Classical mechanics applies only to macroscopic particles. For microscopic "particles" we require a new form of mechanics, called quantum mechanics. We now consider some of the contrasts between classical and quantum mechanics. For simplicity a one-particle, one-dimensional system will be discussed.

In classical mechanics the motion of a particle is governed by Newton's second law:

\(\begin{equation*}F=m a=m \frac{d^{2} x}{d t^{2}} \tag{1.8}\end{equation*}\)

where $F$ is the force acting on the particle, $m$ is its mass, and $t$ is the time; $a$ is the acceleration, given by $a=d v / d t=(d / d t)(d x / d t)=d^{2} x / d t^{2}$, where $v$ is the velocity. Equation (1.8) contains the second derivative of the coordinate $x$ with respect to time. To solve it, we must carry out two integrations. This introduces two arbitrary constants $c_{1}$ and $c_{2}$ into the solution, and

\(\begin{equation*}x=g\left(t, c_{1}, c_{2}\right) \tag{1.9}\end{equation*}\)

where $g$ is some function of time. We now ask: What information must we possess at a given time $t_{0}$ to be able to predict the future motion of the particle? If we know that at $t_{0}$ the particle is at point $x_{0}$, we have

\(\begin{equation*}x_{0}=g\left(t_{0}, c_{1}, c_{2}\right) \tag{1.10}\end{equation*}\)

Since we have two constants to determine, more information is needed. Differentiating (1.9), we have

\(\frac{d x}{d t}=v=\frac{d}{d t} g\left(t, c_{1}, c_{2}\right)\)

If we also know that at time $t_{0}$ the particle has velocity $v_{0}$, then we have the additional relation

\(\begin{equation*}v_{0}=\left.\frac{d}{d t} g\left(t, c_{1}, c_{2}\right)\right|_{t=t_{0}} \tag{1.11}\end{equation*}\)

We may then use (1.10) and (1.11) to solve for $c_{1}$ and $c_{2}$ in terms of $x_{0}$ and $v_{0}$. Knowing $c_{1}$ and $c_{2}$, we can use Eq. (1.9) to predict the exact future motion of the particle.

As an example of Eqs. (1.8) to (1.11), consider the vertical motion of a particle in the earth's gravitational field. Let the $x$ axis point upward. The force on the particle is downward and is $F=-m g$, where $g$ is the gravitational acceleration constant. Newton's second law (1.8) is $-m g=m d^{2} x / d t^{2}$, so $d^{2} x / d t^{2}=-g$. A single integration gives $d x / d t=-g t+c_{1}$. The arbitrary constant $c_{1}$ can be found if we know that at time $t_{0}$ the particle had velocity $v_{0}$. Since $v=d x / d t$, we have $v_{0}=-g t_{0}+c_{1}$ and $c_{1}=v_{0}+g t_{0}$. Therefore, $d x / d t=-g t+g t_{0}+v_{0}$. Integrating a second time, we introduce another arbitrary constant $c_{2}$, which can be evaluated if we know that at time $t_{0}$ the particle had position $x_{0}$. We find (Prob. 1.7) $x=x_{0}-\frac{1}{2} g\left(t-t_{0}\right)^{2}+v_{0}\left(t-t_{0}\right)$. Knowing $x_{0}$ and $v_{0}$ at time $t_{0}$, we can predict the future position of the particle.

The classical-mechanical potential energy $V$ of a particle moving in one dimension is defined to satisfy

\(\begin{equation*}\frac{\partial V(x, t)}{\partial x}=-F(x, t) \tag{1.12}\end{equation*}\)

For example, for a particle moving in the earth's gravitational field, $\partial V / \partial x=-F=m g$ and integration gives $V=m g x+c$, where $c$ is an arbitrary constant. We are free to set the zero level of potential energy wherever we please. Choosing $c=0$, we have $V=m g x$ as the potential-energy function.

The word state in classical mechanics means a specification of the position and velocity of each particle of the system at some instant of time, plus specification of the forces
acting on the particles. According to Newton's second law, given the state of a system at any time, its future state and future motions are exactly determined, as shown by Eqs. (1.9)-(1.11). The impressive success of Newton's laws in explaining planetary motions led many philosophers to use Newton's laws as an argument for philosophical determinism. The mathematician and astronomer Laplace (1749-1827) assumed that the universe consisted of nothing but particles that obeyed Newton's laws. Therefore, given the state of the universe at some instant, the future motion of everything in the universe was completely determined. A super-being able to know the state of the universe at any instant could, in principle, calculate all future motions.


#### Abstract

Although classical mechanics is deterministic, many classical-mechanical systems (for example, a pendulum oscillating under the influence of gravity, friction, and a periodically varying driving force) show chaotic behavior for certain ranges of the systems' parameters. In a chaotic system, the motion is extraordinarily sensitive to the initial values of the particles' positions and velocities and to the forces acting, and two initial states that differ by an experimentally undetectable amount will eventually lead to very different future behavior of the system. Thus, because the accuracy with which one can measure the initial state is limited, prediction of the long-term behavior of a chaotic classical-mechanical system is, in practice, impossible, even though the system obeys deterministic equations. Computer calculations of solar-system planetary orbits over tens of millions of years indicate that the motions of the planets are chaotic [I. Peterson, Newton's Clock: Chaos in the Solar System, Freeman, 1993; J. J. Lissauer, Rev. Mod. Phys., 71, 835 (1999)].


Given exact knowledge of the present state of a classical-mechanical system, we can predict its future state. However, the Heisenberg uncertainty principle shows that we cannot determine simultaneously the exact position and velocity of a microscopic particle, so the very knowledge required by classical mechanics for predicting the future motions of a system cannot be obtained. We must be content in quantum mechanics with something less than complete prediction of the exact future motion.

Our approach to quantum mechanics will be to postulate the basic principles and then use these postulates to deduce experimentally testable consequences such as the energy levels of atoms. To describe the state of a system in quantum mechanics, we postulate the existence of a function $\Psi$ of the particles' coordinates called the state function or wave function (often written as wavefunction). Since the state will, in general, change with time, $\Psi$ is also a function of time. For a one-particle, one-dimensional system, we have $\Psi=\Psi(x, t)$. The wave function contains all possible information about a system, so instead of speaking of "the state described by the wave function $\Psi$," we simply say "the state $\Psi$." Newton's second law tells us how to find the future state of a classicalmechanical system from knowledge of its present state. To find the future state of a quantum-mechanical system from knowledge of its present state, we want an equation that tells us how the wave function changes with time. For a one-particle, one-dimensional system, this equation is postulated to be

\(\begin{equation*}-\frac{\hbar}{i} \frac{\partial \Psi(x, t)}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}+V(x, t) \Psi(x, t) \tag{1.13}\end{equation*}\)

where the constant $\hbar$ (h-bar) is defined as

\(\begin{equation*}\hbar \equiv \frac{h}{2 \pi} \tag{1.14}\end{equation*}\)

The concept of the wave function and the equation governing its change with time were discovered in 1926 by the Austrian physicist Erwin Schrödinger (1887-1961). In this equation, known as the time-dependent Schrödinger equation (or the Schrödinger wave equation), $i=\sqrt{-1}, m$ is the mass of the particle, and $V(x, t)$ is the potentialenergy function of the system. (Many of the historically important papers in quantum mechanics are available at dieumsnh.qfb.umich.mx/archivoshistoricosmq.)

The time-dependent Schrödinger equation contains the first derivative of the wave function with respect to time and allows us to calculate the future wave function (state) at any time, if we know the wave function at time $t_{0}$.

The wave function contains all the information we can possibly know about the system it describes. What information does $\Psi$ give us about the result of a measurement of the $x$ coordinate of the particle? We cannot expect $\Psi$ to involve the definite specification of position that the state of a classical-mechanical system does. The correct answer to this question was provided by Max Born shortly after Schrödinger discovered the Schrödinger equation. Born postulated that for a one-particle, one-dimensional system,

\(\begin{equation*}|\Psi(x, t)|^{2} d x \tag{1.15}\end{equation*}\)

gives the probability at time $t$ of finding the particle in the region of the $x$ axis lying between $x$ and $x+d x$. In (1.15) the bars denote the absolute value and $d x$ is an infinitesimal length on the $x$ axis. The function $|\Psi(x, t)|^{2}$ is the probability density for finding the particle at various places on the $x$ axis. (Probability is reviewed in Section 1.6.) For example, suppose that at some particular time $t_{0}$ the particle is in a state characterized by the wave function $a e^{-b x^{2}}$, where $a$ and $b$ are real constants. If we measure the particle's position at time $t_{0}$, we might get any value of $x$, because the probability density $a^{2} e^{-2 b x^{2}}$ is nonzero everywhere. Values of $x$ in the region around $x=0$ are more likely to be found than other values, since $|\Psi|^{2}$ is a maximum at the origin in this case.

To relate $|\Psi|^{2}$ to experimental measurements, we would take many identical noninteracting systems, each of which was in the same state $\Psi$. Then the particle's position in each system is measured. If we had $n$ systems and made $n$ measurements, and if $d n_{x}$ denotes the number of measurements for which we found the particle between $x$ and $x+d x$, then $d n_{x} / n$ is the probability for finding the particle between $x$ and $x+d x$. Thus

\(\frac{d n_{x}}{n}=|\Psi|^{2} d x\)

and a graph of $(1 / n) d n_{x} / d x$ versus $x$ gives the probability density $|\Psi|^{2}$ as a function of $x$. It might be thought that we could find the probability-density function by taking one system that was in the state $\Psi$ and repeatedly measuring the particle's position. This procedure is wrong because the process of measurement generally changes the state of a system. We saw an example of this in the discussion of the uncertainty principle (Section 1.3).

Quantum mechanics is statistical in nature. Knowing the state, we cannot predict the result of a position measurement with certainty; we can only predict the probabilities of various possible results. The Bohr theory of the hydrogen atom specified the precise path of the electron and is therefore not a correct quantum-mechanical picture.

Quantum mechanics does not say that an electron is distributed over a large region of space as a wave is distributed. Rather, it is the probability patterns (wave functions) used to describe the electron's motion that behave like waves and satisfy a wave equation.

How the wave function gives us information on other properties besides the position is discussed in later chapters.

The postulates of thermodynamics (the first, second, and third laws of thermodynamics) are stated in terms of macroscopic experience and hence are fairly readily understood. The postulates of quantum mechanics are stated in terms of the microscopic world and appear quite abstract. You should not expect to fully understand the postulates of quantum mechanics at first reading. As we treat various examples, understanding of the postulates will increase.

It may bother the reader that we wrote down the Schrödinger equation without any attempt to prove its plausibility. By using analogies between geometrical optics and classical mechanics on the one hand, and wave optics and quantum mechanics on the other hand, one can show the plausibility of the Schrödinger equation. Geometrical optics is an approximation to wave optics, valid when the wavelength of the light is much less than the size of the apparatus. (Recall its use in treating lenses and mirrors.) Likewise, classical mechanics is an approximation to wave mechanics, valid when the particle's wavelength is much less than the size of the apparatus. One can make a plausible guess as to how to get the proper equation for quantum mechanics from classical mechanics based on the known relation between the equations of geometrical and wave optics. Since many chemists are not particularly familiar with optics, these arguments have been omitted. In any case, such analogies can only make the Schrödinger equation seem plausible. They cannot be used to derive or prove this equation. The Schrödinger equation is a postulate of the theory, to be tested by agreement of its predictions with experiment. (Details of the reasoning that led Schrödinger to his equation are given in Jammer, Section 5.3. A reference with the author's name italicized is listed in the Bibliography.)

Quantum mechanics provides the law of motion for microscopic particles. Experimentally, macroscopic objects obey classical mechanics. Hence for quantum mechanics to be a valid theory, it should reduce to classical mechanics as we make the transition from microscopic to macroscopic particles. Quantum effects are associated with the de Broglie wavelength $\lambda=h / m v$. Since $h$ is very small, the de Broglie wavelength of macroscopic objects is essentially zero. Thus, in the limit $\lambda \rightarrow 0$, we expect the time-dependent Schrödinger equation to reduce to Newton's second law. We can prove this to be so (see Prob. 7.59).

A similar situation holds in the relation between special relativity and classical mechanics. In the limit $v / c \rightarrow 0$, where $c$ is the speed of light, special relativity reduces to classical mechanics. The form of quantum mechanics that we will develop will be nonrelativistic. A complete integration of relativity with quantum mechanics has not been achieved.

Historically, quantum mechanics was first formulated in 1925 by Heisenberg, Born, and Jordan using matrices, several months before Schrödinger's 1926 formulation using differential equations. Schrödinger proved that the Heisenberg formulation (called matrix mechanics) is equivalent to the Schrödinger formulation (called wave mechanics). In 1926, Dirac and Jordan, working independently, formulated quantum mechanics in an abstract version called transformation theory that is a generalization of matrix mechanics and wave mechanics. In 1948, Feynman devised the path integral formulation of quantum mechanics [R. P. Feynman, Rev. Mod. Phys., 20, 367 (1948); R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965].

The time-dependent Schrödinger equation (1.13) is formidable looking. Fortunately, many applications of quantum mechanics to chemistry do not use this equation. Instead, the simpler time-independent Schrödinger equation is used. We now derive the
time-independent from the time-dependent Schrödinger equation for the one-particle, one-dimensional case.

We begin by restricting ourselves to the special case where the potential energy $V$ is not a function of time but depends only on $x$. This will be true if the system experiences no time-dependent external forces. The time-dependent Schrödinger equation reads

\(
\begin{equation*}
-\frac{\hbar}{i} \frac{\partial \Psi(x, t)}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}+V(x) \Psi(x, t) \tag{1.16}
\end{equation*}
\)

We now restrict ourselves to looking for those solutions of (1.16) that can be written as the product of a function of time and a function of $x$ :

\(
\begin{equation*}
\Psi(x, t)=f(t) \psi(x) \tag{1.17}
\end{equation*}
\)

Capital psi is used for the time-dependent wave function and lowercase psi for the factor that depends only on the coordinate $x$. States corresponding to wave functions of the form (1.17) possess certain properties (to be discussed shortly) that make them of great interest. [Not all solutions of (1.16) have the form (1.17); see Prob. 3.51.] Taking partial derivatives of (1.17), we have

\(
\frac{\partial \Psi(x, t)}{\partial t}=\frac{d f(t)}{d t} \psi(x), \quad \frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}=f(t) \frac{d^{2} \psi(x)}{d x^{2}}
\)

Substitution into (1.16) gives

\(
\begin{gather*}
-\frac{\hbar}{i} \frac{d f(t)}{d t} \psi(x)=-\frac{\hbar^{2}}{2 m} f(t) \frac{d^{2} \psi(x)}{d x^{2}}+V(x) f(t) \psi(x) \\
-\frac{\hbar}{i} \frac{1}{f(t)} \frac{d f(t)}{d t}=-\frac{\hbar^{2}}{2 m} \frac{1}{\psi(x)} \frac{d^{2} \psi(x)}{d x^{2}}+V(x) \tag{1.18}
\end{gather*}
\)

where we divided by $f \psi$. In general, we expect the quantity to which each side of (1.18) is equal to be a certain function of $x$ and $t$. However, the right side of (1.18) does not depend on $t$, so the function to which each side of (1.18) is equal must be independent of $t$. The left side of (1.18) is independent of $x$, so this function must also be independent of $x$. Since the function is independent of both variables, $x$ and $t$, it must be a constant. We call this constant $E$.

Equating the left side of (1.18) to $E$, we get

\(
\frac{d f(t)}{f(t)}=-\frac{i E}{\hbar} d t
\)

Integrating both sides of this equation with respect to $t$, we have

\(
\ln f(t)=-i E t / \hbar+C
\)

where $C$ is an arbitrary constant of integration. Hence

\(
f(t)=e^{C} e^{-i E t / \hbar}=A e^{-i E t / \hbar}
\)

where the arbitrary constant $A$ has replaced $e^{C}$. Since $A$ can be included as a factor in the function $\psi(x)$ that multiplies $f(t)$ in (1.17), $A$ can be omitted from $f(t)$. Thus

\(
f(t)=e^{-i E t / \hbar}
\)

Equating the right side of (1.18) to $E$, we have

\(
\begin{equation*}
-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi(x)}{d x^{2}}+V(x) \psi(x)=E \psi(x) \tag{1.19}
\end{equation*}
\)

Equation (1.19) is the time-independent Schrödinger equation for a single particle of mass $m$ moving in one dimension.

What is the significance of the constant $E$ ? Since $E$ occurs as $[E-V(x)]$ in (1.19), $E$ has the same dimensions as $V$, so $E$ has the dimensions of energy. In fact, we postulate that $E$ is the energy of the system. (This is a special case of a more general postulate to be discussed in a later chapter.) Thus, for cases where the potential energy is a function of $x$ only, there exist wave functions of the form

\(
\begin{equation*}
\Psi(x, t)=e^{-i E t / \hbar} \psi(x) \tag{1.20}
\end{equation*}
\)

and these wave functions correspond to states of constant energy $E$. Much of our attention in the next few chapters will be devoted to finding the solutions of (1.19) for various systems.

The wave function in (1.20) is complex, but the quantity that is experimentally observable is the probability density $|\Psi(x, t)|^{2}$. The square of the absolute value of a complex quantity is given by the product of the quantity with its complex conjugate, the complex conjugate being formed by replacing $i$ with $-i$ wherever it occurs. (See Section 1.7.) Thus

\(
\begin{equation*}
|\Psi|^{2}=\Psi^{*} \Psi \tag{1.21}
\end{equation*}
\)

where the star denotes the complex conjugate. For the wave function (1.20),

\(
\begin{align*}
|\Psi(x, t)|^{2} & =\left[e^{-i E t / \hbar} \psi(x)\right]^{*} e^{-i E t / \hbar} \psi(x) \\
& =e^{i E t / \hbar} \psi^{*}(x) e^{-i E t / \hbar} \psi(x) \\
& =e^{0} \psi^{*}(x) \psi(x)=\psi^{*}(x) \psi(x) \\
|\Psi(x, t)|^{2} & =|\psi(x)|^{2} \tag{1.22}
\end{align*}
\)

In deriving (1.22), we assumed that $E$ is a real number, so $E=E^{*}$. This fact will be proved in Section 7.2.

Hence for states of the form (1.20), the probability density is given by $|\Psi(x)|^{2}$ and does not change with time. Such states are called stationary states. Since the physically significant quantity is $|\Psi(x, t)|^{2}$, and since for stationary states $|\Psi(x, t)|^{2}=|\psi(x)|^{2}$, the function $\psi(x)$ is often called the wave function, although the complete wave function of a stationary state is obtained by multiplying $\psi(x)$ by $e^{-i E t / \hbar}$. The term stationary state should not mislead the reader into thinking that a particle in a stationary state is at rest. What is stationary is the probability density $|\Psi|^{2}$, not the particle itself.

We will be concerned mostly with states of constant energy (stationary states) and hence will usually deal with the time-independent Schrödinger equation (1.19). For simplicity we will refer to this equation as "the Schrödinger equation." Note that the Schrödinger equation contains two unknowns: the allowed energies $E$ and the allowed wave functions $\psi$. To solve for two unknowns, we need to impose additional conditions (called boundary conditions) on $\psi$ besides requiring that it satisfy (1.19). The boundary conditions determine the allowed energies, since it turns out that only certain values of $E$ allow $\psi$ to satisfy the boundary conditions. This will become clearer when we discuss specific examples in later chapters.

Probability plays a fundamental role in quantum mechanics. This section reviews the mathematics of probability.

There has been much controversy about the proper definition of probability. One definition is the following: If an experiment has $n$ equally probable outcomes, $m$ of which are favorable to the occurrence of a certain event $A$, then the probability that $A$ occurs is $m / n$. Note that this definition is circular, since it specifies equally probable outcomes when probability is what we are trying to define. It is simply assumed that we can recognize equally probable outcomes. An alternative definition is based on actually performing the experiment many times. Suppose that we perform the experiment $N$ times and that in $M$ of these trials the event $A$ occurs. The probability of $A$ occurring is then defined as

\(
\lim _{N \rightarrow \infty} \frac{M}{N}
\)

Thus, if we toss a coin repeatedly, the fraction of heads will approach $1 / 2$ as we increase the number of tosses.

For example, suppose we ask for the probability of drawing a heart when a card is picked at random from a standard 52 -card deck containing 13 hearts. There are 52 cards and hence 52 equally probable outcomes. There are 13 hearts and hence 13 favorable outcomes. Therefore, $m / n=13 / 52=1 / 4$. The probability for drawing a heart is $1 / 4$.

Sometimes we ask for the probability of two related events both occurring. For example, we may ask for the probability of drawing two hearts from a 52 -card deck, assuming we do not replace the first card after it is drawn. There are 52 possible outcomes of the first draw, and for each of these possibilities there are 51 possible second draws. We have $52 \cdot 51$ possible outcomes. Since there are 13 hearts, there are $13 \cdot 12$ different ways to draw two hearts. The desired probability is $(13 \cdot 12) /(52 \cdot 51)=1 / 17$. This calculation illustrates the theorem: The probability that two events $A$ and $B$ both occur is the probability that $A$ occurs, multiplied by the conditional probability that $B$ then occurs, calculated with the assumption that $A$ occurred. Thus, if $A$ is the probability of drawing a heart on the first draw, the probability of $A$ is $13 / 52$. The probability of drawing a heart on the second draw, given that the first draw yielded a heart, is $12 / 51$ since there remain 12 hearts in the deck. The probability of drawing two hearts is then $(13 / 52)(12 / 51)=1 / 17$, as found previously.

In quantum mechanics we must deal with probabilities involving a continuous variable, for example, the $x$ coordinate. It does not make much sense to talk about the probability of a particle being found at a particular point such as $x=0.5000 \ldots$, since there are an infinite number of points on the $x$ axis, and for any finite number of measurements we make, the probability of getting exactly $0.5000 \ldots$ is vanishingly small. Instead we talk of the probability of finding the particle in a tiny interval of the $x$ axis lying between $x$ and $x+d x, d x$ being an infinitesimal element of length. This probability will naturally be proportional to the length of the interval, $d x$, and will vary for different regions of the $x$ axis. Hence the probability that the particle will be found between $x$ and $x+d x$ is equal to $g(x) d x$, where $g(x)$ is some function that tells how the probability varies over the $x$ axis. The function $g(x)$ is called the probability density, since it is a probability per unit length. Since probabilities are real, nonnegative numbers, $g(x)$ must be a real function that is everywhere nonnegative. The wave function $\Psi$ can take on negative and complex values and is not a probability density. Quantum mechanics postulates that the probability density is $|\Psi|^{2}$ [Eq. (1.15)].

What is the probability that the particle lies in some finite region of space $a \leq x \leq b$ ? To find this probability, we sum up the probabilities $|\Psi|^{2} d x$ of finding the particle in all
the infinitesimal regions lying between $a$ and $b$. This is just the definition of the definite integral

\(
\begin{equation*}
\int_{a}^{b}|\Psi|^{2} d x=\operatorname{Pr}(a \leq x \leq b) \tag{1.23}
\end{equation*}
\)

where Pr denotes a probability. A probability of 1 represents certainty. Since it is certain that the particle is somewhere on the $x$ axis, we have the requirement

\(
\begin{equation*}
\int_{-\infty}^{\infty}|\Psi|^{2} d x=1 \tag{1.24}
\end{equation*}
\)

When $\Psi$ satisfies (1.24), it is said to be normalized. For a stationary state, $|\Psi|^{2}=|\psi|^{2}$ and $\int_{-\infty}^{\infty}|\psi|^{2} d x=1$.

## EXAMPLE

A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured. (a) Find the probability that the measured value lies between $x=1.5000 \mathrm{~nm}$ and $x=1.5001 \mathrm{~nm}$. (b) Find the probability that the measured value is between $x=0$ and $x=2 \mathrm{~nm}$. (c) Verify that $\Psi$ is normalized.
(a) In this tiny interval, $x$ changes by only 0.0001 nm , and $\Psi$ goes from $e^{-1.5000} \mathrm{~nm}^{-1 / 2}=0.22313 \mathrm{~nm}^{-1 / 2}$ to $e^{-1.5001} \mathrm{~nm}^{-1 / 2}=0.22311 \mathrm{~nm}^{-1 / 2}$, so $\Psi$ is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given by (1.15) as

\(
\begin{aligned}
|\Psi|^{2} d x=a^{-1} e^{-2|x| / a} d x & =(1 \mathrm{~nm})^{-1} e^{-2(1.5 \mathrm{~nm}) /(1 \mathrm{~nm})}(0.0001 \mathrm{~nm}) \\
& =4.979 \times 10^{-6}
\end{aligned}
\)

(See also Prob. 1.14.)
(b) Use of Eq. (1.23) and $|x|=x$ for $x \geq 0$ gives

\(
\begin{aligned}
\operatorname{Pr}(0 \leq x \leq 2 \mathrm{~nm}) & =\int_{0}^{2 \mathrm{~nm}}|\Psi|^{2} d x=a^{-1} \int_{0}^{2 \mathrm{~nm}} e^{-2 x / a} d x \\
& =-\left.\frac{1}{2} e^{-2 x / a}\right|_{0} ^{2 \mathrm{~nm}}=-\frac{1}{2}\left(e^{-4}-1\right)=0.4908
\end{aligned}
\)

(c) Use of $\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^{0} f(x) d x+\int_{0}^{\infty} f(x) d x,|x|=-x$ for $x \leq 0$, and $|x|=x$ for $x \geq 0$, gives

\(
\begin{aligned}
\int_{-\infty}^{\infty}|\Psi|^{2} d x & =a^{-1} \int_{-\infty}^{0} e^{2 x / a} d x+a^{-1} \int_{0}^{\infty} e^{-2 x / a} d x \\
& =a^{-1}\left(\left.\frac{1}{2} a e^{2 x / a}\right|_{-\infty} ^{0}\right)+a^{-1}\left(-\left.\frac{1}{2} a e^{-2 x / a}\right|_{0} ^{\infty}\right)=\frac{1}{2}+\frac{1}{2}=1
\end{aligned}
\)

EXERCISE For a system whose state function at the time of a position measurement is $\Psi=\left(32 a^{3} / \pi\right)^{1 / 4} x e^{-a x^{2}}$, where $a=1.0000 \mathrm{~nm}^{-2}$, find the probability that the particle is found between $x=1.2000 \mathrm{~nm}$ and 1.2001 nm . Treat the interval as infinitesimal.
(Answer: 0.0000258.)

We have seen that the wave function can be complex, so we now review some properties of complex numbers.

A complex number $z$ is a number of the form

\(
\begin{equation}
z=x+i y, \quad \text { where } i \equiv \sqrt{-1} \tag{1.25}
\end{equation}
\)

and where $x$ and $y$ are real numbers (numbers that do not involve the square root of a negative quantity). If $y=0$ in (1.25), then $z$ is a real number. If $y \neq 0$, then $z$ is an imaginary number. If $x=0$ and $y \neq 0$, then $z$ is a pure imaginary number. For example, 6.83 is a real number, $5.4-3 i$ is an imaginary number, and $0.60 i$ is a pure imaginary number. Real and pure imaginary numbers are special cases of complex numbers. In (1.25), $x$ and $y$ are called the real and imaginary parts of $z$, respectively: $x=\operatorname{Re}(z) ; y=\operatorname{Im}(z)$.

The complex number $z$ can be represented as a point in the complex plane (Fig. 1.3), where the real part of $z$ is plotted on the horizontal axis and the imaginary part on the vertical axis. This diagram immediately suggests defining two quantities that characterize the complex number $z$ : the distance $r$ of the point $z$ from the origin is called the absolute value or modulus of $z$ and is denoted by $|z|$; the angle $\theta$ that the radius vector to the point $z$ makes with the positive horizontal axis is called the phase or argument of $z$. We have

\(
\begin{gather}
|z|=r=\left(x^{2}+y^{2}\right)^{1 / 2}, \quad \tan \theta=y / x \tag{1.26}\
x=r \cos \theta, \quad y=r \sin \theta
\end{gather}
\)

So we may write $z=x+i y$ as

\(
\begin{equation}
z=r \cos \theta+i r \sin \theta=r e^{i \theta} \tag{1.27}
\end{equation}
\)

since (Prob. 4.3)

\(
\begin{equation}
e^{i \theta}=\cos \theta+i \sin \theta \tag{1.28}
\end{equation}
\)

The angle $\theta$ in these equations is in radians.
If $z=x+i y$, the complex conjugate $z^{*}$ of the complex number $z$ is defined as

\(
\begin{equation}
z^{} \equiv x-i y=r e^{-i \theta} \tag{1.29}
\end{equation}
\)

FIGURE 1.3 (a) Plot of a complex number $z=x+i y$. (b) Plot of the number $-2+i$.

If $z$ is a real number, its imaginary part is zero. Thus $z$ is real if and only if $z=z^{}$. Taking the complex conjugate twice, we get $z$ back again, $\left(z^{}\right)^{*}=z$. Forming the product of $z$ and its complex conjugate and using $i^{2}=-1$, we have

\(
\begin{gather}
z z^{}=(x+i y)(x-i y)=x^{2}+i y x-i y x-i^{2} y^{2} \
z z^{}=x^{2}+y^{2}=r^{2}=|z|^{2} \tag{1.30}
\end{gather}
\)

For the product and quotient of two complex numbers $z{1}=r{1} e^{i \theta{1}}$ and $z{2}=r{2} e^{i \theta{2}}$, we have

\(
\begin{equation}
z{1} z{2}=r{1} r{2} e^{i\left(\theta{1}+\theta{2}\right)}, \quad \frac{z{1}}{z{2}}=\frac{r{1}}{r{2}} e^{i\left(\theta{1}-\theta{2}\right)} \tag{1.31}
\end{equation}
\)

It is easy to prove, either from the definition of complex conjugate or from (1.31), that

\(
\begin{equation}
\left(z{1} z{2}\right)^{}=z{1}^{*} z{2}^{} \tag{1.32}
\end{equation}
\)

Likewise,

\(
\begin{equation}
\left(z{1} / z{2}\right)^{}=z{1}^{*} / z{2}^{}, \quad\left(z{1}+z{2}\right)^{}=z{1}^{*}+z{2}^{}, \quad\left(z{1}-z{2}\right)^{}=z{1}^{*}-z{2}^{} \tag{1.33}
\end{equation}
\)

For the absolute values of products and quotients, it follows from (1.31) that

\(
\begin{equation}
\left|z{1} z{2}\right|=\left|z{1}\right|\left|z{2}\right|, \quad\left|\frac{z{1}}{z{2}}\right|=\frac{\left|z{1}\right|}{\left|z{2}\right|} \tag{1.34}
\end{equation}
\)

Therefore, if $\psi$ is a complex wave function, we have

\(
\begin{equation}
\left|\psi^{2}\right|=|\psi|^{2}=\psi^{} \psi \tag{1.35}
\end{equation}
\)

We now obtain a formula for the $n$th roots of the number 1 . We may take the phase of the number 1 to be 0 or $2 \pi$ or $4 \pi$, and so on. Hence $1=e^{i 2 \pi k}$, where $k$ is any integer, zero, negative, or positive. Now consider the number $\omega$, where $\omega \equiv e^{i 2 \pi k / n}, n$ being a positive integer. Using (1.31) $n$ times, we see that $\omega^{n}=e^{i 2 \pi k}=1$. Thus $\omega$ is an $n$th root of unity. There are $n$ different complex $n$th roots of unity, and taking $n$ successive values of the integer $k$ gives us all of them:

\(
\begin{equation}
\omega=e^{i 2 \pi k / n}, \quad k=0,1,2, \ldots, n-1 \tag{1.36}
\end{equation}
\)

Any other value of $k$ besides those in (1.36) gives a number whose phase differs by an integral multiple of $2 \pi$ from one of the numbers in (1.36) and hence is not a different root. For $n=2$ in (1.36), we get the two square roots of 1 ; for $n=3$, the three cube roots of 1 ; and so on.

We now develop the theory of quantum mechanics in a more general way than previously. We begin by writing the one-particle, one-dimensional, time-independent Schrödinger equation (1.19) in the form

\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)\right] \psi(x)=E \psi(x) \tag{3.1}
\end{equation}
\)

The entity in brackets in (3.1) is an operator. Equation (3.1) suggests that we have an energy operator, which, operating on the wave function, gives us the wave function back again, but multiplied by an allowed value of the energy. We therefore discuss operators.

An operator is a rule that transforms a given function into another function. For example, let $\hat{D}$ be the operator that differentiates a function with respect to $x$. We use a circumflex to denote an operator. Provided $f(x)$ is differentiable, the result of operating on $f(x)$ with $\hat{D}$ is $\hat{D} f(x)=f^{\prime}(x)$. For example, $\hat{D}\left(x^{2}+3 e^{2 x}\right)=2 x+6 e^{2 x}$. If $\hat{3}$ is the operator that multiplies a function by 3 , then $\hat{3}\left(x^{2}+3 e^{x}\right)=3 x^{2}+9 e^{x}$. If tan is the operator that takes the tangent of a function, then application of $\tan$ to the function $x^{2}+1$ gives $\tan \left(x^{2}+1\right)$. If the operator $\hat{A}$ transforms the function $f(x)$ into the function $g(x)$, we write $\hat{A} f(x)=g(x)$.

We define the sum and the difference of two operators $\hat{A}$ and $\hat{B}$ by

\(
\begin{align}
& (\hat{A}+\hat{B}) f(x) \equiv \hat{A} f(x)+\hat{B} f(x) \tag{3.2}\
& (\hat{A}-\hat{B}) f(x) \equiv \hat{A} f(x)-\hat{B} f(x)
\end{align}
\)

For example, if $\hat{D} \equiv d / d x$, then

\(
(\hat{D}+\hat{3})\left(x^{3}-5\right) \equiv \hat{D}\left(x^{3}-5\right)+\hat{3}\left(x^{3}-5\right)=3 x^{2}+\left(3 x^{3}-15\right)=3 x^{3}+3 x^{2}-15
\)

An operator can involve more than one variable. For example, the operator $\partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}$ has the following effect:

\(
\left(\partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}\right) g(x, y)=\partial^{2} g / \partial x^{2}+\partial^{2} g / \partial y^{2}
\)

The product of two operators $\hat{A}$ and $\hat{B}$ is defined by

\(
\begin{equation}
\hat{A} \hat{B} f(x) \equiv \hat{A}[\hat{B} f(x)] \tag{3.3}
\end{equation}
\)

In other words, we first operate on $f(x)$ with the operator on the right of the operator product, and then we take the resulting function and operate on it with the operator on the left of the operator product. For example, $\hat{3} \hat{D} f(x)=\hat{3}[\hat{D} f(x)]=\hat{3} f^{\prime}(x)=3 f^{\prime}(x)$.

The operators $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ may not have the same effect. Consider, for example, the operators $d / d x$ and $\hat{x}$ (where $\hat{x}$ means multiplication by $x$ ):

\(
\begin{gather}
\hat{D} \hat{x} f(x)=\frac{d}{d x}[x f(x)]=f(x)+x f^{\prime}(x)=(\hat{1}+\hat{x} \hat{D}) f(x) \tag{3.4}\
\hat{x} \hat{D} f(x)=\hat{x}\left[\frac{d}{d x} f(x)\right]=x f^{\prime}(x)
\end{gather}
\)

Thus $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are different operators in this case.
We can develop an operator algebra as follows. Two operators $\hat{A}$ and $\hat{B}$ are said to be equal if $\hat{A} f=\hat{B} f$ for all functions $f$. Equal operators produce the same result when they operate on a given function. For example, (3.4) shows that

\(
\begin{equation}
\hat{D} \hat{x}=1+\hat{x} \hat{D} \tag{3.5}
\end{equation}
\)

The operator $\hat{1}$ (multiplication by 1) is the unit operator. The operator $\hat{0}$ (multiplication by 0 ) is the null operator. We usually omit the circumflex over operators that are simply multiplication by a constant. We can transfer operators from one side of an operator equation to the other (Prob. 3.7). Thus (3.5) is equivalent to $\hat{D} \hat{x}-\hat{x} \hat{D}-1=0$, where circumflexes over the null and unit operators were omitted.

Operators obey the associative law of multiplication:

\(
\begin{equation}
\hat{A}(\hat{B} \hat{C})=(\hat{A} \hat{B}) \hat{C} \tag{3.6}
\end{equation}
\)

The proof of (3.6) is outlined in Prob. 3.10. As an example, let $\hat{A}=d / d x, \hat{B}=\hat{x}$, and $\hat{C}=3$. Using (3.5), we have

\(
\begin{array}{ll}
(\hat{A} \hat{B})=\hat{D} \hat{x}=1+\hat{x} \hat{D}, & {[(\hat{A} \hat{B}) \hat{C}] f=(1+\hat{x} \hat{D}) 3 f=3 f+3 x f^{\prime}} \
(\hat{B} \hat{C})=3 \hat{x}, & {[\hat{A}(\hat{B} \hat{C})] f=\hat{D}(3 x f)=3 f+3 x f^{\prime}}
\end{array}
\)

A major difference between operator algebra and ordinary algebra is that numbers obey the commutative law of multiplication, but operators do not necessarily do so; $a b=b a$ if $a$ and $b$ are numbers, but $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are not necessarily equal operators. We define the commutator $[\hat{A}, \hat{B}]$ of the operators $\hat{A}$ and $\hat{B}$ as the operator $\hat{A} \hat{B}-\hat{B} \hat{A}$ :

\(
\begin{equation}
[\hat{A}, \hat{B}] \equiv \hat{A} \hat{B}-\hat{B} \hat{A} \tag{3.7}
\end{equation}
\)

If $\hat{A} \hat{B}=\hat{B} \hat{A}$, then $[\hat{A}, \hat{B}]=0$, and we say that $\hat{A}$ and $\hat{B}$ commute. If $\hat{A} \hat{B} \neq \hat{B} \hat{A}$, then $\hat{A}$ and $\hat{B}$ do not commute. Note that $[\hat{A}, \hat{B}] f=\hat{A} \hat{B} f-\hat{B} \hat{A} f$. Since the order in which we apply the operators 3 and $d / d x$ makes no difference, we have

\(
\left[\hat{3}, \frac{d}{d x}\right]=\hat{3} \frac{d}{d x}-\frac{d}{d x} \hat{3}=0
\)

From Eq. (3.5) we have

\(
\begin{equation}
\left[\frac{d}{d x}, \hat{x}\right]=\hat{D} \hat{x}-\hat{x} \hat{D}=1 \tag{3.8}
\end{equation}
\)

The operators $d / d x$ and $\hat{x}$ do not commute.

EXAMPLE

Find $\left[z^{3}, d / d z\right]$.
To find $\left[z^{3}, d / d z\right]$, we apply this operator to an arbitrary function $g(z)$. Using the commutator definition (3.7) and the definitions of the difference and product of two operators, we have

\(
\begin{aligned}
{\left[z^{3}, d / d z\right] g=\left[z^{3}(d / d z)-(d / d z) z^{3}\right] g } & =z^{3}(d / d z) g-(d / d z)\left(z^{3} g\right) \
& =z^{3} g^{\prime}-3 z^{2} g-z^{3} g^{\prime}=-3 z^{2} g
\end{aligned}
\)

Deleting the arbitrary function $g$, we get the operator equation $\left[z^{3}, d / d z\right]=-3 z^{2}$.

EXERCISE Find $\left[d / d x, 5 x^{2}+3 x+4\right]$. (Answer: $10 x+3$.)

The square of an operator is defined as the product of the operator with itself: $\hat{B}^{2}=\hat{B} \hat{B}$. Let us find the square of the differentiation operator:

\(
\begin{aligned}
\hat{D}^{2} f(x) & =\hat{D}(\hat{D} f)=\hat{D} f^{\prime}=f^{\prime \prime} \
\hat{D}^{2} & =d^{2} / d x^{2}
\end{aligned}
\)

As another example, the square of the operator that takes the complex conjugate of a function is equal to the unit operator, since taking the complex conjugate twice gives the original function. The operator $\hat{B}^{n}(n=1,2,3, \ldots)$ is defined to mean applying the operator $\hat{B} n$ times in succession.

It turns out that the operators occurring in quantum mechanics are linear. $\hat{A}$ is a linear operator if and only if it has the following two properties:

\(
\begin{equation}
\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x) \tag{3.9}
\end{equation}
\)

\(
\begin{equation}
\hat{A}[c f(x)]=c \hat{A} f(x) \tag{3.10}
\end{equation}
\)

where $f$ and $g$ are arbitrary functions and $c$ is an arbitrary constant (not necessarily real). Examples of linear operators include $\hat{x}^{2}, d / d x$, and $d^{2} / d x^{2}$. Some nonlinear operators are cos and ()$^{2}$, where ()$^{2}$ squares the function it acts on.

EXAMPLE

Is $d / d x$ a linear operator? Is $\sqrt{ }$ a linear operator?
We have

\(
\begin{gathered}
(d / d x)[f(x)+g(x)]=d f / d x+d g / d x=(d / d x) f(x)+(d / d x) g(x) \
(d / d x)[c f(x)]=c d f(x) / d x
\end{gathered}
\)

so $d / d x$ obeys (3.9) and (3.10) and is a linear operator. However,

\(
\sqrt{f(x)+g(x)} \neq \sqrt{f(x)}+\sqrt{g(x)}
\)

so $\sqrt{ }$ does not obey (3.9) and is nonlinear.
EXERCISE Is the operator $x^{2} \times$ (multiplication by $x^{2}$ ) linear? (Answer: Yes.)

Useful identities in linear-operator manipulations are

\(
\begin{equation}
(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C} \tag{3.11}
\end{equation}
\)

\(
\begin{equation}
\hat{A}(\hat{B}+\hat{C})=\hat{A} \hat{B}+\hat{A} \hat{C} \tag{3.12}
\end{equation}
\)

EXAMPLE

Prove the distributive law (3.11) for linear operators.
A good way to begin a proof is to first write down what is given and what is to be proved. We are given that $\hat{A}, \hat{B}$, and $\hat{C}$ are linear operators. We must prove that $(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C}$.

To prove that the operator $(\hat{A}+\hat{B}) \hat{C}$ is equal to the operator $\hat{A} \hat{C}+\hat{B} \hat{C}$, we must prove that these two operators give the same result when applied to an arbitrary function $f$. Thus we must prove that

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A} \hat{C}+\hat{B} \hat{C}) f
\)

We start with $[(\hat{A}+\hat{B}) \hat{C}] f$. This expression involves the product of the two operators $\hat{A}+\hat{B}$ and $\hat{C}$. The operator-product definition (3.3) with $\hat{A}$ replaced by $\hat{A}+\hat{B}$ and $\hat{B}$ replaced by $\hat{C}$ gives $[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A}+\hat{B})(\hat{C} f)$. The entity $\hat{C} f$ is a function, and use of the definition (3.2) of the sum $\hat{A}+\hat{B}$ of the two operators $\hat{A}$ and $\hat{B}$ gives $(\hat{A}+\hat{B})(\hat{C} f)=\hat{A}(\hat{C} f)+\hat{B}(\hat{C} f)$. Thus

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A}+\hat{B})(\hat{C} f)=\hat{A}(\hat{C} f)+\hat{B}(\hat{C} f)
\)

Use of the operator-product definition (3.3) gives $\hat{A}(\hat{C} f)=\hat{A} \hat{C} f$ and $\hat{B}(\hat{C} f)=\hat{B} \hat{C} f$. Hence

\(
\begin{equation}
[(\hat{A}+\hat{B}) \hat{C}] f=\hat{A} \hat{C} f+\hat{B} \hat{C} f \tag{3.13}
\end{equation}
\)

Use of the operator-sum definition (3.2) with $\hat{A}$ replaced by $\hat{A} \hat{C}$ and $\hat{B}$ replaced by $\hat{B} \hat{C}$ gives $(\hat{A} \hat{C}+\hat{B} \hat{C}) f=\hat{A} \hat{C} f+\hat{B} \hat{C} f$, so (3.13) becomes

\(
[(\hat{A}+\hat{B}) \hat{C}] f=(\hat{A} \hat{C}+\hat{B} \hat{C}) f
\)

which is what we wanted to prove. Hence $(\hat{A}+\hat{B}) \hat{C}=\hat{A} \hat{C}+\hat{B} \hat{C}$.
Note that we did not need to use the linearity of $\hat{A}, \hat{B}$, and $\hat{C}$. Hence (3.11) holds for all operators. However, (3.12) holds only if $\hat{A}$ is linear (see Prob. 3.17).

EXAMPLE

Find the square of the operator $d / d x+\hat{x}$.
To find the effect of $(d / d x+\hat{x})^{2}$, we apply this operator to an arbitrary function $f(x)$. Letting $\hat{D} \equiv d / d x$, we have

\(
\begin{aligned}
(\hat{D}+\hat{x})^{2} f(x)= & (\hat{D}+\hat{x})[(\hat{D}+x) f]=(\hat{D}+\hat{x})\left(f^{\prime}+x f\right) \
= & f^{\prime \prime}+f+x f^{\prime}+x f^{\prime}+x^{2} f=\left(\hat{D}^{2}+2 \hat{x} \hat{D}+\hat{x}^{2}+1\right) f(x) \
& (\hat{D}+\hat{x})^{2}=\hat{D}^{2}+2 \hat{x} \hat{D}+\hat{x}^{2}+1
\end{aligned}
\)

Let us repeat this calculation, using only operator equations:

\(
\begin{aligned}
(\hat{D}+\hat{x})^{2} & =(\hat{D}+\hat{x})(\hat{D}+\hat{x})=\hat{D}(\hat{D}+\hat{x})+\hat{x}(\hat{D}+\hat{x}) \
& =\hat{D}^{2}+\hat{D} \hat{x}+\hat{x} \hat{D}+\hat{x}^{2}=\hat{D}^{2}+\hat{x} \hat{D}+1+\hat{x} \hat{D}+\hat{x}^{2} \
& =\hat{D}^{2}+2 x \hat{D}+x^{2}+1
\end{aligned}
\)

where (3.11), (3.12), and (3.5) have been used and the circumflex over the operator "multiplication by $x$ " has been omitted. Until you have become thoroughly experienced with operators, it is safest when doing operator manipulations always to let the operator operate on an arbitrary function $f$ and then delete $f$ at the end.

EXERCISE Find $\left(d^{2} / d x^{2}+x\right)^{2}$. (Answer: $d^{4} / d x^{4}+2 x d^{2} / d x^{2}+2 d / d x+x^{2}$.)

Suppose that the effect of operating on some function $f(x)$ with the linear operator $\hat{A}$ is simply to multiply $f(x)$ by a certain constant $k$. We then say that $f(x)$ is an eigenfunction of $\hat{A}$ with eigenvalue $k$. (Eigen is a German word meaning characteristic.) As part of the definition, we shall require that the eigenfunction $f(x)$ is not identically zero. By this we mean that, although $f(x)$ may vanish at various points, it is not everywhere zero. We have

\(
\begin{equation}
\hat{A} f(x)=k f(x) \tag{3.14}
\end{equation}
\)

As an example of (3.14), $e^{2 x}$ is an eigenfunction of the operator $d / d x$ with eigenvalue 2:

\(
(d / d x) e^{2 x}=2 e^{2 x}
\)

However, $\sin 2 x$ is not an eigenfunction of $d / d x$, since $(d / d x)(\sin 2 x)=2 \cos 2 x$, which is not a constant times $\sin 2 x$.

EXAMPLE

If $f(x)$ is an eigenfunction of the linear operator $\hat{A}$ and $c$ is any constant, prove that $c f(x)$ is an eigenfunction of $\hat{A}$ with the same eigenvalue as $f(x)$.

A good way to see how to do a proof is to carry out the following steps:

1. Write down the given information and translate this information from words into equations.
2. Write down what is to be proved in the form of an equation or equations.
3. (a) Manipulate the given equations of step 1 so as to transform them to the desired equations of step 2. (b) Alternatively, start with one side of the equation that we want to prove and use the given equations of step 1 to manipulate this side until it is transformed into the other side of the equation to be proved.

We are given three pieces of information: $f$ is an eigenfunction of $\hat{A} ; \hat{A}$ is a linear operator; $c$ is a constant. Translating these statements into equations, we have [see Eqs. (3.14), (3.9), and (3.10)]

\(
\begin{gather}
\hat{A} f=k f \tag{3.15}\
\hat{A}(f+g)=\hat{A} f+\hat{A} g \quad \text { and } \quad \hat{A}(b f)=b \hat{A} f \tag{3.16}\
c=\text { a constant }
\end{gather}
\)

where $k$ and $b$ are constants and $f$ and $g$ are functions.
We want to prove that $c f$ is an eigenfunction of $\hat{A}$ with the same eigenvalue as $f$, which, written as an equation, is

\(
\hat{A}(c f)=k(c f)
\)

Using the strategy of step $3(\mathrm{~b})$, we start with the left side $\hat{A}(c f)$ of this last equation and try to show that it equals $k(c f)$. Using the second equation in the linearity definition (3.16), we have $\hat{A}(c f)=c \hat{A} f$. Using the eigenvalue equation (3.15), we have $c \hat{A} f=c k f$. Hence

\(
\hat{A}(c f)=c \hat{A} f=c k f=k(c f)
\)

which completes the proof.

EXAMPLE

(a) Find the eigenfunctions and eigenvalues of the operator $d / d x$. (b) If we impose the boundary condition that the eigenfunctions remain finite as $x \rightarrow \pm \infty$, find the eigenvalues.
(a) Equation (3.14) with $\hat{A}=d / d x$ becomes

\(
\begin{align}
\frac{d f(x)}{d x} & =k f(x) \tag{3.17}\
\frac{1}{f} d f & =k d x
\end{align}
\)

Integration gives

\(
\begin{align}
\ln f & =k x+\text { constant } \
f & =e^{\text {constant }} e^{k x} \
f & =c e^{k x} \tag{3.18}
\end{align}
\)

The eigenfunctions of $d / d x$ are given by (3.18). The eigenvalues are $k$, which can be any number whatever and (3.17) will still be satisfied. The eigenfunctions contain an arbitrary multiplicative constant $c$. This is true for the eigenfunctions of every linear operator, as was proved in the previous example. Each different value of $k$ in (3.18) gives a different eigenfunction. However, eigenfunctions with the same value of $k$ but different values of $c$ are not independent of each other.
(b) Since $k$ can be complex, we write it as $k=a+i b$, where $a$ and $b$ are real numbers. We then have $f(x)=c e^{a x} e^{i b x}$. If $a>0$, the factor $e^{a x}$ goes to infinity as $x$ goes to infinity. If $a<0$, then $e^{a x} \rightarrow \infty$ in the limit $x \rightarrow-\infty$. Thus the boundary conditions require that $a=0$, and the eigenvalues are $k=i b$, where $b$ is real.

In the first example in Section 3.1, we found that $\left[z^{3}, d / d z\right] g(z)=-3 z^{2} g(z)$ for every function $g$, and we concluded that $\left[z^{3}, d / d z\right]=-3 z^{2}$. In contrast, the eigenvalue equation $\hat{A} f(x)=k f(x)$ [Eq. (3.14)] does not hold for every function $f(x)$, and we cannot conclude from this equation that $\hat{A}=k$. Thus the fact that $(d / d x) e^{2 x}=2 e^{2 x}$ does not mean that the operator $d / d x$ equals multiplication by 2 .

We now examine the relationship between operators and quantum mechanics. Comparing Eq. (3.1) with (3.14), we see that the Schrödinger equation is an eigenvalue problem. The values of the energy $E$ are the eigenvalues. The eigenfunctions are the time-independent wave functions $\psi$. The operator whose eigenfunctions and eigenvalues are desired is $-\left(\hbar^{2} / 2 m\right) d^{2} / d x^{2}+V(x)$. This operator is called the Hamiltonian operator for the system.

Sir William Rowan Hamilton (1805-1865) devised an alternative form of Newton's equations of motion involving a function $H$, the Hamiltonian function for the system. For a system where the potential energy is a function of the coordinates only, the total energy remains constant with time; that is, $E$ is conserved. We shall restrict ourselves to such conservative systems. For conservative systems, the classical-mechanical Hamiltonian function turns out to be simply the total energy expressed in terms of coordinates and conjugate momenta. For Cartesian coordinates $x, y, z$, the conjugate
momenta are the components of linear momentum in the $x, y$, and $z$ directions: $p{x}, p{y}$, and $p_{z}$ :

\(
\begin{equation}
p{x} \equiv m v{x}, \quad p{y} \equiv m v{y}, \quad p{z} \equiv m v{z} \tag{3.19}
\end{equation}
\)

where $v{x}, v{y}$, and $v_{z}$ are the components of the particle's velocity in the $x, y$, and $z$ directions.

Let us find the classical-mechanical Hamiltonian function for a particle of mass $m$ moving in one dimension and subject to a potential energy $V(x)$. The Hamiltonian function is equal to the energy, which is composed of kinetic and potential energies. The familiar form of the kinetic energy, $\frac{1}{2} m v{x}^{2}$, will not do, however, since we must express the Hamiltonian as a function of coordinates and momenta, not velocities. Since $v{x}=p{x} / m$, the form of the kinetic energy we want is $p{x}^{2} / 2 m$. The Hamiltonian function is

\(
\begin{equation}
H=\frac{p_{x}^{2}}{2 m}+V(x) \tag{3.20}
\end{equation}
\)

The time-independent Schrödinger equation (3.1) indicates that, corresponding to the Hamiltonian function (3.20), we have a quantum-mechanical operator

\(
-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)
\)

whose eigenvalues are the possible values of the system's energy. This correspondence between physical quantities in classical mechanics and operators in quantum mechanics is general. It is a fundamental postulate of quantum mechanics that every physical property (for example, the energy, the $x$ coordinate, the momentum) has a corresponding quantummechanical operator. We further postulate that the operator corresponding to the property $B$ is found by writing the classical-mechanical expression for $B$ as a function of Cartesian coordinates and corresponding momenta and then making the following replacements. Each Cartesian coordinate $q$ is replaced by the operator multiplication by that coordinate:

\(
\hat{q}=q \times
\)

Each Cartesian component of linear momentum $p_{q}$ is replaced by the operator

\(
\hat{p}_{q}=\frac{\hbar}{i} \frac{\partial}{\partial q}=-i \hbar \frac{\partial}{\partial q}
\)

where $i=\sqrt{-1}$ and $\partial / \partial q$ is the operator for the partial derivative with respect to the coordinate $q$. Note that $1 / i=i / i^{2}=i /(-1)=-i$.

Consider some examples. The operator corresponding to the $x$ coordinate is multiplication by $x$ :

\(
\begin{equation}
\hat{x}=x \times \tag{3.21}
\end{equation}
\)

Also,

\(
\begin{equation}
\hat{y}=y \times \quad \text { and } \quad \hat{z}=z \times \tag{3.22}
\end{equation}
\)

The operators for the components of linear momentum are

\(
\begin{equation}
\hat{p}{x}=\frac{\hbar}{i} \frac{\partial}{\partial x}, \quad \hat{p}{y}=\frac{\hbar}{i} \frac{\partial}{\partial y}, \quad \hat{p}_{z}=\frac{\hbar}{i} \frac{\partial}{\partial z} \tag{3.23}
\end{equation}
\)

The operator corresponding to $p_{x}^{2}$ is

\(
\begin{equation}
\hat{p}_{x}^{2}=\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right)^{2}=\frac{\hbar}{i} \frac{\partial}{\partial x} \frac{\hbar}{i} \frac{\partial}{\partial x}=-\hbar^{2} \frac{\partial^{2}}{\partial x^{2}} \tag{3.24}
\end{equation}
\)

with similar expressions for $\hat{p}{y}^{2}$ and $\hat{p}{z}^{2}$.
What are the potential-energy and kinetic-energy operators in one dimension? Suppose a system has the potential-energy function $V(x)=a x^{2}$, where $a$ is a constant. Replacing $x$ with $x \times$, we see that the potential-energy operator is simply multiplication by $a x^{2}$; that is, $\hat{V}(x)=a x^{2} \times$. In general, we have for any potential-energy function

\(
\begin{equation}
\hat{V}(x)=V(x) \times \tag{3.25}
\end{equation}
\)

The classical-mechanical expression for the kinetic energy $T$ in (3.20) is

\(
\begin{equation}
T=p_{x}^{2} / 2 m \tag{3.26}
\end{equation}
\)

Replacing $p_{x}$ by the corresponding operator (3.23), we have

\(
\begin{equation}
\hat{T}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}} \tag{3.27}
\end{equation}
\)

where (3.24) has been used, and the partial derivative becomes an ordinary derivative in one dimension. The classical-mechanical Hamiltonian (3.20) is

\(
\begin{equation}
H=T+V=p_{x}^{2} / 2 m+V(x) \tag{3.28}
\end{equation}
\)

The corresponding quantum-mechanical Hamiltonian (or energy) operator is

\(
\begin{equation}
\hat{H}=\hat{T}+\hat{V}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x) \tag{3.29}
\end{equation}
\)

which agrees with the operator in the Schrödinger equation (3.1). Note that all these operators are linear.

How are the quantum-mechanical operators related to the corresponding properties of a system? Each such operator has its own set of eigenfunctions and eigenvalues. Let $\hat{B}$ be the quantum-mechanical operator that corresponds to the physical property $B$. Letting $f{i}$ and $b{i}$ symbolize the eigenfunctions and eigenvalues of $\hat{B}$, we have [Eq. (3.14)]

\(
\begin{equation}
\hat{B} f{i}=b{i} f_{i}, \quad i=1,2,3, \ldots \tag{3.30}
\end{equation}
\)

The operator $\hat{B}$ has many eigenfunctions and eigenvalues, and the subscript $i$ is used to indicate this. $\hat{B}$ is usually a differential operator, and (3.30) is a differential equation whose solutions give the eigenfunctions and eigenvalues. Quantum mechanics postulates that (no matter what the state function of the system happens to be) a measurement of the property $B$ must yield one of the eigenvalues $b{i}$ of the operator $\hat{B}$. For example, the only values that can be found for the energy of a system are the eigenvalues of the energy (Hamiltonian) operator $\hat{H}$. Using $\psi{i}$ to symbolize the eigenfunctions of $\hat{H}$, we have as the eigenvalue equation (3.30)

\(
\begin{equation}
\hat{H} \psi{i}=E{i} \psi_{i} \tag{3.31}
\end{equation}
\)

Using the Hamiltonian (3.29) in (3.31), we obtain for a one-dimensional, one-particle system

\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)\right] \psi{i}=E{i} \psi_{i} \tag{3.32}
\end{equation}
\)

which is the time-independent Schrödinger equation (3.1). Thus our postulates about operators are consistent with our previous work. We shall later further justify the choice (3.23) for the momentum operator by showing that in the limiting transition to classical mechanics this choice yields $p_{x}=m(d x / d t)$, as it should. (See Prob. 7.59.)

In Chapter 1 we postulated that the state of a quantum-mechanical system is specified by a state function $\Psi(x, t)$, which contains all the information we can know about the system. How does $\Psi$ give us information about the property $B$ ? We postulate that if $\Psi$ is an eigenfunction of $\hat{B}$ with eigenvalue $b{k}$, then a measurement of $B$ is certain to yield the value $b{k}$. Consider, for example, the energy. The eigenfunctions of the energy operator are the solutions $\psi(x)$ of the time-independent Schrödinger equation (3.32). Suppose the system is in a stationary state with state function [Eq. (1.20)]

\(
\begin{equation}
\Psi(x, t)=e^{-i E t / \hbar} \psi(x) \tag{3.33}
\end{equation}
\)

Is $\Psi(x, t)$ an eigenfunction of the energy operator $\hat{H}$ ? We have

\(
\hat{H} \Psi(x, t)=\hat{H} e^{-i E t / \hbar} \psi(x)
\)

$\hat{H}$ contains no derivatives with respect to time and therefore does not affect the exponential factor $e^{-i E t / \hbar}$. We have

\(
\begin{align}
\hat{H} \Psi(x, t)=e^{-i E t / \hbar} \hat{H} \psi(x) & =E e^{-i E t / \hbar} \psi(x)=E \Psi(x, t) \
\hat{H} \Psi & =E \Psi \tag{3.34}
\end{align}
\)

where (3.31) was used. Hence, for a stationary state, $\Psi(x, t)$ is an eigenfunction of $\hat{H}$, and we are certain to obtain the value $E$ when we measure the energy.

As an example of another property, consider momentum. The eigenfunctions $g$ of $\hat{p}_{x}$ are found by solving

\(
\begin{align}
\hat{p}_{x} g & =k g \
\frac{\hbar}{i} \frac{d g}{d x} & =k g \tag{3.35}
\end{align}
\)

We find (Prob. 3.29)

\(
\begin{equation}
g=A e^{i k x / \hbar} \tag{3.36}
\end{equation}
\)

where $A$ is an arbitrary constant. To keep $g$ finite for large $|x|$, the eigenvalues $k$ must be real. Thus the eigenvalues of $\hat{p}_{x}$ are all the real numbers

\(
\begin{equation}
-\infty<k<\infty \tag{3.37}
\end{equation}
\)

which is reasonable. Any measurement of $p{x}$ must yield one of the eigenvalues (3.37) of $\hat{p}{x}$. Each different value of $k$ in (3.36) gives a different eigenfunction $g$. It might seem surprising that the operator for the physical property momentum involves the imaginary number $i$. Actually, the presence of $i$ in $\hat{p}_{x}$ ensures that the eigenvalues $k$ are real. Recall that the eigenvalues of $d / d x$ are imaginary (Section 3.2).

Comparing the free-particle wave function (2.30) with the eigenfunctions (3.36) of $\hat{p}_{x}$, we note the following physical interpretation: The first term in (2.30) corresponds to positive momentum and represents motion in the $+x$ direction; the second term in (2.30) corresponds to negative momentum and represents motion in the $-x$ direction.

Now consider the momentum of a particle in a box. The state function for a particle in a stationary state in a one-dimensional box is [Eqs. (3.33), (2.20), and (2.23)]

\(
\begin{equation}
\Psi(x, t)=e^{-i E t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right) \tag{3.38}
\end{equation}
\)

where $E=n^{2} h^{2} / 8 m l^{2}$. Does the particle have a definite value of $p{x}$ ? That is, is $\Psi(x, t)$ an eigenfunction of $\hat{p}{x}$ ? Looking at the eigenfunctions (3.36) of $\hat{p}{x}$, we see that there is no numerical value of the real constant $k$ that will make the exponential function in (3.36) become a sine function, as in (3.38). Hence $\Psi$ is not an eigenfunction of $\hat{p}{x}$. We can verify this directly; we have

\(
\hat{p}_{x} \Psi=\frac{\hbar}{i} \frac{\partial}{\partial x} e^{-i E t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right)=\frac{n \pi \hbar}{i l} e^{-i E t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \cos \left(\frac{n \pi x}{l}\right)
\)

Since $\hat{p}{x} \Psi \neq$ constant $\cdot \Psi$, the state function $\Psi$ is not an eigenfunction of $\hat{p}{x}$.
Note that the system's state function $\Psi$ need not be an eigenfunction $f{i}$ of the operator $\hat{B}$ in (3.30) that corresponds to the physical property $B$ of the system. Thus, the particle-in-a-box stationary-state wave functions are not eigenfunctions of $\hat{p}{x}$. Despite this, we still must get one of the eigenvalues (3.37) of $\hat{p}{x}$ when we measure $p{x}$ for a particle-in-a-box stationary state.

Are the particle-in-a-box stationary-state wave functions eigenfunctions of $\hat{p}_{x}^{2}$ ? We have [Eq. (3.24)]

\(
\begin{align}
& \hat{p}{x}^{2} \Psi=-\hbar^{2} \frac{\partial^{2}}{\partial x^{2}} e^{-i E t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right)=\frac{n^{2} \pi^{2} \hbar^{2}}{l^{2}} e^{-i E t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right) \
& \hat{p}{x}^{2} \Psi=\frac{n^{2} h^{2}}{4 l^{2}} \Psi \tag{3.39}
\end{align}
\)

Hence a measurement of $p_{x}^{2}$ will always give the result $n^{2} h^{2} / 4 l^{2}$ when the particle is in the stationary state with quantum number $n$. This should come as no surprise: The potential energy in the box is zero, and the Hamiltonian operator is

\(
\hat{H}=\hat{T}+\hat{V}=\hat{T}=\hat{p}_{x}^{2} / 2 m
\)

We then have [Eq. (3.34)]

\(
\begin{gather}
\hat{H} \Psi=E \Psi=\frac{\hat{p}{x}^{2}}{2 m} \Psi \
\hat{p}{x}^{2} \Psi=2 m E \Psi=2 m \frac{n^{2} h^{2}}{8 m l^{2}} \Psi=\frac{n^{2} h^{2}}{4 l^{2}} \Psi \tag{3.40}
\end{gather}
\)

in agreement with (3.39). The only possible value for $p_{x}^{2}$ is

\(
\begin{equation}
p_{x}^{2}=n^{2} h^{2} / 4 l^{2} \tag{3.41}
\end{equation}
\)

Equation (3.41) suggests that a measurement of $p{x}$ would necessarily yield one of the two values $\pm \frac{1}{2} n h / l$, corresponding to the particle moving to the right or to the left in the box. This plausible suggestion is not accurate. An analysis using the methods of Chapter 7 shows that there is a high probability that the measured value will be close to one of the two values $\pm \frac{1}{2} n h / l$, but that any value consistent with (3.37) can result from a measurement of $p{x}$ for the particle in a box; see Prob. 7.41.

We postulated that a measurement of the property $B$ must give a result that is one of the eigenvalues of the operator $\hat{B}$. If the state function $\Psi$ happens to be an eigenfunction
of $\hat{B}$ with eigenvalue $b$, we are certain to get $b$ when we measure $B$. Suppose, however, that $\Psi$ is not one of the eigenfunctions of $\hat{B}$. What then? We still assert that we will get one of the eigenvalues of $\hat{B}$ when we measure $B$, but we cannot predict which eigenvalue will be obtained. We shall see in Chapter 7 that the probabilities for obtaining the various eigenvalues of $\hat{B}$ can be predicted.

EXAMPLE

The energy of a particle of mass $m$ in a one-dimensional box of length $l$ is measured. What are the possible values that can result from the measurement if at the time the measurement begins, the particle's state function is (a) $\Psi=\left(30 / l^{5}\right)^{1 / 2} x(l-x)$ for $0 \leq x \leq l ;$ (b) $\Psi=(2 / l)^{1 / 2} \sin (3 \pi x / l)$ for $0 \leq x \leq l$ ?
(a) The possible outcomes of a measurement of the property $E$ are the eigenvalues of the system's energy (Hamiltonian) operator $\hat{H}$. Therefore, the measured value must be one of the numbers $n^{2} h^{2} / 8 m l^{2}$, where $n=1,2,3, \ldots$ Since $\Psi$ is not one of the eigenfunctions $(2 / l)^{1 / 2} \sin (n \pi x / l)$ [Eq. (2.23)] of $\hat{H}$, we cannot predict which one of these eigenvalues will be obtained for this nonstationary state. (The probabilities for obtaining these eigenvalues are found in the last example in Section 7.6.)
(b) Since $\Psi$ is an eigenfunction of $\hat{H}$ with eigenvalue $3^{2} h^{2} / 8 m l^{2}$ [Eq. (2.20)], the measurement must give $9 h^{2} / 8 m l^{2}$.

Up to now we have restricted ourselves to one-dimensional, one-particle systems. The operator formalism developed in the last section allows us to extend our work to threedimensional, many-particle systems. The time-dependent Schrödinger equation for the time development of the state function is postulated to have the form of Eq. (1.13):

\(
\begin{equation}
i \hbar \frac{\partial \Psi}{\partial t}=\hat{H} \Psi \tag{3.42}
\end{equation}
\)

The time-independent Schrödinger equation for the energy eigenfunctions and eigenvalues is

\(
\begin{equation}
\hat{H} \psi=E \psi \tag{3.43}
\end{equation}
\)

which is obtained from (3.42) by taking the potential energy as independent of time and applying the separation-of-variables procedure used to obtain (1.19) from (1.13).

For a one-particle, three-dimensional system, the classical-mechanical Hamiltonian is

\(
\begin{equation}
H=T+V=\frac{1}{2 m}\left(p{x}^{2}+p{y}^{2}+p_{z}^{2}\right)+V(x, y, z) \tag{3.44}
\end{equation}
\)

Introducing the quantum-mechanical operators [Eq. (3.24)], we have for the Hamiltonian operator

\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)+V(x, y, z) \tag{3.45}
\end{equation}
\)

The operator in parentheses in (3.45) is called the Laplacian operator $\nabla^{2}$ (read as "del squared"):

\(
\begin{equation}
\nabla^{2} \equiv \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} \tag{3.46}
\end{equation}
\)

The one-particle, three-dimensional, time-independent Schrödinger equation is then

\(
\begin{equation}
-\frac{\hbar^{2}}{2 m} \nabla^{2} \psi+V \psi=E \psi \tag{3.47}
\end{equation}
\)

Now consider a three-dimensional system with $n$ particles. Let particle $i$ have mass $m{i}$ and coordinates $\left(x{i}, y{i}, z{i}\right)$, where $i=1,2,3, \ldots, n$. The kinetic energy is the sum of the kinetic energies of the individual particles:

\(
T=\frac{1}{2 m{1}}\left(p{x{1}}^{2}+p{y{1}}^{2}+p{z{1}}^{2}\right)+\frac{1}{2 m{2}}\left(p{x{2}}^{2}+p{y{2}}^{2}+p{z{2}}^{2}\right)+\cdots+\frac{1}{2 m{n}}\left(p{x{n}}^{2}+p{y{n}}^{2}+p{z_{n}}^{2}\right)
\)

where $p{x{i}}$ is the $x$ component of the linear momentum of particle $i$, and so on. The kineticenergy operator is

\(
\begin{gather}
\hat{T}=-\frac{\hbar^{2}}{2 m{1}}\left(\frac{\partial^{2}}{\partial x{1}^{2}}+\frac{\partial^{2}}{\partial y{1}^{2}}+\frac{\partial^{2}}{\partial z{1}^{2}}\right)-\cdots-\frac{\hbar^{2}}{2 m{n}}\left(\frac{\partial^{2}}{\partial x{n}^{2}}+\frac{\partial^{2}}{\partial y{n}^{2}}+\frac{\partial^{2}}{\partial z{n}^{2}}\right) \
\hat{T}=-\sum{i=1}^{n} \frac{\hbar^{2}}{2 m{i}} \nabla{i}^{2} \tag{3.48}\
\nabla{i}^{2} \equiv \frac{\partial^{2}}{\partial x{i}^{2}}+\frac{\partial^{2}}{\partial y{i}^{2}}+\frac{\partial^{2}}{\partial z_{i}^{2}} \tag{3.49}
\end{gather}
\)

We shall usually restrict ourselves to cases where the potential energy depends only on the $3 n$ coordinates:

\(
V=V\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)
\)

The Hamiltonian operator for an $n$-particle, three-dimensional system is then

\(
\begin{equation}
\hat{H}=-\sum{i=1}^{n} \frac{\hbar^{2}}{2 m{i}} \nabla{i}^{2}+V\left(x{1}, \ldots, z_{n}\right) \tag{3.50}
\end{equation}
\)

and the time-independent Schrödinger equation is

\(
\begin{equation}
\left[-\sum{i=1}^{n} \frac{\hbar^{2}}{2 m{i}} \nabla{i}^{2}+V\left(x{1}, \ldots, z_{n}\right)\right] \psi=E \psi \tag{3.51}
\end{equation}
\)

where the time-independent wave function is a function of the $3 n$ coordinates of the $n$ particles:

\(
\begin{equation}
\psi=\psi\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right) \tag{3.52}
\end{equation}
\)

The Schrödinger equation (3.51) is a linear partial differential equation.
As an example, consider a system of two particles interacting so that the potential energy is inversely proportional to the distance between them, with $c$ being the proportionality constant. The Schrödinger equation (3.51) becomes

FIGURE 3.1 An infinitesimal box-shaped region located at $x^{\prime}, y^{\prime}, z^{\prime}$.

\(
\begin{gather}
{\left[-\frac{\hbar^{2}}{2 m{1}}\left(\frac{\partial^{2}}{\partial x{1}^{2}}+\frac{\partial^{2}}{\partial y{1}^{2}}+\frac{\partial^{2}}{\partial z{1}^{2}}\right)-\frac{\hbar^{2}}{2 m{2}}\left(\frac{\partial^{2}}{\partial x{2}^{2}}+\frac{\partial^{2}}{\partial y{2}^{2}}+\frac{\partial^{2}}{\partial z{2}^{2}}\right)\right.} \
\left.+\frac{c}{\left[\left(x{1}-x{2}\right)^{2}+\left(y{1}-y{2}\right)^{2}+\left(z{1}-z{2}\right)^{2}\right]^{1 / 2}}\right] \psi=E \psi \tag{3.53}\
\psi=\psi\left(x{1}, y{1}, z{1}, x{2}, y{2}, z{2}\right)
\end{gather}
\)

Although (3.53) looks formidable, we shall solve it in Chapter 6.
For a one-particle, one-dimensional system, the Born postulate [Eq. (1.15)] states that $\left|\Psi\left(x^{\prime}, t\right)\right|^{2} d x$ is the probability of observing the particle between $x^{\prime}$ and $x^{\prime}+d x$ at time $t$, where $x^{\prime}$ is a particular value of $x$. We extend this postulate as follows. For $a$ three-dimensional, one-particle system, the quantity

\(
\begin{equation}
\left|\Psi\left(x^{\prime}, y^{\prime}, z^{\prime}, t\right)\right|^{2} d x d y d z \tag{3.54}
\end{equation}
\)

is the probability of finding the particle in the infinitesimal region of space with its $x$ coordinate lying between $x^{\prime}$ and $x^{\prime}+d x$, its $y$ coordinate lying between $y^{\prime}$ and $y^{\prime}+d y$, and its $z$ coordinate between $z^{\prime}$ and $z^{\prime}+d z$ (Fig. 3.1). Since the total probability of finding the particle is 1 , the normalization condition is

\(
\begin{equation}
\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int_{-\infty}^{\infty}|\Psi(x, y, z, t)|^{2} d x d y d z=1 \tag{3.55}
\end{equation}
\)

For a three-dimensional, n-particle system, we postulate that

\(
\begin{equation}
\left|\Psi\left(x{1}^{\prime}, y{1}^{\prime}, z{1}^{\prime}, x{2}^{\prime}, y{2}^{\prime}, z{2}^{\prime}, \ldots, x{n}^{\prime}, y{n}^{\prime}, z{n}^{\prime}, t\right)\right|^{2} d x{1} d y{1} d z{1} d x{2} d y{2} d z{2} \ldots d x{n} d y{n} d z{n} \tag{3.56}
\end{equation}
\)

is the probability at time $t$ of simultaneously finding particle 1 in the infinitesimal rectangular box-shaped region at $\left(x{1}^{\prime}, y{1}^{\prime}, z{1}^{\prime}\right)$ with edges $d x{1}, d y{1}, d z{1}$, particle 2 in the infinitesimal box-shaped region at $\left(x{2}^{\prime}, y{2}^{\prime}, z{2}^{\prime}\right)$ with edges $d x{2}, d y{2}, d z{2}, \ldots$, and particle $n$ in the infinitesimal box-shaped region at $\left(x{n}^{\prime}, y{n}^{\prime}, z{n}^{\prime}\right)$ with edges $d x{n}, d y{n}, d z{n}$. The total probability of finding all the particles is 1 , and the normalization condition is

\(
\begin{equation}
\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty}|\Psi|^{2} d x{1} d y{1} d z{1} \cdots d x{n} d y{n} d z{n}=1 \tag{3.57}
\end{equation}
\)

It is customary in quantum mechanics to denote integration over the full range of all the coordinates of a system by $\int d \tau$. A shorthand way of writing (3.55) or (3.57) is

\(
\begin{equation}
\int|\Psi|^{2} d \tau=1 \tag{3.58}
\end{equation}
\)

Although (3.58) may look like an indefinite integral, it is understood to be a definite integral. The integration variables and their ranges are understood from the context.

For a stationary state, $|\Psi|^{2}=|\psi|^{2}$, and

\(
\begin{equation}
\int|\psi|^{2} d \tau=1 \tag{3.59}
\end{equation}
\)

For the present, we confine ourselves to one-particle problems. In this section we consider the three-dimensional case of the problem solved in Section 2.2, the particle in a box.

There are many possible shapes for a three-dimensional box. The box we consider is a rectangular parallelepiped with edges of length $a, b$, and $c$. We choose our coordinate system so that one corner of the box lies at the origin and the box lies in the first octant of space (Fig. 3.2). Within the box, the potential energy is zero. Outside the box, it is infinite:

\(
\begin{align}
V(x, y, z) & =0 \text { in the region }\left{\begin{array}{l}
0<x<a \
0<y<b \
0<z<c
\end{array}\right. \tag{3.60}\
V & =\infty \quad \text { elsewhere }
\end{align}
\)

Since the probability for the particle to have infinite energy is zero, the wave function must be zero outside the box. Within the box, the potential-energy operator is zero and the Schrödinger equation (3.47) is

\(
\begin{equation}
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\right)=E \psi \tag{3.61}
\end{equation}
\)

To solve (3.61), we assume that the solution can be written as the product of a function of $x$ alone times a function of $y$ alone times a function of $z$ alone:

\(
\begin{equation}
\psi(x, y, z)=f(x) g(y) h(z) \tag{3.62}
\end{equation}
\)

It might be thought that this assumption throws away solutions that are not of the form (3.62). However, it can be shown that, if we can find solutions of the form (3.62) that satisfy the boundary conditions, then there are no other solutions of the Schrödinger equation that will satisfy the boundary conditions. (For a proof, see G. F. D. Duff and

FIGURE 3.2 Inside the box-shaped region, $V=0$.
D. Naylor, Differential Equations of Applied Mathematics, Wiley, 1966, pp. 257-258.) The method we are using to solve (3.62) is called separation of variables.

From (3.62), we find

\(
\begin{equation}
\frac{\partial^{2} \psi}{\partial x^{2}}=f^{\prime \prime}(x) g(y) h(z), \quad \frac{\partial^{2} \psi}{\partial y^{2}}=f(x) g^{\prime \prime}(y) h(z), \quad \frac{\partial^{2} \psi}{\partial z^{2}}=f(x) g(y) h^{\prime \prime}(z) \tag{3.63}
\end{equation}
\)

Substitution of (3.62) and (3.63) into (3.61) gives

\(
\begin{equation}
-\left(\hbar^{2} / 2 m\right) f^{\prime \prime} g h-\left(\hbar^{2} / 2 m\right) f g^{\prime \prime} h-\left(\hbar^{2} / 2 m\right) f g h^{\prime \prime}-E f g h=0 \tag{3.64}
\end{equation}
\)

Division of this equation by $f g h$ gives

\(
\begin{gather}
-\frac{\hbar^{2} f^{\prime \prime}}{2 m f}-\frac{\hbar^{2} g^{\prime \prime}}{2 m g}-\frac{\hbar^{2} h^{\prime \prime}}{2 m h}-E=0 \tag{3.65}\
-\frac{\hbar^{2} f^{\prime \prime}(x)}{2 m f(x)}=\frac{\hbar^{2} g^{\prime \prime}(y)}{2 m g(y)}+\frac{\hbar^{2} h^{\prime \prime}(z)}{2 m h(z)}+E \tag{3.66}
\end{gather}
\)

Let us define $E_{x}$ as equal to the left side of (3.66):

\(
\begin{equation}
E_{x} \equiv-\hbar^{2} f^{\prime \prime}(x) / 2 m f(x) \tag{3.67}
\end{equation}
\)

The definition (3.67) shows that $E{x}$ is independent of $y$ and $z$. Equation (3.66) shows that $E{x}$ equals $\hbar^{2} g^{\prime \prime}(y) / 2 m g(y)+\hbar^{2} h^{\prime \prime}(z) / 2 m h(z)+E$; therefore, $E{x}$ must be independent of $x$. Being independent of $x, y$, and $z$, the quantity $E{x}$ must be a constant.

Similar to (3.67), we define $E{y}$ and $E{z}$ by

\(
\begin{equation}
E{y} \equiv-\hbar^{2} g^{\prime \prime}(y) / 2 m g(y), \quad E{z} \equiv-\hbar^{2} h^{\prime \prime}(z) / 2 m h(z) \tag{3.68}
\end{equation}
\)

Since $x, y$, and $z$ occur symmetrically in (3.65), the same reasoning that showed $E{x}$ to be a constant shows that $E{y}$ and $E_{z}$ are constants. Substitution of the definitions (3.67) and (3.68) into (3.65) gives

\(
\begin{equation}
E{x}+E{y}+E_{z}=E \tag{3.69}
\end{equation}
\)

Equations (3.67) and (3.68) are

\(
\begin{gather}
\frac{d^{2} f(x)}{d x^{2}}+\frac{2 m}{\hbar^{2}} E{x} f(x)=0 \tag{3.70}\
\frac{d^{2} g(y)}{d y^{2}}+\frac{2 m}{\hbar^{2}} E{y} g(y)=0, \quad \frac{d^{2} h(z)}{d z^{2}}+\frac{2 m}{\hbar^{2}} E_{z} h(z)=0 \tag{3.71}
\end{gather}
\)

We have converted the partial differential equation in three variables into three ordinary differential equations. What are the boundary conditions on (3.70)? Since the wave function vanishes outside the box, continuity of $\psi$ requires that it vanish on the walls of the box. In particular, $\psi$ must be zero on the wall of the box lying in the $y z$ plane, where $x=0$, and it must be zero on the parallel wall of the box, where $x=a$. Therefore, $f(0)=0$ and $f(a)=0$.

Now compare Eq. (3.70) with the Schrödinger equation [Eq. (2.10)] for a particle in a one-dimensional box. The equations are the same in form, with $E_{x}$ in (3.70) corresponding to $E$ in (2.10). Are the boundary conditions the same? Yes, except that we have $x=a$ instead of $x=l$ as the second point where the independent variable vanishes. Thus we can use the work in Section 2.2 to write as the solution [see Eqs. (2.23) and (2.20)]

\(
\begin{aligned}
f(x) & =\left(\frac{2}{a}\right)^{1 / 2} \sin \left(\frac{n{x} \pi x}{a}\right) \
E{x} & =\frac{n{x}^{2} h^{2}}{8 m a^{2}}, \quad n{x}=1,2,3 \ldots
\end{aligned}
\)

The same reasoning applied to the $y$ and $z$ equations gives

\(
\begin{gathered}
g(y)=\left(\frac{2}{b}\right)^{1 / 2} \sin \left(\frac{n{y} \pi y}{b}\right), \quad h(z)=\left(\frac{2}{c}\right)^{1 / 2} \sin \left(\frac{n{z} \pi z}{c}\right) \
E{y}=\frac{n{y}^{2} h^{2}}{8 m b^{2}}, \quad n{y}=1,2,3 \ldots \quad \text { and } \quad E{z}=\frac{n{z}^{2} h^{2}}{8 m c^{2}}, \quad n{z}=1,2,3 \ldots
\end{gathered}
\)

From (3.69), the energy is

\(
\begin{equation}
E=\frac{h^{2}}{8 m}\left(\frac{n{x}^{2}}{a^{2}}+\frac{n{y}^{2}}{b^{2}}+\frac{n_{z}^{2}}{c^{2}}\right) \tag{3.72}
\end{equation}
\)

As with the particle in a one-dimensional box, the ground-state energy is greater than the classical-mechanical, lowest-energy value of zero.

From (3.62), the wave function inside the box is

\(
\begin{equation}
\psi(x, y, z)=\left(\frac{8}{a b c}\right)^{1 / 2} \sin \left(\frac{n{x} \pi x}{a}\right) \sin \left(\frac{n{y} \pi y}{b}\right) \sin \left(\frac{n_{z} \pi z}{c}\right) \tag{3.73}
\end{equation}
\)

The wave function has three quantum numbers, $n{x}, n{y}, n_{z}$. We can attribute this to the three-dimensional nature of the problem. The three quantum numbers vary independently of one another.

In a one-particle, one-dimensional problem such as the particle in a one-dimensional box, the nodes are where $\psi(x)=0$, and solving this equation for $x$, we get points where $\psi=0$. In a one-particle, three-dimensional problem, the nodes are where $\psi(x, y, z)=0$, and solving this equation for $z$, we get solutions of the form $z=f(x, y)$. Each such solution is the equation of a nodal surface in three-dimensional space. For example, for the stationary state with $n{x}=1, n{y}=1, n{z}=2$, the wave function $\psi$ in (3.73) is zero on the surface where $z=c / 2$; this is the equation of a plane that lies parallel to the top and bottom faces of the box and is midway between these faces. Similarly, for the state $n{x}=2$, $n{y}=1, n{z}=1$ the plane $x=a / 2$ is a nodal surface.

Since the $x, y$, and $z$ factors in the wave function are each independently normalized, the wave function is normalized:

\(
\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty}|\psi|^{2} d x d y d z=\int{0}^{a}|f(x)|^{2} d x \int{0}^{b}|g(y)|^{2} d y \int{0}^{c}|h(z)|^{2} d z=1
\)

where we used (Prob. 3.40)

\(
\begin{equation}
\iiint F(x) G(y) H(z) d x d y d z=\int F(x) d x \int G(y) d y \int H(z) d z \tag{3.74}
\end{equation}
\)

What are the dimensions of $\psi(x, y, z)$ in (3.73)? For a one-particle, three-dimensional system, $|\psi|^{2} d x d y d z$ is a probability, and probabilities are dimensionless. Since the dimensions of $d x d y d z$ are length ${ }^{3}, \psi(x, y, z)$ must have dimensions of length ${ }^{-3 / 2}$ to make $|\psi|^{2} d x d y d z$ dimensionless.

Suppose that $a=b=c$. We then have a cube. The energy levels are then

\(
\begin{equation}
E=\left(h^{2} / 8 m a^{2}\right)\left(n{x}^{2}+n{y}^{2}+n_{z}^{2}\right) \tag{3.75}
\end{equation}
\)

Let us tabulate some of the allowed energies of a particle confined to a cube with infinitely strong walls:

$n{x} n{y} n_{z}$	111	211	121	112	122	212	221	113	131	311	222
$E\left(8 m a^{2} / h^{2}\right)$	3	6	6	6	9	9	9	11	11	11	12

FIGURE 3.3 Energies of the lowest few states of a particle in a cubic box.

Note that states with different quantum numbers may have the same energy (Fig. 3.3). For example, the states $\psi{211}, \psi{121}$, and $\psi{112}$ (where the subscripts give the quantum numbers) all have the same energy. However, Eq. (3.73) shows that these three sets of quantum numbers give three different, independent wave functions and therefore do represent different states of the system. When two or more independent wave functions correspond to states with the same energy eigenvalue, the eigenvalue is said to be degenerate. The degree of degeneracy (or, simply, the degeneracy) of an energy level is the number of states that have that energy. Thus the second-lowest energy level of the particle in a cube is threefold degenerate. We got the degeneracy when we made the edges of the box equal. Degeneracy is usually related to the symmetry of the system. Note that the wave functions $\psi{211}, \psi{121}$, and $\psi{112}$ can be transformed into one another by rotating the cubic box. Usually, the bound-state energy levels in one-dimensional problems are nondegenerate.

In the statistical-mechanical evaluation of the molecular partition function of an ideal gas, the translational energy levels of each gas molecule are taken to be the levels of a particle in a three-dimensional rectangular box (the box is the container holding the gas); see Levine, Physical Chemistry, Sections 21.6 and 21.7.

In the free-electron theory of metals, the valence electrons of a nontransition metal are treated as noninteracting particles in a box, the sides of the box being the surfaces of the metal. This approximation, though crude, gives fairly good results for some properties of metals.

For an $n$-fold degenerate energy level, there are $n$ independent wave functions $\psi{1}, \psi{2}, \ldots, \psi_{n}$, each having the same energy eigenvalue $w$ :

\(
\begin{equation}
\hat{H} \psi{1}=w \psi{1}, \quad \hat{H} \psi{2}=w \psi{2}, \quad \ldots, \quad \hat{H} \psi{n}=w \psi{n} \tag{3.76}
\end{equation}
\)

We wish to prove the following important theorem: Every linear combination

\(
\begin{equation}
\phi \equiv c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n} \tag{3.77}
\end{equation}
\)

of the wave functions of a degenerate level with energy eigenvalue $w$ is an eigenfunction of the Hamiltonian operator with eigenvalue $w$. [A linear combination of the functions $\psi{1}, \psi{2}, \ldots, \psi_{n}$ is defined as a function of the form (3.77) where the $c$ 's are constants, some of which might be zero.] To prove this theorem, we must show that $\hat{H} \phi=w \phi$ or

\(
\begin{equation}
\hat{H}\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right)=w\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right) \tag{3.78}
\end{equation}
\)

Since $\hat{H}$ is a linear operator, we can apply Eq. (3.9) $n-1$ times to the left side of (3.78) to get

\(
\hat{H}\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right)=\hat{H}\left(c{1} \psi{1}\right)+\hat{H}\left(c{2} \psi{2}\right)+\cdots+\hat{H}\left(c{n} \psi{n}\right)
\)

Use of Eqs. (3.10) and (3.76) gives

\(
\begin{aligned}
\hat{H}\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right) & =c{1} \hat{H} \psi{1}+c{2} \hat{H} \psi{2}+\cdots+c{n} \hat{H} \psi{n} \
& =c{1} w \psi{1}+c{2} w \psi{2}+\cdots+c{n} w \psi{n} \
\hat{H}\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right) & =w\left(c{1} \psi{1}+c{2} \psi{2}+\cdots+c{n} \psi{n}\right)
\end{aligned}
\)

which completes the proof.
For example, the stationary-state wave functions $\psi{211}, \psi{121}$, and $\psi{112}$ for the particle in a cubic box are degenerate, and the linear combination $c{1} \psi{211}+c{2} \psi{121}+c{3} \psi{112}$ is an eigenfunction of the particle-in-a-cubic-box Hamiltonian with eigenvalue $6 h^{2} / 8 m a^{2}$, the same eigenvalue as for each of $\psi{211}, \psi{121}$, and $\psi{112}$.

Note that the linear combination $c{1} \psi{1}+c{2} \psi{2}$ is not an eigenfunction of $\hat{H}$ if $\psi{1}$ and $\psi{2}$ correspond to different energy eigenvalues $\left(\hat{H} \psi{1}=E{1} \psi{1}\right.$ and $\hat{H} \psi{2}=E{2} \psi{2}$ with $E{1} \neq E{2}$ ).

Since any linear combination of the wave functions corresponding to a degenerate energy level is an eigenfunction of $\hat{H}$ with the same eigenvalue, we can construct an infinite number of different wave functions for any degenerate energy level. Actually, we are only interested in eigenfunctions that are linearly independent. The $n$ functions $f{1}, \ldots, f{n}$ are said to be linearly independent if the equation $c{1} f{1}+\cdots+c{n} f{n}=0$ can only be satisfied with all the constants $c{1}, \ldots, c{n}$ equal to zero. This means that no member of a set of linearly independent functions can be expressed as a linear combination of the remaining members. For example, the functions $f{1}=3 x, f{2}=5 x^{2}-x$, $f{3}=x^{2}$ are not linearly independent, since $f{2}=5 f{3}-\frac{1}{3} f{1}$. The functions $g{1}=1$, $g{2}=x, g_{3}=x^{2}$ are linearly independent, since none of them can be written as a linear combination of the other two.

The degree of degeneracy of an energy level is equal to the number of linearly independent wave functions corresponding to that value of the energy. The one-dimensional free-particle wave functions (2.30) are linear combinations of two linearly independent functions that are each an eigenfunction with the same energy eigenvalue $E$. Thus each such energy eigenvalue (except $E=0$ ) is doubly degenerate (meaning that the degree of degeneracy is two).

It was pointed out in Section 3.3 that, when the state function $\Psi$ is not an eigenfunction of the operator $\hat{B}$, a measurement of $B$ will give one of a number of possible values (the eigenvalues of $\hat{B}$ ). We now consider the average value of the property $B$ for a system whose state is $\Psi$.

To find the average value of $B$ experimentally, we take many identical, noninteracting systems each in the same state $\Psi$ and we measure $B$ in each system. The average value of $B$, symbolized by $\langle B\rangle$, is defined as the arithmetic mean of the observed values $b{1}$, $b{2}, \ldots, b_{N}$ :

\(
\begin{equation}
\langle B\rangle=\frac{\sum{j=1}^{N} b{j}}{N} \tag{3.79}
\end{equation}
\)

where $N$, the number of systems, is extremely large.
Instead of summing over the observed values of $B$, we can sum over all the possible values of $B$, multiplying each possible value by the number of times it is observed, to get the equivalent expression

\(
\begin{equation}
\langle B\rangle=\frac{\sum{b} n{b} b}{N} \tag{3.80}
\end{equation}
\)

where $n_{b}$ is the number of times the value $b$ is observed. An example will make this clear. Suppose a class of nine students takes a quiz that has five questions and the students receive these grades: $0,20,20,60,60,80,80,80,100$. Calculating the average grade according to (3.79), we have

\(
\frac{1}{N} \sum{j=1}^{N} b{j}=\frac{0+20+20+60+60+80+80+80+100}{9}=56
\)

To calculate the average grade according to (3.80), we sum over the possible grades: 0,20 , $40,60,80,100$. We have

\(
\frac{1}{N} \sum{b} n{b} b=\frac{1(0)+2(20)+0(40)+2(60)+3(80)+1(100)}{9}=56
\)

Equation (3.80) can be written as

\(
\langle B\rangle=\sum{b}\left(\frac{n{b}}{N}\right) b
\)

Since $N$ is very large, $n{b} / N$ is the probability $P{b}$ of observing the value $b$, and

\(
\begin{equation}
\langle B\rangle=\sum{b} P{b} b \tag{3.81}
\end{equation}
\)

Now consider the average value of the $x$ coordinate for a one-particle, one-dimensional system in the state $\Psi(x, t)$. The $x$ coordinate takes on a continuous range of values, and the probability of observing the particle between $x$ and $x+d x$ is $|\Psi|^{2} d x$. The summation over the infinitesimal probabilities is equivalent to an integration over the full range of $x$, and (3.81) becomes

\(
\begin{equation}
\langle x\rangle=\int_{-\infty}^{\infty} x|\Psi(x, t)|^{2} d x \tag{3.82}
\end{equation}
\)

For the one-particle, three-dimensional case, the probability of finding the particle in the volume element at point $(x, y, z)$ with edges $d x, d y, d z$ is

\(
\begin{equation}
|\Psi(x, y, z, t)|^{2} d x d y d z \tag{3.83}
\end{equation}
\)

If we want the probability that the particle is between $x$ and $x+d x$, we must integrate (3.83) over all possible values of $y$ and $z$, since the particle can have any values for its $y$ and $z$ coordinates while its $x$ coordinate lies between $x$ and $x+d x$. Hence, in the threedimensional case (3.82) becomes

\(
\begin{align}
& \langle x\rangle=\int{-\infty}^{\infty}\left[\int{-\infty}^{\infty} \int{-\infty}^{\infty}|\Psi(x, y, z, t)|^{2} d y d z\right] x d x \
& \langle x\rangle=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty}|\Psi(x, y, z, t)|^{2} x d x d y d z \tag{3.84}
\end{align}
\)

Now consider the average value of some physical property $B(x, y, z)$ that is a function of the particle's coordinates. An example is the potential energy $V(x, y, z)$. The same reasoning that gave Eq. (3.84) yields

\(
\begin{gather}
\langle B(x, y, z)\rangle=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty}|\Psi(x, y, z, t)|^{2} B(x, y, z) d x d y d z \tag{3.85}\
\langle B(x, y, z)\rangle=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} \Psi B \Psi d x d y d z \tag{3.86}
\end{gather*}
\)

The form (3.86) might seem like a bit of whimsy, since it is no different from (3.85). In a moment we shall see its significance.

In general, the property $B$ depends on both coordinates and momenta:

\(
B=B\left(x, y, z, p{x}, p{y}, p_{z}\right)
\)

for the one-particle, three-dimensional case. How do we find the average value of $B$ ? We postulate that $\langle B\rangle$ for a system in state $\Psi$ is

\(
\begin{align}
& \langle B\rangle=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int_{-\infty}^{\infty} \Psi B\left(x, y, z, \frac{\hbar}{i} \frac{\partial}{\partial x}, \frac{\hbar}{i} \frac{\partial}{\partial y}, \frac{\hbar}{i} \frac{\partial}{\partial z}\right) \Psi d x d y d z \
& \langle B\rangle=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int_{-\infty}^{\infty} \Psi \hat{B} \Psi d x d y d z \tag{3.87}
\end{align}
\)

where $\hat{B}$ is the quantum-mechanical operator for the property $B$. [Later we shall provide some justification for this postulate by using (3.87) to show that the time-dependent Schrödinger equation reduces to Newton's second law in the transition from quantum to classical mechanics; see Prob. 7.59.] For the $n$-particle case, we postulate that

\(
\begin{equation}
\langle B\rangle=\int \Psi \hat{B} \Psi d \tau \tag{3.88}
\end{}
\)

where $\int d \tau$ denotes a definite integral over the full range of the $3 n$ coordinates. The state function in (3.88) must be normalized, since we took $\Psi^{} \Psi$ as the probability density. It is important to have the operator properly sandwiched between $\Psi^{}$ and $\Psi$. The quantities $\hat{B} \Psi^{} \Psi$ and $\Psi^{} \Psi \hat{B}$ are not the same as $\Psi \hat{B} \Psi$, unless $B$ is a function of coordinates only. In $\int \Psi^{} \hat{B} \Psi d \tau$, one first operates on $\Psi$ with $\hat{B}$ to produce a new function $\hat{B} \Psi$, which is then multiplied by $\Psi^{*}$; one then integrates over all space to produce a number, which is $\langle B\rangle$.

For a stationary state, we have [Eq. (1.20)]

\(
\Psi^{} \hat{B} \Psi=e^{i E t / \hbar} \psi^{} \hat{B} e^{-i E t / \hbar} \psi=e^{0} \psi^{} \hat{B} \psi=\psi^{} \hat{B} \psi
\)

since $\hat{B}$ contains no time derivatives and does not affect the time factor in $\Psi$. Hence, for a stationary state,

\(
\begin{equation}
\langle B\rangle=\int \psi^{} \hat{B} \psi d \tau \tag{3.89}
\end{}
\)

Thus, if $\hat{B}$ is time-independent, then $\langle B\rangle$ is time-independent in a stationary state.
Consider the special case where $\Psi$ is an eigenfunction of $\hat{B}$. When $\hat{B} \Psi=k \Psi$, Eq. (3.88) becomes

\(
\langle B\rangle=\int \Psi^{} \hat{B} \Psi d \tau=\int \Psi^{} k \Psi d \tau=k \int \Psi^{*} \Psi d \tau=k
\)

since $\Psi$ is normalized. This result is reasonable, since when $\hat{B} \Psi=k \Psi, k$ is the only possible value we can find for $B$ when we make a measurement (Section 3.3).

The following properties of average values are easily proved from Eq. (3.88) (see Prob. 3.49):

\(
\begin{equation}
\langle A+B\rangle=\langle A\rangle+\langle B\rangle \quad\langle c B\rangle=c\langle B\rangle \tag{3.90}
\end{equation}
\)

where $A$ and $B$ are any two properties and $c$ is a constant. However, the average value of a product need not equal the product of the average values: $\langle A B\rangle \neq\langle A\rangle\langle B\rangle$.

The term expectation value is often used instead of average value. The expectation value is not necessarily one of the possible values we might observe.

EXAMPLE

Find $\langle x\rangle$ and $\left\langle p_{x}\right\rangle$ for the ground stationary state of a particle in a three-dimensional box.
Substitution of the stationary-state wave function $\psi=f(x) g(y) h(z)$ [Eq. (3.62)] into the average-value postulate (3.89) gives

\(
\langle x\rangle=\int \psi^{} \hat{x} \psi d \tau=\int{0}^{c} \int{0}^{b} \int_{0}^{a} f^{} g^{} h^{} x f g h d x d y d z
\)

since $\psi=0$ outside the box. Use of (3.74) gives

\(
\langle x\rangle=\int{0}^{a} x|f(x)|^{2} d x \int{0}^{b}|g(y)|^{2} d y \int{0}^{c}|h(z)|^{2} d z=\int{0}^{a} x|f(x)|^{2} d x
\)

since $g(y)$ and $h(z)$ are each normalized. For the ground state, $n_{x}=1$ and $f(x)=(2 / a)^{1 / 2} \sin (\pi x / a)$. So

\(
\begin{equation}
\langle x\rangle=\frac{2}{a} \int_{0}^{a} x \sin ^{2}\left(\frac{\pi x}{a}\right) d x=\frac{a}{2} \tag{3.91}
\end{equation}
\)

where the Appendix integral (A.3) was used. A glance at Fig. 2.4 shows that this result is reasonable.

Also,

\(
\begin{align}
& \left\langle p_{x}\right\rangle=\int \psi^{} \hat{p}{x} \psi d \tau=\int{0}^{c} \int{0}^{b} \int{0}^{a} f^{} g^{} h^{} \frac{\hbar}{i} \frac{\partial}{\partial x}[f(x) g(y) h(z)] d x d y d z \
& \left\langle p{x}\right\rangle=\frac{\hbar}{i} \int{0}^{a} f^{}(x) f^{\prime}(x) d x \int{0}^{b}|g(y)|^{2} d y \int{0}^{c}|h(z)|^{2} d z \
& \left\langle p{x}\right\rangle=\frac{\hbar}{i} \int{0}^{a} f(x) f^{\prime}(x) d x=\left.\frac{\hbar}{2 i} f^{2}(x)\right|_{0} ^{a}=0 \tag{3.92}
\end{align*}
\)

where the boundary conditions $f(0)=0$ and $f(a)=0$ were used. The result (3.92) is reasonable since the particle is equally likely to be headed in the $+x$ or $-x$ direction.
EXERCISE Find $\left\langle p_{x}^{2}\right\rangle$ for the ground state of a particle in a three-dimensional box. (Answer: $h^{2} / 4 l^{2}$.)

In solving the particle in a box, we required $\psi$ to be continuous. We now discuss other requirements the wave function must satisfy.

Since $|\psi|^{2} d \tau$ is a probability, we want to be able to normalize the wave function by choosing a suitable normalization constant $N$ as a multiplier of the wave function. If $\psi$ is unnormalized and $N \psi$ is normalized, the normalization condition (3.59) gives

\(
\begin{align}
1 & =\int|N \psi|^{2} d \tau=|N|^{2} \int|\psi|^{2} d \tau \
|N| & =\left(\int|\psi|^{2} d \tau\right)^{-1 / 2} \tag{3.93}
\end{align}
\)

FIGURE 3.4 Function (a) is continuous, and its first derivative is continuous. Function (b) is continuous, but its first derivative has a discontinuity. Function (c) is discontinuous.

The definite integral $\int|\psi|^{2} d \tau$ will equal zero only if the function $\psi$ is zero everywhere. However, $\psi$ cannot be zero everywhere (this would mean no particles were present), so this integral is never zero. If $\int|\psi|^{2} d \tau$ is infinite, then the magnitude $|N|$ of the normalization constant is zero and $\psi$ cannot be normalized. We can normalize $\psi$ if and only if $\int|\psi|^{2} d \tau$ has a finite, rather than infinite, value. If the integral over all space $\int|\psi|^{2} d \tau$ is finite, $\psi$ is said to be quadratically integrable. Thus we generally demand that $\psi$ be quadratically integrable. The important exception is a particle that is not bound. Thus the wave functions for the unbound states of the particle in a well (Section 2.4) and for a free particle are not quadratically integrable.

Since $\psi^{} \psi$ is the probability density, it must be single-valued. It would be embarrassing if our theory gave two different values for the probability of finding a particle at a certain point. If we demand that $\psi$ be single-valued, then surely $\psi^{} \psi$ will be single-valued. It is possible to have $\psi$ multivalued [for example, $\psi(q)=-1,+1, i$ ] and still have $\psi^{*} \psi$ single-valued. We will, however, demand single-valuedness for $\psi$.

In addition to demanding that $\psi$ be continuous, we usually also require that all the partial derivatives $\partial \psi / \partial x, \partial \psi / \partial y$, and so on, be continuous. (See Fig. 3.4.) Referring back to Section 2.2, however, we note that for the particle in a box, $d \psi / d x$ is discontinuous at the walls of the box; $\psi$ and $d \psi / d x$ are zero everywhere outside the box; but from Eq. (2.23) we see that $d \psi / d x$ does not become zero at the walls. The discontinuity in $\psi^{\prime}$ is due to the infinite jump in potential energy at the walls of the box. For a box with walls of finite height, $\psi^{\prime}$ is continuous at the walls (Section 2.4).

In line with the requirement of quadratic integrability, it is sometimes stated that the wave function must be finite everywhere, including infinity. However, this is usually a much stronger requirement than quadratic integrability. In fact, it turns out that some of the relativistic wave functions for the hydrogen atom are infinite at the origin but are quadratically integrable. Occasionally, one encounters nonrelativistic wave functions that are infinite at the origin [L. D. Landau and E. M. Lifshitz, Quantum Mechanics, 3rd ed. (1977), Section 35]. Thus the fundamental requirement is quadratic integrability, rather than finiteness.

We require that the eigenfunctions of any operator representing a physical quantity meet the above requirements. A function meeting these requirements is said to be well-behaved.

So far we have considered only cases where the potential energy $V(x)$ is a constant. This makes the Schrödinger equation a second-order linear homogeneous differential equation with constant coefficients, which we know how to solve. For cases in which $V$ varies with $x$, a useful approach is to try a power-series solution of the Schrödinger equation.

To illustrate the method, consider the differential equation

\(
\begin{equation}
y^{\prime \prime}(x)+c^{2} y(x)=0 \tag{4.1}
\end{equation}
\)

where $c^{2}>0$. Of course, this differential equation has constant coefficients, but we can solve it with the power-series method if we want. Let us first find the solution by using the auxiliary equation, which is $s^{2}+c^{2}=0$. We find $s= \pm i c$. Recalling the work in Section 2.2 [Eqs. (2.10) and (4.1) are the same], we get trigonometric solutions when the roots of the auxiliary equation are pure imaginary:

\(
\begin{equation}
y=A \cos c x+B \sin c x \tag{4.2}
\end{equation}
\)

where $A$ and $B$ are the constants of integration. A different form of (4.2) is

\(
\begin{equation}
y=D \sin (c x+e) \tag{4.3}
\end{equation}
\)

where $D$ and $e$ are arbitrary constants. Using the formula for the sine of the sum of two angles, we can show that (4.3) is equivalent to (4.2).

Now let us solve (4.1) using the power-series method. We start by assuming that the solution can be expanded in a Taylor series (see Prob. 4.1) about $x=0$; that is, we assume that

\(
\begin{equation}
y(x)=\sum{n=0}^{\infty} a{n} x^{n}=a{0}+a{1} x+a{2} x^{2}+a{3} x^{3}+\cdots \tag{4.4}
\end{equation}
\)

where the $a$ 's are constant coefficients to be determined so as to satisfy (4.1). Differentiating (4.4), we have

\(
\begin{equation}
y^{\prime}(x)=a{1}+2 a{2} x+3 a{3} x^{2}+\cdots=\sum{n=1}^{\infty} n a_{n} x^{n-1} \tag{4.5}
\end{equation}
\)

where we assumed that term-by-term differentiation is valid for the series. (This is not always true for infinite series.) For $y^{\prime \prime}$, we have

\(
\begin{equation}
y^{\prime \prime}(x)=2 a{2}+3(2) a{3} x+\cdots=\sum{n=2}^{\infty} n(n-1) a{n} x^{n-2} \tag{4.6}
\end{equation}
\)

Substituting (4.4) and (4.6) into (4.1), we get

\(
\begin{equation}
\sum{n=2}^{\infty} n(n-1) a{n} x^{n-2}+\sum{n=0}^{\infty} c^{2} a{n} x^{n}=0 \tag{4.7}
\end{equation}
\)

We want to combine the two sums in (4.7). Provided certain conditions are met, we can add two infinite series term by term to get their sum:

\(
\begin{equation}
\sum{j=0}^{\infty} b{j} x^{j}+\sum{j=0}^{\infty} c{j} x^{j}=\sum{j=0}^{\infty}\left(b{j}+c_{j}\right) x^{j} \tag{4.8}
\end{equation}
\)

To apply (4.8) to the two sums in (4.7), we want the limits in each sum to be the same and the powers of $x$ to be the same. We therefore change the summation index in the first sum in (4.7), defining $k$ as $k \equiv n-2$. The limits $n=2$ to $\infty$ correspond to $k=0$ to $\infty$ and use of $n=k+2$ gives
$\sum{n=2}^{\infty} n(n-1) a{n} x^{n-2}=\sum{k=0}^{\infty}(k+2)(k+1) a{k+2} x^{k}=\sum{n=0}^{\infty}(n+2)(n+1) a{n+2} x^{n}$
The last equality in (4.9) is valid because the summation index is a dummy variable; it makes no difference what letter we use to denote this variable. For example, the sums $\sum{i=1}^{3} c{i} x^{i}$ and $\sum{m=1}^{3} c{m} x^{m}$ are equal because only the dummy variables in the two sums differ. This equality is easy to see if we write out the sums:

\(
\sum{i=1}^{3} c{i} x^{i}=c{1} x+c{2} x^{2}+c{3} x^{3} \quad \text { and } \quad \sum{m=1}^{3} c{m} x^{m}=c{1} x+c{2} x^{2}+c{3} x^{3}
\)

In the last equality in (4.9), we simply changed the symbol denoting the summation index from $k$ to $n$.

The integration variable in a definite integral is also a dummy variable, since the value of a definite integral is unaffected by what letter we use for this variable:

\(
\begin{equation}
\int{a}^{b} f(x) d x=\int{a}^{b} f(t) d t \tag{4.10}
\end{equation}
\)

Using (4.9) in (4.7), we find, after applying (4.8), that

\(
\begin{equation}
\sum{n=0}^{\infty}\left[(n+2)(n+1) a{n+2}+c^{2} a_{n}\right] x^{n}=0 \tag{4.11}
\end{equation}
\)

If (4.11) is to be true for all values of $x$, then the coefficient of each power of $x$ must vanish. To see this, consider the equation

\(
\begin{equation}
\sum{j=0}^{\infty} b{j} x^{j}=0 \tag{4.12}
\end{equation}
\)

Putting $x=0$ in (4.12) shows that $b{0}=0$. Taking the first derivative of (4.12) with respect to $x$ and then putting $x=0$ shows that $b{1}=0$. Taking the $n$th derivative and putting $x=0$ gives $b_{n}=0$. Thus, from (4.11), we have

\(
\begin{gather}
(n+2)(n+1) a{n+2}+c^{2} a{n}=0 \tag{4.13}\
a{n+2}=-\frac{c^{2}}{(n+1)(n+2)} a{n} \tag{4.14}
\end{gather}
\)

Equation (4.14) is a recursion relation. If we know the value of $a{0}$, we can use (4.14) to find $a{2}, a{4}, a{6}, \ldots$ If we know $a{1}$, we can find $a{3}, a{5}, a{7}, \ldots$. Since there is no restriction on the values of $a{0}$ and $a{1}$, they are arbitrary constants, which we denote by $A$ and $B c$ :

\(
\begin{equation}
a{0}=A, \quad a{1}=B c \tag{4.15}
\end{equation}
\)

Using (4.14), we find for the coefficients

\(
\begin{align}
a{0}=A, \quad a{2} & =-\frac{c^{2} A}{1 \cdot 2}, \quad a{4}=\frac{c^{4} A}{4 \cdot 3 \cdot 2 \cdot 1}, \quad a{6}=-\frac{c^{6} A}{6!}, \ldots \
a{2 k} & =(-1)^{k} \frac{c^{2 k} A}{(2 k)!}, \quad k=0,1,2,3, \ldots \tag{4.16}\
a{1}=B c, \quad a{3} & =-\frac{c^{3} B}{2 \cdot 3}, \quad a{5}=\frac{c^{5} B}{5 \cdot 4 \cdot 3 \cdot 2}, \quad a{7}=-\frac{c^{7} B}{7!}, \ldots \
a{2 k+1} & =(-1)^{k} \frac{c^{2 k+1} B}{(2 k+1)!}, \quad k=0,1,2, \ldots \tag{4.17}
\end{align}
\)

From (4.4), (4.16), and (4.17), we have

\(
\begin{align}
& y=\sum{n=0}^{\infty} a{n} x^{n}=\sum{n=0,2,4, \ldots}^{\infty} a{n} x^{n}+\sum{n=1,3,5, \ldots}^{\infty} a{n} x^{n} \tag{4.18}\
& y=A \sum{k=0}^{\infty}(-1)^{k} \frac{c^{2 k} x^{2 k}}{(2 k)!}+B \sum{k=0}^{\infty}(-1)^{k} \frac{c^{2 k+1} x^{2 k+1}}{(2 k+1)!} \tag{4.19}
\end{align}
\)

The two series in (4.19) are the Taylor series for $\cos c x$ and $\sin c x$ (Prob. 4.2). Hence, in agreement with (4.2), we have $y=A \cos c x+B \sin c x$.

In this section we will increase our quantum-mechanical repertoire by solving the Schrödinger equation for the one-dimensional harmonic oscillator. This system is important as a model for molecular vibrations.

Classical-Mechanical Treatment

Before looking at the wave mechanics of the harmonic oscillator, we review the classical treatment. We have a single particle of mass $m$ attracted toward the origin by a force proportional to the particle's displacement from the origin:

\(
\begin{equation}
F_{x}=-k x \tag{4.20}
\end{equation}
\)

The proportionality constant $k$ is called the force constant. $F_{x}$ is the $x$ component of the force on the particle. This is also the total force in this one-dimensional problem. Equation (4.20) is obeyed by a particle attached to a spring, provided the spring is not stretched greatly from its equilibrium position.

Newton's second law, $F=m a$, gives

\(
\begin{equation}
-k x=m \frac{d^{2} x}{d t^{2}} \tag{4.21}
\end{equation}
\)

where $t$ is the time. Equation (4.21) is the same as Eq. (4.1) with $c^{2}=k / m$; hence the solution is [Eq. (4.3) with $c=(k / m)^{1 / 2}$ ]

\(
\begin{equation}
x=A \sin (2 \pi \nu t+b) \tag{4.22}
\end{equation}
\)

where $A$ (the amplitude of the vibration) and $b$ are the integration constants, and the vibration frequency $\nu$ is

\(
\begin{equation}
\nu=\frac{1}{2 \pi}\left(\frac{k}{m}\right)^{1 / 2} \tag{4.23}
\end{equation}
\)

Since the sine function has maximum and minimum values of 1 and -1 , respectively, $x$ in (4.22) oscillates between $A$ and $-A$. The sine function repeats itself every $2 \pi$ radians, and the time needed for one complete oscillation (called the period) is the time it takes for the argument of the sine function to increase by $2 \pi$. At time $t+1 / \nu$, the argument of the sine function is $2 \pi \nu(t+1 / \nu)+b=2 \pi \nu t+2 \pi+b$, which is $2 \pi$ greater than the argument at time $t$, so the period is $1 / \nu$. The reciprocal of the period is the number of vibrations per unit time (the vibrational frequency), and so the frequency is $\nu$.

Now consider the energy. The potential energy $V$ is related to the components of force in the three-dimensional case by

\(
\begin{equation}
F{x}=-\frac{\partial V}{\partial x}, \quad F{y}=-\frac{\partial V}{\partial y}, \quad F_{z}=-\frac{\partial V}{\partial z} \tag{4.24}
\end{equation}
\)

Equation (4.24) is the definition of potential energy. Since this is a one-dimensional problem, we have [Eq. (1.12)]

\(
\begin{equation}
F_{x}=-\frac{d V}{d x}=-k x \tag{4.25}
\end{equation}
\)

Integration of (4.25) gives $V=\int k x d x=\frac{1}{2} k x^{2}+C$, where $C$ is a constant. The potential energy always has an arbitrary additive constant. Choosing $C=0$, we have [Eq. (4.23)]

\(
\begin{gather}
V=\frac{1}{2} k x^{2} \tag{4.26}\
V=2 \pi^{2} \nu^{2} m x^{2} \tag{4.27}
\end{gather}
\)

The graph of $V(x)$ is a parabola (Fig. 4.5). The kinetic energy $T$ is

\(
\begin{equation}
T=\frac{1}{2} m(d x / d t)^{2} \tag{4.28}
\end{equation}
\)

and can be found by differentiating (4.22) with respect to $t$. Adding $T$ and $V$, one finds for the total energy (Prob. 4.4)

\(
\begin{equation}
E=T+V=\frac{1}{2} k A^{2}=2 \pi^{2} \nu^{2} m A^{2} \tag{4.29}
\end{equation}
\)

where the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$ was used.

According to (4.22), the classical harmonic oscillator vibrates back and forth between $x=A$ and $x=-A$. These two points are the turning points for the motion. The particle has zero speed at these points, and the speed increases to a maximum at $x=0$, where the potential energy is zero and the energy is all kinetic energy. The classical harmonic oscillator spends more time in each of the regions near $x=A$ and $x=-A$ (where it is moving the slowest) than it does in the region near $x=0$. Problem 4.18 works out the probability density for finding the classical harmonic oscillator at various locations. (Interestingly, this probability density becomes infinite at the turning points.)

Quantum-Mechanical Treatment

The harmonic-oscillator Hamiltonian operator is [Eqs. (3.27) and (4.27)]

\(
\begin{equation}
\hat{H}=\hat{T}+\hat{V}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+2 \pi^{2} \nu^{2} m x^{2}=-\frac{\hbar^{2}}{2 m}\left(\frac{d^{2}}{d x^{2}}-\alpha^{2} x^{2}\right) \tag{4.30}
\end{equation}
\)

where, to save time in writing, $\alpha$ was defined as

\(
\begin{equation}
\alpha \equiv 2 \pi \nu m / \hbar \tag{4.31}
\end{equation}
\)

The Schrödinger equation $\hat{H} \psi=E \psi$ reads, after multiplication by $2 m / \hbar^{2}$,

\(
\begin{equation}
\frac{d^{2} \psi}{d x^{2}}+\left(2 m E \hbar^{-2}-\alpha^{2} x^{2}\right) \psi=0 \tag{4.32}
\end{equation}
\)

We might now attempt a power-series solution of (4.32). If we do now try a power series for $\psi$ of the form (4.4), we will find that it leads to a three-term recursion relation, which is harder to deal with than a two-term recursion relation like Eq. (4.14). We therefore modify the form of (4.32) so as to get a two-term recursion relation when we try a series solution. A substitution that will achieve this purpose is (see Prob. 4.22) $f(x) \equiv e^{\alpha x^{2} / 2} \psi(x)$. Thus

\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} f(x) \tag{4.33}
\end{equation}
\)

This equation is simply the definition of a new function $f(x)$ that replaces $\psi(x)$ as the unknown function to be solved for. (We can make any substitution we please in a differential equation.) Differentiating (4.33) twice, we have

\(
\begin{equation}
\psi^{\prime \prime}=e^{-\alpha x^{2} / 2}\left(f^{\prime \prime}-2 \alpha x f^{\prime}-\alpha f+\alpha^{2} x^{2} f\right) \tag{4.34}
\end{equation}
\)

Substituting (4.33) and (4.34) into (4.32), we find

\(
\begin{equation}
f^{\prime \prime}(x)-2 \alpha x f^{\prime}(x)+\left(2 m E \hbar^{-2}-\alpha\right) f(x)=0 \tag{4.35}
\end{equation}
\)

Now we try a series solution for $f(x)$ :

\(
\begin{equation}
f(x)=\sum{n=0}^{\infty} c{n} x^{n} \tag{4.36}
\end{equation}
\)

Assuming the validity of term-by-term differentiation of (4.36), we get

\(
\begin{equation}
f^{\prime}(x)=\sum{n=1}^{\infty} n c{n} x^{n-1}=\sum{n=0}^{\infty} n c{n} x^{n-1} \tag{4.37}
\end{equation}
\)

[The first term in the second sum in (4.37) is zero.] Also,

\(
f^{\prime \prime}(x)=\sum{n=2}^{\infty} n(n-1) c{n} x^{n-2}=\sum{j=0}^{\infty}(j+2)(j+1) c{j+2} x^{j}=\sum{n=0}^{\infty}(n+2)(n+1) c{n+2} x^{n}
\)

where we made the substitution $j=n-2$ and then changed the summation index from $j$ to $n$. [See Eq. (4.9).] Substitution into (4.35) gives

\(
\begin{align}
& \sum{n=0}^{\infty}(n+2)(n+1) c{n+2} x^{n}-2 \alpha \sum{n=0}^{\infty} n c{n} x^{n}+\left(2 m E \hbar^{-2}-\alpha\right) \sum{n=0}^{\infty} c{n} x^{n}=0 \
& \sum{n=0}^{\infty}\left[(n+2)(n+1) c{n+2}-2 \alpha n c{n}+\left(2 m E \hbar^{-2}-\alpha\right) c{n}\right] x^{n}=0 \tag{4.38}
\end{align}
\)

Setting the coefficient of $x^{n}$ equal to zero [for the same reason as in Eq. (4.11)], we have

\(
\begin{equation}
c{n+2}=\frac{\alpha+2 \alpha n-2 m E \hbar^{-2}}{(n+1)(n+2)} c{n} \tag{4.39}
\end{equation}
\)

which is the desired two-term recursion relation. Equation (4.39) has the same form as (4.14), in that knowing $c{n}$ we can calculate $c{n+2}$. We thus have two arbitrary constants: $c{0}$ and $c{1}$. If we set $c_{1}$ equal to zero, then we will have as a solution a power series containing only even powers of $x$, multiplied by the exponential factor:

\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} f(x)=e^{-\alpha x^{2} / 2} \sum{n=0,2,4, \ldots}^{\infty} c{n} x^{n}=e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l} x^{2 l} \tag{4.40}
\end{equation}
\)

If we set $c_{0}$ equal to zero, we get another independent solution:

\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} \sum{n=1,3, \ldots}^{\infty} c{n} x^{n}=e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l+1} x^{2 l+1} \tag{4.41}
\end{equation}
\)

The general solution of the Schrödinger equation is a linear combination of these two independent solutions [recall Eq. (2.4)]:

\(
\begin{equation}
\psi=A e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l+1} x^{2 l+1}+B e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l} x^{2 l} \tag{4.42}
\end{equation}
\)

where $A$ and $B$ are arbitrary constants.
We now must see if the boundary conditions on the wave function lead to any restrictions on the solution. To see how the two infinite series behave for large $x$, we examine the ratio of successive coefficients in each series. The ratio of the coefficient of $x^{2 l+2}$ to that of $x^{2 l}$ in the second series is [set $n=2 l$ in (4.39)]

\(
\frac{c{2 l+2}}{c{2 l}}=\frac{\alpha+4 \alpha l-2 m E \hbar^{-2}}{(2 l+1)(2 l+2)}
\)

Assuming that for large values of $x$ the later terms in the series are the dominant ones, we look at this ratio for large values of $l$ :

\(
\begin{equation}
\frac{c{2 l+2}}{c{2 l}} \sim \frac{4 \alpha l}{(2 l)(2 l)}=\frac{\alpha}{l} \quad \text { for } l \text { large } \tag{4.43}
\end{equation}
\)

Setting $n=2 l+1$ in (4.39), we find that for large $l$ the ratio of successive coefficients in the first series is also $\alpha / l$. Now consider the power-series expansion for the function $e^{\alpha x^{2}}$. Using (Prob. 4.3)

\(
\begin{equation}
e^{z}=\sum_{n=0}^{\infty} \frac{z^{n}}{n!}=1+z+\frac{z^{2}}{2!}+\cdots \tag{4.44}
\end{equation}
\)

FIGURE 4.1 Lowest five energy levels for the one-dimensional harmonic oscillator.
we get

\(
e^{\alpha x^{2}}=1+\alpha x^{2}+\cdots+\frac{\alpha^{l} x^{2 l}}{l!}+\frac{\alpha^{l+1} x^{2 l+2}}{(l+1)!}+\cdots
\)

The ratio of the coefficients of $x^{2 l+2}$ and $x^{2 l}$ in this series is

\(
\frac{\alpha^{l+1}}{(l+1)!} \div \frac{\alpha^{l}}{l!}=\frac{\alpha}{l+1} \sim \frac{\alpha}{l} \quad \text { for large } l
\)

Thus the ratio of successive coefficients in each of the infinite series in the solution (4.42) is the same as in the series for $e^{\alpha x^{2}}$ for large $l$. We conclude that, for large $x$, each series behaves as $e^{\alpha x^{2}}$. [This is not a rigorous proof. A proper mathematical derivation is given in H. A. Buchdahl, Am. J. Phys., 42, 47 (1974); see also M. Bowen and J. Coster, Am. J. Phys., 48, 307 (1980).]

If each series behaves as $e^{\alpha x^{2}}$, then (4.42) shows that $\psi$ will behave as $e^{\alpha x^{2} / 2}$ for large $x$. The wave function will become infinite as $x$ goes to infinity and will not be quadratically integrable. If we could somehow break off the series after a finite number of terms, then the factor $e^{-\alpha x^{2} / 2}$ would ensure that $\psi$ went to zero as $x$ became infinite. (Using l'Hôpital's rule, it is easy to show that $x^{p} e^{-\alpha x^{2} / 2}$ goes to zero as $x \rightarrow \infty$, where $p$ is any finite power.) To have one of the series break off after a finite number of terms, the coefficient of $c{n}$ in the recursion relation (4.39) must become zero for some value of $n$, say for $n=v$. This makes $c{v+2}, c{v+4}, \ldots$ all equal to zero, and one of the series in (4.42) will have a finite number of terms. In the recursion relation (4.39), there is one quantity whose value is not yet fixed, but can be adjusted to make the coefficient of $c{v}$ vanish. This quantity is the energy $E$. Setting the coefficient of $c_{v}$ equal to zero in (4.39) and using (4.31) for $\alpha$, we get

\(
\begin{gather}
\alpha+2 \alpha v-2 m E \hbar^{-2}=0 \
2 m E \hbar^{-2}=(2 v+1) 2 \pi \nu m \hbar^{-1} \
E=\left(v+\frac{1}{2}\right) h \nu, \quad v=0,1,2, \ldots \tag{4.45}
\end{gather}
\)

The harmonic-oscillator stationary-state energy levels (4.45) are equally spaced (Fig. 4.1). Do not confuse the quantum number $v$ (vee) with the vibrational frequency $\nu$ (nu).

Substitution of (4.45) into the recursion relation (4.39) gives

\(
\begin{equation}
c{n+2}=\frac{2 \alpha(n-v)}{(n+1)(n+2)} c{n} \tag{4.46}
\end{equation}
\)

By quantizing the energy according to (4.45), we have made one of the series break off after a finite number of terms. To get rid of the other infinite series in (4.42), we must set the arbitrary constant that multiplies it equal to zero. This leaves us with a wave function that is $e^{-\alpha x^{2} / 2}$ times a finite power series containing only even or only odd powers of $x$, depending on whether $v$ is even or odd, respectively. The highest power of $x$ in this power series is $x^{v}$, since we chose $E$ to make $c{v+2}, c{v+4}, \ldots$ all vanish. The wave functions (4.42) are thus

\(
\psi{v}=\left{\begin{array}{lr}
e^{-\alpha x^{2} / 2}\left(c{0}+c{2} x^{2}+\cdots+c{v} x^{v}\right) & \text { for } v \text { even } \tag{4.47}\
e^{-\alpha x^{2} / 2}\left(c{1} x+c{3} x^{3}+\cdots+c_{v} x^{v}\right) & \text { for } v \text { odd }
\end{array}\right.
\)

where the arbitrary constants $A$ and $B$ in (4.42) can be absorbed into $c{1}$ and $c{0}$, respectively, and can therefore be omitted. The coefficients after $c{0}$ and $c{1}$ are found from the recursion relation (4.46). Since the quantum number $v$ occurs in the recursion relation, we get a different set of coefficients $c{i}$ for each different $v$. For example, $c{2}$ in $\psi{4}$ differs from $c{2}$ in $\psi_{2}$.

As in the particle in a box, the requirement that the wave functions be well-behaved forces us to quantize the energy. For values of $E$ that differ from (4.45), $\psi$ is not quadratically integrable. For example, Fig. 4.2 plots $\psi$ of Eq. (4.40) for the values $E / h \nu=0.499,0.500$, and 0.501 , where the recursion relation (4.39) is used to calculate the coefficients $c_{n}$ (see also Prob. 4.23). Figure 4.3 gives an enlarged view of these curves in the region near $\alpha^{1 / 2} x=3$.

The harmonic-oscillator ground-state energy is nonzero. This energy, $\frac{1}{2} h \nu$, is called the zero-point energy. This would be the vibrational energy of a harmonic oscillator in a collection of harmonic oscillators at a temperature of absolute zero. The zero-point energy can be understood from the uncertainty principle. If the lowest state had an energy of zero, both its potential and kinetic energies (which are nonnegative) would have to be zero. Zero kinetic energy would mean that the momentum was exactly zero, and $\Delta p{x}$ would be zero. Zero potential energy would mean that the particle was always located at the origin, and $\Delta x$ would be zero. But we cannot have both $\Delta x$ and $\Delta p{x}$ equal to zero. Hence the need for a nonzero ground-state energy. Similar ideas apply for the particle in a box. The definition of the zero-point energy (ZPE) is $E{\mathrm{ZPE}}=E{\mathrm{gs}}-V{\min }$, where $E{\mathrm{gs}}$ and $V_{\min }$ are the groundstate energy and the minimum value of the potential-energy function.

FIGURE 4.2 Plots of the harmonic-oscillator Schrödinger-equation solution containing only even powers of $x$ for $E=0.499 h \nu, E=0.500 h \nu$, and $E=0.501 h \nu$. In the region around $x=0$ the three curves nearly coincide. For $\left|\alpha^{1 / 2} x\right|>3$ the $E=0.500 h \nu$ curve nearly coincides with the $x$ axis.

FIGURE 4.3 Enlargement of Fig. 4.2 in the region $\alpha^{1 / 2} x=3$

Even and Odd Functions

Before considering the wave functions in detail, we define even and odd functions. If $f(x)$ satisfies

\(
\begin{equation}
f(-x)=f(x) \tag{4.48}
\end{equation}
\)

then $f$ is an even function of $x$. Thus $x^{2}$ and $e^{-b x^{2}}$ are both even functions of $x$ since $(-x)^{2}=x^{2}$ and $e^{-b(-x)^{2}}=e^{-b x^{2}}$. The graph of an even function is symmetric about the $y$ axis (for example, see Fig. 4.4a). Therefore

\(
\begin{equation}
\int{-a}^{+a} f(x) d x=2 \int{0}^{a} f(x) d x \text { for } f(x) \text { even } \tag{4.49}
\end{equation}
\)

If $g(x)$ satisfies

\(
\begin{equation}
g(-x)=-g(x) \tag{4.50}
\end{equation}
\)

then $g$ is an odd function of $x$. Examples are $x, 1 / x$, and $x^{3} e^{x^{2}}$. Setting $x=0$ in (4.50), we see that an odd function must be zero at $x=0$, provided $g(0)$ is defined and singlevalued. The graph of an odd function has the general appearance of Fig. 4.4b. Because positive contributions on one side of the $y$ axis are canceled by corresponding negative contributions on the other side, we have

\(
\begin{equation}
\int_{-a}^{+a} g(x) d x=0 \quad \text { for } g(x) \text { odd } \tag{4.51}
\end{equation}
\)

It is easy to show that the product of two even functions or of two odd functions is an even function, while the product of an even and an odd function is an odd function.

The Harmonic-Oscillator Wave Functions

The exponential factor $e^{-\alpha x^{2} / 2}$ in (4.47) is an even function of $x$. If $v$ is an even number, the polynomial factor contains only even powers of $x$, which makes $\psi{v}$ an even function. If $v$ is odd, the polynomial factor contains only odd powers of $x$, and $\psi{v}$, being the product of an even function and an odd function, is an odd function. Each harmonic-oscillator stationary state $\psi$ is either an even or odd function according to whether the quantum number $v$ is even or odd. In Section 7.5, we shall see that, when the potential energy $V$ is an even function, the wave functions of nondegenerate levels must be either even or odd functions.

We now find the explicit forms of the wave functions of the lowest three levels. For the $v=0$ ground state, Eq. (4.47) gives

\(
\begin{equation}
\psi{0}=c{0} e^{-\alpha x^{2} / 2} \tag{4.52}
\end{equation}
\)

where the subscript on $\psi$ gives the value of $v$. We fix $c_{0}$ by normalization:

\(
1=\int{-\infty}^{\infty}\left|c{0}\right|^{2} e^{-\alpha x^{2}} d x=2\left|c{0}\right|^{2} \int{0}^{\infty} e^{-\alpha x^{2}} d x
\)

where Eq. (4.49) has been used. Using the integral (A.9) in the Appendix, we find $\left|c_{0}\right|=(\alpha / \pi)^{1 / 4}$. Therefore,

\(
\begin{equation}
\psi_{0}=(\alpha / \pi)^{1 / 4} e^{-\alpha x^{2} / 2} \tag{4.53}
\end{equation}
\)

if we choose the phase of the normalization constant to be zero. The wave function (4.53) is a Gaussian function (Fig. 4.4a).

For the $v=1$ state, Eq. (4.47) gives

\(
\begin{equation}
\psi{1}=c{1} x e^{-\alpha x^{2} / 2} \tag{4.54}
\end{equation}
\)

After normalization using the integral in Eq. (A.10), we have

\(
\begin{equation}
\psi_{1}=\left(4 \alpha^{3} / \pi\right)^{1 / 4} x e^{-\alpha x^{2} / 2} \tag{4.55}
\end{equation}
\)

Figure 4.4 b shows $\psi_{1}$.
For $v=2$, Eq. (4.47) gives

\(
\psi{2}=\left(c{0}+c_{2} x^{2}\right) e^{-\alpha x^{2} / 2}
\)

The recursion relation (4.46) with $v=2$, gives

\(
c{2}=\frac{2 \alpha(-2)}{1 \cdot 2} c{0}=-2 \alpha c_{0}
\)

Therefore

\(
\begin{equation}
\psi{2}=c{0}\left(1-2 \alpha x^{2}\right) e^{-\alpha x^{2} / 2} \tag{4.56}
\end{equation}
\)

Evaluating $c_{0}$ by normalization, we find (Prob. 4.10)

\(
\begin{equation}
\psi_{2}=(\alpha / 4 \pi)^{1 / 4}\left(2 \alpha x^{2}-1\right) e^{-\alpha x^{2} / 2} \tag{4.57}
\end{equation}
\)

Note that $c{0}$ in $\psi{2}$ is not the same as $c{0}$ in $\psi{0}$.
The number of nodes in the wave function equals the quantum number $v$. It can be proved (see Messiah, pages 109-110) that for the bound stationary states of a onedimensional problem, the number of nodes interior to the boundary points is zero for the ground-state $\psi$ and increases by one for each successive excited state. The boundary points for the harmonic oscillator are $\pm \infty$. Moreover, one can show that a one-dimensional wave function must change sign as it goes through a node (see Prob. 4.45).

FIGURE 4.4 Harmonicoscillator wave functions. The same scale is used for all graphs. The points marked on the $x$ axes are for $\alpha^{1 / 2} x= \pm 2$.

FIGURE 4.5 The classically
allowed $(-a \leq x \leq a)$ and forbidden ( $x<-a$ and $x>a)$ regions for the harmonic oscillator.

The polynomial factors in the harmonic-oscillator wave functions are well known in mathematics and are called Hermite polynomials, after a French mathematician. (See Prob. 4.21.)

According to the quantum-mechanical solution, there is some probability of finding the particle at any point on the $x$ axis (except at the nodes). Classically, $E=T+V$ and the kinetic energy $T$ cannot be negative: $T \geq 0$. Therefore, $E-V=T \geq 0$ and $V \leq E$. The potential energy $V$ is a function of position, and a classical particle is confined to the region of space where $V \leq E$; that is, where the potential energy does not exceed the total energy. In Fig. 4.5, the horizontal line labeled $E$ gives the energy of a harmonic oscillator, and the parabolic curve gives the potential energy $\frac{1}{2} k x^{2}$. For the regions $x<-a$ and $x>a$, we have $V>E$, and these regions are classically forbidden. The classically allowed region $-a \leq x \leq a$ in Fig. 4.5 is where $V \leq E$.

In quantum mechanics, the stationary-state wave functions are not eigenfunctions of $\hat{T}$ or $\hat{V}$, and we cannot assign definite values to $T$ or $V$ for a stationary state. Instead of the classical equations $E=T+V$ and $T \geq 0$, we have in quantum mechanics that $E=\langle T\rangle+\langle V\rangle$ (Prob. 6.35) and $\langle T\rangle \geq 0$ (Prob. 7.7), so $\langle V\rangle \leq E$ in quantum mechanics, but we cannot write $V \leq E$, and a particle has some probability to be found in classically forbidden regions where $V>E$.

It might seem that, by saying the particle can be found outside the classically allowed region, we are allowing it to have negative kinetic energy. Actually, there is no paradox in the quantum-mechanical view. To verify that the particle is in the classically forbidden region, we must measure its position. This measurement changes the state of the system (Sections 1.3 and 1.4). The interaction of the oscillator with the measuring apparatus transfers enough energy to the oscillator for it to be in the classically forbidden region. An accurate measurement of $x$ introduces a large uncertainty in the momentum and hence in the kinetic energy. Penetration of classically forbidden regions was previously discussed in Sections 2.4 and 2.5.

A harmonic-oscillator stationary state has $E=\left(v+\frac{1}{2}\right) h \nu$ and $V=\frac{1}{2} k x^{2}=2 \pi^{2} \nu^{2} m x^{2}$, so the classically allowed region where $V \leq E$ is where $2 \pi^{2} \nu^{2} m x^{2} \leq\left(v+\frac{1}{2}\right) h \nu$, which gives $x^{2} \leq\left(v+\frac{1}{2}\right) h / 2 \pi^{2} \nu m=(2 v+1) / \alpha$, where $\alpha \equiv 2 \pi \nu m / \hbar$ [Eq. (4.31)]. Therefore, the classically allowed region for the harmonic oscillator is where $-(2 v+1)^{1 / 2} \leq$ $\alpha^{1 / 2} x \leq(2 v+1)^{1 / 2}$.

Note from Fig. 4.4 that $\psi$ oscillates in the classically allowed region and decreases exponentially to zero in the classically forbidden region. We previously saw this behavior for the particle in a rectangular well (Section 2.4).

Figure 4.4 shows that, as we go to higher-energy states of the harmonic oscillator, $\psi$ and $|\psi|^{2}$ tend to have maxima farther and farther from the origin. Since $V=\frac{1}{2} k x^{2}$ increases as we go farther from the origin, the average potential energy $\langle V\rangle=\int{-\infty}^{\infty}|\psi|^{2} V d x$ increases as the quantum number increases. The average kinetic
energy is given by $\langle T\rangle=-\left(\hbar^{2} / 2 m\right) \int{-\infty}^{\infty} \psi^{*} \psi^{\prime \prime} d x$. Integration by parts gives (Prob. 7.7b) $\langle T\rangle=\left(\hbar^{2} / 2 m\right) \int_{-\infty}^{\infty}|d \psi / d x|^{2} d x$. The higher number of nodes in states with a higher quantum number produces a faster rate of change of $\psi$, so $\langle T\rangle$ increases as the quantum number increases.

A classical harmonic oscillator is most likely to be found in the regions near the turning points of the motion, where the oscillator is moving the slowest and $V$ is large. In contrast, for the ground state of a quantum harmonic oscillator, the most probable region is the region around the origin. For high oscillator quantum numbers, one finds that the outer peaks of $|\psi|^{2}$ are larger than the peaks near the origin, and the most probable regions become the regions near the classical turning points, where $V$ is large (see Prob. 4.18). This is an example of the correspondence principle (Section 2.2).

Some online simulations of the quantum harmonic oscillator are available at www.phy. davidson.edu/StuHome/cabellf/energy.html (shows energy levels and wave functions and shows how the wave function diverges when the energy is changed from an allowed value); www.falstad.com/qm1d/ (choose harmonic oscillator from the drop-down menu at the top; double click on one of the small circles at the bottom to show a stationary state; shows energies, wave functions, probability densities; $m$ and $k$ can be varied); demonstrations. wolfram.com/HarmonicOscillatorEigenfunctions (shows $\left|\psi{v}\right|^{2}$ ).

We shall see in Section 13.1 that to an excellent approximation one can treat separately the motions of the electrons and the motions of the nuclei of a molecule. (This is due to the much heavier mass of the nuclei.) One first imagines the nuclei to be held stationary and solves a Schrödinger equation for the electronic energy $U$. ( $U$ also includes the energy of nuclear repulsion.) For a diatomic (two-atom) molecule, the electronic energy $U$ depends on the distance $R$ between the nuclei, $U=U(R)$, and the $U$ versus $R$ curve has the typical appearance of Fig. 13.1.

After finding $U(R)$, one solves a Schrödinger equation for nuclear motion, using $U(R)$ as the potential energy for nuclear motion. For a diatomic molecule, the nuclear Schrödinger equation is a two-particle equation. We shall see in Section 6.3 that, when the potential energy of a two-particle system depends only on the distance between the particles, the energy of the system is the sum of (a) the kinetic energy of translational motion of the entire system through space and (b) the energy of internal motion of the particles relative to each other. The classical expression for the two-particle internalmotion energy turns out to be the sum of the potential energy of interaction between the particles and the kinetic energy of a hypothetical particle whose mass is $m{1} m{2} /\left(m{1}+m{2}\right)$ (where $m{1}$ and $m{2}$ are the masses of the two particles) and whose coordinates are the coordinates of one particle relative to the other. The quantity $m{1} m{2} /\left(m{1}+m{2}\right)$ is called the reduced mass $\mu$.

The internal motion of a diatomic molecule consists of vibration, corresponding to a change in the distance $R$ between the two nuclei, and rotation, corresponding to a change in the spatial orientation of the line joining the nuclei. To a good approximation, one can usually treat the vibrational and rotational motions separately. The rotational energy levels are found in Section 6.4. Here we consider the vibrational levels.

The Schrödinger equation for the vibration of a diatomic molecule has a kineticenergy operator for the hypothetical particle of mass $\mu=m{1} m{2} /\left(m{1}+m{2}\right)$ and a potential-energy term given by $U(R)$. If we place the origin to coincide with the minimum point of the $U$ curve in Fig. 13.1 and take the zero of potential energy at the energy of this minimum point, then the lower portion of the $U(R)$ curve will nearly coincide with

FIGURE 4.6 Potential energy for vibration of a diatomic molecule (solid curve) and for a harmonic oscillator (dashed curve). Also shown are the boundstate vibrational energy levels for the diatomic molecule. In contrast to the harmonic oscillator, a diatomic molecule has only a finite number of bound vibrational levels.

the potential-energy curve of a harmonic oscillator with the appropriate force constant $k$ (see Fig. 4.6 and Prob. 4.28). The minimum in the $U(R)$ curve occurs at the equilibrium distance $R{e}$ between the nuclei. In Fig. 4.6, $x$ is the deviation of the internuclear distance from its equilibrium value: $x \equiv R-R{e}$.

The harmonic-oscillator force constant $k$ in Eq. (4.26) is obtained as $k=d^{2} V / d x^{2}$, and the harmonic-oscillator curve essentially coincides with the $U(R)$ curve at $R=R{e}$, so the molecular force constant is $k=d^{2} U /\left.d R^{2}\right|{R=R_{e}}$ (see also Prob. 4.28). Differences in nuclear mass have virtually no effect on the electronic-energy curve $U(R)$, so different isotopic species of the same molecule have essentially the same force constant $k$.

We expect, therefore, that a reasonable approximation to the vibrational energy levels $E_{\text {vib }}$ of a diatomic molecule would be the harmonic-oscillator vibrational energy levels; Eqs. (4.45) and (4.23) give

\(
\begin{equation}
E{\mathrm{vib}} \approx\left(v+\frac{1}{2}\right) h \nu{e}, \quad v=0,1,2, \ldots \tag{4.58}
\end{equation}
\)

\(
\begin{equation}
\nu{e}=\frac{1}{2 \pi}\left(\frac{k}{\mu}\right)^{1 / 2}, \quad \mu=\frac{m{1} m{2}}{m{1}+m{2}}, \quad k=\left.\frac{d^{2} U}{d R^{2}}\right|{R=R_{e}} \tag{4.59}
\end{equation}
\)

$\nu_{e}$ is called the equilibrium (or harmonic) vibrational frequency. This approximation is best for the lower vibrational levels. As $v$ increases, the nuclei spend more time in regions far from their equilibrium separation. For such regions the potential energy deviates substantially from that of a harmonic oscillator and the harmonic-oscillator approximation is poor. Instead of being equally spaced, one finds that the vibrational levels of a diatomic molecule come closer and closer together as $v$ increases (Fig. 4.6). Eventually, the vibrational energy is large enough to dissociate the diatomic molecule into atoms that are not bound to each other. Unlike the harmonic oscillator, a diatomic molecule has only a finite number of bound-state vibrational levels. A more accurate expression for the molecular vibrational energy that allows for the anharmonicity of the vibration is

\(
\begin{equation}
E{\mathrm{vib}}=\left(v+\frac{1}{2}\right) h \nu{e}-\left(v+\frac{1}{2}\right)^{2} h \nu{e} x{e} \tag{4.60}
\end{equation}
\)

where the anharmonicity constant $\nu{e} x{e}$ is positive in nearly all cases.

Using the time-dependent Schrödinger equation, one finds (Section 9.9) that the most probable vibrational transitions when a diatomic molecule is exposed to electromagnetic radiation are those where $v$ changes by $\pm 1$. Furthermore, for absorption or emission of electromagnetic radiation to occur, the vibration must change the molecule's dipole moment. Hence homonuclear diatomics (such as $\mathrm{H}{2}$ or $\mathrm{N}{2}$ ) cannot undergo transitions between vibrational levels by absorption or emission of radiation. (Such transitions can occur during intermolecular collisions.) The relation $E{\text {upper }}-E{\text {lower }}=h \nu$, the approximate equation (4.58), and the selection rule $\Delta v=1$ for absorption of radiation show that a heteronuclear diatomic molecule whose vibrational frequency is $\nu{e}$ will most strongly absorb light of frequency $\nu{\text {light }}$ given approximately by
$\nu{\text {light }}=\left(E{2}-E{1}\right) / h \approx\left[\left(v{2}+\frac{1}{2}\right) h \nu{e}-\left(v{1}+\frac{1}{2}\right) h \nu{e}\right] / h=\left(v{2}-v{1}\right) \nu{e}=\nu{e}$
The values of $k$ and $\mu$ in (4.59) for diatomic molecules are such that $\nu{\text {light }}$ usually falls in the infrared region of the spectrum. Transitions with $\Delta v=2,3, \ldots$ also occur, but these (called overtones) are much weaker than the $\Delta v=1$ absorption.

Use of the more accurate equation (4.60) gives (Prob. 4.27)

\(
\begin{equation}
\nu{\text {light }}=\nu{e}-2 \nu{e} x{e}\left(v_{1}+1\right) \tag{4.62}
\end{equation}
\)

where $v_{1}$ is the quantum number of the lower level and $\Delta v=1$.
The relative population of two molecular energy levels in a system in thermal equilibrium is given by the Boltzmann distribution law (see any physical chemistry text) as

\(
\begin{equation}
\frac{N{i}}{N{j}}=\frac{g{i}}{g{j}} e^{-\left(E{i}-E{j}\right) / k T} \tag{4.63}
\end{equation}
\)

where energy levels $i$ and $j$ have energies $E{i}$ and $E{j}$ and degeneracies $g{i}$ and $g{j}$ and are populated by $N{i}$ and $N{j}$ molecules, and where $k$ is Boltzmann's constant and $T$ the absolute temperature. For a nondegenerate level, $g_{i}=1$.

The magnitude of $\nu=(1 / 2 \pi)(k / \mu)^{1 / 2}$ is such that for light diatomics (for example, $\mathrm{H}{2}, \mathrm{HCl}, \mathrm{CO}$ ) only the $v=0$ vibrational level is significantly populated at room temperature. For heavy diatomics (for example, $\mathrm{I}{2}$, there is significant room-temperature population of one or more excited vibrational levels.

The vibrational absorption spectrum of a polar diatomic molecule consists of a $v=0 \rightarrow 1$ band, much weaker overtone bands $(v=0 \rightarrow 2,0 \rightarrow 3, \ldots)$, and, if $v>0$ levels are significantly populated, hot bands such as $v=1 \rightarrow 2,2 \rightarrow 3$. Each band corresponding to a particular vibrational transition consists of several closely spaced lines. Each such line corresponds to a different change in rotational state simultaneous with the change in vibrational state. Each line is the result of a vibration-rotation transition.

The SI unit for spectroscopic frequencies is the hertz $(\mathrm{Hz})$, defined by $1 \mathrm{~Hz} \equiv 1 \mathrm{~s}^{-1}$. Multiples such as the megahertz ( MHz ) equal to $10^{6} \mathrm{~Hz}$ and the gigahertz $(\mathrm{GHz})$ equal to $10^{9} \mathrm{~Hz}$ are often used. Infrared (IR) absorption lines are usually specified by giving their wavenumber $\widetilde{\nu}$ defined as

\(
\begin{equation}
\widetilde{\nu} \equiv 1 / \lambda=\nu / c \tag{4.64}
\end{equation}
\)

where $\lambda$ is the wavelength in vacuum.
In the harmonic-oscillator approximation, the quantum-mechanical energy levels of a polyatomic molecule turn out to be $E{\text {vib }}=\sum{i}\left(v{i}+\frac{1}{2}\right) h \nu{i}$, where the $\nu{i}$ 's are the frequencies of the normal modes of vibration of the molecule and $v{i}$ is the vibrational quantum
number of the $i$ th normal mode. Each $v_{i}$ takes on the values $0,1,2, \ldots$ independently of the values of the other vibrational quantum numbers. A linear molecule with $n$ atoms has $3 n-5$ normal modes; a nonlinear molecule has $3 n-6$ normal modes.

To calculate the reduced mass $\mu$ in (4.59), one needs the masses of isotopic species. Some relative isotopic masses are listed in Table A. 3 in the Appendix.

EXAMPLE

The strongest infrared band of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$ occurs at $\widetilde{\nu}=2143 \mathrm{~cm}^{-1}$. Find the force constant of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$. State any approximation made.
The strongest infrared band corresponds to the $v=0 \rightarrow 1$ transition. We approximate the molecular vibration as that of a harmonic oscillator. From (4.61), the equilibrium molecular vibrational frequency is approximately

\(
\nu{e} \approx \nu{\text {light }}=\widetilde{\nu} c=\left(2143 \mathrm{~cm}^{-1}\right)\left(2.9979 \times 10^{10} \mathrm{~cm} / \mathrm{s}\right)=6.424 \times 10^{13} \mathrm{~s}^{-1}
\)

To relate $k$ to $\nu{e}$ in (4.59), we need the reduced mass $\mu=m{1} m{2} /\left(m{1}+m_{2}\right)$. One mole of ${ }^{12} \mathrm{C}$ has a mass of 12 g and contains Avogadro's number of atoms. Hence the mass of one atom of ${ }^{12} \mathrm{C}$ is $(12 \mathrm{~g}) /\left(6.02214 \times 10^{23}\right)$. The reduced mass and force constant are

\(
\begin{gathered}
\mu=\frac{12(15.9949) \mathrm{g}}{27.9949} \frac{1}{6.02214 \times 10^{23}}=1.1385 \times 10^{-23} \mathrm{~g} \
k=4 \pi^{2} \nu_{e}^{2} \mu=4 \pi^{2}\left(6.424 \times 10^{13} \mathrm{~s}^{-1}\right)^{2}\left(1.1385 \times 10^{-26} \mathrm{~kg}\right)=1855 \mathrm{~N} / \mathrm{m}
\end{gathered}
\)

EXERCISE (a) Find the approximate zero-point energy of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$.
(Answer: $2.1 \times 10^{-20} \mathrm{~J}$.) (b) Estimate $\nu_{e}$ of ${ }^{13} \mathrm{C}^{16} \mathrm{O}$. (Answer: $6.28 \times 10^{13} \mathrm{~s}^{-1}$.)

The Numerov Method

We solved the Schrödinger equation exactly for the particle in a box and the harmonic oscillator. For many potential-energy functions $V(x)$, the one-particle, one-dimensional Schrödinger equation cannot be solved exactly. This section presents a numerical method (the Numerov method) for computer solution of the one-particle, one-dimensional Schrödinger equation that allows one to get accurate bound-state eigenvalues and eigenfunctions for an arbitrary $V(x)$.

To solve the Schrödinger equation numerically, we deal with a portion of the $x$ axis that includes the classically allowed region and that extends somewhat into the classically forbidden region at each end of the classically allowed region. We divide this portion of the $x$ axis into small intervals, each of length $s$ (Fig. 4.7). The points $x{0}$ and $x{\text {max }}$ are the endpoints of this portion, and $x{n}$ is the endpoint of the $n$th interval. Let $\psi{n-1}, \psi{n}$, and $\psi{n+1}$ denote the values of $\psi$ at the points $x{n}-s, x{n}$, and $x_{n}+s$, respectively (these are the endpoints of adjacent intervals)

\(
\begin{equation}
\psi{n-1} \equiv \psi\left(x{n}-s\right), \quad \psi{n} \equiv \psi\left(x{n}\right), \quad \psi{n+1} \equiv \psi\left(x{n}+s\right) \tag{4.65}
\end{equation}
\)

Don't be confused by the notation. The subscripts $n-1, n$, and $n+1$ do not label different states but rather indicate values of one particular wave function $\psi$ at points on the $x$ axis separated by the interval $s$. The $n$ subscript means $\psi$ is evaluated at the point $x_{n}$. We write the Schrödinger equation $-\left(\hbar^{2} / 2 m\right) \psi^{\prime \prime}+V \psi=E \psi$ as

\(
\begin{equation}
\psi^{\prime \prime}=G \psi, \quad \text { where } G \equiv m \hbar^{-2}[2 V(x)-2 E] \tag{4.66}
\end{equation}
\)

By expanding $\psi\left(x{n}+s\right)$ and $\psi\left(x{n}-s\right)$ in Taylor series involving powers of $s$, adding these two expansions to eliminate odd powers of $s$, using the Schrödinger equation to express $\psi^{\prime \prime}$ and $\psi^{(\mathrm{iv})}$ in terms of $\psi$, and neglecting terms in $s^{6}$ and higher powers of $s$ (an approximation that will be accurate if $s$ is small), one finds that (Prob. 4.43)

\(
\begin{equation}
\psi{n+1} \approx \frac{2 \psi{n}-\psi{n-1}+5 G{n} \psi{n} s^{2} / 6+G{n-1} \psi{n-1} s^{2} / 12}{1-G{n+1} s^{2} / 12} \tag{4.67}
\end{equation}
\)

where $G{n} \equiv G\left(x{n}\right) \equiv m \hbar^{-2}\left[2 V\left(x{n}\right)-2 E\right]$ [Eqs. (4.66) and (4.65)]. Equation (4.67) allows us to calculate $\psi{n+1}$, the value of $\psi$ at point $x{n}+s$, if we know $\psi{n}$ and $\psi{n-1}$, the values of $\psi$ at the preceding two points $x{n}$ and $x_{n}-s$.

How do we use (4.67) to solve the Schrödinger equation? We first guess a value $E{\text {guess }}$ for an energy eigenvalue. We start at a point $x{0}$ well into the left-hand classically forbidden region (Fig. 4.7), where $\psi$ will be very small, and we approximate $\psi$ as zero at this point: $\psi{0} \equiv \psi\left(x{0}\right)=0$. Also, we pick a point $x{\max }$ well into the right-hand classically forbidden region, where $\psi$ will be very small and we shall demand that $\psi\left(x{\max }\right)=0$. We pick a small value for the interval $s$ between successive points, and we take $\psi$ at $x{0}+s$ as some small number, say, $0.0001: \psi{1} \equiv \psi\left(x{1}\right) \equiv \psi\left(x{0}+s\right)=0.0001$. The value of $\psi{1}$ will not make any difference in the eigenvalues found. If 0.001 were used instead of 0.0001 for $\psi{1}$, Eq. (4.67) shows that this would simply multiply all values of $\psi$ at subsequent points by 10 (Prob. 4.41). This would not affect the eigenvalues [see the example after Eq. (3.14)]. The wave function can be normalized after each eigenvalue is found.

Having chosen values for $\psi{0}$ and $\psi{1}$, we then use (4.67) with $n=1$ to calculate $\psi{2} \equiv \psi\left(x{2}\right) \equiv \psi\left(x{1}+s\right)$, where the $G$ values are calculated using $E{\text {guess }}$. Next, (4.67) with $n=2$ gives $\psi{3}$; and so on. We continue until we reach $x{\max }$. If $E{\text {guess }}$ is not equal to or very close to an eigenvalue, $\psi$ will not be quadratically integrable and $\left|\psi\left(x{\max }\right)\right|$ will be very large. If $\psi\left(x{\max }\right)$ is not found to be close to zero, we start again at $x{0}$ and repeat the process using a new $E{\text {guess }}$. The process is repeated until we find an $E{\text {guess }}$ that makes $\psi\left(x{\max }\right)$ very close to zero. $E{\text {guess }}$ is then essentially equal to an eigenvalue. The systematic way to locate the eigenvalues is to count the nodes in the $\psi$ produced by $E{\text {guess }}$. Recall (Section 4.2) that in a one-dimensional problem, the number of interior nodes is 0 for the ground state, 1 for the first excited state, and so on. Let $E{1}, E{2}, E{3}, \ldots$ denote the energies of the ground state, the first excited state, the second excited state, and so on, respectively. If $\psi{\text {guess }}$ contains no nodes between $x{0}$ and $x{\text {max }}$, then $E{\text {guess }}$ is less than or equal to $E_{1}$; if

FIGURE 4.7 $V$ versus $x$ for a one-particle, onedimensional system.

FIGURE 4.8 The number of nodes in a Numerov-method solution as a function of the energy $E_{\text {guess }}$.

$\psi{\text {guess }}$ contains one interior node, then $E{\text {guess }}$ is between $E{1}$ and $E{2}$ (Fig. 4.8). Examples are given later.

Dimensionless Variables

The Numerov method requires that we guess values of $E$. What should be the order of magnitude of our guesses: $10^{-20} \mathrm{~J}, 10^{-15} \mathrm{~J}, \ldots$ ? To answer this question, we reformulate the Schrödinger equation using dimensionless variables, taking the harmonic oscillator as the example.

The harmonic oscillator has $V=\frac{1}{2} k x^{2}$, and the harmonic-oscillator Schrödinger equation contains the three constants $k, m$, and $\hbar$. We seek to find a dimensionless reduced energy $E{r}$ and a dimensionless reduced $x$ coordinate $x{r}$ that are defined by

\(
\begin{equation}
E{r} \equiv E / A, \quad x{r} \equiv x / B \tag{4.68}
\end{equation}
\)

where the constant $A$ is a combination of $k, m$, and $\hbar$ that has dimensions of energy, and $B$ is a combination with dimensions of length. The dimensions of energy are mass $\times$ length $^{2} \times$ time $^{-2}$, which we write as

\(
\begin{equation}
[E]=\mathrm{ML}^{2} \mathrm{~T}^{-2} \tag{4.69}
\end{equation}
\)

where the brackets around $E$ denote its dimensions, and M, L, and T stand for the dimensions mass, length, and time, respectively. The equation $V=\frac{1}{2} k x^{2}$ shows that $k$ has dimensions of energy $\times$ length $^{-2}$, and (4.69) gives $[k]=\mathrm{MT}^{-2}$. The dimensions of $\hbar$ are energy $\times$ time. Thus

\(
\begin{equation}
[m]=\mathrm{M}, \quad[k]=\mathrm{MT}^{-2}, \quad[\hbar]=\mathrm{ML}^{2} \mathrm{~T}^{-1} \tag{4.70}
\end{equation}
\)

The dimensions of $A$ and $B$ in (4.68) are energy and length, respectively, so

\(
\begin{equation}
[A]=\mathrm{ML}^{2} \mathrm{~T}^{-2}, \quad[B]=\mathrm{L} \tag{4.71}
\end{equation}
\)

Let $A=m^{a} k^{b} \hbar^{c}$, where $a, b$, and $c$ are powers that are determined by the requirement that the dimensions of $A$ must be $\mathrm{ML}^{2} \mathrm{~T}^{-2}$. We have

\(
\begin{equation}
[A]=\left[m^{a} k^{b} \hbar^{c}\right]=\mathbf{M}^{a}\left(\mathrm{MT}^{-2}\right)^{b}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{c}=\mathbf{M}^{a+b+c} \mathrm{~L}^{2 c} \mathrm{~T}^{-2 b-c} \tag{4.72}
\end{equation}
\)

Equating the exponents of each of $\mathrm{M}, \mathrm{L}$, and T in (4.71) and (4.72), we have

\(
a+b+c=1, \quad 2 c=2, \quad-2 b-c=-2
\)

Solving these equations, we get $c=1, b=\frac{1}{2}, a=-\frac{1}{2}$. Therefore,

\(
\begin{equation}
A=m^{-1 / 2} k^{1 / 2} \hbar \tag{4.73}
\end{equation}
\)

Let $B=m^{d} k^{e} \hbar^{f}$. The same dimensional-analysis procedure that gave (4.73) gives (Prob. 4.44)

\(
\begin{equation}
B=m^{-1 / 4} k^{-1 / 4} \hbar^{1 / 2} \tag{4.74}
\end{equation}
\)

From (4.68), (4.73), and (4.74), the reduced variables for the harmonic oscillator are

\(
\begin{equation}
E{r}=E / m^{-1 / 2} k^{1 / 2} \hbar, \quad x{r}=x / m^{-1 / 4} k^{-1 / 4} \hbar^{1 / 2} \tag{4.75}
\end{equation}
\)

Using $k^{1 / 2}=2 \pi \nu m^{1 / 2}$ [Eq. (4.23)] to eliminate $k$ from (4.75) and recalling the definition $\alpha \equiv 2 \pi \nu m / \hbar$ [Eq. (4.31)], we have the alternative expressions

\(
\begin{equation}
E{r}=E / h \nu, \quad x{r}=\alpha^{1 / 2} x \tag{4.76}
\end{equation}
\)

Similar to the equation $E{r} \equiv E / A$ [Eq. (4.68)], we define the reduced potential energy function $V{r}$ as

\(
\begin{equation}
V_{r} \equiv V / A \tag{4.77}
\end{equation}
\)

Since $|\psi(x)|^{2} d x$ is a probability, and probabilities are dimensionless, the normalized $\psi(x)$ must have the dimensions of length ${ }^{-1 / 2}$. We therefore define a reduced normalized wave function $\psi_{r}$ that is dimensionless. From (4.71), $B$ has dimensions of length, so $B^{-1 / 2}$ has units of length ${ }^{-1 / 2}$. Therefore,

\(
\begin{equation}
\psi_{r}=\psi / B^{-1 / 2} \tag{4.78}
\end{equation}
\)

$\psi{r}$ satisfies $\int{-\infty}^{\infty}\left|\psi{r}\right|^{2} d x{r}=1$; this follows from (4.68), (4.78), and $\int{-\infty}^{\infty}|\psi|^{2} d x=1$.
We now rewrite the Schrödinger equation in terms of the reduced variables $x{r}, \psi{r}, V{r}$, and $E_{r}$. We have

\(
\begin{align}
\frac{d^{2} \psi}{d x^{2}}=\frac{d^{2}}{d x^{2}} B^{-1 / 2} \psi{r}=B^{-1 / 2} \frac{d}{d x} \frac{d \psi{r}}{d x} & =B^{-1 / 2} \frac{d}{d x} \frac{d \psi{r}}{d x{r}} \frac{d x{r}}{d x}=B^{-1 / 2} \frac{d\left(d \psi{r} / d x{r}\right)}{d x{r}} \frac{d x{r}}{d x} \frac{d x{r}}{d x} \
\frac{d^{2} \psi}{d x^{2}} & =B^{-5 / 2} \frac{d^{2} \psi{r}}{d x{r}^{2}} \tag{4.79}
\end{align}
\)

since $d x_{r} / d x=B^{-1}$ [Eq. (4.68)]. Substitution of (4.68), (4.77), and (4.79) into the Schrödinger equation $-\left(\hbar^{2} / 2 m\right)\left(d^{2} \psi / d x^{2}\right)+V \psi=E \psi$ gives

\(
\begin{align}
-\frac{\hbar^{2}}{2 m} B^{-5 / 2} \frac{d^{2} \psi{r}}{d x{r}^{2}}+A V{r} B^{-1 / 2} \psi{r} & =A E{r} B^{-1 / 2} \psi{r} \
-\frac{\hbar^{2}}{2 m} \frac{1}{A B^{2}} \frac{d^{2} \psi{r}}{d x{r}^{2}}+V{r} \psi{r} & =E{r} \psi{r} \tag{4.80}
\end{align}
\)

From (4.73) and (4.74), we get $A B^{2}=\hbar^{2} / m$, so $\hbar^{2} / m A B^{2}=1$ for the harmonic oscillator.

More generally, let $V$ contain a single parameter $c$ that is not dimensionless. For example, we might have $V=c x^{4}$ or $V=c x^{2}\left(1+0.05 m^{1 / 2} c^{1 / 2} \hbar^{-1} x^{2}\right)$. (Note that $m^{1 / 2} c^{1 / 2} \hbar^{-1} x^{2}$ is dimensionless, as it must be, since 1 is dimensionless.) The quantity $A B^{2}$ in (4.80) must have the form $A B^{2}=\hbar^{r} m^{s} c^{t}$, where $r, s$, and $t$ are certain powers. Since the term $V{r} \psi{r}$ in (4.80) is dimensionless, the first term is dimensionless. Therefore, $\hbar^{2} / m A B^{2}$ is dimensionless and $A B^{2}$ has the same dimensions as $\hbar^{2} / m$; so $r=2, s=-1$, and $t=0$. With $A B^{2}=\hbar^{2} / m$, Eq. (4.80) gives as the dimensionless Schrödinger equation

\(
\begin{gather}
\frac{d^{2} \psi{r}}{d x{r}^{2}}=\left(2 V{r}-2 E{r}\right) \psi{r} \tag{4.81}\
\psi{r}^{\prime \prime}=G{r} \psi{r}, \quad \text { where } \quad G{r} \equiv 2 V{r}-2 E_{r} \tag{4.82}
\end{gather}
\)

For the harmonic oscillator, $V{r} \equiv V / A=\frac{1}{2} k x^{2} / m^{-1 / 2} k^{1 / 2} \hbar=\frac{1}{2} x{r}^{2}$ [Eqs. (4.73) and (4.75)]:

\(
\begin{equation}
V{r}=\frac{1}{2} x{r}^{2} \tag{4.83}
\end{equation}
\)

Having reduced the harmonic-oscillator Schrödinger equation to the form (4.81) involving only dimensionless quantities, we can expect that the lowest energy eigenvalues will be of the order of magnitude 1 .

The reduced harmonic-oscillator Schrödinger equation (4.82) has the same form as (4.66), so we can use the Numerov formula (4.67) with $\psi, G$, and $s$ replaced by $\psi{r}, G{r}$, and $s{r}$, respectively, where, similar to (4.68), $s{r} \equiv s / B$.

Once numerical values of the reduced energy $E_{r}$ have been found, the energies $E$ are found from (4.75) or (4.76).

Choice of $x{r, 0}, x{r, \text { max }}$, and $s_{r}$

We now need to choose initial and final values of $x{r}$ and the value of the interval $s{r}$ between adjacent points. Suppose we want to find all the harmonic-oscillator eigenvalues and eigenfunctions with $E{r} \leq 5$. We start the solution in the left-hand classically forbidden region, so we first locate the classically forbidden regions for $E{r}=5$. The boundaries between the classically allowed and forbidden regions are where $E{r}=V{r}$. From (4.83), $V{r}=\frac{1}{2} x{r}^{2}$. Thus $E{r}=V{r}$ becomes $5=\frac{1}{2} x{r}^{2}$ and the classically allowed region for $E{r}=5$ is from $x{r}=-(10)^{1 / 2}=-3.16$ to +3.16 . For $E{r}<5$, the classically allowed region is smaller. We want to start the solution at a point well into the left-hand classically forbidden region, where $\psi$ is very small, and we want to end the solution well into the right-hand classically forbidden region. The left-hand classically forbidden region ends at $x{r}=-3.16$ for $E{r}=5$, and a reasonable choice is to start at $x{r}=-5$. [Starting too far into the classically forbidden region can sometimes lead to trouble (see the following), so some trial-and-error might be needed in picking the starting point.] Since $V$ is symmetrical, we shall end the solution at $x{r}=5$.

For reasonable accuracy, one usually needs a minimum of 100 points, so we shall take $s{r}=0.1$ to give us 100 points. As is evident from the derivation of the Numerov method, $s{r}$ must be small. A reasonable rule might be to have $s_{r}$ no greater than 0.1.

If, as is often true, $V \rightarrow \infty$ as $x \rightarrow \pm \infty$, then starting too far into the classically forbidden region can make the denominator $1-G{n+1} s^{2} / 12$ in the Numerov formula (4.67) negative. We have $G{r}=2 V{r}-2 E{r}$, and if we start at a point $x{0}$ where $V{r}$ is extremely large, $G{r}$ at that point might be large enough to make the Numerov denominator negative. The method will then fail to work. We are taking $\psi{0}$ as zero and $\psi{1}$ as a positive number. The Numerov formula (4.67) shows that if the denominator is negative, then $\psi{2}$ will be negative, and we will have produce a spurious node in $\psi$ between $x{1}$ and $x{2}$. To avoid this problem, we can decrease either the step size $s{r}$ or $x{r, \text { max }}-x_{r, 0}$ (see Prob. 4.46).

Computer Program for the Numerov Method

Table 4.1 contains a C ++ computer program that applies the Numerov method to the harmonic-oscillator Schrödinger equation. The fifth and sixth lines declare variables as either integers or double precision. m is the number of intervals between $x{r, 0}$ and $x{r, \text { max }}$ and equals $\left(x{r, \text { max }}-x{r, 0}\right) / s$. cout and cin provide for output to the computer screen and input to the variables of the program. Note the colon at the end of line 13 and the semicolons after most other lines. The three lines beginning with $g[0]=, g[1]=$, and $g[i+1]=$ contain two times the potential-energy function. These lines must be modified if the problem is not the harmonic oscillator. If there is a node between two successive values of $x{r}$, then the $\psi{r}$ values at these two points will have opposite signs (see Prob. 4.45) and the statement $n n=n n+1$ will increase the nodes counter $n n$ by 1 .

You can download a free $\mathrm{C}++$ compiler and integrated development environmentsee the Wikipedia article, List of Compilers. Even simpler, you can enter and run C ++ programs online without downloading anything. The website ideone.com provides this service (after you register) for $\mathrm{C}++$ and other languages. However, at this site, you must enter your complete input into the input area before running the program, since the website does not accept input once the program begins; each input number is separated from its neighbors by a space.

TABLE 4.1 C++ Program for Numerov Solution of the One-Dimensional Schrödinger Equation

#include <iostream>

#include <math.h>

using namespace std;

int main() {<br />
    int m, i, nn;<br />
    double x<span class="hljs-string">[1000]</span>, g<span class="hljs-string">[1000]</span>, p<span class="hljs-string">[1000]</span>, E, s, ss;<br />
    cout << <span class="hljs-string">"Enter initial xr "</span>;<br />
    cin>>x<span class="hljs-string">[0]</span>;<br />
    cout << <span class="hljs-string">"Enter the increment sr "</span>;<br />
    cin>>S;<br />
    cout << <span class="hljs-string">"Enter the number of intervals m "</span>;<br />
    cin>>m;<br />
    label1:<br />
    cout << <span class="hljs-string">"Enter the reduced energy Er (enter 1e10 to quit) "</span>;<br />
    cin>>E;<br />
    if (E > 1e9) {<br />
        cout << <span class="hljs-string">"Quitting"</span>;<br />
        return <span class="hljs-number">0</span>;<br />
    }

    nn=0; p[0]=0;

    p[1]=0.0001;

    x[1]=x[0]+5;

    g[0]=x[0]*x[0]-2*E;

    g[1]=x[1]*x[1]-2*E;

    ss=s*s/12;

    for (i=1; i<=m-1; i=i+1) {<br />
        x<span class="hljs-string">[i+1]</span>=x<span class="hljs-string">[i]</span>+s;<br />
        g<span class="hljs-string">[i+1]</span>=x<span class="hljs-string">[i+1]</span>*x<span class="hljs-string">[i+1]</span>-<span class="hljs-number">2</span>*E;<br />
        p<span class="hljs-string">[i+1]</span>=(-p<span class="hljs-string">[i-1]</span>+<span class="hljs-number">2</span>*p<span class="hljs-string">[i]</span>+<span class="hljs-number">10</span>*g<span class="hljs-string">[i]</span>*p<span class="hljs-string">[i]</span>*ss+g<span class="hljs-string">[i-1]</span>*p<span class="hljs-string">[i-1]</span>*ss)/(<span class="hljs-number">1</span>-g<span class="hljs-string">[i+1]</span>*ss);<br />
        if(p<span class="hljs-string">[i+1]</span>*p<span class="hljs-string">[i]</span><<span class="hljs-number">0</span>)<br />
            nn=nn+<span class="hljs-number">1</span>;<br />
    }

    cout << " Er = " << E << " Nodes = " << nn << " Psir(xm) = " << p[m] << endl;

    goto label1;

}

 

For example, suppose we want the harmonic-oscillator ground-state energy. The program of Table 4.1, with $s{r}=0.1, x{r, 0}=-5$, and $\mathrm{m}=100$ gives the following results. The guess $E{r}=0$ gives a wave function with zero nodes, $\mathrm{nn}=0$, telling us (Fig. 4.8) that the ground-state energy $E{r, 1}$ is above 0 . (Also, the wave function at the rightmost point is found to be $9.94 \times 10^{6}$, very far from 0 . If we now guess 0.9 for $E{r}$, we get a function with one
node, so (Fig. 4.8) 0.9 is between $E{r, 1}$ and $E{r, 2}$. Hence the ground state $E{r}$ is between 0 and 0.9 . Averaging these, we try 0.45 . This value gives a function with no nodes, and so 0.45 is below $E{r, 1}$. Averaging 0.45 and 0.9 , we get 0.675 , which is found to give one node and so is too high. Averaging 0.675 and 0.45 , we try 0.5625 , which gives one node and is too high. We next try 0.50625 , and so on. The program's results show that as we get closer to the true $E{r, 1}, \psi_{r}(5)$ comes closer to zero.

Use of a Spreadsheet to Solve the One-Dimensional Schrödinger Equation

An alternative to a Numerov-method computer program is a spreadsheet.
The following directions for the Excel 2010 spreadsheet apply the Numerov method to solve the harmonic-oscillator Schrödinger equation. (Other versions of Excel can be used with modified directions.)

The columns in the spreadsheet are labeled $\mathrm{A}, \mathrm{B}, \mathrm{C}, \ldots$ and the rows are labeled 1 , $2,3, \ldots$ (see Fig. 4.9 later in this section). A cell's location is specified by its row and column. For example, the cell at the upper left is cell A1. To enter something into a cell, you first select that cell, either by moving the mouse pointer over the desired cell and then clicking the (left) mouse button, or by using the arrow keys to move from the currently selected cell (which has a heavy outline) to the desired cell. After a cell has been selected, type the entry for that cell and press Enter or one of the four arrow keys.

To begin, enter a title in cell A1. Then enter $\mathrm{Er}=$ in cell A3. We shall enter our guesses for $E{r}$ in cell B3. We shall look first for the ground-state (lowest) eigenvalue, pretending that we don't know the answer. The minimum value of $V(x)$ for the harmonic oscillator is zero, so $E{r}$ cannot be negative. We shall take zero as our initial guess for $E_{r}$, so enter 0 in cell B3. Enter $\mathrm{Sr}=$ in cell C3.

Enter 0.1 (the $s{r}$ value chosen earlier) in cell D3. Enter xr in cell A5, Gr in cell B5, and psir in C5. (These entries are labels for the data columns that we shall construct.) Enter -5 (the starting value for $x{r}$ ) in cell A7. Enter $=\mathrm{A} 7+\\( D $\$ 3$ in cell A8. The equal sign at the beginning of the entry tells the spreadsheet that a formula is being entered. This formula tells the spreadsheet to add the numbers in cells A7 and D3. The reason for the $\\) signs in D3 will be explained shortly. When you type a formula, you will see it displayed in the formula bar above the spreadsheet. When you press Enter, the value -4.9 is displayed in cell A8. This is the sum of cells A7 and D3. (If you see a different value in A8 or get an error message, you probably mistyped the formula. To correct the formula, select cell A8 and click in the formula displayed in the formula bar to make the correction.)

Select cell A8 and then click the Home tab and in the Clipboard group click the Copy icon. This copies the contents of cell A8 to a storage area called the Clipboard. Then in the Editing group click the Find \& Select icon and choose Go To. In the Go To box, enter A9:A107 under Reference: and click OK. This selects cells A9 through A107. Then in the Clipboard group click the Paste icon. This pastes the cell A8 formula (which was stored on the Clipboard) into each of cells A9 through A107. To see how this works, click on cell A9. You will see the formula $=A 8+\$ D \$ 3$ in the formula bar. Note that when the cell A8 formula $=A 7+\$ D \$ 3$ was copied one cell down to cell A9, the A7 in the formula was changed to A8. However, the $\\( signs prevented cell D3 in the formula from being changed when it was copied. In a spreadsheet formula, a cell address without $\\) signs is called a relative reference, whereas a cell address with $\$$ signs is called an absolute reference. When a relative reference is copied to the next row in a column, the row number is increased by one; when it is copied two rows below the original row, the row number is increased by two; and so on. Click in some of the other cells in column A to see their formulas. The net result of this copy-and-paste procedure is to fill
the column-A cells with numbers from -5 to 5 in increments of 0.1. (Spreadsheets have faster ways to accomplish this than using Copy and Paste.)

We next fill in the $G{r}$ column. From Eqs. (4.82) and (4.83), $G{r}=x{r}^{2}-2 E{r}$ for the harmonic oscillator. We therefore enter $=A 7 \wedge 2-2 \$ B \$ 3$ in cell B7 (which will contain the value of $G{r}$ at $x{r}=-5$ ). Cell A7 contains the $x_{r}=-5$ value, and the $\wedge$ symbol denotes exponentiation. The denotes multiplication. Cell B3 contains the $E{r}$ value. Next, the rest of the $G{r}$ values are calculated. Select cell B7. Then click Copy in the Clipboard group. Now select cells B8 through B107 by using Find \& Select as before. Then click Paste in the Clipboard group. This fills the cells with the appropriate $G_{r}$ values. (Click on cell B8 or B9 and see how its formula compares with that of B7.)

We now go to the $\psi{r}$ values. Enter 0 in cell C7. Cell C7 contains the value of $\psi{r}$ at $x{r}=-5$. Since this point is well into the classically forbidden region, $\psi{r}$ will be very small here, and we can approximate it as zero. Cell C8 contains the value of $\psi{r}$ at $x{r}=-4.9$. This value will be very small, and we can enter any small number in C 8 without affecting the eigenvalues. Enter $1 \mathrm{E}-4$ in cell C 8 (where E denotes the power of 10 ). Now that we have values in cells C7 and C8 for $\psi{r}$ at the first two points -5.0 and -4.9 [points $x{n-1}$ and $x{n}$ in (4.67)], we use (4.67) to calculate $\psi{r}$ at $x{r}=-4.8$ (point $x{n+1}$ ). Therefore, enter the Eq. (4.67) formula

=(2*C8-C7+5*B8*C8*$D$3^2/6+B7*C7*$D$3^2/12)/(1-B9*$D$3^2/12)

 

in cell C9. After the Enter key is pressed, the value 0.000224 for $\psi{r}$ at $x{r}=-4.8$ appears in cell C9. Select cell C9. Click Copy. Select C10 through C107. Click Paste. Cells C10 through C107 will now be filled with their appropriate $\psi_{r}$ values. As a further check that you entered the C9 formula correctly, verify that cell C10 contains 0.000401 .

Since the number of nodes tells us which eigenvalues our energy guess is between (Fig. 4.8), we want to count and display the number of nodes in $\psi{r}$. To do this, enter into cell D 9 the formula $=\operatorname{IF}(\mathrm{C} 9 * \mathrm{C} 8<0,1,0)$. This formula enters the number 1 into D 9 if C 9 times C 8 is negative and enters 0 into D 9 if C 9 times C 8 is not negative. If there is a node between the $x{r}$ values in A8 and A9, then the $\psi{r}$ values in C8 and C9 will have opposite signs, and the value 1 will be entered into D9. Use Copy and Paste to copy the cell D9 formula into cells D10 through D107. Enter nodes = into cell E2. Enter =SUM(D9:D107) into F2. This formula gives the sum of the values in D9 through D107 and thus gives the number of interior nodes in $\psi{r}$.

Next, we graph $\psi{r}$ versus $x{r}$. Select cells A7 through A107 and C7 through C107 by clicking Find \& Select, clicking Go To, typing A7:A107,C7:C107 in the Reference box, and clicking OK. Then click the Insert tab and in the Chart group click the Scatter icon and then click the chart showing smoothed lines with markers (Scatter with Smooth Lines and Markers). On the chart that appears, right-click Series1 and choose Delete. Right-click the horizontal grid line at $4 \times 10^{6}$ and choose Delete. The spreadsheet will look like Fig. 4.9.

Since $x{r}=5$ is well into the right-hand classically forbidden region, $\psi{r}$ should be very close to zero at this point. However, the graph shows that for our choice $E{r}=0$, the wave function $\psi{r}$ is very large at $x{r}=5$. Therefore, $E{r}=0$ does not give a well-behaved $\psi{r}$, and we must try a different $E{r}$. Cell F2 has a zero, so this $\psi{r}$ has no nodes. Therefore (Fig. 4.8), the guess $E{r}=0$ is less than the true ground-state energy. Let us try $E{r}=2$. Select cell B3 and enter 2 into it. After you press Enter, the spreadsheet will recalculate every column B and column C cell whose value depends on $E{r}$ (all column B and C cells except C 7 and C 8 change) and will then replot the graph. The graph for $E{r}=2$ goes to a very large positive $\psi{r}$ value at $x{r}=5$. Also, cell F2 tells us that $\psi{r}$ for $E{r}=2$ contains two nodes. These are not readily visible on the graph, but the column-C data show that $\psi{r}$ changes sign between the $x_{r}$ values -0.4 and -0.3 and between 1.2 and 1.3 . We are

FIGURE 4.9 Spreadsheet for Numerov-method solution of the harmonic oscillator.

	A	B	C	D	E	F	G	H	I
1	Numerov m	thod for S	rodinger	n, V	$=0.5 \mathrm{kx}$ ^2
2					Nodes=	0
3	Er =	0	$\mathrm{s}=$	0.1
4						1000000
5	xr	Gr	psir			90000
6						80000
7	-5	25	0			70000
8	-4.9	24.01	1.00E-04			60000
9	-4.8	23.04	0.000224	0		50000
10	-4.7	22.09	0.000401	0		40000
11	-4.6	21.16	0.000668	0		30000
12	-4.5	20.25	0.001078	0		20000
13	-4.4	19.36	0.001709	0		100000
14	-4.3	18.49	0.002675	0		,
15	-4.2	17.64	0.004142	0	-6	-4 -2	0	4
16	-4.1	16.81	0.006348	0
17	-4	16	0.009633	0

looking for the ground-state eigenfunction, which does not contain a node, or rather, in our approximation, will contain nodes at -5 and 5 . The presence of the two interior nodes shows (Fig. 4.8) that the value 2 for $E{r}$ is not only too high for the ground state, but is higher than $E{r}$ for the first excited state (whose wave function contains one interior node). We therefore need to try a lower value of $E_{r}$.

Before doing so, let us change the graph scale so as to make the nodes more visible. Double click on the $y$ axis of the graph. In the Format Axis dialog box that appears, click Axis Options at the left and click Fixed next to Minimum and Maximum. Replace the original numbers in the Minimum and Maximum boxes with -10 and 10, respectively. Then click Close. The graph will be redrawn with -10 and 10 as the minimum and maximum $y$-axis values, making the two nodes easily visible.

We now change $E{r}$ to a smaller value, say 1.2. Enter 1.2 in cell B3. We get a $\psi{r}$ that goes to a large negative value on the right and that has only one node. The presence of one node tells us that we are now below the energy of the second-lowest state and above the ground-state energy (see Fig. 4.8). We have bracketed the ground-state energy to lie between 0 and 1.2. Let us average these two numbers and try 0.6 as the energy. When we enter 0.6 into cell B3, we get a function with one node, so we are still above the groundstate energy. Since the maximum on the graph is now off scale, it's a good idea to change the graph scale and reset the $y$ maximum and minimum values to 25 and -25 .

We have found the lowest eigenvalue to be between 0 and 0.6 . Averaging these values, we enter 0.3 into B3. This gives a function that has no nodes, so 0.3 is below the groundstate reduced energy, and $E{r}$ is between 0.3 and 0.6 . Averaging these, we enter 0.45 into B3. We get a function that has no nodes, so we are still below the correct eigenvalue. However, if we rescale the $y$ axis suitably (taking, for example, -15 and 30 as the minimum and maximum values), we see a function that for values of $x{r}$ less than 0.2 resembles closely what we expect for the ground state, so we are getting warm. The eigenvalue is now known to be between 0.45 and 0.60 . Averaging these, we enter 0.525 into B3. We get a function with one node, so we are too high. Averaging 0.45 and 0.525 , we try 0.4875 , which we find to be too low.

Continuing in this manner, we get the successive values 0.50625 (too high), 0.496875 (too low), . . , 0.4999995943548 (low), 0.4999996301176 (high). Thus we have found
0.4999996 as the lowest eigenvalue. Since $E_{r}=E / h \nu$, we should have gotten 0.5 . The slight error is due to the approximations of the Numerov method.

Suppose we want the second-lowest eigenvalue. We previously found this eigenvalue to be below 2.0 , so it lies between 0.5 and 2.0. Averaging these numbers, we enter 1.25 into B3. We get a function that has the desired one node but that goes to a large negative value at the right, rather than returning to the $x$ axis at the right. Therefore, 1.25 is too low (Fig. 4.8). Averaging 1.25 and 2.0 , we next try 1.625 . This gives a function with two nodes, so we are too high. Continuing, we get the successive values 1.4375 (low), 1.53125 (high), 1.484375 (low), 1.5078125 (high), and so on.

To test the accuracy of the eigenvalues found, we can repeat the calculation with $s{r}$ half as large and see if the new eigenvalues differ significantly from those found with the larger $s{r}$. Also, we can start further into the classically forbidden region.

Finding eigenvalues as we have done by trial and error is instructive and fun the first few times, but if you have a lot of eigenvalues to find, you can use a faster method. Most spreadsheets have a built-in program that can adjust the value in one cell so as to yield a desired value in a second cell. This tool is called the Solver in Excel. Click the Data tab. In the Analysis group (at the right), see if there is an icon for Solver. (If the Solver icon is missing, click the File tab and click Options at the left. In the Excel Options box, click Add-Ins at the left. In the Manage box select Excel Add-Ins at the bottom and click the Go button. In the Add-Ins box check Solver Add-In and click OK. After a while, the Solver icon will appear in the Analysis group.) To see how the Solver works, enter 0 into cell B3. Click the Solver icon. The Solver Parameters box opens. The $\psi{r}$ value at $x{r}=5$ is in cell C107, and we want to make this value zero. Therefore, in the Set Objective box of the Solver, enter $\$ C \$ 107$. Next to To: click on Value of and enter 0 . We want to adjust the energy so as to satisfy the boundary condition at $x_{r}=5$, so click in the By Changing Variable Cells box and enter $\$ B \$ 3$. Select the Solver Method as GRG Nonlinear. Then click on Solve. When Solver declares it has found a solution, select Accept Solver solution to close the Solver box. The solution found by Excel has 0.500002 in cell B3 (the formula bar will show the full value if you select cell B3). Cell C 107 will have the value -6.38 . Since this value is not close to the desired value of 0 , click the Solver icon and then click Solve, thereby re-running Solver starting from the 0.500002 value. This time Solver gives 0.4999996089 , similar to the value found by hand, and C 107 has the value $-3.66 \times 10^{-9}$. (The Solver uses either the quasi-Newton or the conjugate-gradient method, both of which are discussed in Section 15.10.)

To find higher eigenvalues using the Solver, start with a value in B3 that is well above the previously found eigenvalue. If the program converges to the previous eigenvalue, start with a still-higher value in B3. You can check which eigenvalue you have found by counting the nodes in the wave function. If the Solver fails to find the desired eigenvalue, use trial and error to find an approximate value and then use the Solver starting from the approximate eigenvalue.

The wave function we have found is unnormalized. The normalization constant is given by Eq. (3.93) as $N=\left[\int{-\infty}^{\infty}\left|\psi{r}\right|^{2} d x{r}\right]^{-1 / 2}$. We have $\int{-\infty}^{\infty}\left|\psi{r}\right|^{2} d x{r} \approx \int{-5}^{5} \psi{r}^{2} d x{r} \approx \sum{i=1}^{100} \psi{r, i}^{2} s{r}$, where the $\psi{r, i}$ values are in column B. Enter npsir in E5. In cell D109, enter the formula $=$ SUMSQ(C8:C107)*\$D\$3. The SUMSQ function adds the squares of a series of numbers. Enter $=C 7 / \$ D \$ 109 \wedge 0.5$ in E7. Copy and paste E7 into E8 through E107. Column E will contain the normalized $\psi{r}$ values if $E_{\text {guess }}$ is equal to an eigenvalue.

Excel is widely used to do statistical and scientific calculations, but studies of Excel 2007 and its predecessors found many significant errors [B. D. McCullough and D. A. Heiser, Comput. Statist. Data Anal., 52, 4570 (2008)]. For example, McCullough and Heiser state: "It has been noted that Excel Solver has a marked tendency to stop at a point that is not a solution and declare that it has found a solution." Therefore one should
always verify the correctness of the Solver's solution. A study of Excel 2010 found that many of the errors present in earlier versions of Excel were fixed, but some problems remain [G. Mélard, "On the Accuracy of Statistical Procedures in Excel 2010," available at homepages.ulb.ac.be/~gmelard/rech/gmelard_csda24.pdf]. Mélard noted that "The conclusion is that Microsoft did not make an attempt to fix all the errors in Excel, and this point needs to be made strongly." Mélard found that the Solver yielded results with zero significant-figure accuracy in a substantial fraction of test cases.

Use of Mathcad to Solve the One-Dimensional Schrödinger Equation

Several programs classified as computer algebra systems do a wide variety of mathematical procedures, including symbolic integration and differentiation, numerical integration, algebraic manipulations, solving systems of equations, graphing, and matrix computations. Examples of such computer algebra systems are Maple, Mathematica, MATLAB, Mathcad, and LiveMath Maker. The Numerov procedure can be performed using these programs. One nice feature of Mathcad is its ability to produce animations ("movies"). With Mathcad one can create a movie showing how $\psi{r}$ changes as $E{r}$ goes through an eigenvalue.

Summary of Numerov-Method Steps

Problems 4.30-4.38 apply the Numerov method to several other one-dimensional problems. In solving these problems, you need to (a) find combinations of the constants in the problem that will give a dimensionless reduced energy and length [Eq. (4.68)]; (b) convert the Schrödinger equation to dimensionless form and find what $G{r}\left(x{r}\right)$ in (4.82) is; (c) decide on a maximum $E{r}$ value, below which you will find all eigenvalues; (d) locate the boundaries between the classically allowed and forbidden regions for this maximum $E{r}$ value and choose $x{r, 0}$ and $x{r, \text { max }}$ values in the classically forbidden regions (for the particle in a box with infinitely high walls, use $x{0}=0$ and $x{\max }=l$ ); and (e) decide on a value for the interval $s_{r}$.

In this chapter we discuss angular momentum, and in the next chapter we show that for the stationary states of the hydrogen atom the magnitude of the electron's angular momentum is constant. As a preliminary, we consider what criterion we can use to decide which properties of a system can be simultaneously assigned definite values.

In Section 3.3 we postulated that if the state function $\Psi$ is an eigenfunction of the operator $\hat{A}$ with eigenvalue $s$, then a measurement of the physical property $A$ is certain to give the result $s$. If $\Psi$ is simultaneously an eigenfunction of the two operators $\hat{A}$ and $\hat{B}$, that is, if $\hat{A} \Psi=s \Psi$ and $\hat{B} \Psi=t \Psi$, then we can simultaneously assign definite values to the physical quantities $A$ and $B$. When will it be possible for $\Psi$ to be simultaneously an eigenfunction of two different operators? In Chapter 7, we shall prove the following two theorems. First, a necessary condition for the existence of a complete set of simultaneous eigenfunctions of two operators is that the operators commute with each other. (The word complete is used here in a certain technical sense, which we won't worry about until Chapter 7.) Conversely, if $\hat{A}$ and $\hat{B}$ are two commuting operators that correspond to physical quantities, then there exists a complete set of functions that are eigenfunctions of both $\hat{A}$ and $\hat{B}$. Thus, if $[\hat{A}, \hat{B}]=0$, then $\Psi$ can be an eigenfunction of both $\hat{A}$ and $\hat{B}$.

Recall that the commutator of $\hat{A}$ and $\hat{B}$ is $[\hat{A}, \hat{B}] \equiv \hat{A} \hat{B}-\hat{B} \hat{A}$ [Eq. (3.7)]. The following identities are helpful in evaluating commutators. These identities are proved by writing out the commutators in detail (Prob. 5.2):

\(
\begin{gather}
{[\hat{A}, \hat{B}]=-[\hat{B}, \hat{A}]} \tag{5.1}\
{\left[\hat{A}, \hat{A}^{n}\right]=0, \quad n=1,2,3, \ldots} \
{[k \hat{A}, \hat{B}]=[\hat{A}, k \hat{B}]=k[\hat{A}, \hat{B}]} \
{[\hat{A}, \hat{B}+\hat{C}]=[\hat{A}, \hat{B}]+[\hat{A}, \hat{C}]} \
\hline[\hat{A}, \hat{B} \hat{C}]=[\hat{A}, \hat{B}] \hat{C}+\hat{B}, \hat{B}[\hat{A}, \hat{C}]=[\hat{A}, \hat{C}]+[\hat{B}, \hat{C}] \
\hline[\hat{A} \hat{B}, \hat{C}]=[\hat{A}, \hat{C}] \hat{B}+\hat{A}[\hat{B}, \hat{C}] \
\hline
\end{gather}
\)

where $k$ is a constant and the operators are assumed to be linear.

EXAMPLE

Starting from $[\partial / \partial x, x]=1$ [Eq. (3.8)], use the commutator identities (5.1)-(5.5) to find (a) $\left[\hat{x}, \hat{p}{x}\right]$; (b) $\left[\hat{x}, \hat{p}{x}^{2}\right]$; and (c) $[\hat{x}, \hat{H}]$ for a one-particle, three-dimensional system.
(a) Use of (5.3), (5.1), and $[\partial / \partial x, x]=1$ gives

\(
\begin{align}
& {\left[\hat{x}, \hat{p}{x}\right]=\left[x, \frac{\hbar}{i} \frac{\partial}{\partial x}\right]=\frac{\hbar}{i}\left[x, \frac{\partial}{\partial x}\right]=-\frac{\hbar}{i}\left[\frac{\partial}{\partial x}, x\right]=-\frac{\hbar}{i}} \
& {\left[\hat{x}, \hat{p}{x}\right]=i \hbar} \tag{5.6}
\end{align}
\)

(b) Use of (5.5) and (5.6) gives

\(
\begin{align}
& {\left[\hat{x}, \hat{p}{x}^{2}\right]=\left[\hat{x}, \hat{p}{x}\right] \hat{p}{x}+\hat{p}{x}\left[\hat{x}, \hat{p}{x}\right]=i \hbar \cdot \frac{\hbar}{i} \frac{\partial}{\partial x}+\frac{\hbar}{i} \frac{\partial}{\partial x} \cdot i \hbar} \
& {\left[\hat{x}, \hat{p}{x}^{2}\right]=2 \hbar^{2} \frac{\partial}{\partial x}} \tag{5.7}
\end{align}
\)

(c) Use of (5.4), (5.3), and (5.7) gives

\(
\begin{align}
{[\hat{x}, \hat{H}] } & =[\hat{x}, \hat{T}+\hat{V}]=[\hat{x}, \hat{T}]+[\hat{x}, \hat{V}(x, y, z)]=[\hat{x}, \hat{T}] \
& =\left[\hat{x},(1 / 2 m)\left(\hat{p}{x}^{2}+\hat{p}{y}^{2}+\hat{p}{z}^{2}\right)\right] \
& =(1 / 2 m)\left[\hat{x}, \hat{p}{x}^{2}\right]+(1 / 2 m)\left[\hat{x}, \hat{p}{y}^{2}\right]+(1 / 2 m)\left[\hat{x}, \hat{p}{z}^{2}\right] \
& =\frac{1}{2 m} 2 \hbar^{2} \frac{\partial}{\partial x}+0+0 \
& \quad[\hat{x}, \hat{H}]=\frac{\hbar^{2}}{m} \frac{\partial}{\partial x}=\frac{i \hbar}{m} \hat{p}_{x} \tag{5.8}
\end{align}
\)

EXERCISE Show that for a one-particle, three-dimensional system,

\(
\begin{equation}
\left[\hat{p}_{x}, \hat{H}\right]=-i \hbar \partial V(x, y, z) / \partial x \tag{5.9}
\end{equation}
\)

These commutators have important physical consequences. Since $\left[\hat{x}, \hat{p}{x}\right] \neq 0$, we cannot expect the state function to be simultaneously an eigenfunction of $\hat{x}$ and of $\hat{p}{x}$. Hence we cannot simultaneously assign definite values to $x$ and $p_{x}$, in agreement with the uncertainty principle. Since $\hat{x}$ and $\hat{H}$ do not commute, we cannot expect to assign definite values to the energy and the $x$ coordinate at the same time. A stationary state (which has a definite energy) shows a spread of possible values for $x$, the probabilities for observing various values of $x$ being given by the Born postulate.

For a state function $\Psi$ that is not an eigenfunction of $\hat{A}$, we get various possible outcomes when we measure $A$ in identical systems. We want some measure of the spread or dispersion in the set of observed values $A{i}$. If $\langle A\rangle$ is the average of these values, then the deviation of each measurement from the average is $A{i}-\langle A\rangle$. If we averaged all the deviations, we would get zero, since positive and negative deviations would cancel. Hence to make all deviations positive, we square them. The average of the squares of the deviations is called the variance of $A$, symbolized in statistics by $\sigma^{2}(A)$ and in quantum mechanics by $(\Delta A)^{2}$ :

\(
\begin{equation}
(\Delta A)^{2} \equiv \sigma^{2}(A) \equiv\left\langle(A-\langle A\rangle)^{2}\right\rangle=\int \Psi^{}(\hat{A}-\langle A\rangle)^{2} \Psi d \tau \tag{5.10}
\end{}
\)

where the average-value expression (3.88) was used. The definition (5.10) is equivalent to (Prob. 5.7)

\(
\begin{equation}
(\Delta A)^{2}=\left\langle A^{2}\right\rangle-\langle A\rangle^{2} \tag{5.11}
\end{equation}
\)

The positive square root of the variance is called the standard deviation, $\sigma(A)$ or $\Delta A$. The standard deviation is the most commonly used measure of spread, and we shall take it as the measure of the "uncertainty" in the property $A$.

Robertson in 1929 proved that the product of the standard deviations of two properties of a quantum-mechanical system whose state function is $\Psi$ must satisfy the inequality

\(
\begin{equation}
\sigma(A) \sigma(B) \equiv \Delta A \Delta B \geq \frac{1}{2}\left|\int \Psi^{}[\hat{A}, \hat{B}] \Psi d \tau\right| \tag{5.12}
\end{}
\)

The proof of (5.12), which follows from the postulates of quantum mechanics, is outlined in Prob. 7.60. If $\hat{A}$ and $\hat{B}$ commute, then the integral in (5.12) is zero, and $\Delta A$ and $\Delta B$ may both be zero, in agreement with the previous discussion.

As an example of (5.12), we find, using (5.6), $\left|z{1} z{2}\right|=\left|z{1}\right|\left|z{2}\right|$ [Eq. (1.34)], and normalization:

\(
\begin{gather}
\Delta x \Delta p_{x} \geq \frac{1}{2}\left|\int \Psi^{}\left[\hat{x}, \hat{p}{x}\right] \Psi d \tau\right|=\frac{1}{2}\left|\int \Psi i \hbar \Psi d \tau\right|=\frac{1}{2} \hbar|i|\left|\int \Psi \Psi d \tau\right| \
\sigma(x) \sigma\left(p{x}\right) \equiv \Delta x \Delta p_{x} \geq \frac{1}{2} \hbar \tag{5.13}
\end{}
\)

Equation (5.13) is usually considered to be the quantitative statement of the Heisenberg uncertainty principle (Section 1.3). However, the meaning of the standard deviations in Eqs. (5.12) and (5.13) is rather different than the meaning of the uncertainties in Section 1.3. To find $\Delta x$ in (5.13) we take a very large number of systems, each of which has the same state function $\Psi$, and we perform one measurement of $x$ in each system. From these measured values, symbolized by $x{i}$, we calculate $\langle x\rangle$ and the squares of the deviations $\left(x{i}-\langle x\rangle\right)^{2}$. We average the squares of the deviations to get the variance and take the square root to get the standard deviation $\sigma(x) \equiv \Delta x$. Then we take many systems, each of which is in the same state $\Psi$ as used to get $\Delta x$, and we do a single measurement of $p{x}$ in each system, calculating $\Delta p{x}$ from these measurements. Thus, the statistical quantities $\Delta x$ and $\Delta p{x}$ in (5.13) are not errors of individual measurements and are not found from simultaneous measurements of $x$ and $p{x}$ (see Ballentine, pp. 225-226).

Let $\varepsilon(x)$ be the typical error in a single measurement of $x$ and let $\eta\left(p{x}\right)$ be the typical disturbance in $p{x}$ caused by the measurement of $x$. In 1927, Heisenberg analyzed specific thought experiments that perform position measurements and concluded that the product $\varepsilon(x) \eta\left(p_{x}\right)$ was of the order of magnitude of $h$ or larger. Heisenberg did not give a precise definition of these quantities. Ozawa rewrote Heisenberg's relation as

\(
\varepsilon(x) \eta\left(p_{x}\right) \geq \frac{1}{2} \hbar
\)

where $\varepsilon(x)$ is defined as the root-mean-square deviation of measured $x$ values from the theoretical value and $\eta\left(p{x}\right)$ is defined as the root-mean-square deviation of the change in $p{x}$ produced by the measurement of $x$. More generally, for any two properties, Ozawa wrote the Heisenberg uncertainty principle as

\(
\begin{equation}
\varepsilon(A) \eta(B) \geq \frac{1}{2}\left|\int \Psi^{}[\hat{A}, \hat{B}] \Psi d \tau\right| \tag{5.14}
\end{}
\)

Ozawa presented arguments that, in certain circumstances, the Heisenberg inequality (5.14) can be violated. Ozawa derived the following relation to replace (5.14) [M. Ozawa, Phys. Rev. A, 67, 042105 (2003); available at arxiv.org/abs/quant-ph/0207121]:

\(
\varepsilon(A) \eta(B)+\varepsilon(A) \sigma(B)+\sigma(A) \eta(B) \geq \frac{1}{2}\left|\int \Psi^{*}[\hat{A}, \hat{B}] \Psi d \tau\right|
\)

where $\sigma(A)$ and $\sigma(B)$ are found from (5.10). In 2012, an experiment that measured components of neutron spin found that the Heisenberg error-disturbance inequality (5.14) was not obeyed for the spin components but that the Ozawa inequality was obeyed [J. Erhart et al., Nature Physics, 8, 185 (2012); arxiv.org/abs/1201.1833].

Another inequality is the Heisenberg uncertainty relation for simultaneous measurement of two properties $A$ and $B$ by an apparatus that measures both $A$ and $B$ :

\(
\varepsilon(A) \varepsilon(B) \geq \frac{1}{2}\left|\int \Psi^{*}[\hat{A}, \hat{B}] \Psi d \tau\right|
\)

where $\varepsilon(A)$ and $\varepsilon(B)$ are the experimental errors in the measured $A$ and $B$ values. This relation has been proven to hold, provided a certain plausible assumption (believed to hold for all currently available measuring devices) is valid (see references 6-12 in the abovecited Ozawa paper).

EXAMPLE

Equations (3.91), (3.92), (3.39), the equation following (3.89), and Prob. 3.48 give for the ground state of the particle in a three-dimensional box

\(
\langle x\rangle=\frac{a}{2}, \quad\left\langle x^{2}\right\rangle=\left(\frac{1}{3}-\frac{1}{2 \pi^{2}}\right) a^{2}, \quad\left\langle p{x}\right\rangle=0, \quad\left\langle p{x}^{2}\right\rangle=\frac{h^{2}}{4 a^{2}}
\)

Use these results to check that the uncertainty principle (5.13) is obeyed.
We have

\(
\begin{gathered}
(\Delta x)^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=\left(\frac{1}{3}-\frac{1}{2 \pi^{2}}\right) a^{2}-\frac{a^{2}}{4}=\frac{\pi^{2}-6}{12 \pi^{2}} a^{2} \
\Delta x=\left(\frac{\pi^{2}-6}{12}\right)^{1 / 2} \frac{a}{\pi} \
\left(\Delta p{x}\right)^{2}=\left\langle p{x}^{2}\right\rangle-\left\langle p{x}\right\rangle^{2}=\frac{h^{2}}{4 a^{2}}, \quad \Delta p{x}=\frac{h}{2 a} \
\Delta x \Delta p_{x}=\left(\frac{\pi^{2}-6}{12}\right)^{1 / 2} \frac{h}{2 \pi}=0.568 \hbar>\frac{1}{2} \hbar
\end{gathered}
\)

There is also an uncertainty relation involving energy and time:

\(
\begin{equation}
\Delta E \Delta t \geq \frac{1}{2} \hbar \tag{5.15}
\end{equation}
\)

Some texts state that (5.15) is derived from (5.12) by taking $i \hbar \partial / \partial t$ as the energy operator and multiplication by $t$ as the time operator. However, the energy operator is the Hamiltonian $\hat{H}$ and not $i \hbar \partial / \partial t$. Moreover, time is not an observable but is a parameter in quantum mechanics. Hence there is no quantum-mechanical time operator. (The noun observable in quantum mechanics means a physically measurable property of a system.) Equation (5.15) must be derived by a special treatment, which we omit.
(See Ballentine, Section 12.3.) The derivation of (5.15) shows that $\Delta t$ is to be interpreted as the lifetime of the state whose energy is uncertain by $\Delta E$. It is often stated that $\Delta t$ in (5.15) is the duration of the energy measurement. However, Aharonov and Bohm have shown that "energy can be measured reproducibly in an arbitrarily short time" [Y. Aharonov and D. Bohm, Phys. Rev., 122, 1649 (1961); 134, B 1417 (1964); see also S. Massar and S. Popescu, Phys. Rev. A, 71, 042106 (2005); P. Busch, The Time-Energy Uncertainty Relation, arxiv.org/abs/quant-ph/0105049].

Now consider the possibility of simultaneously assigning definite values to three physical quantities: $A, B$, and $C$. Suppose

\(
\begin{equation}
[\hat{A}, \hat{B}]=0 \quad \text { and } \quad[\hat{A}, \hat{C}]=0 \tag{5.16}
\end{equation}
\)

Is this enough to ensure that there exist simultaneous eigenfunctions of all three operators? Since $[\hat{A}, \hat{B}]=0$, we can construct a common set of eigenfunctions for $\hat{A}$ and $\hat{B}$. Since $[\hat{A}, \hat{C}]=0$, we can construct a common set of eigenfunctions for $\hat{A}$ and $\hat{C}$. If these two sets of eigenfunctions are the same, then we will have a common set of eigenfunctions for all three operators. Hence we ask: Is the set of eigenfunctions of the linear operator $\hat{A}$ uniquely determined (apart from arbitrary multiplicative constants)? The answer is, in general, no. If there is more than one independent eigenfunction corresponding to an eigenvalue of $\hat{A}$ (that is, degeneracy), then any linear combination of the eigenfunctions of the degenerate eigenvalue is an eigenfunction of $\hat{A}$ (Section 3.6). It might well be that the proper linear combinations needed to give eigenfunctions of $\hat{B}$ would differ from the linear combinations that give eigenfunctions of $\hat{C}$. It turns out that, to have a common complete set of eigenfunctions of all three operators, we require that $[\hat{B}, \hat{C}]=0$ in addition to (5.16). To have a complete set of functions that are simultaneous eigenfunctions of several operators, each operator must commute with every other operator.

In the next section we shall solve the eigenvalue problem for angular momentum, which is a vector property. We therefore first review vectors.

Physical properties (for example, mass, length, energy) that are completely specified by their magnitude are called scalars. Physical properties (for example, force, velocity, momentum) that require specification of both magnitude and direction are called vectors. A vector is represented by a directed line segment whose length and direction give the magnitude and direction of the property.

The sum of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined by the following procedure: Slide the first vector so that its tail touches the head of the second vector, keeping the direction of the first vector fixed. Then draw a new vector from the tail of the second vector to the head of the first vector. See Fig. 5.1. The product of a vector and a scalar, $c \mathbf{A}$, is defined as a vector of length $|c|$ times the length of $\mathbf{A}$ with the same direction as $\mathbf{A}$ if $c$ is positive, or the opposite direction to $\mathbf{A}$ if $c$ is negative.

FIGURE 5.1 Addition of two vectors.

(a)

(b) $\mathbf{C}=\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}$

FIGURE 5.2 Unit vectors $\mathbf{i}, \mathbf{j}$,
$\mathbf{k}$, and components of $\mathbf{A}$.

To obtain an algebraic (as well as geometric) way of representing vectors, we set up Cartesian coordinates in space. We draw a vector of unit length directed along the positive $x$ axis and call it i. (No connection with $i=\sqrt{-1}$.) Unit vectors in the positive $y$ and $z$ directions are called $\mathbf{j}$ and $\mathbf{k}$ (Fig. 5.2). To represent any vector $\mathbf{A}$ in terms of the three unit vectors, we first slide $\mathbf{A}$ so that its tail is at the origin, preserving its direction during this process. We then find the projections of $\mathbf{A}$ on the $x, y$, and $z$ axes: $A{x}, A{y}$, and $A_{z}$. From the definition of vector addition, it follows that (Fig. 5.2)

\(
\begin{equation}
\mathbf{A}=A{x} \mathbf{i}+A{y} \mathbf{j}+A_{z} \mathbf{k} \tag{5.17}
\end{equation}
\)

We can specify $\mathbf{A}$ by specifying its three components: $\left(A{x}, A{y}, A_{z}\right)$. A vector in threedimensional space can therefore be defined as an ordered set of three numbers.

Two vectors $\mathbf{A}$ and $\mathbf{B}$ are equal if and only if all their corresponding components are equal: $A{x}=B{x}, A{y}=B{y}, A{z}=B{z}$. Therefore a vector equation is equivalent to three scalar equations.

To add two vectors, we add corresponding components:

\(
\mathbf{A}+\mathbf{B}=A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}+B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}
\)

\(
\begin{equation}
\mathbf{A}+\mathbf{B}=\left(A{x}+B{x}\right) \mathbf{i}+\left(A{y}+B{y}\right) \mathbf{j}+\left(A{z}+B{z}\right) \mathbf{k} \tag{5.18}
\end{equation}
\)

Also, if $c$ is a scalar, then

\(
\begin{equation}
c \mathbf{A}=c A{x} \mathbf{i}+c A{y} \mathbf{j}+c A_{z} \mathbf{k} \tag{5.19}
\end{equation}
\)

The magnitude of a vector $\mathbf{A}$ is its length and is denoted by $A$ or $|\mathbf{A}|$. The magnitude $A$ is a scalar.

The dot product or scalar product $\mathbf{A} \cdot \mathbf{B}$ of two vectors is defined by

\(
\begin{equation}
\mathbf{A} \cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}| \cos \theta=\mathbf{B} \cdot \mathbf{A} \tag{5.20}
\end{equation}
\)

where $\theta$ is the angle between the vectors. The dot product, being the product of three scalars, is a scalar. Note that $|\mathbf{A}| \cos \theta$ is the projection of $\mathbf{A}$ on $\mathbf{B}$. From the definition of vector addition, it follows that the projection of the vector $\mathbf{A}+\mathbf{B}$ on some vector $\mathbf{C}$ is the sum of the projections of $\mathbf{A}$ and of $\mathbf{B}$ on $\mathbf{C}$. Therefore

\(
\begin{equation}
(\mathbf{A}+\mathbf{B}) \cdot \mathbf{C}=\mathbf{A} \cdot \mathbf{C}+\mathbf{B} \cdot \mathbf{C} \tag{5.21}
\end{equation}
\)

Since the three unit vectors $\mathbf{i}, \mathbf{j}$, and $\mathbf{k}$ are each of unit length and are mutually perpendicular, we have

\(
\begin{equation}
\mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=\cos 0=1, \quad \mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{k} \cdot \mathbf{i}=\cos (\pi / 2)=0 \tag{5.22}
\end{equation}
\)

Using (5.22) and the distributive law (5.21), we have

\(
\begin{gather}
\mathbf{A} \cdot \mathbf{B}=\left(A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}\right) \cdot\left(B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}\right) \
\mathbf{A} \cdot \mathbf{B}=A{x} B{x}+A{y} B{y}+A{z} B{z} \tag{5.23}
\end{gather}
\)

where six of the nine terms in the dot product are zero.
Equation (5.20) gives

\(
\begin{equation}
\mathbf{A} \cdot \mathbf{A}=|\mathbf{A}|^{2} \tag{5.24}
\end{equation}
\)

Using (5.23), we therefore have

\(
\begin{equation}
|\mathbf{A}|=\left(A{x}^{2}+A{y}^{2}+A_{z}^{2}\right)^{1 / 2} \tag{5.25}
\end{equation}
\)

For three-dimensional vectors, there is another type of product. The cross product or vector product $\mathbf{A} \times \mathbf{B}$ is a vector whose magnitude is

\(
\begin{equation}
|\mathbf{A} \times \mathbf{B}|=|\mathbf{A}||\mathbf{B}| \sin \theta \tag{5.26}
\end{equation}
\)

whose line segment is perpendicular to the plane defined by $\mathbf{A}$ and $\mathbf{B}$, and whose direction is such that $\mathbf{A}, \mathbf{B}$, and $\mathbf{A} \times \mathbf{B}$ form a right-handed system (just as the $x, y$, and $z$ axes form a right-handed system). See Fig. 5.3. From the definition it follows that

\(
\mathbf{B} \times \mathbf{A}=-\mathbf{A} \times \mathbf{B}
\)

Also, it can be shown that (Taylor and Mann, Section 10.2)

\(
\begin{equation}
\mathbf{A} \times(\mathbf{B}+\mathbf{C})=\mathbf{A} \times \mathbf{B}+\mathbf{A} \times \mathbf{C} \tag{5.27}
\end{equation}
\)

For the three unit vectors, we have

\(
\begin{gathered}
\mathbf{i} \times \mathbf{i}=\mathbf{j} \times \mathbf{j}=\mathbf{k} \times \mathbf{k}=\sin 0=0 \
\mathbf{i} \times \mathbf{j}=\mathbf{k}, \quad \mathbf{j} \times \mathbf{i}=-\mathbf{k}, \quad \mathbf{j} \times \mathbf{k}=\mathbf{i}, \quad \mathbf{k} \times \mathbf{j}=-\mathbf{i}, \quad \mathbf{k} \times \mathbf{i}=\mathbf{j}, \quad \mathbf{i} \times \mathbf{k}=-\mathbf{j}
\end{gathered}
\)

Using these equations and the distributive property (5.27), we find

\(
\begin{gathered}
\mathbf{A} \times \mathbf{B}=\left(A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}\right) \times\left(B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}\right) \
\mathbf{A} \times \mathbf{B}=\left(A{y} B{z}-A{z} B{y}\right) \mathbf{i}+\left(A{z} B{x}-A{x} B{z}\right) \mathbf{j}+\left(A{x} B{y}-A{y} B{x}\right) \mathbf{k}
\end{gathered}
\)

FIGURE 5.3 Cross product of two vectors.

As a memory aid, we can express the cross product as a determinant (see Section 8.3):

\(
\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \tag{5.28}\
A{x} & A{y} & A{z} \
B{x} & B{y} & B{z}
\end{array}\right|=\mathbf{i}\left|\begin{array}{cc}
A{y} & A{z} \
B{y} & B{z}
\end{array}\right|-\mathbf{j}\left|\begin{array}{cc}
A{x} & A{z} \
B{x} & B{z}
\end{array}\right|+\mathbf{k}\left|\begin{array}{cc}
A{x} & A{y} \
B{x} & B{y}
\end{array}\right|
\)

We define the vector operator del as

\(
\begin{equation}
\nabla \equiv \mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z} \tag{5.29}
\end{equation}
\)

From Eq. (3.23), the operator for the linear-momentum vector is $\hat{\mathbf{p}}=-i \hbar \nabla$.
The gradient of a function $g(x, y, z)$ is defined as the result of operating on $g$ with del:

\(
\begin{equation}
\operatorname{grad} g(x, y, z) \equiv \nabla g(x, y, z) \equiv \mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z} \tag{5.30}
\end{equation}
\)

The gradient of a scalar function is a vector function. The vector $\nabla g(x, y, z)$ represents the spatial rate of change of the function $g$ : The $x$ component of $\nabla g$ is the rate of change of $g$ with respect to $x$, and so on. It can be shown that the vector $\nabla g$ points in the direction in which the rate of change of $g$ is greatest. From Eq. (4.24), the relation between force and potential energy is

\(
\begin{equation}
\mathbf{F}=-\nabla V(x, y, z)=-\mathbf{i} \frac{\partial V}{\partial x}-\mathbf{j} \frac{\partial V}{\partial y}-\mathbf{k} \frac{\partial V}{\partial z} \tag{5.31}
\end{equation}
\)

Suppose that the components of the vector $\mathbf{A}$ are each functions of some parameter $t$; $A{x}=A{x}(t), A{y}=A{y}(t), A{z}=A{z}(t)$. The derivative of $\mathbf{A}$ with respect to $t$ is defined as

\(
\begin{equation}
\frac{d \mathbf{A}}{d t}=\mathbf{i} \frac{d A{x}}{d t}+\mathbf{j} \frac{d A{y}}{d t}+\mathbf{k} \frac{d A_{z}}{d t} \tag{5.32}
\end{equation}
\)

Vector notation is a convenient way to represent the variables of a function. The wave function of a two-particle system can be written as $\psi\left(x{1}, y{1}, z{1}, x{2}, y{2}, z{2}\right)$. If $\mathbf{r}{1}$ is the vector from the origin to particle 1 , then $\mathbf{r}{1}$ has components $x{1}, y{1}, z{1}$ and specification of $\mathbf{r}{1}$ is equivalent to specification of the three coordinates $x{1}, y{1}, z{1}$. The same is true for the vector $\mathbf{r}{2}$ from the origin to particle 2 . Therefore, we can write the wave function as $\psi\left(\mathbf{r}{1}, \mathbf{r}{2}\right)$. Vector notation can be used in integrals. For example, the integral over all space in Eq. (3.57) is often written as $\int \cdots \int\left|\Psi\left(\mathbf{r}{1}, \ldots, \mathbf{r}{n}, t\right)\right|^{2} d \mathbf{r}{1} \cdots d \mathbf{r}{n}$.

Vectors in $n$-Dimensional Space

The definition of a vector can be generalized to more than three dimensions. A vector $\mathbf{A}$ in three-dimensional space can be defined by its magnitude $|\mathbf{A}|$ and its direction, or it can be defined by its three components $\left(A{x}, A{y}, A{z}\right)$ in a Cartesian coordinate system. Therefore, we can define a three-dimensional vector as a set of three real numbers $\left(A{x}, A{y}, A{z}\right)$ in a particular order. A vector $\mathbf{B}$ in an $n$-dimensional real vector "space" (sometimes called a hyperspace) is defined as an ordered set of $n$ real numbers $\left(B{1}, B{2}, \ldots, B{n}\right)$, where $B{1}, B{2}, \ldots, B{n}$ are the components of $\mathbf{B}$. Don't be concerned that you can't visualize vectors in an $n$-dimensional space.

The variables of a function are often denoted using $n$-dimensional vector notation. For example, instead of writing the wave function of a two-particle system as $\psi\left(\mathbf{r}{1}, \mathbf{r}{2}\right)$, we can define a six-dimensional vector $\mathbf{q}$ whose components are $q{1}=x{1}, q{2}=y{1}$,
$q{3}=z{1}, q{4}=x{2}, q{5}=y{2}, q{6}=z{2}$ and write the wave function as $\psi(\mathbf{q})$. For an $n$-particle system, we can define $\mathbf{q}$ to have $3 n$ components and write the wave function as $\psi(\mathbf{q})$ and the normalization integral over all space as $\int|\psi(\mathbf{q})|^{2} d \mathbf{q}$.

The theory of searching for the equilibrium geometry of a molecule uses $n$-dimensional vectors (Section 15.10). The rest of Section 5.2 is relevant to Section 15.10 and need not be read until you study Section 15.10.

Two $n$-dimensional vectors are equal if all their corresponding components are equal; $\mathbf{B}=\mathbf{C}$ if and only if $B{1}=C{1}, B{2}=C{2}, \ldots, B{n}=C{n}$. Therefore, in $n$-dimensional space, a vector equation is equivalent to $n$ scalar equations. The sum of two $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{D}$ is defined as the vector $\left(B{1}+D{1}, B{2}+D{2}, \ldots, B{n}+D{n}\right)$. The difference is defined similarly. The vector $k \mathbf{B}$ is defined as the vector $\left(k B{1}, k B{2}, \ldots, k B{n}\right)$, where $k$ is a scalar. In three-dimensional space, the vectors $k \mathbf{A}$, where $k>0$, all lie in the same direction. In $n$-dimensional space the vectors $k \mathbf{B}$ all lie in the same direction. Just as the numbers $\left(A{x}, A{y}, A{z}\right)$ define a point in three-dimensional space, the numbers $\left(B{1}, B{2}, \ldots, B_{n}\right)$ define a point in $n$-dimensional space.

The length (or magnitude or Euclidean norm) $|\mathbf{B}|$ (sometimes denoted $|\mathbf{B}|$ ) of an $n$-dimensional real vector is defined as

\(
|\mathbf{B}| \equiv(\mathbf{B} \cdot \mathbf{B})^{1 / 2}=\left(B{1}^{2}+B{2}^{2}+\cdots+B_{n}^{2}\right)^{1 / 2}
\)

A vector whose length is 1 is said to be normalized.
The inner product (or scalar product) $\mathbf{B} \cdot \mathbf{G}$ of two real $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{G}$ is defined as the scalar

\(
\mathbf{B} \cdot \mathbf{G} \equiv B{1} G{1}+B{2} G{2}+\cdots+B{n} G{n}
\)

If $\mathbf{B} \cdot \mathbf{G}=0$, the vectors $\mathbf{B}$ and $\mathbf{G}$ are said to be orthogonal. The cosine of the angle $\theta$ between two $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{C}$ is defined by analogy to (5.20) as $\cos \theta \equiv \mathbf{B} \cdot \mathbf{C} /|\mathbf{B} | \mathbf{C}|$. One can show that this definition makes $\cos \theta$ lie in the range -1 to 1 .

In three-dimensional space, the unit vectors $\mathbf{i}=(1,0,0), \mathbf{j}=(0,1,0), \mathbf{k}=(0,0,1)$ are mutually perpendicular. Also, any vector can be written as a linear combination of these three vectors [Eq. (5.17)]. In an $n$-dimensional real vector space, the unit vectors $\mathbf{e}{1} \equiv(1,0,0, \ldots, 0), \mathbf{e}{2} \equiv(0,1,0, \ldots, 0), \ldots, \mathbf{e}{n} \equiv(0,0,0, \ldots, 1)$ are mutually orthogonal. Since the $n$-dimensional vector $\mathbf{B}$ equals $B{1} \mathbf{e}{1}+B{2} \mathbf{e}{2}+\cdots+B{n} \mathbf{e}{n}$, any $n$-dimensional real vector can be written as a linear combination of the $n$ unit vectors $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}{n}$. This set of $n$ vectors is therefore said to be a basis for the $n$-dimensional real vector space. Since the vectors $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}{n}$ are orthogonal and normalized, they are an orthonormal basis for the real vector space. The scalar product $\mathbf{B} \cdot \mathbf{e}{i}$ gives the component of $\mathbf{B}$ in the direction of the basis vector $\mathbf{e}_{i}$. A vector space has many possible basis sets. Any set of $n$ linearly independent real vectors can serve as a basis for the $n$-dimensional real vector space.

A three-dimensional vector can be specified by its three components or by its length and its direction. The direction can be specified by giving the three angles that the vector makes with the positive halves of the $x, y$, and $z$ axes. These angles are the direction angles of the vector and lie in the range 0 to $180^{\circ}$. However, the direction angle with the $z$ axis is fixed once the other two direction angles have been given, so only two direction angles are independent. Thus a three-dimensional vector can be specified by its length and two direction angles. Similarly, in $n$-dimensional space, the direction angles between a vector and each unit vector $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}_{n}$ can be found from the above formula for the cosine of the angle between two vectors. An $n$-dimensional vector can thus be specified by its length and $n-1$ direction angles.

The gradient of a function of three variables is defined by (5.30). The gradient $\nabla f$ of a function $f\left(q{1}, q{2}, \ldots, q_{n}\right)$ of $n$ variables is defined as the $n$-dimensional vector whose components are the first partial derivatives of $f$ :

\(
\nabla f \equiv\left(\partial f / \partial q{1}\right) \mathbf{e}{1}+\left(\partial f / \partial q{2}\right) \mathbf{e}{2}+\cdots+\left(\partial f / \partial q{n}\right) \mathbf{e}{n}
\)

We have considered real, $n$-dimensional vector spaces. Dirac's formulation of quantum mechanics uses a complex, infinite-dimensional vector space, discussion of which is omitted.

In Section 3.3 we found the eigenfunctions and eigenvalues for the linear-momentum operator $\hat{p}_{x}$. In this section we consider the same problem for the angular momentum of a particle. Angular momentum plays a key role in the quantum mechanics of atomic structure. We begin by reviewing the classical mechanics of angular momentum.

Classical Mechanics of One-Particle Angular Momentum

Consider a moving particle of mass $m$. We set up a Cartesian coordinate system that is fixed in space. Let $\mathbf{r}$ be the vector from the origin to the instantaneous position of the particle. We have

\(
\begin{equation}
\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z \tag{5.33}
\end{equation}
\)

where $x, y$, and $z$ are the particle's coordinates at a given instant. These coordinates are functions of time. Defining the velocity vector $\mathbf{v}$ as the time derivative of the position vector, we have [Eq. (5.32)]

\(
\begin{gather}
\mathbf{v} \equiv \frac{d \mathbf{r}}{d t}=\mathbf{i} \frac{d x}{d t}+\mathbf{j} \frac{d y}{d t}+\mathbf{k} \frac{d z}{d t} \tag{5.34}\
v{x}=d x / d t, \quad v{y}=d y / d t, \quad v_{z}=d z / d t
\end{gather}
\)

We define the particle's linear momentum vector $\mathbf{p}$ by

\(
\begin{gather}
\mathbf{p} \equiv m \mathbf{v} \tag{5.35}\
p{x}=m v{x}, \quad p{y}=m v{y}, \quad p{z}=m v{z} \tag{5.36}
\end{gather}
\)

The particle's angular momentum $\mathbf{L}$ with respect to the coordinate origin is defined in classical mechanics as

\(
\begin{gather}
\mathbf{L} \equiv \mathbf{r} \times \mathbf{p} \tag{5.37}\
\mathbf{L}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
x & y & z \
p{x} & p{y} & p{z}
\end{array}\right| \tag{5.38}\
L{x}=y p{z}-z p{y}, \quad L{y}=z p{x}-x p{z}, \quad L{z}=x p{y}-y p{x} \tag{5.39}
\end{gather}
\)

where (5.28) was used. $L{x}, L{y}$, and $L_{z}$ are the components of $\mathbf{L}$ along the $x, y$, and $z$ axes. The angular-momentum vector $\mathbf{L}$ is perpendicular to the plane defined by the particle's position vector $\mathbf{r}$ and its velocity $\mathbf{v}$ (Fig. 5.4).

FIGURE 5.4 L $\equiv \mathbf{r} \times \mathbf{p}$.

The torque $\tau$ acting on a particle is defined as the cross product of $\mathbf{r}$ and the force $\mathbf{F}$ acting on the particle: $\tau \equiv \mathbf{r} \times \mathbf{F}$. One can show that $\tau=d \mathbf{L} / d t$. When no torque acts on a particle, the rate of change of its angular momentum is zero; that is, its angular momentum is constant (or conserved). For a planet orbiting the sun, the gravitational force is radially directed. Since the cross product of two parallel vectors is zero, there is no torque on the planet and its angular momentum is conserved.

One-Particle Orbital-Angular-Momentum Operators

Now let us turn to the quantum-mechanical treatment. In quantum mechanics, there are two kinds of angular momenta. Orbital angular momentum results from the motion of a particle through space, and is the analog of the classical-mechanical quantity $\mathbf{L}$. Spin angular momentum (Chapter 10) is an intrinsic property of many microscopic particles and has no classical-mechanical analog. We are now considering only orbital angular momentum. We get the quantum-mechanical operators for the components of orbital angular momentum of a particle by replacing the coordinates and momenta in the classical equations (5.39) by their corresponding operators [Eqs. (3.21)-(3.23)]. We find

\(
\begin{align}
& \hat{L}{x}=-i \hbar\left(y \frac{\partial}{\partial z}-z \frac{\partial}{\partial y}\right) \tag{5.40}\
& \hat{L}{y}=-i \hbar\left(z \frac{\partial}{\partial x}-x \frac{\partial}{\partial z}\right) \tag{5.41}\
& \hat{L}_{z}=-i \hbar\left(x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}\right) \tag{5.42}
\end{align}
\)

(Since $\hat{y} \hat{p}{z}=\hat{p}{z} \hat{y}$, and so on, we do not run into any problems of noncommutativity in constructing these operators.) Using

\(
\begin{equation}
\hat{L}^{2}=|\hat{\mathbf{L}}|^{2}=\hat{\mathbf{L}} \cdot \hat{\mathbf{L}}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}_{z}^{2} \tag{5.43}
\end{equation}
\)

we can construct the operator for the square of the angular-momentum magnitude from the operators in (5.40)-(5.42).

Since the commutation relations determine which physical quantities can be simultaneously assigned definite values, we investigate these relations for angular momentum. Operating on some function $f(x, y, z)$ with $\hat{L}_{y}$, we have

\(
\hat{L}_{y} f=-i \hbar\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right)
\)

Operating on this last equation with $\hat{L}_{x}$, we get

\(
\begin{equation}
\hat{L}{x} \hat{L}{y} f=-\hbar^{2}\left(y \frac{\partial f}{\partial x}+y z \frac{\partial^{2} f}{\partial z \partial x}-y x \frac{\partial^{2} f}{\partial z^{2}}-z^{2} \frac{\partial^{2} f}{\partial y \partial x}+z x \frac{\partial^{2} f}{\partial y \partial z}\right) \tag{5.44}
\end{equation}
\)

Similarly,

\(
\begin{gather}
\hat{L}{x} f=-i \hbar\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \
\hat{L}{y} \hat{L}_{x} f=-\hbar^{2}\left(z y \frac{\partial^{2} f}{\partial x \partial z}-z^{2} \frac{\partial^{2} f}{\partial x \partial y}-x y \frac{\partial^{2} f}{\partial z^{2}}+x \frac{\partial f}{\partial y}+x z \frac{\partial^{2} f}{\partial z \partial y}\right) \tag{5.45}
\end{gather}
\)

Subtracting (5.45) from (5.44), we have

\(
\begin{align}
\hat{L}{x} \hat{L}{y} f-\hat{L}{y} \hat{L}{x} f & =-\hbar^{2}\left(y \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial y}\right) \
{\left[\hat{L}{x}, \hat{L}{y}\right] } & =i \hbar \hat{L}_{z} \tag{5.46}
\end{align}
\)

where we used relations such as

\(
\begin{equation}
\frac{\partial^{2} f}{\partial z \partial x}=\frac{\partial^{2} f}{\partial x \partial z} \tag{5.47}
\end{equation}
\)

which are true for well-behaved functions. We could use the same procedure to find [ $\left.\hat{L}{y}, \hat{L}{z}\right]$ and $\left[\hat{L}{z}, \hat{L}{x}\right]$, but we can save time by noting a certain kind of symmetry in (5.40)-(5.42). By a cyclic permutation of $x, y$, and $z$, we mean replacing $x$ by $y$, replacing $y$ by $z$, and replacing $z$ by $x$. If we carry out a cyclic permutation in $\hat{L}{x}$, we get $\hat{L}{y}$; a cyclic permutation in $\hat{L}{y}$ gives $\hat{L}{z}$; and $\hat{L}{z}$ is transformed into $\hat{L}{x}$ by a cyclic permutation. Hence, by carrying out two successive cyclic permutations on (5.46), we get

\(
\begin{equation}
\left[\hat{L}{y}, \hat{L}{z}\right]=i \hbar \hat{L}{x}, \quad\left[\hat{L}{z}, \hat{L}{x}\right]=i \hbar \hat{L}{y} \tag{5.48}
\end{equation}
\)

We next evaluate the commutators of $\hat{L}^{2}$ with each of its components, using commutator identities of Section 5.1.

\(
\begin{align}
{\left[\hat{L}^{2}, \hat{L}{x}\right]=} & {\left[\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{x}^{2}, \hat{L}{x}\right]+\left[\hat{L}{y}^{2}, \hat{L}{x}\right]+\left[\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{y}^{2}, \hat{L}{x}\right]+\left[\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{y}, \hat{L}{x}\right] \hat{L}{y}+\hat{L}{y}\left[\hat{L}{y}, \hat{L}{x}\right]+\left[\hat{L}{z}, \hat{L}{x}\right] \hat{L}{z}+\hat{L}{z}\left[\hat{L}{z}, \hat{L}{x}\right] } \
= & -i \hbar \hat{L}{z} \hat{L}{y}-i \hbar \hat{L}{y} \hat{L}{z}+i \hbar \hat{L}{y} \hat{L}{z}+i \hbar \hat{L}{z} \hat{L}{y} \
& \quad\left[\hat{L}^{2}, \hat{L}{x}\right]=0 \tag{5.49}
\end{align}
\)

Since a cyclic permutation of $x, y$, and $z$ leaves $\hat{L}^{2}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}_{z}^{2}$ unchanged, if we carry out two such permutations on (5.49), we get

\(
\begin{equation}
\left[\hat{L}^{2}, \hat{L}{y}\right]=0, \quad\left[\hat{L}^{2}, \hat{L}{z}\right]=0 \tag{5.50}
\end{equation}
\)

To which of the quantities $L^{2}, L{x}, L{y}, L_{z}$ can we assign definite values simultaneously? Because $\hat{L}^{2}$ commutes with each of its components, we can specify an exact value for $L^{2}$

FIGURE 5.5 Spherical coordinates.

and any one component. However, no two components of $\hat{\mathbf{L}}$ commute with each other, so we cannot specify more than one component simultaneously. (There is one exception to this statement, which will be discussed shortly.) It is traditional to take $L_{z}$ as the component of angular momentum that will be specified along with $L^{2}$. Note that in specifying $L^{2}=|\mathbf{L}|^{2}$ we are not specifying the vector $\mathbf{L}$, only its magnitude. A complete specification of $\mathbf{L}$ requires simultaneous specification of each of its three components, which we usually cannot do. In classical mechanics when angular momentum is conserved, each of its three components has a definite value. In quantum mechanics when angular momentum is conserved, only its magnitude and one of its components are specifiable.

We could now try to find the eigenvalues and common eigenfunctions of $\hat{L}^{2}$ and $\hat{L}_{z}$ by using the forms for these operators in Cartesian coordinates. However, we would find that the partial differential equations obtained would not be separable. Therefore we transform these operators to spherical coordinates (Fig. 5.5). The coordinate $r$ is the distance from the origin to the point $(x, y, z)$. The angle $\theta$ is the angle the vector $\mathbf{r}$ makes with the positive $z$ axis. The angle that the projection of $\mathbf{r}$ in the $x y$ plane makes with the positive $x$ axis is $\phi$. (Mathematics texts often interchange $\theta$ and $\phi$.) A little trigonometry gives

\(
\begin{align}
x=r \sin \theta \cos \phi, \quad y & =r \sin \theta \sin \phi, \quad z=r \cos \theta \tag{5.51}\
r^{2}=x^{2}+y^{2}+z^{2}, \quad \cos \theta & =\frac{z}{\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}, \quad \tan \phi=\frac{y}{x} \tag{5.52}
\end{align}
\)

To transform the angular-momentum operators to spherical coordinates, we must transform $\partial / \partial x, \partial / \partial y$, and $\partial / \partial z$ into these coordinates. [This transformation may be skimmed if desired. Begin reading again after Eq. (5.64).]

To perform this transformation, we use the chain rule. Suppose we have a function of $r, \theta$, and $\phi: f(r, \theta, \phi)$. If we change the independent variables by substituting

\(
r=r(x, y, z), \quad \theta=\theta(x, y, z), \quad \phi=\phi(x, y, z)
\)

into $f$, we transform it into a function of $x, y$, and $z$ :

\(
f[r(x, y, z), \theta(x, y, z), \phi(x, y, z)]=g(x, y, z)
\)

For example, suppose that $f(r, \theta, \phi)=3 r \cos \theta+2 \tan ^{2} \phi$. Using (5.52), we have $g(x, y, z)=3 z+2 y^{2} x^{-2}$.

The chain rule tells us how the partial derivatives of $g(x, y, z)$ are related to those of $f(r, \theta, \phi)$. In fact,

\(
\begin{align}
\left(\frac{\partial g}{\partial x}\right){y, z} & =\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial x}\right){y, z}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial x}\right){y, z}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial x}\right){y, z} \tag{5.53}\
\left(\frac{\partial g}{\partial y}\right){x, z} & =\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial y}\right){x, z}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial y}\right){x, z}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial y}\right){x, z} \tag{5.54}
\end{align}
\)

\(
\begin{equation}
\left(\frac{\partial g}{\partial z}\right){x, y}=\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial z}\right){x, y}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial z}\right){x, y}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial z}\right)_{x, y} \tag{5.55}
\end{equation}
\)

To convert these equations to operator equations, we delete $f$ and $g$ to give

\(
\begin{equation}
\frac{\partial}{\partial x}=\left(\frac{\partial r}{\partial x}\right){y, z} \frac{\partial}{\partial r}+\left(\frac{\partial \theta}{\partial x}\right){y, z} \frac{\partial}{\partial \theta}+\left(\frac{\partial \phi}{\partial x}\right)_{y, z} \frac{\partial}{\partial \phi} \tag{5.56}
\end{equation}
\)

with similar equations for $\partial / \partial y$ and $\partial / \partial z$. The task now is to evaluate the partial derivatives such as $(\partial r / \partial x)_{y, z}$. Taking the partial derivative of the first equation in (5.52) with respect to $x$ at constant $y$ and $z$, we have

\(
\begin{align}
2 r\left(\frac{\partial r}{\partial x}\right){y, z} & =2 x=2 r \sin \theta \cos \phi \
\left(\frac{\partial r}{\partial x}\right){y, z} & =\sin \theta \cos \phi \tag{5.57}
\end{align}
\)

Differentiating $r^{2}=x^{2}+y^{2}+z^{2}$ with respect to $y$ and with respect to $z$, we find

\(
\begin{equation}
\left(\frac{\partial r}{\partial y}\right){x, z}=\sin \theta \sin \phi, \quad\left(\frac{\partial r}{\partial z}\right){x, y}=\cos \theta \tag{5.58}
\end{equation}
\)

From the second equation in (5.52), we find

\(
\begin{align}
& -\sin \theta\left(\frac{\partial \theta}{\partial x}\right){y, z}=-\frac{x z}{r^{3}} \
& \left(\frac{\partial \theta}{\partial x}\right){y, z}=\frac{\cos \theta \cos \phi}{r} \tag{5.59}
\end{align}
\)

Also,

\(
\begin{equation}
\left(\frac{\partial \theta}{\partial y}\right){x, z}=\frac{\cos \theta \sin \phi}{r}, \quad\left(\frac{\partial \theta}{\partial z}\right){x, y}=-\frac{\sin \theta}{r} \tag{5.60}
\end{equation}
\)

From $\tan \phi=y / x$, we find

\(
\begin{equation}
\left(\frac{\partial \phi}{\partial x}\right){y, z}=-\frac{\sin \phi}{r \sin \theta}, \quad\left(\frac{\partial \phi}{\partial y}\right){x, z}=\frac{\cos \phi}{r \sin \theta}, \quad\left(\frac{\partial \phi}{\partial z}\right)_{x, y}=0 \tag{5.61}
\end{equation}
\)

Substituting (5.57), (5.59), and (5.61) into (5.56), we find

\(
\begin{equation}
\frac{\partial}{\partial x}=\sin \theta \cos \phi \frac{\partial}{\partial r}+\frac{\cos \theta \cos \phi}{r} \frac{\partial}{\partial \theta}-\frac{\sin \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \tag{5.62}
\end{equation}
\)

Similarly,

\(
\begin{gather}
\frac{\partial}{\partial y}=\sin \theta \sin \phi \frac{\partial}{\partial r}+\frac{\cos \theta \sin \phi}{r} \frac{\partial}{\partial \theta}+\frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \tag{5.63}\
\frac{\partial}{\partial z}=\cos \theta \frac{\partial}{\partial r}-\frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \tag{5.64}
\end{gather}
\)

At last, we are ready to express the angular-momentum components in spherical coordinates. Substituting (5.51), (5.63), and (5.64) into (5.40), we have

\(
\begin{align}
\hat{L}{x}= & -i \hbar\left[r \sin \theta \sin \phi\left(\cos \theta \frac{\partial}{\partial r}-\frac{\sin \theta}{r} \frac{\partial}{\partial \theta}\right)\right. \
& \left.-r \cos \theta\left(\sin \theta \sin \phi \frac{\partial}{\partial r}+\frac{\cos \theta \sin \phi}{r} \frac{\partial}{\partial \theta}+\frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi}\right)\right] \
\hat{L}{x}= & i \hbar\left(\sin \phi \frac{\partial}{\partial \theta}+\cot \theta \cos \phi \frac{\partial}{\partial \phi}\right) \tag{5.65}
\end{align}
\)

Also, we find

\(
\begin{gather}
\hat{L}{y}=-i \hbar\left(\cos \phi \frac{\partial}{\partial \theta}-\cot \theta \sin \phi \frac{\partial}{\partial \phi}\right) \tag{5.66}\
\hat{L}{z}=-i \hbar \frac{\partial}{\partial \phi} \tag{5.67}
\end{gather}
\)

By squaring each of $\hat{L}{x}, \hat{L}{y}$, and $\hat{L}{z}$ and then adding their squares, we can construct $\hat{L}^{2}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}{z}^{2}[$ Eq. (5.43)]. The result is (Prob. 5.17)

\(
\begin{equation}
\hat{L}^{2}=-\hbar^{2}\left(\frac{\partial^{2}}{\partial \theta^{2}}+\cot \theta \frac{\partial}{\partial \theta}+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right) \tag{5.68}
\end{equation}
\)

Although the angular-momentum operators depend on all three Cartesian coordinates, $x, y$, and $z$, they involve only the two spherical coordinates $\theta$ and $\phi$.

One-Particle Orbital-Angular-Momentum Eigenfunctions and Eigenvalues

We now find the common eigenfunctions of $\hat{L}^{2}$ and $\hat{L}{z}$, which we denote by $Y$. Since these operators involve only $\theta$ and $\phi, Y$ is a function of these two coordinates: $Y=Y(\theta, \phi)$. (Of course, since the operators are linear, we can multiply $Y$ by an arbitrary function of $r$ and still have an eigenfunction of $\hat{L}^{2}$ and $\hat{L}{z}$.) We must solve

\(
\begin{align}
& \hat{L}_{z} Y(\theta, \phi)=b Y(\theta, \phi) \tag{5.69}\
& \hat{L}^{2} Y(\theta, \phi)=c Y(\theta, \phi) \tag{5.70}
\end{align}
\)

where $b$ and $c$ are the eigenvalues of $\hat{L}{z}$ and $\hat{L}^{2}$.
Using the $\hat{L}{z}$ operator, we have

\(
\begin{equation}
-i \hbar \frac{\partial}{\partial \phi} Y(\theta, \phi)=b Y(\theta, \phi) \tag{5.71}
\end{equation}
\)

Since the operator in (5.71) does not involve $\theta$, we try a separation of variables, writing

\(
\begin{equation}
Y(\theta, \phi)=S(\theta) T(\phi) \tag{5.72}
\end{equation}
\)

Equation (5.71) becomes

\(
\begin{align}
-i \hbar \frac{\partial}{\partial \phi}[S(\theta) T(\phi)] & =b S(\theta) T(\phi) \
-i \hbar S(\theta) \frac{d T(\phi)}{d \phi} & =b S(\theta) T(\phi) \
\frac{d T(\phi)}{T(\phi)} & =\frac{i b}{\hbar} d \phi \
T(\phi) & =A e^{i b \phi / \hbar} \tag{5.73}
\end{align}
\)

where $A$ is an arbitrary constant.

Is $T$ suitable as an eigenfunction? The answer is no, since it is not, in general, a singlevalued function. If we add $2 \pi$ to $\phi$, we will still be at the same point in space, and hence we want no change in $T$ when this is done. For $T$ to be single-valued, we have the restriction

\(
\begin{align}
T(\phi+2 \pi) & =T(\phi) \
A e^{i b \phi / \hbar} e^{i b 2 \pi / \hbar} & =A e^{i b \phi / \hbar} \
e^{i b 2 \pi / \hbar} & =1 \tag{5.74}
\end{align}
\)

To satisfy $e^{i \alpha}=\cos \alpha+i \sin \alpha=1$, we must have $\alpha=2 \pi m$, where

\(
m=0, \pm 1, \pm 2, \pm \cdots
\)

Therefore, (5.74) gives

\(
\begin{gather}
2 \pi b / \hbar=2 \pi m \
b=m \hbar, \quad m=\ldots-2,-1,0,1,2, \ldots \tag{5.75}
\end{gather}
\)

and (5.73) becomes

\(
\begin{equation}
T(\phi)=A e^{i m \phi}, \quad m=0, \pm 1, \pm 2, \ldots \tag{5.76}
\end{equation}
\)

The eigenvalues for the $z$ component of angular momentum are quantized.
We fix $A$ by normalizing $T$. First let us consider normalizing some function $F$ of $r, \theta$, and $\phi$. The ranges of the independent variables are (see Fig. 5.5)

\(
\begin{equation}
0 \leq r \leq \infty, \quad 0 \leq \theta \leq \pi, \quad 0 \leq \phi \leq 2 \pi \tag{5.77}
\end{equation}
\)

The infinitesimal volume element in spherical coordinates is (Taylor and Mann, Section 13.9)

\(
\begin{equation}
d \tau=r^{2} \sin \theta d r d \theta d \phi \tag{5.78}
\end{equation}
\)

The quantity (5.78) is the volume of an infinitesimal region of space for which the spherical coordinates lie in the ranges $r$ to $r+d r, \theta$ to $\theta+d \theta$, and $\phi$ to $\phi+d \phi$. The normalization condition for $F$ in spherical coordinates is therefore

\(
\begin{equation}
\int{0}^{\infty}\left[\int{0}^{\pi}\left[\int_{0}^{2 \pi}|F(r, \theta, \phi)|^{2} d \phi\right] \sin \theta d \theta\right] r^{2} d r=1 \tag{5.79}
\end{equation}
\)

If $F$ happens to have the form

\(
F(r, \theta, \phi)=R(r) S(\theta) T(\phi)
\)

then use of the integral identity (3.74) gives for (5.79)

\(
\int{0}^{\infty}|R(r)|^{2} r^{2} d r \int{0}^{\pi}|S(\theta)|^{2} \sin \theta d \theta \int_{0}^{2 \pi}|T(\phi)|^{2} d \phi=1
\)

and it is convenient to normalize each factor of $F$ separately:

\(
\begin{equation}
\int{0}^{\infty}|R|^{2} r^{2} d r=1, \quad \int{0}^{\pi}|S|^{2} \sin \theta d \theta=1, \quad \int_{0}^{2 \pi}|T|^{2} d \phi=1 \tag{5.80}
\end{equation}
\)

Therefore,

\(
\begin{gathered}
\int{0}^{2 \pi}\left(A e^{i m \phi}\right) * A e^{i m \phi} d \phi=1=|A|^{2} \int{0}^{2 \pi} d \phi \
|A|=(2 \pi)^{-1 / 2}
\end{gathered}
\)

\(
\begin{equation}
T(\phi)=\frac{1}{\sqrt{2 \pi}} e^{i m \phi}, \quad m=0, \pm 1, \pm 2, \ldots \tag{5.81}
\end{equation}
\)

We now solve $\hat{L}^{2} Y=c Y$ [Eq. (5.70)] for the eigenvalues $c$ of $\hat{L}^{2}$. Using (5.68) for $\hat{L}^{2}$, (5.72) for $Y$, and (5.81), we have

\(
\begin{gather}
-\hbar^{2}\left(\frac{\partial^{2}}{\partial \theta^{2}}+\cot \theta \frac{\partial}{\partial \theta}+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right)\left(S(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi}\right)=c S(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi} \
\frac{d^{2} S}{d \theta^{2}}+\cot \theta \frac{d S}{d \theta}-\frac{m^{2}}{\sin ^{2} \theta} S=-\frac{c}{\hbar^{2}} S \tag{5.82}
\end{gather}
\)

To solve (5.82), we carry out some tedious manipulations, which may be skimmed if desired. Begin reading again at Eq. (5.91). First, for convenience, we change the independent variable by making the substitution

\(
\begin{equation}
w=\cos \theta \tag{5.83}
\end{equation}
\)

This transforms $S$ into some new function of $w$ :

\(
\begin{equation}
S(\theta)=G(w) \tag{5.8}
\end{equation}
\)

The chain rule gives

\(
\begin{equation}
\frac{d S}{d \theta}=\frac{d G}{d w} \frac{d w}{d \theta}=-\sin \theta \frac{d G}{d w}=-\left(1-w^{2}\right)^{1 / 2} \frac{d G}{d w} \tag{5.85}
\end{equation}
\)

Similarly, we find (Prob. 5.25)

\(
\begin{equation}
\frac{d^{2} S}{d \theta^{2}}=\left(1-w^{2}\right) \frac{d^{2} G}{d w^{2}}-w \frac{d G}{d w} \tag{5.86}
\end{equation}
\)

Using (5.86), (5.85), and $\cot \theta=\cos \theta / \sin \theta=w /\left(1-w^{2}\right)^{1 / 2}$, we find that (5.82) becomes

\(
\begin{equation}
\left(1-w^{2}\right) \frac{d^{2} G}{d w^{2}}-2 w \frac{d G}{d w}+\left[\frac{c}{\hbar^{2}}-\frac{m^{2}}{1-w^{2}}\right] G(w)=0 \tag{5.87}
\end{equation}
\)

The range of $w$ is $-1 \leq w \leq 1$.
To get a two-term recursion relation when we try a power-series solution, we make the following change of dependent variable:

\(
\begin{equation}
G(w)=\left(1-w^{2}\right)^{|m| / 2} H(w) \tag{5.88}
\end{equation}
\)

Differentiating (5.88), we evaluate $G^{\prime}$ and $G^{\prime \prime}$, and (5.87) becomes, after we divide by $\left(1-w^{2}\right)^{|m| / 2}$,

\(
\begin{equation}
\left(1-w^{2}\right) H^{\prime \prime}-2(|m|+1) w H^{\prime}+\left[c \hbar^{-2}-|m|(|m|+1)\right] H=0 \tag{5.89}
\end{equation}
\)

We now try a power series for $H$ :

\(
\begin{equation}
H(w)=\sum{j=0}^{\infty} a{j} w^{j} \tag{5.90}
\end{equation}
\)

Differentiating [compare Eqs. (4.36)-(4.38)], we have

\(
\begin{aligned}
& H^{\prime}(w)=\sum{j=0}^{\infty} j a{j} w^{j-1} \
& H^{\prime \prime}(w)=\sum{j=0}^{\infty} j(j-1) a{j} w^{j-2}=\sum{j=0}^{\infty}(j+2)(j+1) a{j+2} w^{j}
\end{aligned}
\)

Substitution of these power series into (5.89) yields, after combining sums,

\(
\sum{j=0}^{\infty}\left[(j+2)(j+1) a{j+2}+\left(-j^{2}-j-2|m| j+\frac{c}{\hbar^{2}}-|m|^{2}-|m|\right) a_{j}\right] w^{j}=0
\)

Setting the coefficient of $w^{j}$ equal to zero, we get the recursion relation

\(
\begin{equation}
a{j+2}=\frac{(j+|m|)(j+|m|+1)-c / \hbar^{2}}{(j+1)(j+2)} a{j} \tag{5.91}
\end{equation}
\)

Just as in the harmonic-oscillator case, the general solution of (5.89) is an arbitrary linear combination of a series of even powers (whose coefficients are determined by $a{0}$ ) and a series of odd powers (whose coefficients are determined by $a{1}$ ). It can be shown that the infinite series defined by the recursion relation (5.91) does not give well-behaved eigenfunctions. [Many texts point out that the infinite series diverges at $w= \pm 1$. However, this is not sufficient cause to reject the infinite series, since the eigenfunctions might be quadratically integrable, even though infinite at two points. For a careful discussion, see M. Whippman, Am. J. Phys., 34, 656 (1966).] Hence, as in the harmonic-oscillator case, we must cause one of the series to break off, its last term being $a{k} w^{k}$. We eliminate the other series by setting $a{0}$ or $a_{1}$ equal to zero, depending on whether $k$ is odd or even.

Setting the coefficient of $a_{k}$ in (5.91) equal to zero, we have

\(
\begin{equation}
c=\hbar^{2}(k+|m|)(k+|m|+1), \quad k=0,1,2, \ldots \tag{5.92}
\end{equation}
\)

Since $|m|$ takes on the values $0,1,2, \ldots$, the quantity $k+|m|$ takes on the values $0,1,2, \ldots$ We therefore define the quantum number $l$ as

\(
\begin{equation}
l \equiv k+|m| \tag{5.93}
\end{equation}
\)

and the eigenvalues for the square of the magnitude of angular momentum are

\(
\begin{equation}
c=l(l+1) \hbar^{2}, \quad l=0,1,2, \ldots \tag{5.94}
\end{equation}
\)

The magnitude of the orbital angular momentum of a particle is

\(
\begin{equation}
|\mathbf{L}|=[l(l+1)]^{1 / 2} \hbar \tag{5.95}
\end{equation}
\)

From (5.93), it follows that $|m| \leq l$. The possible values for $m$ are thus

\(
\begin{equation}
m=-l,-l+1,-l+2, \ldots,-1,0,1, \ldots, l-2, l-1, l \tag{5.96}
\end{equation}
\)

Let us examine the angular-momentum eigenfunctions. From (5.83), (5.84), (5.88), (5.90), and (5.93), the theta factor in the eigenfunctions is

\(
\begin{equation}
S{l, m}(\theta)=\sin ^{|m|} \theta \sum{\substack{j=1,3, \ldots \ \text { or } j=0,2, \ldots}}^{l-|m|} a_{j} \cos ^{j} \theta \tag{5.97}
\end{equation}
\)

where the sum is over even or odd values of $j$, depending on whether $l-|m|$ is even or odd. The coefficients $a_{j}$ satisfy the recursion relation (5.91), which, using (5.94), becomes

\(
\begin{equation}
a{j+2}=\frac{(j+|m|)(j+|m|+1)-l(l+1)}{(j+1)(j+2)} a{j} \tag{5.98}
\end{equation}
\)

The $\hat{L}^{2}$ and $\hat{L}_{z}$ eigenfunctions are given by Eqs. (5.72) and (5.81) as

\(
\begin{equation}
Y{l}^{m}(\theta, \phi)=S{l, m}(\theta) T(\phi)=\frac{1}{\sqrt{2 \pi}} S_{l, m}(\theta) e^{i m \phi} \tag{5.99}
\end{equation}
\)

The $m$ in $Y_{l}^{m}$ is a label, and not an exponent.

EXAMPLE

Find $Y{l}^{m}(\theta, \phi)$ and the $\hat{L}^{2}$ and $\hat{L}{z}$ eigenvalues for (a) $l=0$; (b) $l=1$.
(a) For $l=0$, Eq. (5.96) gives $m=0$, and (5.97) becomes

\(
\begin{equation}
S{0,0}(\theta)=a{0} \tag{5.100}
\end{equation}
\)

The normalization condition (5.80) gives

\(
\begin{aligned}
\int{0}^{\pi}\left|a{0}\right|^{2} \sin \theta d \theta=1 & =2\left|a{0}\right|^{2} \
\left|a{0}\right| & =2^{-1 / 2}
\end{aligned}
\)

Equation (5.99) gives

\(
\begin{equation}
Y_{0}^{0}(\theta, \phi)=\frac{1}{\sqrt{4 \pi}} \tag{5.101}
\end{equation}
\)

[Obviously, (5.101) is an eigenfunction of $\hat{L}^{2}, \hat{L}{x}, \hat{L}{y}$, and $\hat{L}_{z}$, Eqs. (5.65)-(5.68).] For $l=0$, there is no angular dependence in the eigenfunction; we say that the eigenfunctions are spherically symmetric for $l=0$.

For $l=0$ and $m=0$, Eqs. (5.69), (5.70), (5.75), and (5.94) give the $\hat{L}^{2}$ eigenvalue as $c=0$ and the $\hat{L}_{z}$ eigenvalue as $b=0$.
(b) For $l=1$, the possible values for $m$ in (5.96) are $-1,0$, and 1 . For $|m|=1$, (5.97) gives

\(
\begin{equation}
S{1, \pm 1}(\theta)=a{0} \sin \theta \tag{5.102}
\end{equation}
\)

$a{0}$ in (5.102) is not necessarily the same as $a{0}$ in (5.100). Normalization gives

\(
\begin{gathered}
1=\left|a{0}^{2}\right| \int{0}^{\pi} \sin ^{2} \theta \sin \theta d \theta=\left|a{0}^{2}\right| \int{-1}^{1}\left(1-w^{2}\right) d w \
\left|a_{0}\right|=\sqrt{3} / 2
\end{gathered}
\)

where the substitution $w=\cos \theta$ was made. Thus $S_{1, \pm 1}=\left(3^{1 / 2} / 2\right) \sin \theta$ and (5.99) gives

\(
\begin{equation}
Y{1}^{1}=(3 / 8 \pi)^{1 / 2} \sin \theta e^{i \phi}, \quad Y{1}^{-1}=(3 / 8 \pi)^{1 / 2} \sin \theta e^{-i \phi} \tag{5.103}
\end{equation}
\)

For $l=1$ and $m=0$, we find (see the following exercise) $S{1,0}=(3 / 2)^{1 / 2} \cos \theta$ and $Y{1}^{0}=(3 / 4 \pi)^{1 / 2} \cos \theta$.

For $l=1$, (5.94) gives the $\hat{L}^{2}$ eigenvalue as $2 \hbar^{2}$; for $m=-1,0$, and 1 , (5.75) gives the $\hat{L}_{z}$ eigenvalues as $-\hbar, 0$, and $\hbar$, respectively.

EXERCISE Verify the expressions for $S{1,0}$ and $Y{1}^{0}$.

The functions $S{l, m}(\theta)$ are well known in mathematics and are associated Legendre functions multiplied by a normalization constant. The associated Legendre functions are defined in Prob. 5.34. Table 5.1 gives the $S{l, m}(\theta)$ functions for $l \leq 3$.

The angular-momentum eigenfunctions $Y_{l}^{m}$ in (5.99) are called spherical harmonics (or surface harmonics).

In summary, the one-particle orbital angular-momentum eigenfunctions and eigenvalues are [Eqs. (5.69), (5.70), (5.75), and (5.94)]

TABLE $5.1 \quad \boldsymbol{S}_{\boldsymbol{l}, \boldsymbol{m}}(\boldsymbol{\theta})$

\(
\begin{array}{ll}
l=0: & S{0,0}=\frac{1}{2} \sqrt{2} \
l=1: & S{1,0}=\frac{1}{2} \sqrt{6} \cos \theta \
& S{1, \pm 1}=\frac{1}{2} \sqrt{3} \sin \theta \
l=2: & S{2,0}=\frac{1}{4} \sqrt{10}\left(3 \cos ^{2} \theta-1\right) \
& S{2, \pm 1}=\frac{1}{2} \sqrt{15} \sin \theta \cos \theta \
& S{2, \pm 2}=\frac{1}{4} \sqrt{15} \sin ^{2} \theta \
l=3: & S{3,0}=\frac{3}{4} \sqrt{14}\left(\frac{5}{3} \cos ^{3} \theta-\cos \theta\right) \
& S{3, \pm 1}=\frac{1}{8} \sqrt{42} \sin \theta\left(5 \cos ^{2} \theta-1\right) \
& S{3, \pm 2}=\frac{1}{4} \sqrt{105} \sin ^{2} \theta \cos \theta \
& S{3, \pm 3}=\frac{1}{8} \sqrt{70} \sin ^{3} \theta
\end{array}
\)

\(
\begin{equation}
\hat{L}^{2} Y{l}^{m}(\theta, \phi)=l(l+1) \hbar^{2} Y{l}^{m}(\theta, \phi), \quad l=0,1,2, \ldots \tag{5.104}
\end{equation}
\)

\(
\begin{equation}
\hat{L}{z} Y{l}^{m}(\theta, \phi)=m \hbar Y_{l}^{m}(\theta, \phi), \quad m=-l,-l+1, \ldots, l-1, l \tag{5.105}
\end{equation}
\)

where the eigenfunctions are given by (5.99). Often the symbol $m{l}$ is used instead of $m$ for the $L{z}$ quantum number. We shall later see that the spherical harmonics are orthogonal functions [Eq. (7.27)].

Since $l \geq|m|$, the magnitude $[l(l+1)]^{1 / 2} \hbar$ of the orbital angular momentum $\mathbf{L}$ is greater than the magnitude $|m| \hbar$ of its $z$ component $L{z}$, except for $l=0$. If it were possible to have the angular-momentum magnitude equal to its $z$ component, this would mean that the $x$ and $y$ components were zero, and we would have specified all three components of $\mathbf{L}$. However, since the components of angular momentum do not commute with each other, we cannot do this. The one exception is when $l$ is zero. In this case, $|\mathbf{L}|^{2}=L{x}^{2}+L{y}^{2}+L{z}^{2}$ has zero for its eigenvalue, and it must be true that all three components $L{x}, L{y}$, and $L_{z}$ have zero eigenvalues. From Eq. (5.12), the uncertainties in angular-momentum components satisfy

\(
\begin{equation}
\Delta L{x} \Delta L{y} \geq \frac{1}{2}\left|\int \Psi^{}\left[\hat{L}{x}, \hat{L}{y}\right] \Psi d \tau\right|=\frac{\hbar}{2}\left|\int \Psi^{} \hat{L}_{z} \Psi d \tau\right| \tag{5.106}
\end{equation}
\)

and two similar equations obtained by cyclic permutation. When the eigenvalues of $\hat{L}{z}, \hat{L}{x}$, and $\hat{L}{y}$ are zero, $\hat{L}{x} \Psi=0, \hat{L}{y} \Psi=0, \hat{L}{z} \Psi=0$, the right-hand sides of (5.106) and the two similar equations are zero, and having $\Delta L{x}=\Delta L{y}=\Delta L{z}=0$ is permitted. But what about the statement in Section 5.1 that to have simultaneous eigenfunctions of two operators the operators must commute? The answer is that this theorem refers to the possibility of having a complete set of eigenfunctions of one operator be eigenfunctions of the other operator. Thus, even though $\hat{L}{x}$ and $\hat{L}{z}$ do not commute, it is possible to have some of the eigenfunctions of $\hat{L}{z}$ (those with $l=0=m$ ) be eigenfunctions of $\hat{L}{x}$. However, it is impossible to have all the $\hat{L}{z}$ eigenfunctions also be eigenfunctions of $\hat{L}_{x}$.

FIGURE 5.6 Orientation of L.

FIGURE 5.7 Orientations of $\mathbf{L}$ with respect to the $z$ axis for $l=1$.

Since we cannot specify $L{x}$ and $L{y}$, the vector $\mathbf{L}$ can lie anywhere on the surface of a cone whose axis is the $z$ axis, whose altitude is $m \hbar$, and whose slant height is $\sqrt{l(l+1)} \hbar$ (Fig. 5.6). The possible orientations of $\mathbf{L}$ with respect to the $z$ axis for the case $l=1$ are shown in Fig. 5.7. For each eigenvalue of $\hat{L}^{2}$, there are $2 l+1$ different eigenfunctions $Y_{l}^{m}$, corresponding to the $2 l+1$ values of $m$. We say that the $\hat{L}^{2}$ eigenvalues are $(2 l+1)$-fold degenerate. The term degeneracy is applicable to the eigenvalues of any operator, not just the Hamiltonian.

Of course, there is nothing special about the $z$ axis. All directions of space are equivalent. If we had chosen to specify $L^{2}$ and $L{x}$ (rather than $L{z}$ ), we would have gotten the same eigenvalues for $L{x}$ as we found for $L{z}$. However, it is easier to solve the $\hat{L}{z}$ eigenvalue equation because $\hat{L}{z}$ has a simple form in spherical coordinates, which involve the angle of rotation $\phi$ about the $z$ axis.

We found the eigenvalues of $\hat{L}^{2}$ and $\hat{L}_{z}$ by expressing these orbital angular-momentum operators as differential operators and solving the resulting differential equations. We now show that these eigenvalues can be found using only the operator commutation relations. The work in this section applies to any operators that satisfy the angular-momentum commutation relations. In particular, it applies to spin angular momentum (Chapter 10) as well as orbital angular momentum.

We used the letter $L$ for orbital angular momentum. Here we will use the letter $M$ to indicate that we are dealing with any kind of angular momentum. We have three linear
operators $\hat{M}{x}, \hat{M}{y}$, and $\hat{M}_{z}$, and all we know about them is that they obey the commutation relations [similar to (5.46) and (5.48)]

\(
\begin{equation}
\left[\hat{M}{x}, \hat{M}{y}\right]=i \hbar \hat{M}{z}, \quad\left[\hat{M}{y}, \hat{M}{z}\right]=i \hbar \hat{M}{x}, \quad\left[\hat{M}{z}, \hat{M}{x}\right]=i \hbar \hat{M}_{y} \tag{5.107}
\end{equation}
\)

We define the operator $\hat{M}^{2}$ as

\(
\begin{equation}
\hat{M}^{2}=\hat{M}{x}^{2}+\hat{M}{y}^{2}+\hat{M}_{z}^{2} \tag{5.108}
\end{equation}
\)

Our problem is to find the eigenvalues of $\hat{M}^{2}$ and $\hat{M}_{z}$.
We begin by evaluating the commutators of $\hat{M}^{2}$ with its components, using Eqs. (5.107) and (5.108). The work is identical with that used to derive Eqs. (5.49) and (5.50), and we have

\(
\begin{equation}
\left[\hat{M}^{2}, \hat{M}{x}\right]=\left[\hat{M}^{2}, \hat{M}{y}\right]=\left[\hat{M}^{2}, \hat{M}_{z}\right]=0 \tag{5.109}
\end{equation}
\)

Hence we can have simultaneous eigenfunctions of $\hat{M}^{2}$ and $\hat{M}{z}$.
Next we define two new operators, the raising operator $\hat{M}{+}$and the lowering operator $\hat{M}_{-}$:

\(
\begin{align}
& \hat{M}{+} \equiv \hat{M}{x}+i \hat{M}{y} \tag{5.110}\
& \hat{M}{-} \equiv \hat{M}{x}-i \hat{M}{y} \tag{5.111}
\end{align}
\)

These are examples of ladder operators. The reason for the terminology will become clear shortly. We have

\(
\begin{gather}
\hat{M}{+} \hat{M}{-}=\left(\hat{M}{x}+i \hat{M}{y}\right)\left(\hat{M}{x}-i \hat{M}{y}\right)=\hat{M}{x}\left(\hat{M}{x}-i \hat{M}{y}\right)+i \hat{M}{y}\left(\hat{M}{x}-i \hat{M}{y}\right) \
=\hat{M}{x}^{2}-i \hat{M}{x} \hat{M}{y}+i \hat{M}{y} \hat{M}{x}+\hat{M}{y}^{2}=\hat{M}^{2}-\hat{M}{z}^{2}+i\left[\hat{M}{y}, \hat{M}{x}\right] \
\hat{M}{+} \hat{M}{-}=\hat{M}^{2}-\hat{M}{z}^{2}+\hbar \hat{M}_{z} \tag{5.112}
\end{gather}
\)

Similarly, we find

\(
\begin{equation}
\hat{M}{-} \hat{M}{+}=\hat{M}^{2}-\hat{M}{z}^{2}-\hbar \hat{M}{z} \tag{5.113}
\end{equation}
\)

For the commutators of these operators with $\hat{M}_{z}$, we have

\(
\begin{gather}
{\left[\hat{M}{+}, \hat{M}{z}\right]=\left[\hat{M}{x}+i \hat{M}{y}, \hat{M}{z}\right]=\left[\hat{M}{x}, \hat{M}{z}\right]+i\left[\hat{M}{y}, \hat{M}{z}\right]=-i \hbar \hat{M}{y}-\hbar \hat{M}{x}} \
{\left[\hat{M}{+}, \hat{M}{z}\right]=-\hbar \hat{M}{+}} \
\hat{M}{+} \hat{M}{z}=\hat{M}{z} \hat{M}{+}-\hbar \hat{M}_{+} \tag{5.114}
\end{gather}
\)

where (5.107) was used. Similarly, we find

\(
\begin{equation}
\hat{M}{-} \hat{M}{z}=\hat{M}{z} \hat{M}{-}+\hbar \hat{M}_{-} \tag{5.115}
\end{equation}
\)

Using $Y$ for the common eigenfunctions of $\hat{M}^{2}$ and $\hat{M}_{z}$, we have

\(
\begin{align}
& \hat{M}^{2} Y=c Y \tag{5.116}\
& \hat{M}_{z} Y=b Y \tag{5.117}
\end{align}
\)

where $c$ and $b$ are the eigenvalues. Operating on Eq. (5.117) with $\hat{M}_{+}$, we get

\(
\hat{M}{+} \hat{M}{z} Y=\hat{M}_{+} b Y
\)

Using Eq. (5.114) and the fact that $\hat{M}_{+}$is linear, we have

\(
\begin{gather}
\left(\hat{M}{z} \hat{M}{+}-\hbar \hat{M}{+}\right) Y=b \hat{M}{+} Y \
\hat{M}{z}\left(\hat{M}{+} Y\right)=(b+\hbar)\left(\hat{M}_{+} Y\right) \tag{5.118}
\end{gather}
\)

This last equation says that the function $\hat{M}{+} Y$ is an eigenfunction of $\hat{M}{z}$ with eigenvalue $b+\hbar$. In other words, operating on the eigenfunction $Y$ with the raising operator $\hat{M}{+}$ converts $Y$ into another eigenfunction of $\hat{M}{z}$ with eigenvalue $\hbar$ higher than the eigenvalue of $Y$. If we now apply the raising operator to (5.118) and use (5.114) again, we find similarly

\(
\hat{M}{z}\left(\hat{M}{+}^{2} Y\right)=(b+2 \hbar)\left(\hat{M}_{+}^{2} Y\right)
\)

Repeated application of the raising operator gives

\(
\begin{equation}
\hat{M}{z}\left(\hat{M}{+}^{k} Y\right)=(b+k \hbar)\left(\hat{M}_{+}^{k} Y\right), \quad k=0,1,2, \ldots \tag{5.119}
\end{equation}
\)

If we operate on (5.117) with the lowering operator and apply (5.115), we find in the same manner

\(
\begin{gather}
\hat{M}{z}\left(\hat{M}{-} Y\right)=(b-\hbar)\left(\hat{M}{-} Y\right) \tag{5.120}\
\hat{M}{z}\left(\hat{M}{-}^{k} Y\right)=(b-k \hbar)\left(\hat{M}{-}^{k} Y\right) \tag{5.121}
\end{gather}
\)

Thus by using the raising and lowering operators on the eigenfunction with the eigenvalue $b$, we generate a ladder of eigenvalues, the difference from step to step being $\hbar$ :

\(
\ldots \quad b-2 \hbar, \quad b-\hbar, \quad b, \quad b+\hbar, \quad b+2 \hbar, \quad \ldots
\)

The functions $\hat{M}{ \pm}^{k} Y$ are eigenfunctions of $\hat{M}{z}$ with eigenvalues $b \pm k \hbar$ [Eqs. (5.119) and (5.121)]. We now show that these functions are also eigenfunctions of $\hat{M}^{2}$, all with the same eigenvalue $c$ :

\(
\begin{gather}
\hat{M}{z} \hat{M}{ \pm}^{k} Y=(b \pm k \hbar) \hat{M}{ \pm}^{k} Y \tag{5.122}\
\hat{M}^{2} \hat{M}{ \pm}^{k} Y=c \hat{M}_{ \pm}^{k} Y, \quad k=0,1,2, \ldots \tag{5.123}
\end{gather}
\)

To prove (5.123), we first show that $\hat{M}^{2}$ commutes with $\hat{M}{+}$and $\hat{M}{-}$:

\(
\left[\hat{M}^{2}, \hat{M}{ \pm}\right]=\left[\hat{M}^{2}, \hat{M}{x} \pm i \hat{M}{y}\right]=\left[\hat{M}^{2}, \hat{M}{x}\right] \pm i\left[\hat{M}^{2}, \hat{M}_{y}\right]=0 \pm 0=0
\)

We also have

\(
\left[\hat{M}^{2}, \hat{M}{ \pm}^{2}\right]=\left[\hat{M}^{2}, \hat{M}{ \pm}\right] \hat{M}{ \pm}+\hat{M}{ \pm}\left[\hat{M}^{2}, \hat{M}_{ \pm}\right]=0+0=0
\)

and it follows by induction that

\(
\begin{equation}
\left[\hat{M}^{2}, \hat{M}{ \pm}^{k}\right]=0 \quad \text { or } \quad \hat{M}^{2} \hat{M}{ \pm}^{k}=\hat{M}_{ \pm}^{k} \hat{M}^{2}, \quad k=0,1,2, \ldots \tag{5.124}
\end{equation}
\)

If we operate on (5.116) with $\hat{M}_{ \pm}^{k}$ and use (5.124), we get

\(
\begin{align}
\hat{M}{ \pm}^{k} \hat{M}^{2} Y & =\hat{M}{ \pm}^{k} c Y \
\hat{M}^{2}\left(\hat{M}{ \pm}^{k} Y\right) & =c\left(\hat{M}{ \pm}^{k} Y\right) \tag{5.125}
\end{align}
\)

which is what we wanted to prove.
Next we show that the set of eigenvalues of $\hat{M}{z}$ generated using the ladder operators must be bounded. For the particular eigenfunction $Y$ with $\hat{M}{z}$ eigenvalue $b$, we have

\(
\hat{M}_{z} Y=b Y
\)

and for the set of eigenfunctions and eigenvalues generated by the ladder operators, we have

\(
\begin{equation}
\hat{M}{z} Y{k}=b{k} Y{k} \tag{5.126}
\end{equation}
\)

where

\(
\begin{gather}
Y{k}=\hat{M}{ \pm}^{k} Y \tag{5.127}\
b_{k}=b \pm k \hbar \tag{5.128}
\end{gather}
\)

(Application of $\hat{M}{+}$or $\hat{M}{-}$destroys the normalization of $Y$, so $Y_{k}$ is not normalized. For the normalization constant, see Prob. 10.27.)

Operating on (5.126) with $\hat{M}_{z}$, we have

\(
\begin{align}
\hat{M}{z}^{2} Y{k} & =b{k} \hat{M}{z} Y{k} \
\hat{M}{z}^{2} Y{k} & =b{k}^{2} Y_{k} \tag{5.129}
\end{align}
\)

Now subtract (5.129) from (5.123), and use (5.127) and (5.108):

\(
\begin{align}
\hat{M}^{2} Y{k}-\hat{M}{z}^{2} Y{k} & =c Y{k}-b{k}^{2} Y{k} \
\left(\hat{M}{x}^{2}+\hat{M}{y}^{2}\right) Y{k} & =\left(c-b{k}^{2}\right) Y_{k} \tag{5.130}
\end{align}
\)

The operator $\hat{M}{x}^{2}+\hat{M}{y}^{2}$ corresponds to a nonnegative physical quantity and hence has nonnegative eigenvalues. (This is proved in Prob. 7.11.) Therefore, (5.130) implies that $c-b{k}^{2} \geq 0$ and $c^{1 / 2} \geq\left|b{k}\right|$. Thus

\(
\begin{equation}
c^{1 / 2} \geq b_{k} \geq-c^{1 / 2}, \quad k=0, \pm 1, \pm 2, \ldots \tag{5.131}
\end{equation}
\)

Since $c$ remains constant as $k$ varies, (5.131) shows that the set of eigenvalues $b{k}$ is bounded above and below. Let $b{\max }$ and $b{\text {min }}$ denote the maximum and minimum values of $b{k} . Y{\max }$ and $Y{\min }$ are the corresponding eigenfunctions:

\(
\begin{align}
\hat{M}{z} Y{\max } & =b{\max } Y{\max } \tag{5.132}\
\hat{M}{z} Y{\min } & =b{\min } Y{\min } \tag{5.133}
\end{align}
\)

Now operate on (5.132) with the raising operator and use (5.114):

\(
\begin{gather}
\hat{M}{+} \hat{M}{z} Y{\max }=b{\max } \hat{M}{+} Y{\max } \
\hat{M}{z}\left(\hat{M}{+} Y{\max }\right)=\left(b{\max }+\hbar\right)\left(\hat{M}{+} Y{\max }\right) \tag{5.134}
\end{gather}
\)

This last equation seems to contradict the statement that $b{\text {max }}$ is the largest eigenvalue of $\hat{M}{z}$, since it says that $\hat{M}{+} Y{\text {max }}$ is an eigenfunction of $\hat{M}{z}$ with eigenvalue $b{\text {max }}+\hbar$. The only way out of this contradiction is to have $\hat{M}{+} Y{\max }$ vanish. (We always reject zero as an eigenfunction on physical grounds.) Thus

\(
\begin{equation}
\hat{M}{+} Y{\max }=0 \tag{5.135}
\end{equation}
\)

Operating on (5.135) with the lowering operator and using (5.113), (5.132), and (5.116), we have

\(
\begin{gather}
0=\hat{M}{-} \hat{M}{+} Y{\max }=\left(\hat{M}^{2}-\hat{M}{z}^{2}-\hbar \hat{M}{z}\right) Y{\max }=\left(c-b{\max }^{2}-\hbar b{\max }\right) Y{\max } \
c-b{\max }^{2}-\hbar b{\max }=0 \
c=b{\max }^{2}+\hbar b_{\max } \tag{5.136}
\end{gather}
\)

A similar argument shows that

\(
\begin{equation}
\hat{M}{-} Y{\min }=0 \tag{5.137}
\end{equation}
\)

and by applying the raising operator to this equation and using (5.112), we find

\(
c=b{\min }^{2}-\hbar b{\min }
\)

Subtracting this last equation from (5.136), we have

\(
b{\max }^{2}+\hbar b{\max }+\left(\hbar b{\min }-b{\min }^{2}\right)=0
\)

This is a quadratic equation in the unknown $b_{\max }$, and using the usual formula (it still works in quantum mechanics), we find

\(
b{\max }=-b{\min }, \quad b{\max }=b{\min }-\hbar
\)

The second root is rejected, since it says that $b{\max }$ is less than $b{\text {min }}$. So

\(
\begin{equation}
b{\min }=-b{\max } \tag{5.138}
\end{equation}
\)

Moreover, (5.128) says that $b{\max }$ and $b{\text {min }}$ differ by an integral multiple of $\hbar$ :

\(
\begin{equation}
b{\max }-b{\min }=n \hbar, \quad n=0,1,2, \ldots \tag{5.139}
\end{equation}
\)

Substituting (5.138) in (5.139), we have for the $\hat{M}_{z}$ eigenvalues

\(
\begin{gather}
b{\max }=\frac{1}{2} n \hbar \
b{\max }=j \hbar, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots \tag{5.140}\
b_{\min }=-j \hbar \
b=-j \hbar,(-j+1) \hbar,(-j+2) \hbar, \ldots,(j-2) \hbar,(j-1) \hbar, j \hbar \tag{5.141}
\end{gather}
\)

and from (5.136) we find as the $\hat{M}^{2}$ eigenvalues

\(
\begin{equation}
c=j(j+1) \hbar^{2}, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, \ldots \tag{5.142}
\end{equation}
\)

Thus

\(
\begin{gather}
\hat{M}^{2} Y=j(j+1) \hbar^{2} Y, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots \tag{5.143}\
\hat{M}{z} Y=m{j} \hbar Y, \quad m_{j}=-j,-j+1, \ldots, j-1, j \tag{5.144}
\end{gather}
\)

We have found the eigenvalues of $\hat{M}^{2}$ and $\hat{M}_{z}$ using just the commutation relations. However, comparison of (5.143) and (5.144) with (5.104) and (5.105) shows that in addition to integral values for the angular-momentum quantum number $(l=0,1,2, \ldots)$ we now also have the possibility for half-integral values $\left(j=0, \frac{1}{2}, 1, \frac{3}{2}, \ldots\right)$. This perhaps suggests that there might be another kind of angular momentum besides orbital angular momentum. In Chapter 10 we shall see that spin angular momentum can have half-integral, as well as integral, quantum numbers. For orbital angular momentum, the boundary condition of single-valuedness of the $T(\phi)$ eigenfunctions [see the equation following (5.73)] eliminates the half-integral values of the angular-momentum quantum numbers. [Not everyone accepts single-valuedness as a valid boundary condition on wave functions, and many other reasons have been given for rejecting half-integral orbital-angular-momentum quantum numbers; see C. G. Gray, Am. J. Phys., 37, 559 (1969); M. L. Whippman, Am. J. Phys., 34, 656 (1966).]

The ladder-operator method can be used to solve other eigenvalue problems; see Prob. 5.36.

Before studying the hydrogen atom, we shall consider the more general problem of a single particle moving under a central force. The results of this section will apply to any central-force problem. Examples are the hydrogen atom (Section 6.5) and the isotropic three-dimensional harmonic oscillator (Prob. 6.3).

A central force is one derived from a potential-energy function that is spherically symmetric, which means that it is a function only of the distance of the particle from the origin: $V=V(r)$. The relation between force and potential energy is given by (5.31) as

\(
\begin{equation}
\mathbf{F}=-\nabla V(x, y, z)=-\mathbf{i}(\partial V / \partial x)-\mathbf{j}(\partial V / \partial y)-\mathbf{k}(\partial V / \partial z) \tag{6.1}
\end{equation}
\)

The partial derivatives in (6.1) can be found by the chain rule [Eqs. (5.53)-(5.55)]. Since $V$ in this case is a function of $r$ only, we have $(\partial V / \partial \theta){r, \phi}=0$ and $(\partial V / \partial \phi){r, \theta}=0$. Therefore,

\(
\begin{gather}
\left(\frac{\partial V}{\partial x}\right){y, z}=\frac{d V}{d r}\left(\frac{\partial r}{\partial x}\right){y, z}=\frac{x}{r} \frac{d V}{d r} \tag{6.2}\
\left(\frac{\partial V}{\partial y}\right){x, z}=\frac{y}{r} \frac{d V}{d r}, \quad\left(\frac{\partial V}{\partial z}\right){x, y}=\frac{z}{r} \frac{d V}{d r} \tag{6.3}
\end{gather}
\)

where Eqs. (5.57) and (5.58) have been used. Equation (6.1) becomes

\(
\begin{equation}
\mathbf{F}=-\frac{1}{r} \frac{d V}{d r}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=-\frac{d V(r)}{d r} \frac{\mathbf{r}}{r} \tag{6.4}
\end{equation}
\)

where (5.33) for $\mathbf{r}$ was used. The quantity $\mathbf{r} / r$ in (6.4) is a unit vector in the radial direction. A central force is radially directed.

Now we consider the quantum mechanics of a single particle subject to a central force. The Hamiltonian operator is

\(
\begin{equation}
\hat{H}=\hat{T}+\hat{V}=-\left(\hbar^{2} / 2 m\right) \nabla^{2}+V(r) \tag{6.5}
\end{equation}
\)

where $\nabla^{2} \equiv \partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}+\partial^{2} / \partial z^{2}$ [Eq. (3.46)]. Since $V$ is spherically symmetric, we shall work in spherical coordinates. Hence we want to transform the Laplacian operator to these coordinates. We already have the forms of the operators $\partial / \partial x, \partial / \partial y$, and $\partial / \partial z$ in these coordinates [Eqs. (5.62)-(5.64)], and by squaring each of these operators and
then adding their squares, we get the Laplacian. This calculation is left as an exercise. The result is (Prob. 6.4)

\(
\begin{equation}
\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2}} \cot \theta \frac{\partial}{\partial \theta}+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}} \tag{6.6}
\end{equation}
\)

Looking back to (5.68), which gives the operator $\hat{L}^{2}$ for the square of the magnitude of the orbital angular momentum of a single particle, we see that

\(
\begin{equation}
\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}-\frac{1}{r^{2} \hbar^{2}} \hat{L}^{2} \tag{6.7}
\end{equation}
\)

The Hamiltonian (6.5) becomes

\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2}+V(r) \tag{6.8}
\end{equation}
\)

In classical mechanics a particle subject to a central force has its angular momentum conserved (Section 5.3). In quantum mechanics we might ask whether we can have states with definite values for both the energy and the angular momentum. To have the set of eigenfunctions of $\hat{H}$ also be eigenfunctions of $\hat{L}^{2}$, the commutator $\left[\hat{H}, \hat{L}^{2}\right]$ must vanish. We have

\(
\begin{gather}
{\left[\hat{H}, \hat{L}^{2}\right]=\left[\hat{T}, \hat{L}^{2}\right]+\left[\hat{V}, \hat{L}^{2}\right]} \
{\left[\hat{T}, \hat{L}^{2}\right]=\left[-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2}, \hat{L}^{2}\right]} \
{\left[\hat{T}, \hat{L}^{2}\right]=-\frac{\hbar^{2}}{2 m}\left[\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}, \hat{L}^{2}\right]+\frac{1}{2 m}\left[\frac{1}{r^{2}} \hat{L}^{2}, \hat{L}^{2}\right]} \tag{6.9}
\end{gather}
\)

Recall that $\hat{L}^{2}$ involves only $\theta$ and $\phi$ and not $r$ [Eq. (5.68)]. Hence it commutes with every operator that involves only $r$. [To reach this conclusion, we must use relations like (5.47) with $x$ and $z$ replaced by $r$ and $\theta$.] Thus the first commutator in (6.9) is zero. Moreover, since any operator commutes with itself, the second commutator in (6.9) is zero. Therefore, $\left[\hat{T}, \hat{L}^{2}\right]=0$. Also, since $\hat{L}^{2}$ does not involve $r$, and $V$ is a function of $r$ only, we have $\left[\hat{V}, \hat{L}^{2}\right]=0$. Therefore,

\(
\begin{equation}
\left[\hat{H}, \hat{L}^{2}\right]=0 \quad \text { if } V=V(r) \tag{6.10}
\end{equation}
\)

$\hat{H}$ commutes with $\hat{L}^{2}$ when the potential-energy function is independent of $\theta$ and $\phi$.
Now consider the operator $\hat{L}{z}=-i \hbar \partial / \partial \phi[E q . ~(5.67)]$. Since $\hat{L}{z}$ does not involve $r$ and since it commutes with $\hat{L}^{2}$ [Eq. (5.50)], it follows that $\hat{L}_{z}$ commutes with the Hamiltonian (6.8):

\(
\begin{equation}
\left[\hat{H}, \hat{L}_{z}\right]=0 \quad \text { if } V=V(r) \tag{6.11}
\end{equation}
\)

We can therefore have a set of simultaneous eigenfunctions of $\hat{H}, \hat{L}^{2}$, and $\hat{L}_{z}$ for the central-force problem. Let $\psi$ denote these common eigenfunctions:

\(
\begin{gather}
\hat{H} \psi=E \psi \tag{6.12}\
\hat{L}^{2} \psi=l(l+1) \hbar^{2} \psi, \quad l=0,1,2, \ldots \tag{6.13}\
\hat{L}_{z} \psi=m \hbar \psi, \quad m=-l, \quad-l+1, \ldots, l \tag{6.14}
\end{gather}
\)

where Eqs. (5.104) and (5.105) were used.

Using (6.8) and (6.13), we have for the Schrödinger equation (6.12)

\(
\begin{gather}
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2} \psi+V(r) \psi=E \psi \
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}\right)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} \psi+V(r) \psi=E \psi \tag{6.15}
\end{gather}
\)

The eigenfunctions of $\hat{L}^{2}$ are the spherical harmonics $Y{l}^{m}(\theta, \phi)$, and since $\hat{L}^{2}$ does not involve $r$, we can multiply $Y{l}^{m}$ by an arbitrary function of $r$ and still have eigenfunctions of $\hat{L}^{2}$ and $\hat{L}_{z}$. Therefore,

\(
\begin{equation}
\psi=R(r) Y_{l}^{m}(\theta, \phi) \tag{6.16}
\end{equation}
\)

Using (6.16) in (6.15), we then divide both sides by $Y_{l}^{m}$ to get an ordinary differential equation for the unknown function $R(r)$ :

\(
\begin{equation}
-\frac{\hbar^{2}}{2 m}\left(R^{\prime \prime}+\frac{2}{r} R^{\prime}\right)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} R+V(r) R=E R(r) \tag{6.17}
\end{equation}
\)

We have shown that, for any one-particle problem with a spherically symmetric potential-energy function $V(r)$, the stationary-state wave functions are $\psi=R(r) Y_{l}^{m}(\theta, \phi)$, where the radial factor $R(r)$ satisfies (6.17). By using a specific form for $V(r)$ in (6.17), we can solve it for a particular problem.

Up to this point, we have solved only one-particle quantum-mechanical problems. The hydrogen atom is a two-particle system, and as a preliminary to dealing with the H atom, we first consider a simpler case, that of two noninteracting particles.

Suppose that a system is composed of the noninteracting particles 1 and 2. Let $q{1}$ and $q{2}$ symbolize the coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$ of particles 1 and 2 . Because the particles exert no forces on each other, the classical-mechanical energy of the system is the sum of the energies of the two particles: $E=E{1}+E{2}=T{1}+V{1}+T{2}+V{2}$, and the classical Hamiltonian is the sum of Hamiltonians for each particle: $H=H{1}+H{2}$. Therefore, the Hamiltonian operator is

\(
\hat{H}=\hat{H}{1}+\hat{H}{2}
\)

where $\hat{H}{1}$ involves only the coordinates $q{1}$ and the momentum operators $\hat{p}{1}$ that correspond to $q{1}$. The Schrödinger equation for the system is

\(
\begin{equation}
\left(\hat{H}{1}+\hat{H}{2}\right) \psi\left(q{1}, q{2}\right)=E \psi\left(q{1}, q{2}\right) \tag{6.18}
\end{equation}
\)

We try a solution of (6.18) by separation of variables, setting

\(
\begin{equation}
\psi\left(q{1}, q{2}\right)=G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.19}
\end{equation}
\)

We have

\(
\begin{equation}
\hat{H}{1} G{1}\left(q{1}\right) G{2}\left(q{2}\right)+\hat{H}{2} G{1}\left(q{1}\right) G{2}\left(q{2}\right)=E G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.20}
\end{equation}
\)

Since $\hat{H}{1}$ involves only the coordinate and momentum operators of particle 1 , we have $\hat{H}{1}\left[G{1}\left(q{1}\right) G{2}\left(q{2}\right)\right]=G{2}\left(q{2}\right) \hat{H}{1} G{1}\left(q{1}\right)$, since, as far as $\hat{H}{1}$ is concerned, $G{2}$ is a constant. Using this equation and a similar equation for $\hat{H}{2}$, we find that (6.20) becomes

\(
\begin{equation}
G{2}\left(q{2}\right) \hat{H}{1} G{1}\left(q{1}\right)+G{1}\left(q{1}\right) \hat{H}{2} G{2}\left(q{2}\right)=E G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.21}
\end{equation}
\)

\(
\begin{equation}
\frac{\hat{H}{1} G{1}\left(q{1}\right)}{G{1}\left(q{1}\right)}+\frac{\hat{H}{2} G{2}\left(q{2}\right)}{G{2}\left(q{2}\right)}=E \tag{6.22}
\end{equation}
\)

Now, by the same arguments used in connection with Eq. (3.65), we conclude that each term on the left in (6.22) must be a constant. Using $E{1}$ and $E{2}$ to denote these constants, we have

\(
\begin{gather}
\frac{\hat{H}{1} G{1}\left(q{1}\right)}{G{1}\left(q{1}\right)}=E{1}, \quad \frac{\hat{H}{2} G{2}\left(q{2}\right)}{G{2}\left(q{2}\right)}=E{2} \
E=E{1}+E{2} \tag{6.23}
\end{gather}
\)

Thus, when the system is composed of two noninteracting particles, we can reduce the two-particle problem to two separate one-particle problems by solving

\(
\begin{equation}
\hat{H}{1} G{1}\left(q{1}\right)=E{1} G{1}\left(q{1}\right), \quad \hat{H}{2} G{2}\left(q{2}\right)=E{2} G{2}\left(q{2}\right) \tag{6.24}
\end{equation}
\)

which are separate Schrödinger equations for each particle.
Generalizing this result to $n$ noninteracting particles, we have

\(
\begin{gather}
\hat{H}=\hat{H}{1}+\hat{H}{2}+\cdots+\hat{H}{n} \
\psi\left(q{1}, q{2}, \ldots, q{n}\right)=G{1}\left(q{1}\right) G{2}\left(q{2}\right) \cdots G{n}\left(q{n}\right) \tag{6.25}\
E=E{1}+E{2}+\cdots+E{n} \tag{6.26}\
\hat{H}{i} G{i}=E{i} G_{i}, \quad i=1,2, \ldots, n \tag{6.27}
\end{gather}
\)

For a system of noninteracting particles, the energy is the sum of the individual energies of each particle and the wave function is the product of wave functions for each particle. The wave function $G{i}$ of particle $i$ is found by solving a Schrödinger equation for particle $i$ using the Hamiltonian $\hat{H}{i}$.

These results also apply to a single particle whose Hamiltonian is the sum of separate terms for each coordinate:

\(
\hat{H}=\hat{H}{x}\left(\hat{x}, \hat{p}{x}\right)+\hat{H}{y}\left(\hat{y}, \hat{p}{y}\right)+\hat{H}{z}\left(\hat{z}, \hat{p}{z}\right)
\)

In this case, we conclude that the wave functions and energies are

\(
\begin{gathered}
\psi(x, y, z)=F(x) G(y) K(z), \quad E=E{x}+E{y}+E{z} \
\hat{H}{x} F(x)=E{x} F(x), \quad \hat{H}{y} G(y)=E{y} G(y), \quad \hat{H}{z} K(z)=E_{z} K(z)
\end{gathered}
\)

Examples include the particle in a three-dimensional box (Section 3.5), the three-dimensional free particle (Prob. 3.42), and the three-dimensional harmonic oscillator (Prob. 4.20).

The hydrogen atom contains two particles, the proton and the electron. For a system of two particles 1 and 2 with coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$, the potential energy of interaction between the particles is usually a function of only the relative coordinates $x{2}-x{1}, y{2}-y{1}$, and $z{2}-z{1}$ of the particles. In this case the two-particle problem can be simplified to two separate one-particle problems, as we now prove.

Consider the classical-mechanical treatment of two interacting particles of masses $m{1}$ and $m{2}$. We specify their positions by the radius vectors $\mathbf{r}{1}$ and $\mathbf{r}{2}$ drawn from the origin

FIGURE 6.1 A two-particle system with center of mass at $C$.

of a Cartesian coordinate system (Fig. 6.1). Particles 1 and 2 have coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$. We draw the vector $\mathbf{r}=\mathbf{r}{2}-\mathbf{r}{1}$ from particle 1 to 2 and denote the components of $\mathbf{r}$ by $x, y$, and $z$ :

\(
\begin{equation}
x=x{2}-x{1}, \quad y=y{2}-y{1}, \quad z=z{2}-z{1} \tag{6.28}
\end{equation}
\)

The coordinates $x, y$, and $z$ are called the relative or internal coordinates.
We now draw the vector $\mathbf{R}$ from the origin to the system's center of mass, point $C$, and denote the coordinates of $C$ by $X, Y$, and $Z$ :

\(
\begin{equation}
\mathbf{R}=\mathbf{i} X+\mathbf{j} Y+\mathbf{k} Z \tag{6.29}
\end{equation}
\)

The definition of the center of mass of this two-particle system gives

\(
\begin{equation}
X=\frac{m{1} x{1}+m{2} x{2}}{m{1}+m{2}}, \quad Y=\frac{m{1} y{1}+m{2} y{2}}{m{1}+m{2}}, \quad Z=\frac{m{1} z{1}+m{2} z{2}}{m{1}+m{2}} \tag{6.30}
\end{equation}
\)

These three equations are equivalent to the vector equation

\(
\begin{equation}
\mathbf{R}=\frac{m{1} \mathbf{r}{1}+m{2} \mathbf{r}{2}}{m{1}+m{2}} \tag{6.31}
\end{equation}
\)

We also have

\(
\begin{equation}
\mathbf{r}=\mathbf{r}{2}-\mathbf{r}{1} \tag{6.32}
\end{equation}
\)

We regard (6.31) and (6.32) as simultaneous equations in the two unknowns $\mathbf{r}{1}$ and $\mathbf{r}{2}$ and solve for them to get

\(
\begin{equation}
\mathbf{r}{1}=\mathbf{R}-\frac{m{2}}{m{1}+m{2}} \mathbf{r}, \quad \mathbf{r}{2}=\mathbf{R}+\frac{m{1}}{m{1}+m{2}} \mathbf{r} \tag{6.33}
\end{equation}
\)

Equations (6.31) and (6.32) represent a transformation of coordinates from $x{1}, y{1}, z{1}, x{2}, y{2}, z{2}$ to $X, Y, Z, x, y, z$. Consider what happens to the Hamiltonian under this transformation. Let an overhead dot indicate differentiation with respect to time. The velocity of particle 1 is [Eq. (5.34)] $\mathbf{v}{1}=d \mathbf{r}{1} / d t=\dot{\mathbf{r}}_{1}$. The kinetic energy is the sum of the kinetic energies of the two particles:

\(
\begin{equation}
T=\frac{1}{2} m{1}\left|\dot{\mathbf{r}}{1}\right|^{2}+\frac{1}{2} m{2}\left|\dot{\mathbf{r}}{2}\right|^{2} \tag{6.34}
\end{equation}
\)

Introducing the time derivatives of Eqs. (6.33) into (6.34), we have

\(
\begin{aligned}
T= & \frac{1}{2} m{1}\left(\dot{\mathbf{R}}-\frac{m{2}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \cdot\left(\dot{\mathbf{R}}-\frac{m{2}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \
& +\frac{1}{2} m{2}\left(\dot{\mathbf{R}}+\frac{m{1}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \cdot\left(\dot{\mathbf{R}}+\frac{m{1}}{m{1}+m{2}} \dot{\mathbf{r}}\right)
\end{aligned}
\)

where $|\mathbf{A}|^{2}=\mathbf{A} \cdot \mathbf{A}[$ Eq. (5.24)] has been used. Using the distributive law for the dot products, we find, after simplifying,

\(
\begin{equation}
T=\frac{1}{2}\left(m{1}+m{2}\right)|\dot{\mathbf{R}}|^{2}+\frac{1}{2} \frac{m{1} m{2}}{m{1}+m{2}}|\dot{\mathbf{r}}|^{2} \tag{6.35}
\end{equation}
\)

Let $M$ be the total mass of the system:

\(
\begin{equation}
M \equiv m{1}+m{2} \tag{6.36}
\end{equation}
\)

We define the reduced mass $\mu$ of the two-particle system as

\(
\begin{equation}
\mu \equiv \frac{m{1} m{2}}{m{1}+m{2}} \tag{6.37}
\end{equation}
\)

Then

\(
\begin{equation}
T=\frac{1}{2} M|\dot{\mathbf{R}}|^{2}+\frac{1}{2} \mu|\dot{\mathbf{r}}|^{2} \tag{6.38}
\end{equation}
\)

The first term in (6.38) is the kinetic energy due to translational motion of the whole system of mass $M$. Translational motion is motion in which each particle undergoes the same displacement. The quantity $\frac{1}{2} M|\dot{\mathbf{R}}|^{2}$ would be the kinetic energy of a hypothetical particle of mass $M$ located at the center of mass. The second term in (6.38) is the kinetic energy of internal (relative) motion of the two particles. This internal motion is of two types. The distance $r$ between the two particles can change (vibration), and the direction of the $\mathbf{r}$ vector can change (rotation). Note that $|\dot{\mathbf{r}}|=|d \mathbf{r} / d t| \neq d|\mathbf{r}| / d t$.

Corresponding to the original coordinates $x{1}, y{1}, z{1}, x{2}, y{2}, z{2}$, we have six linear momenta:

\(
\begin{equation}
p{x{1}}=m{1} \dot{x}{1}, \quad \ldots, \quad p{z{2}}=m{2} \dot{z}{2} \tag{6.39}
\end{equation}
\)

Comparing Eqs. (6.34) and (6.38), we define the six linear momenta for the new coordinates $X, Y, Z, x, y, z$ as

\(
\begin{array}{rlrl}
p{X} & \equiv M \dot{X}, & p{Y} \equiv M \dot{Y}, & p{Z} \equiv M \dot{Z} \
p{x} \equiv \mu \dot{x}, & p{y} \equiv \mu \dot{y}, & p{z} \equiv \mu \dot{z}
\end{array}
\)

We define two new momentum vectors as

\(
\mathbf{p}{M} \equiv \mathbf{i} M \dot{X}+\mathbf{j} M \dot{Y}+\mathbf{k} M \dot{Z} \quad \text { and } \quad \mathbf{p}{\mu} \equiv \mathbf{i} \mu \dot{x}+\mathbf{j} \mu \dot{y}+\mathbf{k} \mu \dot{z}
\)

Introducing these momenta into (6.38), we have

\(
\begin{equation}
T=\frac{\left|\mathbf{p}{M}\right|^{2}}{2 M}+\frac{\left|\mathbf{p}{\mu}\right|^{2}}{2 \mu} \tag{6.40}
\end{equation}
\)

Now consider the potential energy. We make the restriction that $V$ is a function only of the relative coordinates $x, y$, and $z$ of the two particles:

\(
\begin{equation}
V=V(x, y, z) \tag{6.41}
\end{equation}
\)

An example of (6.41) is two charged particles interacting according to Coulomb's law [see Eq. (3.53)]. With this restriction on $V$, the Hamiltonian function is

\(
\begin{equation}
H=\frac{p{M}^{2}}{2 M}+\left[\frac{p{\mu}^{2}}{2 \mu}+V(x, y, z)\right] \tag{6.42}
\end{equation}
\)

Now suppose we had a system composed of a particle of mass $M$ subject to no forces and a particle of mass $\mu$ subject to the potential-energy function $V(x, y, z)$. Further suppose that there was no interaction between these particles. If $(X, Y, Z)$ are the coordinates of the particle of mass $M$, and $(x, y, z)$ are the coordinates of the particle of mass $\mu$, what is the Hamiltonian of this hypothetical system? Clearly, it is identical with (6.42).

The Hamiltonian (6.42) can be viewed as the sum of the Hamiltonians $p{M}^{2} / 2 M$ and $p{\mu}^{2} / 2 \mu+V(x, y, z)$ of two hypothetical noninteracting particles with masses $M$ and $\mu$. Therefore, the results of Section 6.2 show that the system's quantummechanical energy is the sum of energies of the two hypothetical particles [Eq. (6.23)]: $E=E{M}+E{\mu}$. From Eqs. (6.24) and (6.42), the translational energy $E{M}$ is found by solving the Schrödinger equation $\left(\hat{p}{M}^{2} / 2 M\right) \psi{M}=E{M} \psi{M}$. This is the Schrödinger equation for a free particle of mass $M$, so its possible eigenvalues are all nonnegative numbers: $E{M} \geq 0$ [Eq. (2.31)]. From (6.24) and (6.42), the energy $E_{\mu}$ is found by solving the Schrödinger equation

\(
\begin{equation}
\left[\frac{\hat{p}{\mu}^{2}}{2 \mu}+V(x, y, z)\right] \psi{\mu}(x, y, z)=E{\mu} \psi{\mu}(x, y, z) \tag{6.43}
\end{equation}
\)

We have thus separated the problem of two particles interacting according to a potential-energy function $V(x, y, z)$ that depends on only the relative coordinates $x, y, z$ into two separate one-particle problems: (1) the translational motion of the entire system of mass $M$, which simply adds a nonnegative constant energy $E_{M}$ to the system's energy, and (2) the relative or internal motion, which is dealt with by solving the Schrödinger equation (6.43) for a hypothetical particle of mass $\mu$ whose coordinates are the relative coordinates $x, y, z$ and that moves subject to the potential energy $V(x, y, z)$.

For example, for the hydrogen atom, which is composed of an electron (e) and a proton $(p)$, the atom's total energy is $E=E{M}+E{\mu}$, where $E{M}$ is the translational energy of motion through space of the entire atom of mass $M=m{e}+m{p}$, and where $E{\mu}$ is found by solving (6.43) with $\mu=m{e} m{p} /\left(m{e}+m{p}\right)$ and $V$ being the Coulomb's law potential energy of interaction of the electron and proton; see Section 6.5.

Before solving the Schrödinger equation for the hydrogen atom, we will first deal with the two-particle rigid rotor. This is a two-particle system with the particles held at a fixed distance from each other by a rigid massless rod of length $d$. For this problem, the vector $\mathbf{r}$ in Fig. 6.1 has the constant magnitude $|\mathbf{r}|=d$. Therefore (see Section 6.3), the kinetic energy of internal motion is wholly rotational energy. The energy of the rotor is entirely kinetic, and

\(
\begin{equation}
V=0 \tag{6.44}
\end{equation}
\)

Equation (6.44) is a special case of Eq. (6.41), and we may therefore use the results of the last section to separate off the translational motion of the system as a whole. We will concern ourselves only with the rotational energy. The Hamiltonian operator for the rotation is given by the terms in brackets in (6.43) as

\(
\begin{equation}
\hat{H}=\frac{\hat{p}{\mu}^{2}}{2 \mu}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}, \quad \mu=\frac{m{1} m{2}}{m{1}+m_{2}} \tag{6.45}
\end{equation}
\)

where $m{1}$ and $m{2}$ are the masses of the two particles. The coordinates of the fictitious particle of mass $\mu$ are the relative coordinates of $m{1}$ and $m{2}$ [Eq. (6.28)].

Instead of the relative Cartesian coordinates $x, y, z$, it will prove more fruitful to use the relative spherical coordinates $r, \theta, \phi$. The $r$ coordinate is equal to the magnitude of
the $\mathbf{r}$ vector in Fig. 6.1, and since $m{1}$ and $m{2}$ are constrained to remain a fixed distance apart, we have $r=d$. Thus the problem is equivalent to a particle of mass $\mu$ constrained to move on the surface of a sphere of radius $d$. Because the radial coordinate is constant, the wave function will be a function of $\theta$ and $\phi$ only. Therefore the first two terms of the Laplacian operator in (6.8) will give zero when operating on the wave function and may be omitted. Looking at things in a slightly different way, we note that the operators in (6.8) that involve $r$ derivatives correspond to the kinetic energy of radial motion, and since there is no radial motion, the $r$ derivatives are omitted from $\hat{H}$.

Since $V=0$ is a special case of $V=V(r)$, the results of Section 6.1 tell us that the eigenfunctions are given by (6.16) with the $r$ factor omitted:

\(
\begin{equation}
\psi=Y_{J}^{m}(\theta, \phi) \tag{6.46}
\end{equation}
\)

where $J$ rather than $l$ is used for the rotational angular-momentum quantum number.
The Hamiltonian operator is given by Eq. (6.8) with the $r$ derivatives omitted and $V(r)=0$. Thus

\(
\hat{H}=\left(2 \mu d^{2}\right)^{-1} \hat{L}^{2}
\)

Use of (6.13) gives

\(
\begin{align}
\hat{H} \psi & =E \psi \
\left(2 \mu d^{2}\right)^{-1} \hat{L}^{2} Y{J}^{m}(\theta, \phi) & =E Y{J}^{m}(\theta, \phi) \
\left(2 \mu d^{2}\right)^{-1} J(J+1) \hbar^{2} Y{J}^{m}(\theta, \phi) & =E Y{J}^{m}(\theta, \phi) \
E=\frac{J(J+1) \hbar^{2}}{2 \mu d^{2}}, \quad J & =0,1,2 \cdots \tag{6.47}
\end{align}
\)

The moment of inertia $I$ of a system of $n$ particles about some particular axis in space as defined as

\(
\begin{equation}
I \equiv \sum{i=1}^{n} m{i} \rho_{i}^{2} \tag{6.48}
\end{equation}
\)

where $m{i}$ is the mass of the $i$ th particle and $\rho{i}$ is the perpendicular distance from this particle to the axis. The value of $I$ depends on the choice of axis. For the two-particle rigid rotor, we choose our axis to be a line that passes through the center of mass and is perpendicular to the line joining $m{1}$ and $m{2}$ (Fig. 6.2). If we place the rotor so that the center of mass, point $C$, lies at the origin of a Cartesian coordinate system and the line joining $m{1}$ and $m{2}$ lies on the $x$ axis, then $C$ will have the coordinates $(0,0,0), m{1}$ will have the coordinates $\left(-\rho{1}, 0,0\right)$, and $m{2}$ will have the coordinates $\left(\rho{2}, 0,0\right)$. Using these coordinates in (6.30), we find

\(
\begin{equation}
m{1} \rho{1}=m{2} \rho{2} \tag{6.49}
\end{equation}
\)

FIGURE 6.2 Axis (dashed line) for calculating the moment of inertia of a two-particle rigid rotor. $C$ is the center of mass.

The moment of inertia of the rotor about the axis we have chosen is

\(
\begin{equation}
I=m{1} \rho{1}^{2}+m{2} \rho{2}^{2} \tag{6.50}
\end{equation}
\)

Using (6.49), we transform Eq. (6.50) to (see Prob. 6.14)

\(
\begin{equation}
I=\mu d^{2} \tag{6.51}
\end{equation}
\)

where $\mu \equiv m{1} m{2} /\left(m{1}+m{2}\right)$ is the reduced mass of the system and $d \equiv \rho{1}+\rho{2}$ is the distance between $m{1}$ and $m{2}$. The allowed energy levels (6.47) of the two-particle rigid rotor are

\(
\begin{equation}
E=\frac{J(J+1) \hbar^{2}}{2 I}, \quad J=0,1,2, \ldots \tag{6.52}
\end{equation}
\)

The lowest level is $E=0$, so there is no zero-point rotational energy. Having zero rotational energy and therefore zero angular momentum for the rotor does not violate the uncertainty principle; recall the discussion following Eq. (5.105). Note that $E$ increases as $J^{2}+J$, so the spacing between adjacent rotational levels increases as $J$ increases.

Are the rotor energy levels (6.52) degenerate? The energy depends on $J$ only, but the wave function (6.46) depends on $J$ and $m$, where $m \hbar$ is the $z$ component of the rotor's angular momentum. For each value of $J$, there are $2 J+1$ values of $m$, ranging from $-J$ to $J$. Hence the levels are $(2 J+1)$-fold degenerate. The states of a degenerate level have different orientations of the angular-momentum vector of the rotor about a space-fixed axis.

The angles $\theta$ and $\phi$ in the wave function (6.46) are relative coordinates of the two point masses. If we set up a Cartesian coordinate system with the origin at the rotor's center of mass, $\theta$ and $\phi$ will be as shown in Fig. 6.3. This coordinate system undergoes the same translational motion as the rotor's center of mass but does not rotate in space.

The rotational angular momentum $\left[J(J+1) \hbar^{2}\right]^{1 / 2}$ is the angular momentum of the two particles with respect to an origin at the system's center of mass $C$.

The rotational levels of a diatomic molecule can be well approximated by the two-particle rigid-rotor energies (6.52). It is found (Levine, Molecular Spectroscopy, Section 4.4) that when a diatomic molecule absorbs or emits radiation, the allowed pure-rotational transitions are given by the selection rule

\(
\begin{equation}
\Delta J= \pm 1 \tag{6.53}
\end{equation}
\)

In addition, a molecule must have a nonzero dipole moment in order to show a pure-rotational spectrum. A pure-rotational transition is one where only the rotational

FIGURE 6.3 Coordinate system for the two-particle rigid rotor.

quantum number changes. [Vibration-rotation transitions (Section 4.3) involve changes in both vibrational and rotational quantum numbers.] The spacing between adjacent low-lying rotational levels is significantly less than that between adjacent vibrational levels, and the pure-rotational spectrum falls in the microwave (or the far-infrared) region. The frequencies of the pure-rotational spectral lines of a diatomic molecule are then (approximately)

\(
\begin{gather}
\nu=\frac{E{J+1}-E{J}}{h}=\frac{[(J+1)(J+2)-J(J+1)] h}{8 \pi^{2} I}=2(J+1) B \tag{6.54}\
B \equiv h / 8 \pi^{2} I, \quad J=0,1,2, \ldots \tag{6.55}
\end{gather}
\)

$B$ is called the rotational constant of the molecule.
The spacings between the diatomic rotational levels (6.52) for low and moderate values of $J$ are generally less than or of the same order of magnitude as $k T$ at room temperature, so the Boltzmann distribution law (4.63) shows that many rotational levels are significantly populated at room temperature. Absorption of radiation by diatomic molecules having $J=0$ (the $J=0 \rightarrow 1$ transition) gives a line at the frequency $2 B$; absorption by molecules having $J=1$ (the $J=1 \rightarrow 2$ transition) gives a line at $4 B$; absorption by $J=2$ molecules gives a line at $6 B$; and so on. See Fig. 6.4.

Measurement of the rotational absorption frequencies allows $B$ to be found. From $B$, we get the molecule's moment of inertia $I$, and from $I$ we get the bond distance $d$. The value of $d$ found is an average over the $v=0$ vibrational motion. Because of the asymmetry of the potential-energy curve in Figs. 4.6 and 13.1, $d$ is very slightly longer than the equilibrium bond length $R_{e}$ in Fig. 13.1.

As noted in Section 4.3, isotopic species such as ${ }^{1} \mathrm{H}^{35} \mathrm{Cl}$ and ${ }^{1} \mathrm{H}^{37} \mathrm{Cl}$ have virtually the same electronic energy curve $U(R)$ and so have virtually the same equilibrium bond distance. However, the different isotopic masses produce different moments of inertia and hence different rotational absorption frequencies.

Because molecules are not rigid, the rotational energy levels for diatomic molecules differ slightly from rigid-rotor levels. From (6.52) and (6.55), the two-particle rigid-rotor levels are $E{\mathrm{rot}}=B h J(J+1)$. Because of the anharmonicity of molecular vibration (Fig. 4.6), the average internuclear distance increases with increasing vibrational quantum number $v$, so as $v$ increases, the moment of inertia $I$ increases and the rotational constant $B$ decreases. To allow for the dependence of $B$ on $v$, one replaces $B$ in $E{\text {rot }}$ by $B{v}$. The mean rotational constant $B{v}$ for vibrational level $v$ is $B{v}=B{e}-\alpha{e}(v+1 / 2)$, where $B{e}$ is calculated using the equilibrium internuclear separation $R_{e}$ at the bottom

FIGURE 6.4 Two-particle rigid-rotor absorption transitions.
of the potential-energy curve in Fig. 4.6, and the vibration-rotation coupling constant $\alpha{e}$ is a positive constant (different for different molecules) that is much smaller than $B{e}$. Also, as the rotational energy increases, there is a very slight increase in average internuclear distance (a phenomenon called centrifugal distortion). This adds the term $-h D J^{2}(J+1)^{2}$ to $E{\text {rot }}$, where the centrifugal-distortion constant $D$ is an extremely small positive constant, different for different molecules. For example, for ${ }^{12} \mathrm{C}^{16} \mathrm{O}$, $B{0}=57636 \mathrm{MHz}, \alpha{e}=540 \mathrm{MHz}$, and $D=0.18 \mathrm{MHz}$. As noted in Section 4.3 , for lighter diatomic molecules, nearly all the molecules are in the ground $v=0$ vibrational level at room temperature, and the observed rotational constant is $B{0}$.

For more discussion of nuclear motion in diatomic molecules, see Section 13.2. For the rotational energies of polyatomic molecules, see Townes and Schawlow, chaps. 2-4.

EXAMPLE

The lowest-frequency pure-rotational absorption line of ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$ occurs at 48991.0 MHz . Find the bond distance in ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$.

The lowest-frequency rotational absorption is the $J=0 \rightarrow 1$ line. Equations (1.4), (6.52), and (6.51) give

\(
h \nu=E{\text {upper }}-E{\text {lower }}=\frac{1(2) \hbar^{2}}{2 \mu d^{2}}-\frac{0(1) \hbar^{2}}{2 \mu d^{2}}
\)

which gives $d=\left(h / 4 \pi^{2} \nu \mu\right)^{1 / 2}$. Table A. 3 in the Appendix gives

\(
\mu=\frac{m{1} m{2}}{m{1}+m{2}}=\frac{12(31.97207)}{(12+31.97207)} \frac{1}{6.02214 \times 10^{23}} \mathrm{~g}=1.44885 \times 10^{-23} \mathrm{~g}
\)

The SI unit of mass is the kilogram, and

\(
\begin{aligned}
d=\frac{1}{2 \pi}\left(\frac{h}{\nu_{0 \rightarrow 1} \mu}\right)^{1 / 2} & =\frac{1}{2 \pi}\left[\frac{6.62607 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(48991.0 \times 10^{6} \mathrm{~s}^{-1}\right)\left(1.44885 \times 10^{-26} \mathrm{~kg}\right)}\right]^{1 / 2} \
& =1.5377 \times 10^{-10} \mathrm{~m}=1.5377 \AA
\end{aligned}
\)

EXERCISE The $J=1$ to $J=2$ pure-rotational transition of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$ occurs at $230.538 \mathrm{GHz} .\left(1 \mathrm{GHz}=10^{9} \mathrm{~Hz}\right.$. ) Find the bond distance in this molecule. (Answer: $1.1309 \times 10^{-10} \mathrm{~m}$.)

The hydrogen atom consists of a proton and an electron. If $e$ symbolizes the charge on the proton $\left(e=+1.6 \times 10^{-19} \mathrm{C}\right)$, then the electron's charge is $-e$.

A few scientists have speculated that the proton and electron charges might not be exactly equal in magnitude. Experiments show that the magnitudes of the electron and proton charges are equal to within one part in $10^{21}$. See G. Bressi et al., Phys. Rev. A, 83, 052101 (2011) (available online at arxiv.org/abs/1102.2766).

We shall assume the electron and proton to be point masses whose interaction is given by Coulomb's law. In discussing atoms and molecules, we shall usually be considering isolated systems, ignoring interatomic and intermolecular interactions.

Instead of treating just the hydrogen atom, we consider a slightly more general problem: the hydrogenlike atom, which consists of one electron and a nucleus of charge $Z e$. For $Z=1$, we have the hydrogen atom; for $Z=2$, the $\mathrm{He}^{+}$ion; for $Z=3$, the $\mathrm{Li}^{2+}$ ion; and so on. The hydrogenlike atom is the most important system in quantum chemistry. An exact solution of the Schrödinger equation for atoms with more than one electron cannot be obtained because of the interelectronic repulsions. If, as a crude first approximation, we ignore these repulsions, then the electrons can be treated independently. (See Section 6.2.) The atomic wave function will be approximated by a product of one-electron functions, which will be hydrogenlike wave functions. A one-electron wave function (whether or not it is hydrogenlike) is called an orbital. (More precisely, an orbital is a one-electron spatial wave function, where the word spatial means that the wave function depends on the electron's three spatial coordinates $x, y$, and $z$ or $r, \theta$, and $\phi$. We shall see in Chapter 10 that the existence of electron spin adds a fourth coordinate to a one-electron wave function, giving what is called a spin-orbital.) An orbital for an electron in an atom is called an atomic orbital. We shall use atomic orbitals to construct approximate wave functions for atoms with many electrons (Chapter 11). Orbitals are also used to construct approximate wave functions for molecules.

For the hydrogenlike atom, let $(x, y, z)$ be the coordinates of the electron relative to the nucleus, and let $\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z$. The Coulomb's law force on the electron in the hydrogenlike atom is [see Eq. (1.37)]

\(
\begin{equation}
\mathbf{F}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r^{2}} \frac{\mathbf{r}}{r} \tag{6.56}
\end{equation}
\)

where $\mathbf{r} / r$ is a unit vector in the $\mathbf{r}$ direction. The minus sign indicates an attractive force.
The possibility of small deviations from Coulomb's law has been considered. Experiments have shown that if the Coulomb's-law force is written as being proportional to $r^{-2+s}$, then $|s|<10^{-16}$. A deviation from Coulomb's law can be shown to imply a nonzero photon rest mass. No evidence exists for a nonzero photon rest mass, and data indicate that any such mass must be less than $10^{-51} \mathrm{~g}$; A. S. Goldhaber and M. M. Nieto, Rev. Mod. Phys., 82, 939 (2010) (arxiv.org/abs/0809.1003); G. Spavieri et al., Eur. Phys. J. D, 61, 531 (2011) (link.springer.com/content/pdf/10.1140/epjd/ e2011-10508-7).

The force in (6.56) is central, and comparison with Eq. (6.4) gives $d V(r) / d r=$ $Z e^{2} / 4 \pi \varepsilon_{0} r^{2}$. Integration gives

\(
\begin{equation}
V=\frac{Z e^{2}}{4 \pi \varepsilon{0}} \int \frac{1}{r^{2}} d r=-\frac{Z e^{2}}{4 \pi \varepsilon{0} r} \tag{6.57}
\end{equation}
\)

where the integration constant has been taken as 0 to make $V=0$ at infinite separation between the charges. For any two charges $Q{1}$ and $Q{2}$ separated by distance $r_{12}$, Eq. (6.57) becomes

\(
\begin{equation}
V=\frac{Q{1} Q{2}}{4 \pi \varepsilon{0} r{12}} \tag{6.58}
\end{equation}
\)

Since the potential energy of this two-particle system depends only on the relative coordinates of the particles, we can apply the results of Section 6.3 to reduce the problem to two one-particle problems. The translational motion of the atom as a whole simply adds some constant to the total energy, and we shall not concern ourselves

FIGURE 6.5 Relative spherical coordinates.

with it. To deal with the internal motion of the system, we introduce a fictitious particle of mass

\(
\begin{equation}
\mu=\frac{m{e} m{N}}{m{e}+m{N}} \tag{6.59}
\end{equation}
\)

where $m{e}$ and $m{N}$ are the electronic and nuclear masses. The particle of reduced mass $\mu$ moves subject to the potential-energy function (6.57), and its coordinates $(r, \theta, \phi)$ are the spherical coordinates of one particle relative to the other (Fig. 6.5).

The Hamiltonian for the internal motion is [Eq. (6.43)]

\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r} \tag{6.60}
\end{equation}
\)

Since $V$ is a function of the $r$ coordinate only, we have a one-particle central-force problem, and we may apply the results of Section 6.1. Using Eqs. (6.16) and (6.17), we have for the wave function

\(
\begin{equation}
\psi(r, \theta, \phi)=R(r) Y_{l}^{m}(\theta, \phi), \quad l=0,1,2, \ldots, \quad|m| \leq l \tag{6.61}
\end{equation}
\)

where $Y_{l}^{m}$ is a spherical harmonic, and the radial function $R(r)$ satisfies

\(
\begin{equation}
-\frac{\hbar^{2}}{2 \mu}\left(R^{\prime \prime}+\frac{2}{r} R^{\prime}\right)+\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}} R-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r} R=E R(r) \tag{6.62}
\end{equation}
\)

To save time in writing, we define the constant $a$ as

\(
\begin{equation}
a \equiv \frac{4 \pi \varepsilon_{0} \hbar^{2}}{\mu e^{2}} \tag{6.63}
\end{equation}
\)

and (6.62) becomes

\(
\begin{equation}
R^{\prime \prime}+\frac{2}{r} R^{\prime}+\left[\frac{8 \pi \varepsilon_{0} E}{a e^{2}}+\frac{2 Z}{a r}-\frac{l(l+1)}{r^{2}}\right] R=0 \tag{6.64}
\end{equation}
\)

Solution of the Radial Equation

We could now try a power-series solution of (6.64), but we would get a three-term rather than a two-term recursion relation. We therefore seek a substitution that will lead to a twoterm recursion relation. It turns out that the proper substitution can be found by examining the behavior of the solution for large values of $r$. For large $r$, (6.64) becomes

\(
\begin{equation}
R^{\prime \prime}+\frac{8 \pi \varepsilon_{0} E}{a e^{2}} R=0, \quad r \text { large } \tag{6.65}
\end{equation}
\)

which may be solved using the auxiliary equation (2.7). The solutions are

\(
\begin{equation}
\exp \left[ \pm\left(-8 \pi \varepsilon_{0} E / a e^{2}\right)^{1 / 2} r\right] \tag{6.66}
\end{equation}
\)

Suppose that $E$ is positive. The quantity under the square-root sign in (6.66) is negative, and the factor multiplying $r$ is imaginary:

\(
\begin{equation}
R(r) \sim e^{ \pm i \sqrt{2 \mu E} r / \hbar}, \quad E \geq 0 \tag{6.67}
\end{equation}
\)

where (6.63) was used. The symbol $\sim$ in (6.67) indicates that we are giving the behavior of $R(r)$ for large values of $r$; this is called the asymptotic behavior of the function. Note the resemblance of (6.67) to Eq. (2.30), the free-particle wave function. Equation (6.67) does not give the complete radial factor in the wave function for positive energies. Further study (Bethe and Salpeter, pages 21-24) shows that the radial function for $E \geq 0$ remains finite for all values of $r$, no matter what the value of $E$. Thus, just as for the free particle, all nonnegative energies of the hydrogen atom are allowed. Physically, these eigenfunctions correspond to states in which the electron is not bound to the nucleus; that is, the atom is ionized. (A classical-mechanical analogy is a comet moving in a hyperbolic orbit about the sun. The comet is not bound and makes but one visit to the solar system.) Since we get continuous rather than discrete allowed values for $E \geq 0$, the positive-energy eigenfunctions are called continuum eigenfunctions. The angular part of a continuum wave function is a spherical harmonic. Like the free-particle wave functions, the continuum eigenfunctions are not normalizable in the usual sense.

We now consider the bound states of the hydrogen atom, with $E<0$. (For a bound state, $\psi \rightarrow 0$ as $x \rightarrow \pm \infty$.) In this case, the quantity in parentheses in (6.66) is positive. Since we want the wave functions to remain finite as $r$ goes to infinity, we prefer the minus sign in (6.66), and in order to get a two-term recursion relation, we make the substitution

\(
\begin{gather}
R(r)=e^{-C r} K(r) \tag{6.68}\
C \equiv\left(-\frac{8 \pi \varepsilon_{0} E}{a e^{2}}\right)^{1 / 2} \tag{6.69}
\end{gather}
\)

where $e$ in (6.68) stands for the base of natural logarithms, and not the proton charge. Use of the substitution (6.68) will guarantee nothing about the behavior of the wave function for large $r$. The differential equation we obtain from this substitution will still have two linearly independent solutions. We can make any substitution we please in a differential equation; in fact, we could make the substitution $R(r)=e^{+C r} J(r)$ and still wind up with the correct eigenfunctions and eigenvalues. The relation between $J$ and $K$ would naturally be $J(r)=e^{-2 C r} K(r)$.

Proceeding with (6.68), we evaluate $R^{\prime}$ and $R^{\prime \prime}$, substitute into (6.64), multiply by $r^{2} e^{C r}$, and use (6.69) to get the following differential equation for $K(r)$ :

\(
\begin{equation}
r^{2} K^{\prime \prime}+\left(2 r-2 C r^{2}\right) K^{\prime}+\left[\left(2 Z a^{-1}-2 C\right) r-l(l+1)\right] K=0 \tag{6.70}
\end{equation}
\)

We could now substitute a power series of the form

\(
\begin{equation}
K=\sum{k=0}^{\infty} c{k} r^{k} \tag{6.71}
\end{equation}
\)

for $K$. If we did we would find that, in general, the first few coefficients in (6.71) are zero. If $c_{s}$ is the first nonzero coefficient, (6.71) can be written as

\(
\begin{equation}
K=\sum{k=s}^{\infty} c{k} r^{k}, \quad c_{s} \neq 0 \tag{6.72}
\end{equation}
\)

Letting $j \equiv k-s$, and then defining $b{j}$ as $b{j} \equiv c_{j+s}$, we have

\(
\begin{equation}
K=\sum{j=0}^{\infty} c{j+s} r^{j+s}=r^{s} \sum{j=0}^{\infty} b{j} r^{j}, \quad b_{0} \neq 0 \tag{6.73}
\end{equation}
\)

(Although the various substitutions we are making might seem arbitrary, they are standard procedure in solving differential equations by power series.) The integer $s$ is evaluated by substitution into the differential equation. Equation (6.73) is

\(
\begin{gather}
K(r)=r^{s} M(r) \tag{6.74}\
M(r)=\sum{j=0}^{\infty} b{j} r^{j}, \quad b_{0} \neq 0 \tag{6.75}
\end{gather}
\)

Evaluating $K^{\prime}$ and $K^{\prime \prime}$ from (6.74) and substituting into (6.70), we get

\(
\begin{equation}
r^{2} M^{\prime \prime}+\left[(2 s+2) r-2 C r^{2}\right] M^{\prime}+\left[s^{2}+s+\left(2 Z a^{-1}-2 C-2 C s\right) r-l(l+1)\right] M=0 \tag{6.76}
\end{equation}
\)

To find $s$, we look at (6.76) for $r=0$. From (6.75), we have

\(
\begin{equation}
M(0)=b{0}, \quad M^{\prime}(0)=b{1}, \quad M^{\prime \prime}(0)=2 b_{2} \tag{6.77}
\end{equation}
\)

Using (6.77) in (6.76), we find for $r=0$

\(
\begin{equation}
b_{0}\left(s^{2}+s-l^{2}-l\right)=0 \tag{6.78}
\end{equation}
\)

Since $b_{0}$ is not zero, the terms in parentheses must vanish: $s^{2}+s-l^{2}-l=0$. This is a quadratic equation in the unknown $s$, with the roots

\(
\begin{equation}
s=l, \quad s=-l-1 \tag{6.79}
\end{equation}
\)

These roots correspond to the two linearly independent solutions of the differential equation. Let us examine them from the standpoint of proper behavior of the wave function. From Eqs. (6.68), (6.74), and (6.75), we have

\(
\begin{equation}
R(r)=e^{-C r} r^{s} \sum{j=0}^{\infty} b{j} r^{j} \tag{6.80}
\end{equation}
\)

Since $e^{-C r}=1-C r+\ldots$, the function $R(r)$ behaves for small $r$ as $b_{0} r^{s}$. For the root $s=l, R(r)$ behaves properly at the origin. However, for $s=-l-1, R(r)$ is proportional to

\(
\begin{equation}
\frac{1}{r^{l+1}} \tag{6.81}
\end{equation}
\)

for small $r$. Since $l=0,1,2, \ldots$, the root $s=-l-1$ makes the radial factor in the wave function infinite at the origin. Many texts take this as sufficient reason for rejecting this root. However, this is not a good argument, since for the relativistic hydrogen atom, the $l=0$ eigenfunctions are infinite at $r=0$. Let us therefore look at (6.81) from the standpoint of quadratic integrability, since we certainly require the bound-state eigenfunctions to be normalizable.

The normalization integral [Eq. (5.80)] for the radial functions that behave like (6.81) looks like

\(
\begin{equation}
\int{0}|R|^{2} r^{2} d r \approx \int{0} \frac{1}{r^{2 l}} d r \tag{6.82}
\end{equation}
\)

for small $r$. The behavior of the integral at the lower limit of integration is

\(
\begin{equation}
\left.\frac{1}{r^{2 l-1}}\right|_{r=0} \tag{6.83}
\end{equation}
\)

For $l=1,2,3, \ldots,(6.83)$ is infinite, and the normalization integral is infinite. Hence we must reject the root $s=-l-1$ for $l \geq 1$. However, for $l=0$, (6.83) is finite, and there is no trouble with quadratic integrability. Thus there is a quadratically integrable solution to the radial equation that behaves as $r^{-1}$ for small $r$.

Further study of this solution shows that it corresponds to an energy value that the experimental hydrogen-atom spectrum shows does not exist. Thus the $r^{-1}$ solution must be rejected, but there is some dispute over the reason for doing so. One view is that the $1 / r$ solution satisfies the Schrödinger equation everywhere in space except at the origin and hence must be rejected [Dirac, page 156; B. H. Armstrong and E. A. Power, Am. J. Phys., 31, 262 (1963)]. A second view is that the $1 / r$ solution must be rejected because the Hamiltonian operator is not Hermitian with respect to it (Merzbacher, Section 10.5). (In Chapter 7 we shall define Hermitian operators and show that quantum-mechanical operators are required to be Hermitian.) Further discussion is given in A. A. Khelashvili and T. P. Nadareishvili, Am. J.Phys., 79, 668 (2011) (see arxiv.org/abs/1102.1185) and in Y. C. Cantelaube, arxiv.org/abs/1203.0551.

Taking the first root in (6.79), we have for the radial factor (6.80)

\(
\begin{equation}
R(r)=e^{-C r_{r}^{l} M(r)} \tag{6.84}
\end{equation}
\)

With $s=l$, Eq. (6.76) becomes

\(
\begin{equation}
r M^{\prime \prime}+(2 l+2-2 C r) M^{\prime}+\left(2 Z a^{-1}-2 C-2 C l\right) M=0 \tag{6.85}
\end{equation}
\)

From (6.75), we have

\(
\begin{gather}
M(r)=\sum{j=0}^{\infty} b{j} r^{j} \tag{6.86}\
M^{\prime}=\sum{j=0}^{\infty} j b{j} r^{j-1}=\sum{j=1}^{\infty} j b{j} r^{j-1}=\sum{k=0}^{\infty}(k+1) b{k+1} r^{k}=\sum{j=0}^{\infty}(j+1) b{j+1} r^{j} \
M^{\prime \prime}=\sum{j=0}^{\infty} j(j-1) b{j} r^{j-2}=\sum{j=1}^{\infty} j(j-1) b{j} r^{j-2}=\sum{k=0}^{\infty}(k+1) k b{k+1} r^{k-1} \
M^{\prime \prime}=\sum{j=0}^{\infty}(j+1) j b{j+1} r^{j-1} \tag{6.87}
\end{gather}
\)

Substituting these expressions in (6.85) and combining sums, we get

\(
\sum{j=0}^{\infty}\left[j(j+1) b{j+1}+2(l+1)(j+1) b{j+1}+\left(\frac{2 Z}{a}-2 C-2 C l-2 C j\right) b{j}\right] r^{j}=0
\)

Setting the coefficient of $r^{j}$ equal to zero, we get the recursion relation

\(
\begin{equation}
b{j+1}=\frac{2 C+2 C l+2 C j-2 Z a^{-1}}{j(j+1)+2(l+1)(j+1)} b{j} \tag{6.88}
\end{equation}
\)

We now must examine the behavior of the infinite series (6.86) for large $r$. The result of the same procedure used to examine the harmonic-oscillator power series in (4.42) suggests that for large $r$ the infinite series (6.86) behaves like $e^{2 C r}$. (See Prob. 6.20.) For large $r$, the radial function (6.84) behaves like

\(
\begin{equation}
R(r) \sim e^{-C r} r^{l} e^{2 C r}=r^{l} e^{C r} \tag{6.89}
\end{equation}
\)

Therefore, $R(r)$ will become infinite as $r$ goes to infinity and will not be quadratically integrable. The only way to avoid this "infinity catastrophe" (as in the harmonicoscillator case) is to have the series terminate after a finite number of terms, in which case the $e^{-C r}$ factor will ensure that the wave function goes to zero as $r$ goes to infinity. Let the
last term in the series be $b{k} r^{k}$. Then, to have $b{k+1}, b{k+2}, \ldots$ all vanish, the fraction multiplying $b{j}$ in the recursion relation (6.88) must vanish when $j=k$. We have

\(
\begin{equation}
2 C(k+l+1)=2 Z a^{-1}, \quad k=0,1,2, \ldots \tag{6.90}
\end{equation}
\)

$k$ and $l$ are integers, and we now define a new integer $n$ by

\(
\begin{equation}
n \equiv k+l+1, \quad n=1,2,3, \ldots \tag{6.91}
\end{equation}
\)

From (6.91) the quantum number $l$ must satisfy

\(
\begin{equation}
l \leq n-1 \tag{6.92}
\end{equation}
\)

Hence $l$ ranges from 0 to $n-1$.

Energy Levels

Use of (6.91) in (6.90) gives

\(
\begin{equation}
C n=Z a^{-1} \tag{6.93}
\end{equation}
\)

Substituting $C \equiv\left(-8 \pi \varepsilon_{0} E / a e^{2}\right)^{1 / 2}$ [Eq. (6.69)] into (6.93) and solving for $E$, we get

\(
\begin{equation}
E=-\frac{Z^{2}}{n^{2}} \frac{e^{2}}{8 \pi \varepsilon{0} a}=-\frac{Z^{2} \mu e^{4}}{8 \varepsilon{0}^{2} n^{2} h^{2}} \tag{6.94}
\end{equation}
\)

where $a \equiv 4 \pi \varepsilon_{0} \hbar^{2} / \mu e^{2}$ [Eq. (6.63)]. These are the bound-state energy levels of the hydrogenlike atom, and they are discrete. Figure 6.6 shows the potential-energy curve [Eq. (6.57)] and some of the allowed energy levels for the hydrogen atom $(Z=1)$. The crosshatching indicates that all positive energies are allowed.

It turns out that all changes in $n$ are allowed in light absorption and emission. The wavenumbers [Eq. (4.64)] of H -atom spectral lines are then

\(
\begin{equation}
\widetilde{\nu} \equiv \frac{1}{\lambda}=\frac{v}{c}=\frac{E{2}-E{1}}{h c}=\frac{e^{2}}{8 \pi \varepsilon{0} a h c}\left(\frac{1}{n{1}^{2}}-\frac{1}{n{2}^{2}}\right) \equiv R{\mathrm{H}}\left(\frac{1}{n{1}^{2}}-\frac{1}{n{2}^{2}}\right) \tag{6.95}
\end{equation}
\)

where $R_{\mathrm{H}}=109677.6 \mathrm{~cm}^{-1}$ is the Rydberg constant for hydrogen.

Degeneracy

Are the hydrogen-atom energy levels degenerate? For the bound states, the energy (6.94) depends only on $n$. However, the wave function (6.61) depends on all three quantum numbers $n, l$, and $m$, whose allowed values are [Eqs. (6.91), (6.92), (5.104), and (5.105)]

FIGURE 6.6 Energy levels of the hydrogen atom.

\(
\begin{gather}
n=1,2,3, \ldots \tag{6.96}\
l=0,1,2, \ldots, n-1 \tag{6.97}\
m=-l,-l+1, \ldots, 0, \ldots, l-1, l \tag{6.98}
\end{gather}
\)

Hydrogen-atom states with different values of $l$ or $m$, but the same value of $n$, have the same energy. The energy levels are degenerate, except for $n=1$, where $l$ and $m$ must both be 0 . For a given value of $n$, we can have $n$ different values of $l$. For each of these values of $l$, we can have $2 l+1$ values of $m$. The degree of degeneracy of an H -atom bound-state level is found to equal $n^{2}$ (spin considerations being omitted); see Prob. 6.16. For the continuum levels, it turns out that for a given energy there is no restriction on the maximum value of $l$; hence these levels are infinity-fold degenerate.

The radial equation for the hydrogen atom can also be solved by the use of ladder operators (also known as factorization); see Z. W. Salsburg, Am. J. Phys., 33, 36 (1965).

The Radial Factor

Using (6.93), we have for the recursion relation (6.88)

\(
\begin{equation}
b{j+1}=\frac{2 Z}{n a} \frac{j+l+1-n}{(j+1)(j+2 l+2)} b{j} \tag{6.99}
\end{equation}
\)

The discussion preceding Eq. (6.91) shows that the highest power of $r$ in the polynomial $M(r)=\Sigma{j} b{j} r^{j}$ [Eq. (6.86)] is $k=n-l-1$. Hence use of $C=Z / n a$ [Eq. (6.93)] in $R(r)=e^{-C r} r l M(r)$ [Eq. (6.84)] gives the radial factor in the hydrogen-atom $\psi$ as

\(
\begin{equation}
R{n l}(r)=r^{l} e^{-Z r / n a} \sum{j=0}^{n-l-1} b_{j} r^{j} \tag{6.100}
\end{equation}
\)

where $a \equiv 4 \pi \varepsilon_{0} \hbar^{2} / \mu e^{2}$ [Eq. (6.63)]. The complete hydrogenlike bound-state wave functions are [Eq. (6.61)]

\(
\begin{equation}
\psi{n l m}=R{n l}(r) Y{l}^{m}(\theta, \phi)=R{n l}(r) S_{l m}(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi} \tag{6.101}
\end{equation}
\)

where the first few theta functions are given in Table 5.1.
How many nodes does $R(r)$ have? The radial function is zero at $r=\infty$, at $r=0$ for $l \neq 0$, and at values of $r$ that make $M(r)$ vanish. $M(r)$ is a polynomial of degree $n-l-1$, and it can be shown that the roots of $M(r)=0$ are all real and positive. Thus, aside from the origin and infinity, there are $n-l-1$ nodes in $R(r)$. The nodes of the spherical harmonics are discussed in Prob. 6.41.

Ground-State Wave Function and Energy

For the ground state of the hydrogenlike atom, we have $n=1, l=0$, and $m=0$. The radial factor (6.100) is

\(
\begin{equation}
R{10}(r)=b{0} e^{-Z r / a} \tag{6.102}
\end{equation}
\)

The constant $b_{0}$ is determined by normalization [Eq. (5.80)]:

\(
\left|b{0}\right|^{2} \int{0}^{\infty} e^{-2 Z r / a} r^{2} d r=1
\)

Using the Appendix integral (A.8), we find

\(
\begin{equation}
R_{10}(r)=2\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \tag{6.103}
\end{equation}
\)

Multiplying by $Y_{0}^{0}=1 /(4 \pi)^{1 / 2}$, we have as the ground-state wave function

\(
\begin{equation}
\psi_{100}=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \tag{6.104}
\end{equation}
\)

The hydrogen-atom energies and wave functions involve the reduced mass, given by (6.59) as

\(
\begin{equation}
\mu{\mathrm{H}}=\frac{m{e} m{p}}{m{e}+m{p}}=\frac{m{e}}{1+m{e} / m{p}}=\frac{m{e}}{1+0.000544617}=0.9994557 m{e} \tag{6.105}
\end{equation}
\)

where $m{p}$ is the proton mass and $m{e} / m_{p}$ was found from Table A.1. The reduced mass is very close to the electron mass. Because of this, some texts use the electron mass instead of the reduced mass in the H atom Schrödinger equation. This corresponds to assuming that the proton mass is infinite compared with the electron mass in (6.105) and that all the internal motion is motion of the electron. The error introduced by using the electron mass for the reduced mass is about 1 part in 2000 for the hydrogen atom. For heavier atoms, the error introduced by assuming an infinitely heavy nucleus is even less than this. Also, for many-electron atoms, the form of the correction for nuclear motion is quite complicated. For these reasons we shall assume in the future an infinitely heavy nucleus and simply use the electron mass in writing the Schrödinger equation for atoms.

If we replace the reduced mass of the hydrogen atom by the electron mass, the quantity $a$ defined by (6.63) becomes

\(
\begin{equation}
a{0}=\frac{4 \pi \varepsilon{0} \hbar^{2}}{m_{e} e^{2}}=0.529177 \AA \tag{6.106}
\end{equation}
\)

where the subscript zero indicates use of the electron mass instead of the reduced mass. $a_{0}$ is called the Bohr radius, since it was the radius of the circle in which the electron moved in the ground state of the hydrogen atom in the Bohr theory. Of course, since the ground-state wave function (6.104) is nonzero for all finite values of $r$, there is some probability of finding the electron at any distance from the nucleus. The electron is certainly not confined to a circle.

A convenient unit for electronic energies is the electronvolt (eV), defined as the kinetic energy acquired by an electron accelerated through a potential difference of 1 volt (V). Potential difference is defined as energy per unit charge. Since $e=1.6021766 \times 10^{-19} \mathrm{C}$ and $1 \mathrm{VC}=1 \mathrm{~J}$, we have

\(
\begin{equation}
1 \mathrm{eV}=1.6021766 \times 10^{-19} \mathrm{~J} \tag{6.107}
\end{equation}
\)

EXAMPLE

Calculate the ground-state energy of the hydrogen atom using SI units and convert the result to electronvolts.

The H atom ground-state energy is given by (6.94) with $n=1$ and $Z=1$ as $E=-\mu e^{4} / 8 h^{2} \varepsilon_{0}^{2}$. Use of (6.105) for $\mu$ gives

\(
\begin{aligned}
E & =-\frac{0.9994557\left(9.109383 \times 10^{-31} \mathrm{~kg}\right)\left(1.6021766 \times 10^{-19} \mathrm{C}\right)^{4}}{8\left(6.626070 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^{2}\left(8.8541878 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N}-\mathrm{m}^{2}\right)^{2}} \frac{n^{2}}{n^{2}} \
E & =-\left(2.178686 \times 10^{-18} \mathrm{~J}\right)\left(Z^{2} / n^{2}\right)\left[(1 \mathrm{eV}) /\left(1.6021766 \times 10^{-19} \mathrm{~J}\right)\right]
\end{aligned}
\)

\(
\begin{equation}
E=-(13.598 \mathrm{eV})\left(Z^{2} / n^{2}\right)=-13.598 \mathrm{eV} \tag{6.108}
\end{equation}
\)

a number worth remembering. The minimum energy needed to ionize a ground-state hydrogen atom is 13.598 eV .
EXERCISE Find the $n=2$ energy of $\mathrm{Li}^{2+}$ in eV ; do the minimum amount of calculation needed. (Answer: -30.60 eV .)

EXAMPLE

Find $\langle T\rangle$ for the hydrogen-atom ground state.
Equations (3.89) for $\langle T\rangle$ and (6.7) for $\nabla^{2} \psi$ give

\(
\begin{gathered}
\langle T\rangle=\int \psi^{} \hat{T} \psi d \tau=-\frac{\hbar^{2}}{2 \mu} \int \psi^{} \nabla^{2} \psi d \tau \
\nabla^{2} \psi=\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}-\frac{1}{r^{2} \hbar^{2}} \hat{L}^{2} \psi=\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}
\end{gathered}
\)

since $\hat{L}^{2} \psi=l(l+1) \hbar^{2} \psi$ and $l=0$ for an $s$ state. From (6.104) with $Z=1$, we have $\psi=\pi^{-1 / 2} a^{-3 / 2} e^{-r / a}$, so $\partial \psi / \partial r=-\pi^{-1 / 2} a^{-5 / 2} e^{-r / a}$ and $\partial^{2} \psi / \partial r^{2}=\pi^{-1 / 2} a^{-7 / 2} e^{-r / a}$.
Using $d \tau=r^{2} \sin \theta d r d \theta d \phi$ [Eq. (5.78)], we have

\(
\begin{gathered}
\langle T\rangle=-\frac{\hbar^{2}}{2 \mu} \frac{1}{\pi a^{4}} \int{0}^{2 \pi} \int{0}^{\pi} \int{0}^{\infty}\left(\frac{1}{a} e^{-2 r / a}-\frac{2}{r} e^{-2 r / a}\right) r^{2} \sin \theta d r d \theta d \phi \
=-\frac{\hbar^{2}}{2 \mu \pi a^{4}} \int{0}^{2 \pi} d \phi \int{0}^{\pi} \sin \theta d \theta \int{0}^{\infty}\left(\frac{r^{2}}{a} e^{-2 r / a}-2 r e^{-2 r / a}\right) d r=\frac{\hbar^{2}}{2 \mu a^{2}}=\frac{e^{2}}{8 \pi \varepsilon_{0} a}
\end{gathered}
\)

where Appendix integral A. 8 and $a=4 \pi \varepsilon{0} \hbar^{2} / \mu e^{2}$ were used. From (6.94), $e^{2} / 8 \pi \varepsilon{0} a$ is minus the ground-state H -atom energy, and (6.108) gives $\langle T\rangle=13.598 \mathrm{eV}$. (See also Sec. 14.4.)

EXERCISE Find $\langle T\rangle$ for the hydrogen-atom $2 p{0}$ state using (6.113).
(Answer: $e^{2} / 32 \pi \varepsilon{0} a=(13.598 \mathrm{eV}) / 4=3.40 \mathrm{eV}$.)
Let us examine a significant property of the ground-state wave function (6.104). We have $r=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}$. For points on the $x$ axis, where $y=0$ and $z=0$, we have $r=\left(x^{2}\right)^{1 / 2}=|x|$, and

\(
\begin{equation}
\psi_{100}(x, 0,0)=\pi^{-1 / 2}(Z / a)^{3 / 2} e^{-Z|x| / a} \tag{6.109}
\end{equation}
\)

Figure 6.7 shows how (6.109) varies along the $x$ axis. Although $\psi_{100}$ is continuous at the origin, the slope of the tangent to the curve is positive at the left of the origin but negative

FIGURE 6.7 Cusp in the hydrogen-atom groundstate wave function.
at its right. Thus $\partial \psi / \partial x$ is discontinuous at the origin. We say that the wave function has a cusp at the origin. The cusp is present because the potential energy $V=-Z e^{2} / 4 \pi \varepsilon_{0} r$ becomes infinite at the origin. Recall the discontinuous slope of the particle-in-a-box wave functions at the walls of the box.

We denoted the hydrogen-atom bound-state wave functions by three subscripts that give the values of $n, l$, and $m$. In an alternative notation, the value of $l$ is indicated by a letter:

Letter	$s$	$p$	$d$	$f$	$g$	$h$	$i$	$k$	$\ldots$
$l$	0	1	2	3	4	5	6	7	$\ldots$

The letters $s, p, d, f$ are of spectroscopic origin, standing for sharp, principal, diffuse, and fundamental. After these we go alphabetically, except that $j$ is omitted. Preceding the code letter for $l$, we write the value of $n$. Thus the ground-state wave function $\psi{100}$ is called $\psi{1 s}$ or, more simply, $1 s$.

Wave Functions for $\boldsymbol{n}=\mathbf{2}$

For $n=2$, we have the states $\psi{200}, \psi{21-1}, \psi{210}$, and $\psi{211}$. We denote $\psi{200}$ as $\psi{2 s}$ or simply as $2 s$. To distinguish the three $2 p$ functions, we use a subscript giving the $m$ value and denote them as $2 p{1}, 2 p{0}$, and $2 p{-1}$. The radial factor in the wave function depends on $n$ and $l$, but not on $m$, as can be seen from (6.100). Each of the three $2 p$ wave functions thus has the same radial factor. The $2 s$ and $2 p$ radial factors may be found in the usual way from (6.100) and (6.99), followed by normalization. The results are given in Table 6.1. Note that the exponential factor in the $n=2$ radial functions is not the same as in the $R{1 s}$ function. The complete wave function is found by multiplying the radial factor by the appropriate spherical harmonic. Using (6.101), Table 6.1, and Table 5.1, we have

\(
\begin{align}
2 s & =\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{2 a}\right)^{3 / 2}\left(1-\frac{Z r}{2 a}\right) e^{-Z r / 2 a} \tag{6.111}\
2 p_{-1} & =\frac{1}{8 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta e^{-i \phi} \tag{6.112}
\end{align}
\)

TABLE 6.1 Radial Factors in the Hydrogenlike-Atom

Wave Functions\(
\begin{aligned}
R{1 s} & =2\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \
R{2 s} & =\frac{1}{\sqrt{2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(1-\frac{Z r}{2 a}\right) e^{-Z r / 2 a} \
R{2 p} & =\frac{1}{2 \sqrt{6}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \
R{3 s} & =\frac{2}{3 \sqrt{3}}\left(\frac{Z}{a}\right)^{3 / 2}\left(1-\frac{2 Z r}{3 a}+\frac{2 Z^{2} r^{2}}{27 a^{2}}\right) e^{-Z r / 3 a} \
R{3 p} & =\frac{8}{27 \sqrt{6}}\left(\frac{Z}{a}\right)^{3 / 2}\left(\frac{Z r}{a}-\frac{Z^{2} r^{2}}{6 a^{2}}\right) e^{-Z r / 3 a} \
R{3 d} & =\frac{4}{81 \sqrt{30}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a}
\end{aligned}
\)

\(
\begin{align}
& 2 p{0}=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{2 a}\right)^{5 / 2} r e^{-Z r / 2 a} \cos \theta \tag{6.113}\
& 2 p{1}=\frac{1}{8 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta e^{i \phi} \tag{6.114}
\end{align}
\)

Table 6.1 lists some of the normalized radial factors in the hydrogenlike wave functions. Figure 6.8 graphs some of the radial functions. The $r^{l}$ factor makes the radial functions zero at $r=0$, except for $s$ states.

The Radial Distribution Function

The probability of finding the electron in the region of space where its coordinates lie in the ranges $r$ to $r+d r, \theta$ to $\theta+d \theta$, and $\phi$ to $\phi+d \phi$ is [Eq. (5.78)]

\(
\begin{equation}
|\psi|^{2} d \tau=\left[R{n l}(r)\right]^{2}\left|Y{l}^{m}(\theta, \phi)\right|^{2} r^{2} \sin \theta d r d \theta d \phi \tag{6.115}
\end{equation}
\)

We now ask: What is the probability of the electron having its radial coordinate between $r$ and $r+d r$ with no restriction on the values of $\theta$ and $\phi$ ? We are asking for the probability of finding the electron in a thin spherical shell centered at the origin, of inner radius $r$ and outer radius $r+d r$. We must thus add up the infinitesimal probabilities (6.115) for all

FIGURE 6.8 Graphs of the radial factor $R_{n l}(r)$ in the hydrogen-atom ( $Z=1$ ) wave functions. The same scale is used in all graphs. (In some texts, these functions are not properly drawn to scale.)
possible values of $\theta$ and $\phi$, keeping $r$ fixed. This amounts to integrating (6.115) over $\theta$ and $\phi$. Hence the probability of finding the electron between $r$ and $r+d r$ is

\(
\begin{equation}
\left[R{n l}(r)\right]^{2} r^{2} d r \int{0}^{2 \pi} \int{0}^{\pi}\left|Y{l}^{m}(\theta, \phi)\right|^{2} \sin \theta d \theta d \phi=\left[R_{n l}(r)\right]^{2} r^{2} d r \tag{6.116}
\end{equation}
\)

since the spherical harmonics are normalized:

\(
\begin{equation}
\int{0}^{2 \pi} \int{0}^{\pi}\left|Y_{l}^{m}(\theta, \phi)\right|^{2} \sin \theta d \theta d \phi=1 \tag{6.117}
\end{equation}
\)

as can be seen from (5.72) and (5.80). The function $R^{2}(r) r^{2}$, which determines the probability of finding the electron at a distance $r$ from the nucleus, is called the radial distribution function; see Fig. 6.9.

For the $1 s$ ground state of H , the probability density $|\psi|^{2}$ is from Eq. (6.104) equal to $e^{-2 r / a}$ times a constant, and so $\left|\psi{1 s}\right|^{2}$ is a maximum at $r=0$ (see Fig. 6.14). However, the radial distribution function $\left[R{1 s}(r)\right]^{2} r^{2}$ is zero at the origin and is a maximum at $r=a$ (Fig. 6.9). These two facts are not contradictory. The probability density $|\psi|^{2}$ is proportional to the probability of finding the electron in an infinitesimal box of volume $d x d y d z$, and this probability is a maximum at the nucleus. The radial distribution function is proportional to the probability of finding the electron in a thin spherical shell of inner and outer radii $r$ and $r+d r$, and this probability is a maximum at $r=a$. Since $\psi{1 s}$ depends only on $r$, the $1 s$ probability density is essentially constant in the thin spherical shell. If we imagine the thin shell divided up into a huge number of infinitesimal boxes each of volume $d x d y d z$, we can sum up the probabilities $\left|\psi{1 s}\right|^{2} d x d y d z$ of being in

FIGURE 6.9 Plots of the radial distribution function $\left[R_{n \prime}(r)\right]^{2} r^{2}$ for the hydrogen atom.

each tiny box in the thin shell to get the probability of finding the electron in the thin shell as being $\left|\psi{1 s}\right|^{2} V{\text {shell }}$. The volume $V_{\text {shell }}$ of the thin shell is

\(
\frac{4}{3} \pi(r+d r)^{3}-\frac{4}{3} \pi r^{3}=4 \pi r^{2} d r
\)

where terms in $(d r)^{2}$ and $(d r)^{3}$ are negligible compared with the $d r$ term. Therefore the probability of being in the thin shell is

\(
\left|\psi{1 s}\right|^{2} V{\text {shell }}=R{1 s}^{2}\left(Y{0}^{0}\right)^{2} 4 \pi r^{2} d r=R{1 s}^{2}\left[(4 \pi)^{-1 / 2}\right]^{2} 4 \pi r^{2} d r=R{1 s}^{2} r^{2} d r
\)

in agreement with (6.116). The $1 s$ radial distribution function is zero at $r=0$ because the volume $4 \pi r^{2} d r$ of the thin spherical shell becomes zero as $r$ goes to zero. As $r$ increases from zero, the probability density $\left|\psi{1 s}\right|^{2}$ decreases and the volume $4 \pi r^{2} d r$ of the thin shell increases. Their product $\left|\psi{1 s}\right|^{2} 4 \pi r^{2} d r$ is a maximum at $r=a$.

EXAMPLE

Find the probability that the electron in the ground-state H atom is less than a distance $a$ from the nucleus.

We want the probability that the radial coordinate lies between 0 and $a$. This is found by taking the infinitesimal probability (6.116) of being between $r$ and $r+d r$ and summing it over the range from 0 to $a$. This sum of infinitesimal quantities is the definite integral

\(
\begin{aligned}
\int{0}^{a} R{n l}^{2} r^{2} d r & =\frac{4}{a^{3}} \int{0}^{a} e^{-2 r / a} r^{2} d r=\left.\frac{4}{a^{3}} e^{-2 r / a}\left(-\frac{r^{2} a}{2}-\frac{2 r a^{2}}{4}-\frac{2 a^{3}}{8}\right)\right|{0} ^{a} \
& =4\left[e^{-2}(-5 / 4)-(-1 / 4)\right]=0.323
\end{aligned}
\)

where $R{10}$ was taken from Table 6.1 and the Appendix integral A. 7 was used.
EXERCISE Find the probability that the electron in a $2 p{1} \mathrm{H}$ atom is less than a distance $a$ from the nucleus. Use a table of integrals or the website integrals.wolfram.com.
(Answer: 0.00366.)

Real Hydrogenlike Functions

The factor $e^{i m \phi}$ makes the spherical harmonics complex, except when $m=0$. Instead of working with complex wave functions such as (6.112) and (6.114), chemists often use real hydrogenlike wave functions formed by taking linear combinations of the complex functions. The justification for this procedure is given by the theorem of Section 3.6: Any linear combination of eigenfunctions of a degenerate energy level is an eigenfunction of the Hamiltonian with the same eigenvalue. Since the energy of the hydrogen atom does not depend on $m$, the $2 p{1}$ and $2 p{-1}$ states belong to a degenerate energy level. Any linear combination of them is an eigenfunction of the Hamiltonian with the same energy eigenvalue.

One way to combine these two functions to obtain a real function is

\(
\begin{equation}
2 p{x} \equiv \frac{1}{\sqrt{2}}\left(2 p{-1}+2 p_{1}\right)=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \cos \phi \tag{6.118}
\end{equation}
\)

where we used (6.112), (6.114), and $e^{ \pm i \phi}=\cos \phi \pm i \sin \phi$. The $1 / \sqrt{2}$ factor normalizes $2 p_{x}$ :

\(
\begin{aligned}
\int\left|2 p{x}\right|^{2} d \tau & =\frac{1}{2}\left(\int\left|2 p{-1}\right|^{2} d \tau+\int\left|2 p{1}\right|^{2} d \tau+\int\left(2 p{-1}\right) 2 p{1} d \tau+\int\left(2 p{1}\right) 2 p_{-1} d \tau\right) \
& =\frac{1}{2}(1+1+0+0)=1
\end{aligned}
\)

Here we used the fact that $2 p{1}$ and $2 p{-1}$ are normalized and are orthogonal to each other, since

\(
\int{0}^{2 \pi}\left(e^{-i \phi}\right) * e^{i \phi} d \phi=\int{0}^{2 \pi} e^{2 i \phi} d \phi=0
\)

The designation $2 p_{x}$ for (6.118) becomes clearer if we note that (5.51) gives

\(
\begin{equation}
2 p_{x}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} x e^{-Z r / 2 a} \tag{6.119}
\end{equation}
\)

A second way of combining the functions is

\(
\begin{gather}
2 p{y} \equiv \frac{1}{i \sqrt{2}}\left(2 p{1}-2 p{-1}\right)=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} r \sin \theta \sin \phi e^{-Z r / 2 a} \tag{6.120}\
2 p{y}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} y e^{-Z r / 2 a} \tag{6.121}
\end{gather}
\)

The function $2 p_{0}$ is real and is often denoted by

\(
\begin{equation}
2 p{0}=2 p{z}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{2 a}\right)^{5 / 2} z e^{-Z r / 2 a} \tag{6.122}
\end{equation}
\)

where capital $Z$ stands for the number of protons in the nucleus, and small $z$ is the $z$ coordinate of the electron. The functions $2 p{x}, 2 p{y}$, and $2 p{z}$ are mutually orthogonal (Prob. 6.42). Note that $2 p{z}$ is zero in the $x y$ plane, positive above this plane, and negative below it.

The functions $2 p{-1}$ and $2 p{1}$ are eigenfunctions of $\hat{L}^{2}$ with the same eigenvalue: $2 \hbar^{2}$. The reasoning of Section 3.6 shows that the linear combinations (6.118) and (6.120) are also eigenfunctions of $\hat{L}^{2}$ with eigenvalue $2 \hbar^{2}$. However, $2 p{-1}$ and $2 p{1}$ are eigenfunctions of $\hat{L}{z}$ with different eigenvalues: $-\hbar$ and $+\hbar$. Therefore, $2 p{x}$ and $2 p{y}$ are not eigenfunctions of $\hat{L}{z}$.

We can extend this procedure to construct real wave functions for higher states. Since $m$ ranges from $-l$ to $+l$, for each complex function containing the factor $e^{-i|m| \phi}$ there is a function with the same value of $n$ and $l$ but having the factor $e^{+i|m| \phi}$. Addition and subtraction of these functions gives two real functions, one with the factor $\cos (|m| \phi)$, the other with the factor $\sin (|m| \phi)$. Table 6.2 lists these real wave functions for the hydrogenlike atom. The subscripts on these functions come from similar considerations as for the $2 p{x}, 2 p{y}$, and $2 p{z}$ functions. For example, the $3 d{x y}$ function is proportional to $x y$ (Prob. 6.37).

The real hydrogenlike functions are derived from the complex functions by replacing $e^{i m \phi} /(2 \pi)^{1 / 2}$ with $\pi^{-1 / 2} \sin (|m| \phi)$ or $\pi^{-1 / 2} \cos (|m| \phi)$ for $m \neq 0$; for $m=0$ the $\phi$ factor is $1 /(2 \pi)^{1 / 2}$ for both real and complex functions.

In dealing with molecules, the real hydrogenlike orbitals are more useful than the complex ones. For example, we shall see in Section 15.5 that the real atomic orbitals $2 p{x}, 2 p{y}$, and $2 p{z}$ of the oxygen atom have the proper symmetry to be used in constructing a wave function for the $\mathrm{H}{2} \mathrm{O}$ molecule, whereas the complex $2 p$ orbitals do not.

TABLE 6.2 Real Hydrogenlike Wave Functions

$1 s=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a}$
$2 s=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a}$
$2 p{z}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \cos \theta$
$2 p{x}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \cos \phi$
$2 p{y}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \sin \phi$
$3 s=\frac{1}{81(3 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(27-18 \frac{Z r}{a}+2 \frac{Z^{2} r^{2}}{a^{2}}\right) e^{-Z r / 3 a}$
$3 p{z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \cos \theta$
$3 p{x}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \sin \theta \cos \phi$
$3 p{y}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \sin \theta \sin \phi$
$3 d{z^{2}}=\frac{1}{81(6 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a}\left(3 \cos ^{2} \theta-1\right)$
$3 d{x z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin \theta \cos \theta \cos \phi$
$3 d{y z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin \theta \cos \theta \sin \phi$
$3 d{x^{2}-y^{2}}=\frac{1}{81(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin ^{2} \theta \cos 2 \phi$
$3 d_{x y}=\frac{1}{81(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin ^{2} \theta \sin 2 \phi$

The hydrogenlike wave functions are one-electron spatial wave functions and so are hydrogenlike orbitals (Section 6.5). These functions have been derived for a one-electron atom, and we cannot expect to use them to get a truly accurate representation of the wave function of a many-electron atom. The use of the orbital concept to approximate manyelectron atomic wave functions is discussed in Chapter 11. For now we restrict ourselves to one-electron atoms.

There are two fundamentally different ways of depicting orbitals. One way is to draw graphs of the functions; a second way is to draw contour surfaces of constant probability density.

First consider drawing graphs. To graph the variation of $\psi$ as a function of the three independent variables $r, \theta$, and $\phi$, we need four dimensions. The three-dimensional nature of our world prevents us from drawing such a graph. Instead, we draw graphs of the

FIGURE 6.10 Polar graphs of the $\theta$ factors in the $s$ and $p_{z}$ hydrogen-atom wave functions.

factors in $\psi$. Graphing $R(r)$ versus $r$, we get the curves of Fig. 6.8, which contain no information on the angular variation of $\psi$.

Now consider graphs of $S(\theta)$. We have (Table 5.1)

\(
S{0,0}=1 / \sqrt{2}, \quad S{1,0}=\frac{1}{2} \sqrt{6} \cos \theta
\)

We can graph these functions using two-dimensional Cartesian coordinates, plotting $S$ on the vertical axis and $\theta$ on the horizontal axis. $S{0,0}$ gives a horizontal straight line, and $S{1,0}$ gives a cosine curve. More commonly, $S$ is graphed using plane polar coordinates. The variable $\theta$ is the angle with the positive $z$ axis, and $S(\theta)$ is the distance from the origin to the point on the graph. For $S{0,0}$, we get a circle; for $S{1,0}$ we obtain two tangent circles (Fig. 6.10). The negative sign on the lower circle of the graph of $S{1,0}$ indicates that $S{1,0}$ is negative for $\frac{1}{2} \pi<\theta \leq \pi$. Strictly speaking, in graphing $\cos \theta$ we only get the upper circle, which is traced out twice; to get two tangent circles, we must graph $|\cos \theta|$.

Instead of graphing the angular factors separately, we can draw a single graph that plots $|S(\theta) T(\phi)|$ as a function of $\theta$ and $\phi$. We will use spherical coordinates, and the distance from the origin to a point on the graph will be $|S(\theta) T(\phi)|$. For an $s$ state, $S T$ is independent of the angles, and we get a sphere of radius $1 /(4 \pi)^{1 / 2}$ as the graph. For a $p{z}$ state, $S T=\frac{1}{2}(3 / \pi)^{1 / 2} \cos \theta$, and the graph of $|S T|$ consists of two spheres with centers on the $z$ axis and tangent at the origin (Fig. 6.11). No doubt Fig. 6.11 is familiar. Some texts say this gives the shape of a $p{z}$ orbital, which is wrong. Figure 6.11 is simply a graph of the angular factor in a $p{z}$ wave function. Graphs of the $p{x}$ and $p_{y}$ angular factors give tangent spheres lying on the $x$ and $y$ axes, respectively. If we graph $S^{2} T^{2}$ in spherical coordinates, we get surfaces with the familiar figure-eight cross sections; to repeat, these are graphs, not orbital shapes.

FIGURE 6.11 Graph of $\left|Y{1}^{0}(\theta, \phi)\right|$, the angular factor in a $p{z}$ wave function.

Now consider drawing contour surfaces of constant probability density. We shall draw surfaces in space, on each of which the value of $|\psi|^{2}$, the probability density, is constant. Naturally, if $|\psi|^{2}$ is constant on a given surface, $|\psi|$ is also constant on that surface. The contour surfaces for $|\psi|^{2}$ and for $|\psi|$ are identical.

For an $s$ orbital, $\psi$ depends only on $r$, and a contour surface is a surface of constant $r$, that is, a sphere centered at the origin. To pin down the size of an orbital, we take a contour surface within which the probability of finding the electron is, say, $95 \%$; thus we want $\int_{V}|\psi|^{2} d \tau=0.95$, where $V$ is the volume enclosed by the orbital contour surface.

Let us obtain the cross section of the $2 p_{y}$ hydrogenlike orbital in the $y z$ plane. In this plane, $\phi=\pi / 2$ (Fig. 6.5), and $\sin \phi=1$; hence Table 6.2 gives for this orbital in the $y z$ plane

\(
\begin{equation}
\left|2 p_{y}\right|=k^{5 / 2} \pi^{-1 / 2} r e^{-k r}|\sin \theta| \tag{6.123}
\end{equation}
\)

where $k=Z / 2 a$. To find the orbital cross section, we use plane polar coordinates to plot (6.123) for a fixed value of $\psi ; r$ is the distance from the origin, and $\theta$ is the angle with the $z$ axis. The result for a typical contour (Prob. 6.44) is shown in Fig. 6.12. Since $y e^{-k r}=y \exp \left[-k\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]$, we see that the $2 p{y}$ orbital is a function of $y$ and $\left(x^{2}+z^{2}\right)$. Hence, on a circle centered on the $y$ axis and parallel to the $x z$ plane, $2 p{y}$ is constant. Thus a three-dimensional contour surface may be developed by rotating the cross section in Fig. 6.12 about the $y$ axis, giving a pair of distorted ellipsoids. The shape of a real $2 p$ orbital is two separated, distorted ellipsoids, and not two tangent spheres.

Now consider the shape of the two complex orbitals $2 p_{ \pm 1}$. We have

\(
\begin{gather}
2 p{ \pm 1}=k^{5 / 2} \pi^{-1 / 2} r e^{-k r} \sin \theta e^{ \pm i \phi} \
\left|2 p{ \pm 1}\right|=k^{5 / 2} \pi^{-1 / 2} e^{-k r} r|\sin \theta| \tag{6.124}
\end{gather}
\)

and these two orbitals have the same shape. Since the right sides of (6.124) and (6.123) are identical, we conclude that Fig. 6.12 also gives the cross section of the $2 p_{ \pm 1}$ orbitals in the $y z$ plane. Since [Eq. (5.51)]

\(
e^{-k r} r|\sin \theta|=\exp \left[-k\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right]\left(x^{2}+y^{2}\right)^{1 / 2}
\)

we see that $2 p_{ \pm 1}$ is a function of $z$ and $x^{2}+y^{2}$; so we get the three-dimensional orbital shape by rotating Fig. 6.12 about the $z$ axis. This gives a doughnut-shaped surface.

Some hydrogenlike orbital surfaces are shown in Fig. 6.13. The $2 s$ orbital has a spherical node, which is not visible; the $3 s$ orbital has two such nodes. The $3 p_{z}$ orbital has a spherical node (indicated by a dashed line) and a nodal plane (the $x y$ plane).

FIGURE 6.12 Contour of a
$2 p_{y}$ orbital.

FIGURE 6.13 Shapes of some hydrogen-atom orbitals.

The $3 d{z^{2}}$ orbital has two nodal cones. The $3 d{x^{2}-y^{2}}$ orbital has two nodal planes. Note that the view shown is not the same for the various orbitals. The relative signs of the wave functions are indicated. The other three real $3 d$ orbitals in Table 6.2 have the same shape as the $3 d{x^{2}-y^{2}}$ orbital but have different orientations. The $3 d{x y}$ orbital has its lobes lying between the $x$ and $y$ axes and is obtained by rotating the $3 d{x^{2}-y^{2}}$ orbital by $45^{\circ}$ about the $z$ axis. The $3 d{y z}$ and $3 d_{x z}$ orbitals have their lobes between the $y$ and $z$ axes and between the $x$ and $z$ axes, respectively. (Online three-dimensional views of the real hydrogenlike orbitals are at www.falstad.com/qmatom; these can be rotated using a mouse.)

FIGURE 6.14 Probability

densities for some hydrogenatom states. [For accurate stereo plots, see D. T. Cromer, J. Chem. Educ., 45, 626 (1968).]

Figure 6.14 represents the probability density in the $y z$ plane for various orbitals. The number of dots in a given region is proportional to the value of $|\psi|^{2}$ in that region. Rotation of these diagrams about the vertical $(z)$ axis gives the three-dimensional probability density. The $2 s$ orbital has a constant for its angular factor and hence has no angular nodes; for this orbital, $n-l-1=1$, indicating one radial node. The sphere on which $\psi_{2 s}=0$ is evident in Fig. 6.14.

Schrödinger's original interpretation of $|\psi|^{2}$ was that the electron is "smeared out" into a charge cloud. If we consider an electron passing from one medium to another, we find that $|\psi|^{2}$ is nonzero in both mediums. According to the charge-cloud interpretation, this would mean that part of the electron was reflected and part transmitted. However, experimentally one never detects a fraction of an electron; electrons behave as indivisible entities. This difficulty is removed by the Born interpretation, according to which the values of $|\psi|^{2}$ in the two mediums give the probabilities for reflection and transmission. The orbital shapes we have drawn give the regions of space in which the total probability of finding the electron is $95 \%$.

In 1896, Zeeman observed that application of an external magnetic field caused a splitting of atomic spectral lines. We shall consider this Zeeman effect for the hydrogen atom. We begin by reviewing magnetism.

Magnetic fields arise from moving electric charges. A charge $Q$ with velocity $\mathbf{v}$ gives rise to a magnetic field $\mathbf{B}$ at point $P$ in space, such that

\(
\begin{equation}
\mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{Q \mathbf{v} \times \mathbf{r}}{r^{3}} \tag{6.125}
\end{equation}
\)

where $\mathbf{r}$ is the vector from $Q$ to point P and where $\mu_{0}$ (called the permeability of vacuum or the magnetic constant) is defined as $4 \pi \times 10^{-7} \mathrm{~N} \mathrm{C}^{-2} \mathrm{~s}^{2}$. [Equation (6.125) is valid only for a nonaccelerated charge moving with a speed much less than the speed of light.] The vector $\mathbf{B}$ is called the magnetic induction or magnetic flux density. (It was formerly believed that the vector $\mathbf{H}$ was the fundamental magnetic field vector, so $\mathbf{H}$ was called the magnetic field strength. It is now known that $\mathbf{B}$ is the fundamental magnetic vector.) Equation (6.125) is in SI units with $Q$ in coulombs and $\mathbf{B}$ in teslas (T), where $1 \mathrm{~T}=1 \mathrm{NC}^{-1} \mathrm{~m}^{-1} \mathrm{~s}$.

Two electric charges $+Q$ and $-Q$ separated by a small distance $b$ constitute an electric dipole. The electric dipole moment is defined as a vector from $-Q$ to $+Q$ with magnitude $Q b$. For a small planar loop of electric current, it turns out that the magnetic field generated by the moving charges of the current is given by the same mathematical
expression as that giving the electric field due to an electric dipole, except that the electric dipole moment is replaced by the magnetic dipole moment $\mathbf{m} ; \mathbf{m}$ is a vector of magnitude $I A$, where $I$ is the current flowing in a loop of area $A$. The direction of $\mathbf{m}$ is perpendicular to the plane of the current loop.

Consider the magnetic (dipole) moment associated with a charge $Q$ moving in a circle of radius $r$ with speed $v$. The current is the charge flow per unit time. The circumference of the circle is $2 \pi r$, and the time for one revolution is $2 \pi r / v$. Hence $I=Q v / 2 \pi r$. The magnitude of $\mathbf{m}$ is

\(
\begin{equation}
|\mathbf{m}|=I A=(Q v / 2 \pi r) \pi r^{2}=Q v r / 2=Q r p / 2 m \tag{6.126}
\end{equation}
\)

where $m$ is the mass of the charged particle and $p$ is its linear momentum. Since the radius vector $\mathbf{r}$ is perpendicular to $\mathbf{p}$, we have

\(
\begin{equation}
\mathbf{m}_{L}=\frac{Q \mathbf{r} \times \mathbf{p}}{2 m}=\frac{Q}{2 m} \mathbf{L} \tag{6.127}
\end{equation}
\)

where the definition of orbital angular momentum $\mathbf{L}$ was used and the subscript on $\mathbf{m}$ indicates that it arises from the orbital motion of the particle. Although we derived (6.127) for the special case of circular motion, its validity is general. For an electron, $Q=-e$, and the magnetic moment due to its orbital motion is

\(
\begin{equation}
\mathbf{m}{L}=-\frac{e}{2 m{e}} \mathbf{L} \tag{6.128}
\end{equation}
\)

The magnitude of $\mathbf{L}$ is given by (5.95), and the magnitude of the orbital magnetic moment of an electron with orbital-angular-momentum quantum number $l$ is

\(
\begin{equation}
\left|\mathbf{m}{L}\right|=\frac{e \hbar}{2 m{e}}[l(l+1)]^{1 / 2}=\mu_{\mathrm{B}}[l(l+1)]^{1 / 2} \tag{6.129}
\end{equation}
\)

The constant $e \hbar / 2 m{e}$ is called the Bohr magneton $\mu{\mathrm{B}}$ :

\(
\begin{equation}
\mu{\mathrm{B}} \equiv e \hbar / 2 m{e}=9.2740 \times 10^{-24} \mathrm{~J} / \mathrm{T} \tag{6.130}
\end{equation}
\)

Now consider applying an external magnetic field to the hydrogen atom. The energy of interaction between a magnetic dipole $\mathbf{m}$ and an external magnetic field $\mathbf{B}$ can be shown to be

\(
\begin{equation}
E_{B}=-\mathbf{m} \cdot \mathbf{B} \tag{6.131}
\end{equation}
\)

Using Eq. (6.128), we have

\(
\begin{equation}
E{B}=\frac{e}{2 m{e}} \mathbf{L} \cdot \mathbf{B} \tag{6.132}
\end{equation}
\)

We take the $z$ axis along the direction of the applied field: $\mathbf{B}=B \mathbf{k}$, where $\mathbf{k}$ is a unit vector in the $z$ direction. We have

\(
E{B}=\frac{e}{2 m{e}} B\left(L{x} \mathbf{i}+L{y} \mathbf{j}+L{z} \mathbf{k}\right) \cdot \mathbf{k}=\frac{e}{2 m{e}} B L{z}=\frac{\mu{\mathrm{B}}}{\hbar} B L_{z}
\)

where $L{z}$ is the $z$ component of orbital angular momentum. We now replace $L{z}$ by the operator $\hat{L}_{z}$ to give the following additional term in the Hamiltonian operator, resulting from the external magnetic field:

\(
\begin{equation}
\hat{H}{B}=\mu{\mathrm{B}} B \hbar^{-1} \hat{L}_{z} \tag{6.133}
\end{equation}
\)

The Schrödinger equation for the hydrogen atom in a magnetic field is

\(
\begin{equation}
\left(\hat{H}+\hat{H}_{B}\right) \psi=E \psi \tag{6.134}
\end{equation}
\)

where $\hat{H}$ is the hydrogen-atom Hamiltonian in the absence of an external field. We readily verify that the solutions of Eq. (6.134) are the complex hydrogenlike wave functions (6.61):

\(
\begin{equation}
\left(\hat{H}+\hat{H}{B}\right) R(r) Y{l}^{m}(\theta, \phi)=\hat{H} R Y{l}^{m}+\mu{\mathrm{B}} \hbar^{-1} B \hat{L}{z} R Y{l}^{m}=\left(-\frac{Z^{2}}{n^{2}} \frac{e^{2}}{8 \pi \varepsilon{0} a}+\mu{\mathrm{B}} B m\right) R Y_{l}^{m} \tag{6.135}
\end{equation}
\)

where Eqs. (6.94) and (5.105) were used. Thus there is an additional term $\mu_{\mathrm{B}} B m$ in the energy, and the external magnetic field removes the $m$ degeneracy. For obvious reasons, $m$ is often called the magnetic quantum number. Actually, the observed energy shifts do not match the predictions of Eq. (6.135) because of the existence of electron spin magnetic moment (Chapter 10 and Section 11.7).

In Chapter 5 we found that in quantum mechanics $\mathbf{L}$ lies on the surface of a cone. A classical-mechanical treatment of the motion of $\mathbf{L}$ in an applied magnetic field shows that the field exerts a torque on $\mathbf{m}{L}$, causing $\mathbf{L}$ to revolve about the direction of $\mathbf{B}$ at a constant frequency given by $\left|\mathbf{m}{L}\right| B / 2 \pi|\mathrm{~L}|$, while maintaining a constant angle with $\mathbf{B}$. This gyroscopic motion is called precession. In quantum mechanics, a complete specification of $\mathbf{L}$ is impossible. However, one finds that $\langle\mathbf{L}\rangle$ precesses about the field direction (Dicke and Wittke, Section 12-3).

For a one-particle central-force problem, the wave function is given by (6.16) as $\psi=R(r) Y_{l}^{m}(\theta, \phi)$ and the radial factor $R(r)$ is found by solving the radial equation (6.17). The Numerov method of Section 4.4 applies to differential equations of the form $\psi^{\prime \prime}=G(x) \psi(x)$ [Eq. (4.66)], so we need to eliminate the first derivative $R^{\prime}$ in (6.17). Let us define $F(r)$ by $F(r) \equiv r R(r)$, so

\(
\begin{equation}
R(r)=r^{-1} F(r) \tag{6.136}
\end{equation}
\)

Then $R^{\prime}=-r^{-2} F+r^{-1} F^{\prime}$ and $R^{\prime \prime}=2 r^{-3} F-2 r^{-2} F^{\prime}+r^{-1} F^{\prime \prime}$. Substitution in (6.17) transforms the radial equation to

\(
\begin{gather}
-\frac{\hbar^{2}}{2 m} F^{\prime \prime}(r)+\left[V(r)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}}\right] F(r)=E F(r) \tag{6.137}\
F^{\prime \prime}(r)=G(r) F(r), \quad \text { where } G(r) \equiv \frac{m}{\hbar^{2}}(2 V-2 E)+\frac{l(l+1)}{r^{2}} \tag{6.138}
\end{gather}
\)

which has the form needed for the Numerov method. In solving (6.137) numerically, one deals separately with each value of $l$. Equation (6.137) resembles the one-dimensional Schrödinger equation $-\left(\hbar^{2} / 2 m\right) \psi^{\prime \prime}(x)+V(x) \psi(x)=E \psi(x)$, except that $r$ (whose range is 0 to $\infty$ ) replaces $x$ (whose range is $-\infty$ to $\infty$ ), $F(r) \equiv r R(r)$ replaces $\psi$, and $V(r)+l(l+1) \hbar^{2} / 2 m r^{2}$ replaces $V(x)$. We can expect that for each value of $l$, the lowest-energy solution will have 0 interior nodes (that is, nodes with $0<r<\infty$ ), the next lowest will have 1 interior node, and so on.

Recall from the discussion after (6.81) that if $R(r)$ behaves as $1 / r^{b}$ near the origin, then if $b>1, R(r)$ is not quadratically integrable; also, the value $b=1$ is not allowed, as noted after (6.83). Hence $F(r) \equiv r R(r)$ must be zero at $r=0$.

For $l \neq 0, G(r)$ in (6.138) is infinite at $r=0$, which upsets most computers. To avoid this problem, one starts the solution at an extremely small value of $r$ (for example, $10^{-15}$ for the dimensionless $r_{r}$ ) and approximates $F(r)$ as zero at this point.

As an example, we shall use the Numerov method to solve for the lowest boundstate H -atom energies. Here, $V=-e^{2} / 4 \pi \varepsilon{0} r=-e^{\prime 2} / r$, where $e^{\prime} \equiv e /\left(4 \pi \varepsilon{0}\right)^{1 / 2}$. The radial equation (6.62) contains the three constants $e^{\prime}, \mu$, and $\hbar$, where $e^{\prime} \equiv e /\left(4 \pi \varepsilon_{0}\right)^{1 / 2}$ has SI units of $\mathrm{m} \mathrm{N}^{1 / 2}$ (see Table A. 1 of the Appendix) and hence has the dimensions $\left[e^{\prime}\right]=\mathrm{L}^{3 / 2} \mathrm{M}^{1 / 2} \mathrm{~T}^{-1}$. Following the procedure used to derive Eq. (4.73), we find the H -atom reduced energy and reduced radial coordinate to be (Prob. 6.47)

\(
\begin{equation}
E{r}=E / \mu e^{\prime 4} \hbar^{-2}, \quad r{r}=r / B=r / \hbar^{2} \mu^{-1} e^{\prime-2} \tag{6.139}
\end{equation}
\)

Use of (6.139) and (4.76) and (4.77) with $\psi$ replaced by $F$ and $B=\hbar^{2} \mu^{-1} e^{\prime-2}$ transforms (6.137) for the H atom to (Prob. 6.47)

\(
\begin{equation}
F{r}^{\prime \prime}=G{r} F{r}, \quad \text { where } G{r}=l(l+1) / r{r}^{2}-2 / r{r}-2 E_{r} \tag{6.140}
\end{equation}
\)

and where $F{r}=F / B^{-1 / 2}$.
The bound-state H -atom energies are all less than zero. Suppose we want to find the H -atom bound-state eigenvalues with $E{r} \leq-0.04$. Equating this energy to $V{r}$, we have (Prob. 6.47) $-0.04=-1 / r{r}$ and the classically allowed region for this energy value extends from $r{r}=0$ to $r{r}=25$. Going two units into the classically forbidden region, we take $r{r, \max }=27$ and require that $F{r}(27)=0$. We shall take $s{r}=0.1$, giving 270 points from 0 to 27 (more precisely, from $10^{-15}$ to $27+10^{-15}$ ).
$G{r}$ in (6.140) contains the parameter $l$, so the program of Table 4.1 has to be modified to input the value of $l$. When setting up a spreadsheet, enter the $l$ value in some cell and refer to this cell when you type the formula for cell B7 (Fig. 4.9) that defines $G{r}$. Start column A at $r{r}=1 \times 10^{-15}$. Column C of the spreadsheet will contain $F{r}$ values instead of $\psi{r}$ values, and $F{r}$ will differ negligibly from zero at $r{r}=1 \times 10^{-15}$, and will be taken as zero at this point.

With these choices, we find (Prob. 6.48a) the lowest three H -atom eigenvalues for $l=0$ to be $E{r}=-0.4970,-0.1246$, and -0.05499 ; the lowest two $l=1$ eigenvalues found are -0.1250 and -0.05526 . The true values [Eqs. (6.94) and (6.139)] are $-0.5000,-0.1250$, and -0.05555 . The mediocre accuracy can be attributed mainly to the rapid variation of $G(r)$ near $r=0$. If $s{r}$ is taken as 0.025 instead of 0.1 (giving 1080 points), the $l=0$ eigenvalues are improved to $-0.4998,-0.12497$, and -0.05510 . See also Prob. 6.48b.

The Schrödinger equation for the one-electron atom (Chapter 6) is exactly solvable. However, because of the interelectronic-repulsion terms in the Hamiltonian, the Schrödinger equation for many-electron atoms and molecules cannot be solved exactly. Hence we must seek approximate methods of solution. The two main approximation methods, the variation method and perturbation theory, will be presented in Chapters 8 and 9 . To derive these methods, we must develop further the theory of quantum mechanics, which is what is done in this chapter.

Before starting, we introduce some notation. The definite integral over all space of an operator sandwiched between two functions occurs often, and various abbreviations are used:

\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau \equiv\left\langle f{m}\right| \hat{A}\left|f_{n}\right\rangle \equiv\langle m| \hat{A}|n\rangle \tag{7.1}
\end{}
\)

where $f{m}$ and $f{n}$ are two functions. If it is clear what functions are meant, we can use just the indexes, as indicated in (7.1). The above notation, introduced by Dirac, is called bracket notation. Another notation is

\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau \equiv A{m n} \tag{7.2}
\end{}
\)

The notations $A_{m n}$ and $\langle m| \hat{A}|n\rangle$ imply that we use the complex conjugate of the function whose letter appears first. The definite integral $\langle m| \hat{A}|n\rangle$ is called a matrix element of the operator $\hat{A}$. Matrices are rectangular arrays of numbers and obey certain rules of combination (see Section 7.10).

For the definite integral over all space between two functions, we write

\(
\begin{equation}
\int f_{m}^{} f{n} d \tau \equiv\left\langle f{m} \mid f{n}\right\rangle \equiv\left(f{m}, f_{n}\right) \equiv\langle m \mid n\rangle \tag{7.3}
\end{}
\)

Note that

\(
\langle f| \hat{B}|g\rangle=\langle f \mid \hat{B} g\rangle
\)

where $f$ and $g$ are functions. Since $\left(\int f{m}^{*} f{n} d \tau\right)^{}=\int f_{n}^{} f_{m} d \tau$, we have the identity

\(
\begin{equation}
\langle m \mid n\rangle^{}=\langle n \mid m\rangle \tag{7.4}
\end{}
\)

Since the complex conjugate of $f_{m}$ in (7.1) is taken, it follows that

\(
\begin{equation}
\langle c f| \hat{B}|g\rangle=c^{}\langle f| \hat{B}|g\rangle \quad \text { and } \quad\langle f| \hat{B}|c g\rangle=c\langle f| \hat{B}|g\rangle \tag{7.5}
\end{}
\)

where $\hat{B}$ is a linear operator and $c$ is a constant.

The quantum-mechanical operators that represent physical quantities are linear (Section 3.1). These operators must meet an additional requirement, which we now discuss.

Definition of Hermitian Operators

Let $\hat{A}$ be the linear operator representing the physical property $A$. The average value of $A$ is [Eq. (3.88)]

\(
\langle A\rangle=\int \Psi^{*} \hat{A} \Psi d \tau
\)

where $\Psi$ is the state function of the system. Since the average value of a physical quantity must be a real number, we demand that

\(
\begin{gather}
\langle A\rangle=\langle A\rangle^{} \
\int \Psi^{} \hat{A} \Psi d \tau=\left[\int \Psi^{} \hat{A} \Psi d \tau\right]^{}=\int\left(\Psi^{}\right)^{}(\hat{A} \Psi)^{} d \tau \
\int \Psi^{} \hat{A} \Psi d \tau=\int \Psi(\hat{A} \Psi)^{} d \tau \tag{7.6}
\end{gather*}
\)

Equation (7.6) must hold for every function $\Psi$ that can represent a possible state of the system; that is, it must hold for all well-behaved functions $\Psi$. A linear operator that satisfies (7.6) for all well-behaved functions is called a Hermitian operator (after the mathematician Charles Hermite).

Many texts define a Hermitian operator as a linear operator that satisfies

\(
\begin{equation}
\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau \tag{7.7}
\end{equation}
\)

for all well-behaved functions $f$ and $g$. Note especially that on the left side of (7.7) $\hat{A}$ operates on $g$, but on the right side $\hat{A}$ operates on $f$. For the special case $f=g$, (7.7) reduces to (7.6). Equation (7.7) is apparently a more stringent requirement than (7.6), but we shall prove that (7.7) is a consequence of (7.6). Therefore the two definitions of a Hermitian operator are equivalent.

We begin the proof by setting $\Psi=f+c g$ in (7.6), where $c$ is an arbitrary constant. This gives

\(
\begin{aligned}
& \int(f+c g)^{} \hat{A}(f+c g) d \tau=\int(f+c g)[\hat{A}(f+c g)] d \tau \
& \int\left(f^{}+c^{} g^{}\right) \hat{A} f d \tau+\int\left(f^{}+c^{} g^{}\right) \hat{A} c g d \tau \
& =\int(f+c g)(\hat{A} f)^{} d \tau+\int(f+c g)(\hat{A} c g)^{} d \tau \
& \int f^{} \hat{A} f d \tau+c^{} \int g^{} \hat{A} f d \tau+c \int f^{} \hat{A} g d \tau+c^{} c \int g^{} \hat{A} g d \tau \
& =\int f(\hat{A} f)^{} d \tau+c \int g(\hat{A} f)^{} d \tau+c^{} \int f(\hat{A} g)^{} d \tau+c c^{} \int g(\hat{A} g)^{} d \tau
\end{aligned}
\)

By virtue of (7.6), the first terms on each side of this last equation are equal to each other; likewise, the last terms on each side are equal. Therefore

\(
\begin{equation}
c^{} \int g^{} \hat{A} f d \tau+c \int f^{} \hat{A} g d \tau=c \int g(\hat{A} f)^{} d \tau+c^{} \int f(\hat{A} g)^{} d \tau \tag{7.8}
\end{equation}
\)

Setting $c=1$ in (7.8), we have

\(
\begin{equation}
\int g^{} \hat{A} f d \tau+\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau+\int f(\hat{A} g)^{} d \tau \tag{7.9}
\end{equation}
\)

Setting $c=i$ in (7.8), we have, after dividing by $i$,

\(
\begin{equation}
-\int g^{} \hat{A} f d \tau+\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau-\int f(\hat{A} g)^{} d \tau \tag{7.10}
\end{equation}
\)

We now add (7.9) and (7.10) to get (7.7). This completes the proof.
Therefore, a Hermitian operator $\hat{A}$ is a linear operator that satisfies

\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau=\int f{n}\left(\hat{A} f_{m}\right)^{} d \tau \tag{7.11}
\end{equation}
\)

where $f{m}$ and $f{n}$ are arbitrary well-behaved functions and the integrals are definite integrals over all space. Using the bracket and matrix-element notations, we write

\(
\begin{align}
\left\langle f{m}\right| \hat{A}\left|f{n}\right\rangle & =\left\langle f{n}\right| \hat{A}\left|f{m}\right\rangle^{} \tag{7.12}\
\langle m| \hat{A}|n\rangle & =\langle n| \hat{A}|m\rangle^{} \tag{7.13}\
A{m n} & =\left(A{n m}\right)^{} \tag{7.14}
\end{align*}
\)

The two sides of (7.12) differ by having the functions interchanged and the complex conjugate taken.

Examples of Hermitian Operators

Let us show that some of the operators we have been using are indeed Hermitian. For simplicity, we shall work in one dimension. To prove that an operator is Hermitian, it suffices to show that it satisfies (7.6) for all well-behaved functions. However, we shall make things a bit harder by proving that (7.11) is satisfied.

First consider the one-particle, one-dimensional potential-energy operator. The right side of (7.11) is

\(
\begin{equation}
\int{-\infty}^{\infty} f{n}(x)\left[V(x) f_{m}(x)\right] d x \tag{7.15}
\end{}
\)

We have $V^{*}=V$, since the potential energy is a real function. Moreover, the order of the factors in (7.15) does not matter. Therefore,

\(
\int{-\infty}^{\infty} f{n}\left(V f{m}\right)^{*} d x=\int{-\infty}^{\infty} f{n} V^{*} f{m}^{} d x=\int{-\infty}^{\infty} f{m}^{} V f_{n} d x
\)

which proves that $V$ is Hermitian.
The operator for the $x$ component of linear momentum is $\hat{p}_{x}=-i \hbar d / d x$ [Eq. (3.23)]. For this operator, the left side of (7.11) is

\(
-i \hbar \int{-\infty}^{\infty} f{m}^{*}(x) \frac{d f_{n}(x)}{d x} d x
\)

Now we use the formula for integration by parts:

\(
\begin{equation}
\int{a}^{b} u(x) \frac{d v(x)}{d x} d x=\left.u(x) v(x)\right|{a} ^{b}-\int_{a}^{b} v(x) \frac{d u(x)}{d x} d x \tag{7.16}
\end{equation}
\)

Let

\(
u(x) \equiv-i \hbar f{m}^{*}(x), \quad v(x) \equiv f{n}(x)
\)

Then

\(
\begin{equation}
-i \hbar \int{-\infty}^{\infty} f{m}^{} \frac{d f{n}}{d x} d x=-\left.i \hbar f{m}^{} f{n}\right|{-\infty} ^{\infty}+i \hbar \int{-\infty}^{\infty} f{n}(x) \frac{d f_{m}^{}(x)}{d x} d x \tag{7.17}
\end{}
\)

Because $f{m}$ and $f{n}$ are well-behaved functions, they vanish at $x= \pm \infty$. (If they didn't vanish at infinity, they wouldn't be quadratically integrable.) Therefore, (7.17) becomes

\(
\int{-\infty}^{\infty} f{m}^{}\left(-i \hbar \frac{d f{n}}{d x}\right) d x=\int{-\infty}^{\infty} f{n}\left(-i \hbar \frac{d f{m}}{d x}\right)^{} d x
\)

which is the same as (7.11) and proves that $\hat{p}_{x}$ is Hermitian. The proof that the kineticenergy operator is Hermitian is left to the reader. The sum of two Hermitian operators can be shown to be Hermitian. Hence the Hamiltonian operator $\hat{H}=\hat{T}+\hat{V}$ is Hermitian.

Theorems about Hermitian Operators

We now prove some important theorems about the eigenvalues and eigenfunctions of Hermitian operators.

Since the eigenvalues of the operator $\hat{A}$ corresponding to the physical quantity $A$ are the possible results of a measurement of $A$ (Section 3.3), these eigenvalues should all be real numbers. We now prove that the eigenvalues of a Hermitian operator are real numbers.

We are given that $\hat{A}$ is Hermitian. Translating these words into an equation, we have [Eq. (7.11)]

\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau=\int f{n}\left(\hat{A} f_{m}\right)^{} d \tau \tag{7.18}
\end{equation}
\)

for all well-behaved functions $f{m}$ and $f{n}$. We want to prove that every eigenvalue of $\hat{A}$ is a real number. Translating this into equations, we want to show that $a{i}=a{i}^{*}$, where the eigenvalues $a{i}$ satisfy $\hat{A} g{i}=a{i} g{i}$; the functions $g_{i}$ are the eigenfunctions.

To introduce the eigenvalues $a{i}$ into (7.18), we write (7.18) for the special case where $f{m}=g{i}$ and $f{n}=g_{i}$ :

\(
\int g{i}^{*} \hat{A} g{i} d \tau=\int g{i}\left(\hat{A} g{i}\right)^{*} d \tau
\)

Use of $\hat{A} g{i}=a{i} g_{i}$ gives

\(
\begin{gather}
a{i} \int g{i}^{} g{i} d \tau=\int g{i}\left(a{i} g{i}\right)^{} d \tau=a_{i}^{} \int g{i} g{i}^{} d \tau \
\left(a{i}-a{i}^{}\right) \int\left|g_{i}\right|^{2} d \tau=0 \tag{7.19}
\end{gather*}
\)

Since the integrand $\left|g{i}\right|^{2}$ is never negative, the only way the integral in (7.19) could be zero would be if $g{i}$ were zero for all values of the coordinates. However, we always reject $g{i}=0$ as an eigenfunction on physical grounds. Hence the integral in (7.19) cannot be zero. Therefore, $\left(a{i}-a{i}^{*}\right)=0$, and $a{i}=a_{i}^{*}$. We have proved:

THEOREM 1. The eigenvalues of a Hermitian operator are real numbers.
To help become familiar with bracket notation, we shall repeat the proof of Theorem 1 using bracket notation. We begin by setting $m=i$ and $n=i$ in (7.13) to get $\langle i| \hat{A}|i\rangle=\langle i| \hat{A}|i\rangle^{}$. Choosing the function with index $i$ to be an eigenfunction of $\hat{A}$ and using the eigenvalue equation $\hat{A} g{i}=a{i} g{i}$, we have $\langle i| a{i}|i\rangle=\langle i| a_{i}|i\rangle^{}$. Therefore $a{i}\langle i \mid i\rangle=a{i}^{}\langle i \mid i\rangle^{}=a{i}^{*}\langle i \mid i\rangle$ and $\left(a{i}-a{i}^{*}\right)\langle i \mid i\rangle=0$. So $a{i}=a_{i}^{*}$, where (7.4) with $m=n$ was used.

We showed that two different particle-in-a-box energy eigenfunctions $\psi{i}$ and $\psi{j}$ are orthogonal, meaning that $\int{-\infty}^{\infty} \psi{i}^{*} \psi{j} d x=0$ for $i \neq j$ [Eq. (2.26)]. Two functions $f{1}$ and $f_{2}$ of the same set of coordinates are said to be orthogonal if

\(
\begin{equation}
\int f_{1}^{} f_{2} d \tau=0 \tag{7.20}
\end{}
\)

where the integral is a definite integral over the full range of the coordinates. We now prove the general theorem that the eigenfunctions of a Hermitian operator are, or can be chosen to be, mutually orthogonal. Given that

\(
\begin{equation}
\hat{B} F=s F, \quad \hat{B} G=t G \tag{7.21}
\end{equation}
\)

where $F$ and $G$ are two linearly independent eigenfunctions of the Hermitian operator $\hat{B}$, we want to prove that

\(
\int F^{*} G d \tau \equiv\langle F \mid G\rangle=0
\)

We begin with Eq. (7.12), which expresses the Hermitian nature of $\hat{B}$ :

\(
\langle F| \hat{B}|G\rangle=\langle G| \hat{B}|F\rangle^{*}
\)

Using (7.21), we have

\(
\begin{aligned}
\langle F| t|G\rangle & =\langle G| s|F\rangle^{} \
t\langle F \mid G\rangle & =s^{}\langle G \mid F\rangle^{*}
\end{aligned}
\)

Since eigenvalues of Hermitian operators are real (Theorem 1), we have $s^{}=s$. Use of $\langle G \mid F\rangle^{}=\langle F \mid G\rangle$ [Eq. (7.4)] gives

\(
\begin{gathered}
t\langle F \mid G\rangle=s\langle F \mid G\rangle \
(t-s)\langle F \mid G\rangle=0
\end{gathered}
\)

If $s \neq t$, then

\(
\begin{equation}
\langle F \mid G\rangle=0 \tag{7.22}
\end{equation}
\)

We have proved that two eigenfunctions of a Hermitian operator that correspond to different eigenvalues are orthogonal. The question now is: Can we have two independent eigenfunctions that have the same eigenvalue? The answer is yes. In the case of degeneracy, we have the same eigenvalue for more than one independent eigenfunction. Therefore, we can only be certain that two independent eigenfunctions of a Hermitian operator are orthogonal to each other if they do not correspond to a degenerate eigenvalue. We now show that in the case of degeneracy we may construct eigenfunctions that will be orthogonal to one another. We shall use the theorem proved in Section 3.6, that any linear combination of eigenfunctions corresponding to a degenerate eigenvalue is an eigenfunction with the same eigenvalue. Let us therefore suppose that $F$ and $G$ are independent eigenfunctions that have the same eigenvalue:

\(
\hat{B} F=s F, \quad \hat{B} G=s G
\)

We take linear combinations of $F$ and $G$ to form two new eigenfunctions $g{1}$ and $g{2}$ that will be orthogonal to each other. We choose

\(
g{1} \equiv F, \quad g{2} \equiv G+c F
\)

where the constant $c$ will be chosen to ensure orthogonality. We want

\(
\begin{gathered}
\int g{1}^{*} g{2} d \tau=0 \
\int F^{}(G+c F) d \tau=\int F^{} G d \tau+c \int F^{*} F d \tau=0
\end{gathered}
\)

Hence choosing

\(
\begin{equation}
c=-\int F^{} G d \tau / \int F^{} F d \tau \tag{7.23}
\end{equation}
\)

we have two orthogonal eigenfunctions $g{1}$ and $g{2}$ corresponding to the degenerate eigenvalue. This procedure (called Schmidt or Gram-Schmidt orthogonalization) can be extended to the case of $n$-fold degeneracy to give $n$ linearly independent orthogonal eigenfunctions corresponding to the degenerate eigenvalue.

Thus, although there is no guarantee that the eigenfunctions of a degenerate eigenvalue are orthogonal, we can always choose them to be orthogonal, if we desire, by using the Schmidt (or some other) orthogonalization method. In fact, unless stated otherwise, we shall always assume that we have chosen the eigenfunctions to be orthogonal:

\(
\begin{equation}
\int g_{i}^{} g_{k} d \tau=0, \quad i \neq k \tag{7.24}
\end{}
\)

where $g{i}$ and $g{k}$ are independent eigenfunctions of a Hermitian operator. We have proved:
THEOREM 2. Two eigenfunctions of a Hermitian operator $\hat{B}$ that correspond to different eigenvalues are orthogonal. Eigenfunctions of $\hat{B}$ that belong to a degenerate eigenvalue can always be chosen to be orthogonal.

An eigenfunction can usually be multiplied by a constant to normalize it, and we shall assume, unless stated otherwise, that all eigenfunctions are normalized:

\(
\begin{equation}
\int g_{i}^{} g_{i} d \tau=1 \tag{7.25}
\end{}
\)

The exception is where the eigenvalues form a continuum, rather than a discrete set of values. In this case, the eigenfunctions are not quadratically integrable. Examples are the linear-momentum eigenfunctions, the free-particle energy eigenfunctions, and the hydrogenatom continuum energy eigenfunctions.

Using the Kronecker delta, defined by $\delta{i k} \equiv 1$ if $i=k$ and $\delta{i k} \equiv 0$ if $i \neq k$ [Eq. (2.28)], we can combine (7.24) and (7.25) into one equation:

\(
\begin{equation}
\int g_{i}^{} g{k} d \tau=\langle i \mid k\rangle=\delta{i k} \tag{7.26}
\end{}
\)

where $g{i}$ and $g{k}$ are eigenfunctions of some Hermitian operator.
As an example, consider the spherical harmonics. We shall prove that

\(
\begin{equation}
\int{0}^{2 \pi} \int{0}^{\pi}\left[Y_{l}^{m}(\theta, \phi)\right] Y{l^{\prime}}^{m^{\prime}}(\theta, \phi) \sin \theta d \theta d \phi=\delta{l, l^{\prime}} \delta_{m, m^{\prime}} \tag{7.27}
\end{}
\)

where the $\sin \theta$ factor comes from the volume element in spherical coordinates, (5.78). The spherical harmonics are eigenfunctions of the Hermitian operator $\hat{L}^{2}$ [Eq. (5.104)]. Since eigenfunctions of a Hermitian operator belonging to different eigenvalues are orthogonal, we conclude that the integral in (7.27) is zero unless $l=l^{\prime}$. Similarly, since the $Y{l}^{m}$ functions are eigenfunctions of $\hat{L}{z}$ [Eq. (5.105)], we conclude that the integral in (7.27) is zero unless $m=m^{\prime}$. Also, the multiplicative constant in $Y_{l}^{m}$ [Eq. (5.147) of Prob. 5.34] has been chosen so that the spherical harmonics are normalized [Eq. (6.117)]. Therefore (7.27) is valid.

The integral $\langle f| \hat{B}|g\rangle$ can be simplified if either $f$ or $g$ is an eigenfunction of the Hermitian operator $\hat{B}$. If $\hat{B} g=c g$, where $c$ is a constant, then

\(
\langle f| \hat{B}|g\rangle=\langle f \mid \hat{B} g\rangle=\langle f \mid c g\rangle=c\langle f \mid g\rangle
\)

If $\hat{B} f=k f$, where $k$ is a constant, then use of the Hermitian property of $\hat{B}$ gives

\(
\langle f| \hat{B}|g\rangle=\langle g| \hat{B}|f\rangle^{}=\langle g \mid \hat{B} f\rangle^{}=\langle g \mid k f\rangle^{}=k^{}\langle g \mid f\rangle^{*}=k\langle f \mid g\rangle
\)

since the eigenvalue $k$ is real. The relation $\langle f| \hat{B}|g\rangle=k\langle f \mid g\rangle$ shows that the Hermitian operator $\hat{B}$ can act to the left in $\langle f| \hat{B}|g\rangle$.

A proof of the uncertainty principle is outlined in Prob. 7.60.

In the previous section, we proved the orthogonality of the eigenfunctions of a Hermitian operator. We now discuss another important property of these functions; this property allows us to expand an arbitrary well-behaved function in terms of these eigenfunctions.

We have used the Taylor-series expansion (Prob. 4.1) of a function as a linear combination of the nonnegative integral powers of $(x-a)$. Can we expand a function as a linear combination of some other set of functions besides $1,(x-a),(x-a)^{2}, \ldots$ ? The answer is yes, as was first shown by Fourier in 1807. A Fourier series is an expansion of a function as a linear combination of an infinite number of sine and cosine functions. We shall not go into detail about Fourier series, but shall simply look at one example.

Expansion of a Function Using Particle-in-a-Box Wave Functions

Let us consider expanding a function in terms of the particle-in-a-box stationary-state wave functions, which are [Eq. (2.23)]

\(
\begin{equation}
\psi_{n}=\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right), \quad n=1,2,3, \ldots \tag{7.28}
\end{equation}
\)

for $x$ between 0 and $l$. What are our chances for representing an arbitrary function $f(x)$ in the interval $0 \leq x \leq l$ by a series of the form

\(
\begin{equation}
f(x)=\sum{n=1}^{\infty} a{n} \psi{n}=\left(\frac{2}{l}\right)^{1 / 2} \sum{n=1}^{\infty} a_{n} \sin \left(\frac{n \pi x}{l}\right), \quad 0 \leq x \leq l \tag{7.29}
\end{equation}
\)

where the $a{n}$ 's are constants? Substitution of $x=0$ and $x=l$ in (7.29) gives the restrictions that $f(0)=0$ and $f(l)=0$. In other words, $f(x)$ must satisfy the same boundary conditions as the $\psi{n}$ functions. We shall also assume that $f(x)$ is finite, single-valued, and continuous, but not necessarily differentiable. With these assumptions it can be shown that the expansion (7.29) is valid. We shall not prove (7.29) but will simply illustrate its use to represent a function.

Before we can apply (7.29) to a specific $f(x)$, we must derive an expression for the expansion coefficients $a{n}$. We start by multiplying (7.29) by $\psi{m}^{*}$ :

\(
\begin{equation}
\psi_{m}^{} f(x)=\sum{n=1}^{\infty} a{n} \psi{m}^{*} \psi{n}=\left(\frac{2}{l}\right) \sum{n=1}^{\infty} a{n} \sin \left(\frac{n \pi x}{l}\right) \sin \left(\frac{m \pi x}{l}\right) \tag{7.30}
\end{}
\)

Now we integrate this equation from 0 to $l$. Assuming the validity of interchanging the integration and the infinite summation, we have

\(
\int{0}^{l} \psi{m}^{} f(x) d x=\sum{n=1}^{\infty} a{n} \int{0}^{l} \psi{m}^{} \psi{n} d x=\sum{n=1}^{\infty} a{n}\left(\frac{2}{l}\right) \int{0}^{l} \sin \left(\frac{n \pi x}{l}\right) \sin \left(\frac{m \pi x}{l}\right) d x
\)

We proved the orthonormality of the particle-in-a-box wave functions [Eq. (2.27)]. Therefore, the last equation becomes

\(
\begin{equation}
\int{0}^{l} \psi{m}^{} f(x) d x=\sum{n=1}^{\infty} a{n} \delta_{m n} \tag{7.31}
\end{}
\)

The type of sum in (7.31) occurs often. Writing it in detail, we have

\(
\begin{gather}
\sum{n=1}^{\infty} a{n} \delta{m n}=a{1} \delta{m, 1}+a{2} \delta{m, 2}+\cdots+a{m} \delta{m, m}+a{m+1} \delta{m, m+1}+\cdots \
=0+0+\cdots+a{m}+0+\cdots \
\sum{n=1}^{\infty} a{n} \delta{m n}=a{m} \tag{7.32}
\end{gather}
\)

Thus, since $\delta_{m n}$ is zero except when the summation index $n$ is equal to $m$, all terms but one vanish, and (7.31) becomes

\(
\begin{equation}
a{m}=\int{0}^{l} \psi_{m}^{} f(x) d x \tag{7.33}
\end{}
\)

which is the desired expression for the expansion coefficients.

FIGURE 7.1 Function to be expanded in terms of particle-in-a-box functions.

Changing $m$ to $n$ in (7.33) and substituting it into (7.29), we have

\(
\begin{equation}
f(x)=\sum{n=1}^{\infty}\left[\int{0}^{l} \psi_{n}^{} f(x) d x\right] \psi_{n}(x) \tag{7.34}
\end{}
\)

This is the desired expression for the expansion of an arbitrary well-behaved function $f(x)(0 \leq x \leq l)$ as a linear combination of the particle-in-a-box wave functions $\psi{n}$. Note that the definite integral $\int{0}^{l} \psi_{n}^{*} f(x) d x$ is a number and not a function of $x$.

We now use (7.29) to represent a specific function, the function of Fig. 7.1, which is defined by

\(
\begin{array}{lrl}
f(x) & =x \quad \text { for } \quad 0 \leq x \leq \frac{1}{2} l \
f(x)=l-x \quad \text { for } \quad \frac{1}{2} l \leq x \leq l \tag{7.35}
\end{array}
\)

To find the expansion coefficients $a_{n}$, we substitute (7.28) and (7.35) into (7.33):

\(
\begin{aligned}
& a{n}=\int{0}^{l} \psi{n}^{*} f(x) d x=\left(\frac{2}{l}\right)^{1 / 2} \int{0}^{l} \sin \left(\frac{n \pi x}{l}\right) f(x) d x \
& a{n}=\left(\frac{2}{l}\right)^{1 / 2} \int{0}^{l / 2} x \sin \left(\frac{n \pi x}{l}\right) d x+\left(\frac{2}{l}\right)^{1 / 2} \int_{l / 2}^{l}(l-x) \sin \left(\frac{n \pi x}{l}\right) d x
\end{aligned}
\)

Using the Appendix integral (A.1), we find

\(
\begin{equation}
a_{n}=\frac{(2 l)^{3 / 2}}{n^{2} \pi^{2}} \sin \left(\frac{n \pi}{2}\right) \tag{7.36}
\end{equation}
\)

Using (7.36) in the expansion (7.29), we have [since $\sin (n \pi / 2)$ equals zero for $n$ even and equals +1 or -1 for $n$ odd]

\(
\begin{align}
& f(x)=\frac{4 l}{\pi^{2}}\left[\sin \left(\frac{\pi x}{l}\right)-\frac{1}{3^{2}} \sin \left(\frac{3 \pi x}{l}\right)+\frac{1}{5^{2}} \sin \left(\frac{5 \pi x}{l}\right)-\cdots\right] \
& f(x)=\frac{4 l}{\pi^{2}} \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(2 n-1)^{2}} \sin \left[(2 n-1) \frac{\pi x}{l}\right] \tag{7.37}
\end{align}
\)

where $f(x)$ is given by (7.35). Let us check (7.37) at $x=\frac{1}{2} l$. We have

\(
\begin{equation}
f\left(\frac{l}{2}\right)=\frac{4 l}{\pi^{2}}\left(1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots\right) \tag{7.38}
\end{equation}
\)

FIGURE 7.2 Plots of (a) the error and (b) the percent error in the expansion of the function of Fig. 7.1 in terms of particle-in-a-box wave functions when 1 and 5 terms are taken in the expansion.

Tabulating the right side of (7.38) as a function of the number of terms we take in the infinite series, we get:

Number of terms	1	2	3	4	5	20	100
Right side of (7.38)	$0.405 l$	$0.450 l$	$0.467 l$	$0.475 l$	$0.480 l$	$0.495 l$	$0.499 l$

If we take an infinite number of terms, the series should sum to $\frac{1}{2} l$, which is the value of $f\left(\frac{1}{2} l\right)$. Assuming the validity of the series, we have the interesting result that the infinite sum in parentheses in (7.38) equals $\pi^{2} / 8$. Figure 7.2 plots $f(x)-\sum{n=1}^{k} a{n} \psi{n}$ [where $f, a{n}$, and $\psi_{n}$ are given by (7.35), (7.36), and (7.28)] for $k$ values of 1 and 5 . As $k$, the number of terms in the expansion, increases, the series comes closer to $f(x)$, and the difference between $f$ and the series goes to zero.

Expansion of a Function in Terms of Eigenfunctions

We have seen an example of the expansion of a function in terms of a set of functionsthe particle-in-a-box energy eigenfunctions. Many different sets of functions can be used to expand an arbitrary function. A set of functions $g{1}, g{2}, \ldots, g{i}, \ldots$ is said to be a complete set if every well-behaved function $f$ that obeys the same boundary conditions as the $g{i}$ functions can be expanded as a linear combination of the $g_{i}$ 's according to

\(
\begin{equation}
f=\sum{i} a{i} g_{i} \tag{7.39}
\end{equation}
\)

where the $a{i}$ 's are constants. Of course, it is understood that $f$ and the $g{i}$ 's are all functions of the same set of variables. The limits have been omitted from the sum in (7.39). It is understood that this sum goes over all members of the complete set. By virtue of theorems of Fourier analysis (which we have not proved), the particle-in-a-box energy eigenfunctions can be shown to be a complete set.

We now postulate that the set of eigenfunctions of every Hermitian operator that represents a physical quantity is a complete set. (Completeness of the eigenfunctions can be proved in many cases, but must be postulated in the general case.) Thus, every wellbehaved function that satisfies the same boundary conditions as the set of eigenfunctions can be expanded according to (7.39). Equation (7.29) is an example of (7.39).

The harmonic-oscillator wave functions are given by a Hermite polynomial $H_{v}$ times an exponential factor [Eq. (4.86) of Prob. 4.21c]. By virtue of the expansion postulate, any well-behaved function $f(x)$ can be expanded as a linear combination of harmonicoscillator energy eigenfunctions:

\(
f(x)=\sum{n=0}^{\infty} a{n}\left(2^{n} n!\right)^{-1 / 2}(\alpha / \pi)^{1 / 4} H_{n}\left(\alpha^{1 / 2} x\right) e^{-\alpha x^{2} / 2}
\)

How about using the hydrogen-atom bound-state wave functions to expand an arbitrary function $f(r, \theta, \phi)$ ? The answer is that these functions do not form a complete set, and we cannot expand $f$ using them. To have a complete set, we must use all the eigenfunctions of a particular Hermitian operator. In addition to the bound-state eigenfunctions of the hydrogen-atom Hamiltonian, we have the continuum eigenfunctions, corresponding to ionized states. If the continuum eigenfunctions are included along with the bound-state eigenfunctions, then we have a complete set. (For the particle in a box and the harmonic oscillator, there are no continuum functions.) Equation (7.39) implies an integration over the continuum eigenfunctions, if there are any. Thus, if $\psi{n l m}(r, \theta, \phi)$ is a bound-state wave function of the hydrogen atom and $\psi{E l m}(r, \theta, \phi)$ is a continuum eigenfunction, then (7.39) becomes

\(
f(r, \theta, \phi)=\sum{n=1}^{\infty} \sum{l=0}^{n-1} \sum{m=-l}^{l} a{n l m} \psi{n l m}(r, \theta, \phi)+\sum{l=0}^{\infty} \sum{m=-l}^{l} \int{0}^{\infty} a{l m}(E) \psi{E l m}(r, \theta, \phi) d E
\)

As another example, consider the eigenfunctions of $\hat{p}_{x}$ [Eq. (3.36)]:

\(
g_{k}=e^{i k x / \hbar}, \quad-\infty<k<\infty
\)

Here the eigenvalues are all continuous, and the eigenfunction expansion (7.39) of an arbitrary function $f$ becomes

\(
f(x)=\int_{-\infty}^{\infty} a(k) e^{i k x / \hbar} d k
\)

The reader with a good mathematical background may recognize this integral as very nearly the Fourier transform of $a(k)$.

Let us evaluate the expansion coefficients in $f=\sum{i} a{i} g{i}$ [Eq. (7.39)], where the $g{i}$ functions are the complete set of eigenfunctions of a Hermitian operator. The procedure is the same as that used to derive (7.33). We multiply $f=\sum{i} a{i} g{i}$ by $g{k}^{*}$ and integrate over all space:

\(
\begin{gather}
g_{k}^{} f=\sum{i} a{i} g{k}^{*} g{i} \
\int g{k}^{*} f d \tau=\sum{i} a{i} \int g{k}^{} g{i} d \tau=\sum{i} a{i} \delta{i k}=a{k} \
a{k}=\int g_{k}^{} f d \tau \tag{7.40}
\end{gather*}
\)

where we used the orthonormality of the eigenfunctions of a Hermitian operator: $\int g{k}^{*} g{i} d \tau=\delta{i k}$ [Eq. (7.26)]. The procedure that led to (7.40) will be used often and is worth remembering. Substitution of (7.40) for $a{i}$ in $f=\sum{i} a{i} g_{i}$ gives

\(
\begin{equation}
f=\sum{i}\left[\int g{i}^{} f d \tau\right] g{i}=\sum{i}\left\langle g{i} \mid f\right\rangle g{i} \tag{7.41}
\end{}
\)

EXAMPLE

Let $F(x)=x(l-x)$ for $0 \leq x \leq l$ and $F(x)=0$ elsewhere. Expand $F$ in terms of the particle-in-a-box energy eigenfunctions $\psi_{n}=(2 / l)^{1 / 2} \sin (n \pi x / l)$ for $0 \leq x \leq l$.

We begin by noting that $F(0)=0$ and $F(l)=0$, so $F$ obeys the same boundary conditions as the $\psi{n}$ 's and can be expanded using the $\psi{n}$ 's. The expansion is $F=\sum{n=1}^{\infty} a{n} \psi{n}$, where $a{n}=\int \psi_{n}^{*} F d \tau$ [Eqs. (7.39) and (7.40)]. Thus

\(
a{n}=\int \psi{n}^{*} F d \tau=\left(\frac{2}{l}\right)^{1 / 2} \int_{0}^{l}\left(\sin \frac{n \pi x}{l}\right) x(l-x) d x=\frac{2^{3 / 2} l^{5 / 2}}{n^{3} \pi^{3}}\left[1-(-1)^{n}\right]
\)

where details of the integral evaluation are left as a problem (Prob. 7.18). The expansion $F=\sum{n=1}^{\infty} a{n} \psi_{n}$ is

\(
x(l-x)=\frac{4 l^{2}}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n^{3}} \sin \frac{n \pi x}{l}, \quad \text { for } 0 \leq x \leq l
\)

EXERCISE Let $G(x)=1$ for $0 \leq x \leq l$ and $G(x)=0$ elsewhere. Expand $G$ in terms of the particle-in-a-box energy eigenfunctions. Since $G$ is not zero at 0 and at $l$, the expansion will not represent $G$ at these points but will represent $G$ elsewhere. Use the first 7 nonzero terms of the expansion to calculate $G$ at $x=\frac{1}{4} l$. Repeat this using the first 70 nonzero terms (use a programmable calculator). (Answers: 0.1219, 0.9977.)

A useful theorem is the following:
THEOREM 3. Let the functions $g{1}, g{2}, \ldots$ be the complete set of eigenfunctions of the Hermitian operator $\hat{A}$, and let the function $F$ be an eigenfunction of $\hat{A}$ with eigenvalue $k$ (that is, $\hat{A} F=k F$ ). Then if $F$ is expanded as $F=\sum{i} a{i} g{i}$, the only nonzero coefficients $a{i}$ are those for which $g{i}$ has the eigenvalue $k$. (Because of degeneracy, several $g{i}$ 's may have the same eigenvalue $k$.)

Thus in the expansion of $F$, we include only those eigenfunctions that have the same eigenvalue as $F$. The proof of Theorem 3 follows at once from $a{k}=\int g{k}^{*} F d \tau$ [Eq. (7.40)]; if $F$ and $g{k}$ correspond to different eigenvalues of the Hermitian operator $\hat{A}$, they will be orthogonal [Eq. (7.22)] and $a{k}$ will vanish.

We shall occasionally use a notation (called ket notation) in which the function $f$ is denoted by the symbol $|f\rangle$. There doesn't seem to be any point to this notation, but in advanced formulations of quantum mechanics, it takes on a special significance. In ket notation, Eq. (7.41) reads

\(
\begin{equation}
|f\rangle=\sum{i}\left|g{i}\right\rangle\left\langle g{i} \mid f\right\rangle=\sum{i}|i\rangle\langle i \mid f\rangle \tag{7.42}
\end{equation}
\)

Ket notation is conveniently used to specify eigenfunctions by listing their eigenvalues. For example, the hydrogen-atom wave function with quantum numbers $n, l, m$ is denoted by $\psi_{n l m}=|n l m\rangle$.

The contents of Sections 7.2 and 7.3 can be summarized by the statement that the eigenfunctions of a Hermitian operator form a complete, orthonormal set, and the eigenvalues are real.

If the state function $\Psi$ is simultaneously an eigenfunction of the two operators $\hat{A}$ and $\hat{B}$ with eigenvalues $a{j}$ and $b{j}$, respectively, then a measurement of the physical property $A$ will yield the result $a{j}$ and a measurement of $B$ will yield $b{j}$. Hence the two properties $A$ and $B$ have definite values when $\Psi$ is simultaneously an eigenfunction of $\hat{A}$ and $\hat{B}$.

In Section 5.1, some statements were made about simultaneous eigenfunctions of two operators. We now prove these statements.

First, we show that if there exists a common complete set of eigenfunctions for two linear operators then these operators commute. Let $\hat{A}$ and $\hat{B}$ denote two linear operators that have a common complete set of eigenfunctions $g{1}, g{2}, \ldots$ :

\(
\begin{equation}
\hat{A} g{i}=a{i} g{i}, \quad \hat{B} g{i}=b{i} g{i} \tag{7.43}
\end{equation}
\)

where $a{i}$ and $b{i}$ are the eigenvalues. We must prove that

\(
\begin{equation}
[\hat{A}, \hat{B}]=\hat{0} \tag{7.44}
\end{equation}
\)

Equation (7.44) is an operator equation. For two operators to be equal, the results of operating with either of them on an arbitrary well-behaved function $f$ must be the same. Hence we must show that

\(
(\hat{A} \hat{B}-\hat{B} \hat{A}) f=\hat{0} f=0
\)

where $f$ is an arbitrary function. We begin the proof by expanding $f$ (assuming that it obeys the proper boundary conditions) in terms of the complete set of eigenfunctions $g_{i}$ :

\(
f=\sum{i} c{i} g_{i}
\)

Operating on each side of this last equation with $\hat{A} \hat{B}-\hat{B} \hat{A}$, we have

\(
(\hat{A} \hat{B}-\hat{B} \hat{A}) f=(\hat{A} \hat{B}-\hat{B} \hat{A}) \sum{i} c{i} g_{i}
\)

Since the products $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are linear operators (Prob. 3.16), we have

\(
(\hat{A} \hat{B}-\hat{B} \hat{A}) f=\sum{i} c{i}(\hat{A} \hat{B}-\hat{B} \hat{A}) g{i}=\sum{i} c{i}\left[\hat{A}\left(\hat{B} g{i}\right)-\hat{B}\left(\hat{A} g_{i}\right)\right]
\)

where the definitions of the sum and the product of operators were used. Use of the eigenvalue equations (7.43) gives

\(
(\hat{A} \hat{B}-\hat{B} \hat{A}) f=\sum{i} c{i}\left[\hat{A}\left(b{i} g{i}\right)-\hat{B}\left(a{i} g{i}\right)\right]=\sum{i} c{i}\left(b{i} a{i} g{i}-a{i} b{i} g{i}\right)=0
\)

This completes the proof of:
THEOREM 4. If the linear operators $\hat{A}$ and $\hat{B}$ have a common complete set of eigenfunctions, then $\hat{A}$ and $\hat{B}$ commute.

It is sometimes erroneously stated that if a common eigenfunction of $\hat{A}$ and $\hat{B}$ exists, then they commute. An example that shows this statement to be false is the fact that the
spherical harmonic $Y{0}^{0}$ is an eigenfunction of both $\hat{L}{z}$ and $\hat{L}_{x}$ even though these two operators do not commute (Section 5.3). It is instructive to examine the so-called proof that is given for this erroneous statement. Let $g$ be the common eigenfunction: $\hat{A} g=a g$ and $\hat{B} g=b g$. We have

\(
\begin{array}{cl}
\hat{A} \hat{B} g=\hat{A} b g=a b g \quad \text { and } \hat{B} \hat{A} g=\hat{B} a g=b a g=a b g \
& \hat{A} \hat{B} g=\hat{B} \hat{A} g \tag{7.45}
\end{array}
\)

The "proof" is completed by canceling $g$ from each side of (7.45) to get

\(
\begin{equation}
\hat{A} \hat{B}=\hat{B} \hat{A}(?) \tag{7.46}
\end{equation}
\)

It is in going from (7.45) to (7.46) that the error occurs. Just because the two operators $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ give the same result when acting on the single function $g$ is no reason to conclude that $\hat{A} \hat{B}=\hat{B} \hat{A}$. (For example, $d / d x$ and $d^{2} / d x^{2}$ give the same result when operating on $e^{x}$, but $d / d x$ is certainly not equal to $d^{2} / d x^{2}$.) The two operators must give the same result when acting on every well-behaved function before we can conclude that they are equal. Thus, even though $\hat{A}$ and $\hat{B}$ do not commute, one or more common eigenfunctions of $\hat{A}$ and $\hat{B}$ might exist. However, we cannot have a common complete set of eigenfunctions of two noncommuting operators, as we proved earlier in this section.

We have shown that, if there exists a common complete set of eigenfunctions of the linear operators $\hat{A}$ and $\hat{B}$, then they commute. We now prove the following:

THEOREM 5. If the Hermitian operators $\hat{A}$ and $\hat{B}$ commute, we can select a common complete set of eigenfunctions for them.

The proof is as follows. Let the functions $g{i}$ and the numbers $a{i}$ be the eigenfunctions and eigenvalues of $\hat{A}$ :

\(
\hat{A} g{i}=a{i} g_{i}
\)

Operating on both sides of this equation with $\hat{B}$, we have

\(
\hat{B} \hat{A} g{i}=\hat{B}\left(a{i} g_{i}\right)
\)

Since $\hat{A}$ and $\hat{B}$ commute and since $\hat{B}$ is linear, we have

\(
\begin{equation}
\hat{A}\left(\hat{B} g{i}\right)=a{i}\left(\hat{B} g_{i}\right) \tag{7.47}
\end{equation}
\)

This equation states that the function $\hat{B} g{i}$ is an eigenfunction of the operator $\hat{A}$ with the same eigenvalue $a{i}$ as the eigenfunction $g{i}$. Suppose the eigenvalues of $\hat{A}$ are nondegenerate, so that for any given eigenvalue $a{i}$ one and only one linearly independent eigenfunction exists. If this is so, then the two eigenfunctions $g{i}$ and $\hat{B} g{i}$, which correspond to the same eigenvalue $a_{i}$, must be linearly dependent; that is, one function must be simply a multiple of the other:

\(
\begin{equation}
\hat{B} g{i}=k{i} g_{i} \tag{7.48}
\end{equation}
\)

where $k{i}$ is a constant. This equation states that the functions $g{i}$ are eigenfunctions of $\hat{B}$, which is what we wanted to prove. In Section 7.3, we postulated that the eigenfunctions of any operator that represents a physical quantity form a complete set. Hence the $g_{i}$ 's form a complete set.

We have just proved the desired theorem for the nondegenerate case, but what about the degenerate case? Let the eigenvalue $a{i}$ be $n$-fold degenerate. We know from Eq. (7.47) that $\hat{B} g{i}$ is an eigenfunction of $\hat{A}$ with eigenvalue $a{i}$. Hence, Theorem 3 of Section 7.3 tells us that, if the function $\hat{B} g{i}$ is expanded in terms of the complete set of eigenfunctions of $\hat{A}$,
then all the expansion coefficients will be zero except those for which the $\hat{A}$ eigenfunction has the eigenvalue $a{i}$. In other words, $\hat{B} g{i}$ must be a linear combination of the $n$ linearly independent $\hat{A}$ eigenfunctions that correspond to the eigenvalue $a_{i}$ :

\(
\begin{equation}
\hat{B} g{i}=\sum{k=1}^{n} c{k} g{k}, \quad \text { where } \quad \hat{A} g{k}=a{i} g_{k} \quad \text { for } k=1 \text { to } n \tag{7.49}
\end{equation}
\)

where $g{1}, \ldots, g{n}$ denote those $\hat{A}$ eigenfunctions that have the degenerate eigenvalue $a{i}$. Equation (7.49) shows that $g{i}$ is not necessarily an eigenfunction of $\hat{B}$. However, by taking suitable linear combinations of the $n$ linearly independent $\hat{A}$ eigenfunctions corresponding to the degenerate eigenvalue $a_{i}$, one can construct a new set of $n$ linearly independent eigenfunctions of $\hat{A}$ that will also be eigenfunctions of $\hat{B}$. Proof of this statement is given in Merzbacher, Section 8.5 .

Thus, when $\hat{A}$ and $\hat{B}$ commute, it is always possible to select a common complete set of eigenfunctions for them. For example, consider the hydrogen atom, where the operators $\hat{L}{z}$ and $\hat{H}$ were shown to commute. If we desired, we could take the phi factor in the eigenfunctions of $\hat{H}$ as $\sin m \phi$ and $\cos m \phi$ (Section 6.6). If we did this, we would not have eigenfunctions of $\hat{L}{z}$, except for $m=0$. However, the linear combinations

\(
R(r) S(\theta)(\cos m \phi+i \sin m \phi)=R S e^{i m \phi}, \quad m=-l, \ldots, l
\)

give us eigenfunctions of $\hat{L}_{z}$ that are still eigenfunctions of $\hat{H}$ by virtue of the theorem in Section 3.6.

Extension of the above proofs to the case of more than two operators shows that for a set of Hermitian operators $\hat{A}, \hat{B}, \hat{C}, \ldots$ there exists a common complete set of eigenfunctions if and only if every operator commutes with every other operator.

A useful theorem that is related to Theorem 5 is:
THEOREM 6. If $g{m}$ and $g{n}$ are eigenfunctions of the Hermitian operator $\hat{A}$ with different eigenvalues (that is, if $\hat{\hat{A}} g{m}=a{m} g{m}$ and $\hat{A} g{n}=a{n} g{n}$ with $a{m} \neq a{n}$ ), and if the linear operator $\hat{B}$ commutes with $\hat{A}$, then

\(
\begin{equation}
\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle=0 \quad \text { for } a{n} \neq a{m} \tag{7.50}
\end{equation}
\)

To prove (7.50), we start with

\(
\begin{equation}
\left\langle g{n}\right| \hat{A} \hat{B}\left|g{m}\right\rangle=\left\langle g{n}\right| \hat{B} \hat{A}\left|g{m}\right\rangle \tag{7.51}
\end{equation}
\)

Use of the Hermitian property of $\hat{A}$ gives the left side of (7.51) as

\(
\begin{aligned}
\left\langle g{n}\right| \hat{A} \hat{B}\left|g{m}\right\rangle=\left\langle g{n}\right| \hat{A}\left|\hat{B} g{m}\right\rangle=\left\langle\hat{B} g{m}\right| \hat{A}\left|g{n}\right\rangle & =\left\langle\hat{B} g{m}\right| a{n}\left|g_{n}\right\rangle^{} \
& =a{n}^{*}\left\langle g{n} \mid \hat{B} g{m}\right\rangle=a{n}\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle
\end{aligned}
\)

The right side of (7.51) is

\(
\left\langle g{n}\right| \hat{B} \hat{A}\left|g{m}\right\rangle=\left\langle g{n}\right| \hat{B}\left|\hat{A} g{m}\right\rangle=a{m}\left\langle g{n}\right| \hat{B}\left|g_{m}\right\rangle
\)

Equating the final expressions for the left and right sides of (7.51), we get

\(
\begin{gathered}
a{n}\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle=a{m}\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle \
\left(a{n}-a{m}\right)\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle=0
\end{gathered}
\)

Since $a{m} \neq a{n}$, we have $\left\langle g{n}\right| \hat{B}\left|g{m}\right\rangle=0$, which completes the proof.

Certain quantum-mechanical operators have no classical analog. An example is the parity operator. Recall that the harmonic-oscillator wave functions are either even or odd. We shall show how this property is related to the parity operator.

The parity operator $\hat{\Pi}$ is defined in terms of its effect on an arbitrary function $f$ :

\(
\begin{equation<em>}
\hat{\Pi} f(x, y, z)=f(-x,-y,-z) \tag{7.52}
\end{equation</em>}
\)

The parity operator replaces each Cartesian coordinate with its negative. For example, $\hat{\Pi}\left(x^{2}-z e^{a y}\right)=x^{2}+z e^{-a y}$.

As with any quantum-mechanical operator, we are interested in the eigenvalues $c{i}$ and the eigenfunctions $g{i}$ of the parity operator:

\(
\begin{equation<em>}
\hat{\Pi} g{i}=c{i} g_{i} \tag{7.53}
\end{equation</em>}
\)

The key to the problem is to calculate the square of $\hat{\Pi}$ :

\(
\hat{\Pi}^{2} f(x, y, z)=\hat{\Pi}[\hat{\Pi} f(x, y, z)]=\hat{\Pi}[f(-x,-y,-z)]=f(x, y, z)
\)

Since $f$ is arbitrary, we conclude that $\hat{\Pi}^{2}$ equals the unit operator:

\(
\begin{equation<em>}
\hat{\Pi}^{2}=\hat{1} \tag{7.54}
\end{equation</em>}
\)

We now operate on (7.53) with $\hat{\Pi}$ to get $\hat{\Pi} \hat{\Pi} g{i}=\hat{\Pi} c{i} g{i}$. Since $\hat{\Pi}$ is linear (Prob. 7.26), we have $\hat{\Pi}^{2} g{i}=c{i} \hat{\Pi} g{i}$, which becomes

\(
\begin{equation<em>}
\hat{\Pi}^{2} g{i}=c{i}^{2} g_{i} \tag{7.55}
\end{equation</em>}
\)

where the eigenvalue equation (7.53) was used. Since $\hat{\Pi}^{2}$ is the unit operator, the left side of (7.55) is simply $g_{i}$, and

\(
g{i}=c{i}^{2} g_{i}
\)

The function $g{i}$ cannot be zero everywhere (zero is always rejected as an eigenfunction on physical grounds). We can therefore divide by $g{i}$ to get $c_{i}^{2}=1$ and

\(
\begin{equation<em>}
c_{i}= \pm 1 \tag{7.56}
\end{equation</em>}
\)

The eigenvalues of $\hat{\Pi}$ are +1 and -1 . Note that this derivation applies to any operator whose square is the unit operator.

What are the eigenfunctions $g_{i}$ ? The eigenvalue equation (7.53) reads

\(
\begin{gathered}
\hat{\Pi} g{i}(x, y, z)= \pm g{i}(x, y, z) \
g{i}(-x,-y,-z)= \pm g{i}(x, y, z)
\end{gathered}
\)

If the eigenvalue is +1 , then $g{i}(-x,-y,-z)=g{i}(x, y, z)$ and $g{i}$ is an even function. If the eigenvalue is -1 , then $g{i}$ is odd: $g{i}(-x,-y,-z)=-g{i}(x, y, z)$. Hence, the eigenfunctions of the parity operator $\hat{\Pi}$ are all possible well-behaved even and odd functions.

When the parity operator commutes with the Hamiltonian operator $\hat{H}$, we can select a common set of eigenfunctions for these operators, as proved in Section 7.4. The eigenfunctions of $\hat{H}$ are the stationary-state wave functions $\psi_{i}$. Hence when

\(
\begin{equation<em>}
[\hat{\Pi}, \hat{H}]=0 \tag{7.57}
\end{equation</em>}
\)

the wave functions $\psi_{i}$ can be chosen to be eigenfunctions of $\hat{\Pi}$. We just proved that the eigenfunctions of $\hat{\Pi}$ are either even or odd. Hence, when (7.57) holds, each wave function can be chosen to be either even or odd. Let us find out when the parity and Hamiltonian operators commute.

We have, for a one-particle system,

\(
[\hat{H}, \hat{\Pi}]=[\hat{T}, \hat{\Pi}]+[\hat{V}, \hat{\Pi}]=-\frac{\hbar^{2}}{2 m}\left(\left[\frac{\partial^{2}}{\partial x^{2}}, \hat{\Pi}\right]+\left[\frac{\partial^{2}}{\partial y^{2}}, \hat{\Pi}\right]+\left[\frac{\partial^{2}}{\partial z^{2}}, \hat{\Pi}\right]\right)+[\hat{V}, \hat{\Pi}]
\)

Since

\(
\begin{aligned}
\hat{\Pi}\left[\frac{\partial^{2}}{\partial x^{2}} f(x, y, z)\right]=\frac{\partial}{\partial(-x)} \frac{\partial}{\partial(-x)} f(-x,-y,-z) & =\frac{\partial^{2}}{\partial x^{2}} f(-x,-y,-z) \
& =\frac{\partial^{2}}{\partial x^{2}} \hat{\Pi} f(x, y, z)
\end{aligned}
\)

where $f$ is any function, we conclude that

\(
\left[\frac{\partial^{2}}{\partial x^{2}}, \hat{\Pi}\right]=0
\)

Similar equations hold for the $y$ and $z$ coordinates, and $[\hat{H}, \hat{\Pi}]$ becomes

\(
\begin{equation<em>}
[\hat{H}, \hat{\Pi}]=[\hat{V}, \hat{\Pi}] \tag{7.58}
\end{equation</em>}
\)

Now

\(
\begin{equation<em>}
\hat{\Pi}[V(x, y, z) f(x, y, z)]=V(-x,-y,-z) f(-x,-y,-z) \tag{7.59}
\end{equation</em>}
\)

If the potential energy is an even function, that is, if $V(-x,-y,-z)=V(x, y, z)$, then (7.59) becomes

\(
\hat{\Pi}[V(x, y, z) f(x, y, z)]=V(x, y, z) f(-x,-y,-z)=V(x, y, z) \hat{\Pi} f(x, y, z)
\)

so $[\hat{V}, \hat{\Pi}]=0$. Hence, when the potential energy is an even function, the parity operator commutes with the Hamiltonian:

\(
\begin{equation<em>}
[\hat{H}, \hat{\Pi}]=0 \quad \text { if } V \text { is even } \tag{7.60}
\end{equation</em>}
\)

These results are easily extended to the $n$-particle case. For an $n$-particle system, the parity operator is defined by

\(
\begin{equation<em>}
\hat{\Pi} f\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)=f\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right) \tag{7.61}
\end{equation</em>}
\)

It is easy to see that (7.57) holds when

\(
\begin{equation<em>}
V\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)=V\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right) \tag{7.62}
\end{equation</em>}
\)

If $V$ satisfies this equation, $V$ is said to be an even function of the $3 n$ coordinates. In summary, we have:

THEOREM 7. When the potential energy $V$ is an even function, we can choose the stationary-state wave functions so that each $\psi_{i}$ is either an even function or an odd function.

A function that is either even or odd is said to be of definite parity.
If the energy levels are all nondegenerate (as is usually true in one-dimensional problems), then only one independent wave function corresponds to each energy eigenvalue and there is no element of choice (apart from an arbitrary multiplicative constant) in the wave functions. Thus, for the nondegenerate case, the stationary-state wave functions must be of definite parity when $V$ is an even function. For example, the one-dimensional harmonic oscillator has $V=\frac{1}{2} k x^{2}$, which is an even function, and the wave functions have definite parity.

The hydrogen-atom potential-energy function is even, and the hydrogenlike orbitals can be chosen to have definite parity (Probs. 7.22 and 7.29).

For the degenerate case, we have an element of choice in the wave functions, since an arbitrary linear combination of the functions corresponding to the degenerate level is an eigenfunction of $\hat{H}$. For a degenerate energy level, by taking appropriate linear combinations we can choose wave functions that are of definite parity, but there is no necessity that they be of definite parity.

Parity aids in evaluating integrals. We showed that $\int_{-\infty}^{\infty} f(x) d x=0$ when $f(x)$ is an odd function [Eq. (4.51)]. Let us extend this result to the $3 n$-dimensional case. An odd function of $3 n$ variables satisfies

\(
\begin{equation<em>}
g\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right)=-g\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right) \tag{7.63}
\end{equation</em>}
\)

If $g$ is an odd function of the $3 n$ variables, then

\(
\begin{equation<em>}
\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} g\left(x{1}, \ldots, z{n}\right) d x{1} \cdots d z{n}=0 \tag{7.64}
\end{equation</em>}
\)

where the integration is over the $3 n$ coordinates. This equation holds because the contribution to the integral from the value of $g$ at $\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)$ is canceled by the contribution from ( $-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}$ ). Equation (7.64) also holds when the integrand is an odd function of some (but not necessarily all) of the variables. See Prob. 7.30.

7.6 Measurement and the Superposition of States

Quantum mechanics can be regarded as a scheme for calculating the probabilities of the various possible outcomes of a measurement. For example, if we know the state function $\Psi(x, t)$, then the probability that a measurement at time $t$ of the particle's position yields a value between $x$ and $x+d x$ is given by $|\Psi(x, t)|^{2} d x$. We now consider measurement of the general property $B$. Our aim is to find out how to use $\Psi$ to calculate the probabilities for each possible result of a measurement of $B$. The results of this section, which tell us what information is contained in the state function $\Psi$, lie at the heart of quantum mechanics.

We shall deal with an $n$-particle system and use $q$ to symbolize the $3 n$ coordinates. We have postulated that the eigenvalues $b{i}$ of the operator $\hat{B}$ are the only possible results of a measurement of the property $B$. Using $g{i}$ for the eigenfunctions of $\hat{B}$, we have

\(
\begin{equation<em>}
\hat{B} g{i}(q)=b{i} g_{i}(q) \tag{7.65}
\end{equation</em>}
\)

We postulated in Section 7.3 that the eigenfunctions of any Hermitian operator that represents a physically observable property form a complete set. Since the $g_{i}$ 's form a complete set, we can expand the state function $\Psi$ as

\(
\begin{equation<em>}
\Psi(q, t)=\sum{i} c{i}(t) g_{i}(q) \tag{7.66}
\end{equation</em>}
\)

To allow for the change of $\Psi$ with time, the expansion coefficients $c_{i}$ vary with time.
Since $|\Psi|^{2}$ is a probability density, we require that

\(
\begin{equation<em>}
\int \Psi^{</em>} \Psi d \tau=1 \tag{7.67}
\end{equation*}
\)

Substituting (7.66) into the normalization condition and using (1.33) and (1.32), we get

\(
\begin{equation<em>}
1=\int \sum{i} c{i}^{</em>} g{i}^{*} \sum{i} c{i} g{i} d \tau=\int \sum{i} c{i}^{<em>} g_{i}^{</em>} \sum{k} c{k} g{k} d \tau=\int \sum{i} \sum{k} c{i}^{<em>} c{k} g{i}^{</em>} g_{k} d \tau \tag{7.68}
\end{equation*}
\)

Since the summation indexes in the two sums in (7.68) need not have the same value, different symbols must be used for these two dummy indexes. For example, consider the following product of two sums:

\(
\sum{i=1}^{2} s{i} \sum{i=1}^{2} t{i}=\left(s{1}+s{2}\right)\left(t{1}+t{2}\right)=s{1} t{1}+s{1} t{2}+s{2} t{1}+s{2} t{2}
\)

If we carelessly write

\(
\sum{i=1}^{2} s{i} \sum{i=1}^{2} t{i} \stackrel{(\text { wrong) })}{=} \sum{i=1}^{2} \sum{i=1}^{2} s{i} t{i}=\sum{i=1}^{2}\left(s{1} t{1}+s{2} t{2}\right)=2\left(s{1} t{1}+s{2} t_{2}\right)
\)

we get the wrong answer. The correct way to write the product is

\(
\sum{i=1}^{2} s{i} \sum{i=1}^{2} t{i}=\sum{i=1}^{2} s{i} \sum{k=1}^{2} t{k}=\sum{i=1}^{2} \sum{k=1}^{2} s{i} t{k}=\sum{i=1}^{2}\left(s{i} t{1}+s{i} t{2}\right)=s{1} t{1}+s{1} t{2}+s{2} t{1}+s{2} t_{2}
\)

which gives the right answer.
Assuming the validity of interchanging the infinite summation and the integration in (7.68), we have

\(
\sum{i} \sum{k} c{i}^{*} c{k} \int g{i}^{*} g{k} d \tau=1
\)

Since $\hat{B}$ is Hermitian, its eigenfunctions $g_{i}$ are orthonormal [Eq. (7.26)]; hence

\(
\begin{align<em>}
\sum{i} \sum{k} c_{i}^{</em>} c{k} \delta{i k} & =1 \
\sum{i}\left|c{i}\right|^{2} & =1 \tag{7.69}
\end{align*}
\)

We shall point out the significance of (7.69) shortly.
Recall the postulate (Section 3.7) that, if $\Psi$ is the normalized state function of a system, then the average value of the property $B$ is

\(
\langle B\rangle=\int \Psi^{*}(q, t) \hat{B} \Psi(q, t) d \tau
\)

Using the expansion (7.66) in the average-value expression, we have

\(
\langle B\rangle=\int \sum{i} c{i}^{<em>} g_{i}^{</em>} \hat{B} \sum{k} c{k} g{k} d \tau=\sum{i} \sum{k} c{i}^{<em>} c{k} \int g{i}^{</em>} \hat{B} g_{k} d \tau
\)

where the linearity of $\hat{B}$ was used. Use of $\hat{B} g{k}=b{k} g_{k}[E q$. (7.65)] gives

\(
\begin{gather<em>}
\langle B\rangle=\sum{i} \sum{k} c_{i}^{</em>} c{k} b{k} \int g{i}^{*} g{k} d \tau=\sum{i} \sum{k} c{i}^{*} c{k} b{k} \delta{i k} \
\langle B\rangle=\sum{i}\left|c{i}\right|^{2} b_{i} \tag{7.70}
\end{gather*}
\)

How do we interpret (7.70)? We postulated in Section 3.3 that the eigenvalues of an operator are the only possible numbers we can get when we measure the property that the operator represents. In any measurement of $B$, we get one of the values $b_{i}$ (assuming there is no experimental error). Now recall Eq. (3.81):

\(
\begin{equation<em>}
\langle B\rangle=\sum{b</em>{i}} P\left(b{i}\right) b{i} \tag{7.71}
\end{equation</em>}
\)

where $P\left(b{i}\right)$ is the probability of getting $b{i}$ in a measurement of $B$. The sum in (7.71) goes over the different eigenvalues $b{i}$, whereas the sum in (7.70) goes over the different eigenfunctions $g{i}$, since the expansion (7.66) is over the $g{i}$ 's. If there is only one independent eigenfunction for each eigenvalue, then a sum over eigenfunctions is the same as a sum over eigenvalues, and comparison of (7.71) and (7.70) shows that, when there is no degeneracy in the $\hat{B}$ eigenvalues, $\left|c{i}\right|^{2}$ is the probability of getting the value $b{i}$ in a measurement of the property $B$. Note that the $\left|c{i}\right|^{2}$ values sum to 1 , as probabilities should [Eq. (7.69)]. Suppose the eigenvalue $b{i}$ is degenerate. From (7.71), $P\left(b{i}\right)$ is given by the quantity that multiplies $b{i}$. With degeneracy, more than one term in (7.70) contains $b{i}$, so the probability $P\left(b{i}\right)$ of getting $b{i}$ in a measurement is found by adding the $\left|c{i}\right|^{2}$ values for those eigenfunctions that have the same eigenvalue $b{i}$. We have proved the following:

THEOREM 8. If $b{m}$ is a nondegenerate eigenvalue of the operator $\hat{B}$ and $g{m}$ is the corresponding normalized eigenfunction $\left(\hat{B} g{m}=b{m} g{m}\right)$, then, when the property $B$ is measured in a quantum-mechanical system whose state function at the time of the measurement is $\Psi$, the probability of getting the result $b{m}$ is given by $\left|c{m}\right|^{2}$, where $c{m}$ is the coefficient of $g{m}$ in the expansion $\Psi=\sum{i} c{i} g{i}$. If the eigenvalue $b{m}$ is degenerate, the probability of obtaining $b{m}$ when $B$ is measured is found by adding the $\left|c{i}\right|^{2}$ values for those eigenfunctions whose eigenvalue is $b{m}$.

When can the result of a measurement of $B$ be predicted with certainty? We can do this if all the coefficients in the expansion $\Psi=\sum{i} c{i} g{i}$ are zero, except one: $c{i}=0$ for all $i \neq k$ and $c{k} \neq 0$. For this case, Eq. (7.69) gives $\left|c{k}\right|^{2}=1$ and we are certain to find the result $b{k}$. In this case, the state function $\Psi=\sum{i} c{i} g{i}$ is given by $\Psi=g{k}$. When $\Psi$ is an eigenfunction of $\hat{B}$ with eigenvalue $b{k}$, we are certain to get the value $b_{k}$ when we measure $B$.

We can thus view the expansion $\Psi=\sum{i} c{i} g{i}$ [Eq. (7.66)] as expressing the general state $\Psi$ as a superposition of the eigenstates $g{i}$ of the operator $\hat{B}$. Each eigenstate $g{i}$ corresponds to the value $b{i}$ for the property $B$. The degree to which any eigenfunction $g{i}$ occurs in the expansion of $\Psi$, as measured by $\left|c{i}\right|^{2}$, determines the probability of getting the value $b_{i}$ in a measurement of $B$.

How do we calculate the expansion coefficients $c{i}$ so that we can get the probabilities $\left|c{i}\right|^{2}$ ? We multiply $\Psi=\sum{i} c{i} g{i}$ by $g{j}^{*}$, integrate over all space, and use the orthonormality of the eigenfunctions of the Hermitian operator $\hat{B}$ to get

\(
\begin{gather<em>}
\int g_{j}^{</em>} \Psi d \tau=\sum{i} c{i} \int g{j}^{*} g{i} d \tau=\sum{i} c{i} \delta{i j} \
c{j}=\int g{j}^{*} \Psi d \tau=\left\langle g{j} \mid \Psi\right\rangle \tag{7.72}
\end{gather*}
\)

The probability of finding the nondegenerate eigenvalue $b_{j}$ in a measurement of $B$ is

\(
\begin{equation<em>}
\left|c{j}\right|^{2}=\left|\int g{j}^{</em>} \Psi d \tau\right|^{2}=\left|\left\langle g_{j} \mid \Psi\right\rangle\right|^{2} \tag{7.73}
\end{equation*}
\)

where $\hat{B} g{j}=b{j} g{j}$. The quantity $\left\langle g{j} \mid \Psi\right\rangle$ is called a probability amplitude.
Thus, if we know the state of the system as determined by the state function $\Psi$, we can use (7.73) to predict the probabilities of the various possible outcomes of a measurement of any property $B$. Determination of the eigenfunctions $g{j}$ and eigenvalues $b{j}$ of $\hat{B}$ is a mathematical problem.

To determine experimentally the probability of finding $g{j}$ when $B$ is measured, we take a very large number $n$ of identical, noninteracting systems, each in the same state $\Psi$, and measure the property $B$ in each system. If $n{j}$ of the measurements yield $b{j}$, then $P\left(b{j}\right)=n{j} / n=\left|\left\langle g{j} \mid \Psi\right\rangle\right|^{2}$.

We can restate the first part of Theorem 8 as follows:
THEOREM 9. If the property $B$ is measured in a quantum-mechanical system whose state function at the time of the measurement is $\Psi$, then the probability of observing the nondegenerate $\hat{B}$ eigenvalue $b{j}$ is $\left|\left\langle g{j} \mid \Psi\right\rangle\right|^{2}$, where $g{j}$ is the normalized eigenfunction corresponding to the eigenvalue $b{j}$.

The integral $\left\langle g{j} \mid \Psi\right\rangle=\int g{j}^{<em>} \Psi d \tau$ will have a substantial absolute value if the normalized functions $g{j}$ and $\Psi$ resemble each other closely and so have similar magnitudes in each region of space. If $g{j}$ and $\Psi$ do not resemble each other, then in regions where $g{j}$ is large $\Psi$ will be small (and vice versa), so the product $g{j}^{</em>} \Psi$ will always be small and the absolute value of the integral $\int g{j}^{*} \Psi d \tau$ will be small; the probability $\left|\left\langle g{j} \mid \Psi\right\rangle\right|^{2}$ of getting $b_{j}$ will then be small.

EXAMPLE

Suppose that we measure $L{z}$ of the electron in a hydrogen atom whose state at the time the measurement begins is the $2 p{x}$ state. Give the possible outcomes of the measurement and give the probability of each outcome. Use these probabilities to calculate $\left\langle L{z}\right\rangle$ for the $2 p{x}$ state.

From (6.118),

\(
\Psi=2 p{x}=2^{-1 / 2}\left(2 p{1}\right)+2^{-1 / 2}\left(2 p_{-1}\right)
\)

This equation is the expansion of $\Psi$ as a linear combination of $\hat{L}{z}$ eigenfunctions. The only nonzero coefficients are for $2 p{1}$ and $2 p{-1}$, which are eigenfunctions of $\hat{L}{z}$ with eigenvalues $\hbar$ and $-\hbar$, respectively. (Recall that the 1 and -1 subscripts give the $m$ quantum number and that the $\hat{L}{z}$ eigenvalues are $m \hbar$.) Using Theorem 8 , we take the squares of the absolute values of the coefficients in the expansion of $\Psi$ to get the probabilities. Hence the probability for getting $\hbar$ when $L{z}$ is measured is $\left|2^{-1 / 2}\right|^{2}=0.5$, and the probability for getting $-\hbar$ is $\left|2^{-1 / 2}\right|^{2}=0.5$. To find $\left\langle L_{z}\right\rangle$ from the probabilities, we use Eq. (7.71):

\(
\left\langle L{z}\right\rangle=\sum{L<em>{z i i}} P\left(L{z, i}\right) L_{z, i}=0.5 \hbar+0.5(-\hbar)=0
\)

where $P\left(L{z, i}\right)$ is the probability of observing the value $L{z, i}$, and all probabilities except two are zero. Of course, $\left\langle L{z}\right\rangle$ can also be calculated from the average-value equation (3.88) as $\left\langle 2 p{x}\right| \hat{L}{z}\left|2 p{x}\right\rangle$.

EXERCISE Write down a hydrogen-atom wave function for which the probability of getting the $n=2$ energy if $E$ is measured is 1 , the probability of getting $2 \hbar^{2}$ if $L^{2}$ is measured is 1 , and there are equal probabilities for getting $-\hbar, 0$, and $\hbar$ if $L_{z}$ is measured. Is there only one possible answer? (Partial answer: No.)

EXAMPLE

Suppose that the energy $E$ is measured for a particle in a box of length $l$ and that at the time of the measurement the particle is in the nonstationary state $\Psi=30^{1 / 2} l^{-5 / 2} x(l-x)$ for $0 \leq x \leq l$. Give the possible outcomes of the measurement and give the probability of each possible outcome.

The possible outcomes are given by postulate (c) of Section 3.9 as the eigenvalues of the energy operator $\hat{H}$. The eigenvalues of the particle-in-a-box Hamiltonian are $E=n^{2} h^{2} / 8 m l^{2}(n=1,2,3, \ldots)$, and these are nondegenerate. The probabilities are found by expanding $\Psi$ in terms of the eigenfunctions $\psi{n}$ of $\hat{H} ; \Psi=\sum{n=1}^{\infty} c{n} \psi{n}$, where $\psi{n}=(2 / l)^{1 / 2} \sin (n \pi x / l)$. In the example after Eq. (7.41), the function $x(l-x)$ was expanded in terms of the particle-in-a-box energy eigenfunctions. The state function $30^{1 / 2} l^{-5 / 2} x(l-x)$ equals $x(l-x)$ multiplied by the normalization constant $30^{1 / 2} l^{-5 / 2}$. Hence the expansion coefficients $c{n}$ are found by multiplying the $a_{n}$ coefficients in the earlier example by $30^{1 / 2} l^{-5 / 2}$ to get

\(
c_{n}=\frac{(240)^{1 / 2}}{n^{3} \pi^{3}}\left[1-(-1)^{n}\right]
\)

The probability $P\left(E{n}\right)$ of observing the value $E{n}=n^{2} h^{2} / 8 m l^{2}$ equals $\left|c_{n}\right|^{2}$ :

\(
\begin{equation<em>}
P\left(E_{n}\right)=\frac{240}{n^{6} \pi^{6}}\left[1-(-1)^{n}\right]^{2} \tag{7.74}
\end{equation</em>}
\)

The first few probabilities are

$n$	1	2	3	4	5
$E_{n}$	$h^{2} / 8 m l^{2}$	$4 h^{2} / 8 m l^{2}$	$9 h^{2} / 8 m l^{2}$	$16 h^{2} / 8 m l^{2}$	$25 h^{2} / 8 m l^{2}$
$P\left(E_{n}\right)$	0.99855	0	0.001370	0	0.000064

The very high probability of finding the $n=1$ energy is related to the fact that the parabolic state function $30^{1 / 2} l^{-5 / 2} x(l-x)$ closely resembles the $n=1$ particle-in-a-box wave function $(2 / l)^{1 / 2} \sin (\pi x / l)$ (Fig. 7.3). The zero probabilities for $n=2,4,6, \ldots$

FIGURE 7.3 Plots of $\Psi=(30)^{1 / 2} I^{-5 / 2} x(I-x)$ and the $n=1$ particle-in-a-box wave function.

are due to the fact that, if the origin is put at the center of the box, the state function $\Psi=30^{1 / 2} l^{-5 / 2} x(l-x)$ is an even function, whereas the $n=2,4,6, \ldots$ functions are odd functions (Fig. 2.3) and so cannot contribute to the expansion of $\Psi$. The integral $\left\langle g_{n} \mid \Psi\right\rangle$ vanishes when the integrand is an odd function.

If the property $B$ has a continuous range of eigenvalues (for example, position; Section 7.7), the summation in the expansion (7.66) of $\Psi$ is replaced by an integration over the values of $b$ :

\(
\begin{equation<em>}
\Psi=\int c{b} g{b}(q) d b \tag{7.75}
\end{equation</em>}
\)

and $\left|\left\langle g_{b}(q) \mid \Psi\right\rangle\right|^{2}$ is interpreted as a probability density; that is, the probability of finding a value of $B$ between $b$ and $b+d b$ for a system in the state $\Psi$ is

\(
\begin{equation<em>}
\left|\left\langle g_{b}(q) \mid \Psi(q, t)\right\rangle\right|^{2} d b \tag{7.76}
\end{equation</em>}
\)

We derived the eigenfunctions of the linear-momentum and angular-momentum operators. We now ask: What are the eigenfunctions of the position operator $\hat{x}$ ?

The operator $\hat{x}$ is multiplication by $x$. Denoting the position eigenfunctions by $g_{a}(x)$, we write

\(
\begin{equation}
x g{a}(x)=a g{a}(x) \tag{7.77}
\end{equation}
\)

where $a$ symbolizes the possible eigenvalues. It follows that

\(
\begin{equation}
(x-a) g_{a}(x)=0 \tag{7.78}
\end{equation}
\)

We conclude from (7.78) that

\(
\begin{equation}
g_{a}(x)=0 \quad \text { for } x \neq a \tag{7.79}
\end{equation}
\)

Moreover, since an eigenfunction that is zero everywhere is unacceptable, we have

\(
\begin{equation}
g_{a}(x) \neq 0 \quad \text { for } x=a \tag{7.80}
\end{equation}
\)

These conclusions make sense. If the state function is an eigenfunction of $\hat{x}$ with eigenvalue $a, \Psi=g_{a}(x)$, we know (Section 7.6) that a measurement of $x$ is certain to give the value $a$. This can be true only if the probability density $|\Psi|^{2}$ is zero for $x \neq a$, in agreement with (7.79).

Before considering further properties of $g_{a}(x)$, we define the Heaviside step function $H(x)$ by (see Fig. 7.4)

\(
\begin{array}{ll}
H(x)=1 & \text { for } x>0 \
H(x)=0 & \text { for } x<0 \tag{7.81}\
H(x)=\frac{1}{2} & \text { for } x=0
\end{array}
\)

We next define the Dirac delta function $\delta(x)$ as the derivative of the Heaviside step function:

\(
\begin{equation}
\delta(x) \equiv \frac{d H(x)}{d x} \tag{7.82}
\end{equation}
\)

From (7.81) and (7.82), we have at once (see also Fig. 7.4)

\(
\begin{equation}
\delta(x)=0 \quad \text { for } x \neq 0 \tag{7.83}
\end{equation}
\)

FIGURE 7.4 The Heaviside step function.

Since $H(x)$ makes a sudden jump at $x=0$, its derivative is infinite at the origin:

\(
\begin{equation}
\delta(x)=\infty \quad \text { for } x=0 \tag{7.84}
\end{equation}
\)

We can generalize these equations slightly by setting $x=t-a$ and then changing the symbol $t$ to $x$. Equations (7.81) to (7.84) become

\(
\begin{gather}
H(x-a)=1, \quad x>a \tag{7.85}\
H(x-a)=0, \quad x<a \tag{7.86}\
H(x-a)=\frac{1}{2}, \quad x=a \tag{7.87}\
\delta(x-a)=d H(x-a) / d x \tag{7.88}\
\delta(x-a)=0, \quad x \neq a \quad \text { and } \quad \delta(x-a)=\infty, \quad x=a \tag{7.89}
\end{gather}
\)

Now consider the following integral:

\(
\int_{-\infty}^{\infty} f(x) \delta(x-a) d x
\)

We evaluate it using integration by parts:

\(
\begin{gathered}
\int u d v=u v-\int v d u \
u=f(x), \quad d v=\delta(x-a) d x
\end{gathered}
\)

Using (7.88), we have

\(
\begin{gather}
d u=f^{\prime}(x) d x, \quad v=H(x-a) \
\int{-\infty}^{\infty} f(x) \delta(x-a) d x=\left.f(x) H(x-a)\right|{-\infty} ^{\infty}-\int{-\infty}^{\infty} H(x-a) f^{\prime}(x) d x \
\int{-\infty}^{\infty} f(x) \delta(x-a) d x=f(\infty)-\int_{-\infty}^{\infty} H(x-a) f^{\prime}(x) d x \tag{7.90}
\end{gather}
\)

where (7.85) and (7.86) were used. Since $H(x-a)$ vanishes for $x<a$, (7.90) becomes

\(
\begin{align}
& \int{-\infty}^{\infty} f(x) \delta(x-a) d x=f(\infty)-\int{a}^{\infty} H(x-a) f^{\prime}(x) d x \
&=f(\infty)-\int{a}^{\infty} f^{\prime}(x) d x=f(\infty)-\left.f(x)\right|{a} ^{\infty} \
& \int_{-\infty}^{\infty} f(x) \delta(x-a) d x=f(a) \tag{7.91}
\end{align}
\)

Comparing (7.91) with the equation $\sum{k} c{k} \delta{i k}=c{i}$, we see that the Dirac delta function plays the same role in an integral that the Kronecker delta plays in a sum. The special case of (7.91) with $f(x)=1$ is

\(
\int_{-\infty}^{\infty} \delta(x-a) d x=1
\)

The properties (7.89) of the Dirac delta function agree with the properties (7.79) and (7.80) of the position eigenfunctions $g_{a}(x)$. We therefore tentatively set

\(
\begin{equation}
g_{a}(x)=\delta(x-a) \tag{7.92}
\end{equation}
\)

To verify (7.92), we now show it to be in accord with the Born postulate that $|\Psi(a, t)|^{2} d a$ is the probability of observing a value of $x$ between $a$ and $a+d a$. According to (7.76), this probability is given by

\(
\begin{equation}
\left|\left\langle g{a}(x) \mid \Psi(x, t)\right\rangle\right|^{2} d a=\left|\int{-\infty}^{\infty} g_{a}^{}(x) \Psi(x, t) d x\right|^{2} d a \tag{7.93}
\end{}
\)

Using (7.92) and then (7.91), we have for (7.93)

\(
\left|\int_{-\infty}^{\infty} \delta(x-a) \Psi(x, t) d x\right|^{2} d a=|\Psi(a, t)|^{2} d a
\)

which completes the proof.
Since the quantity $a$ in $\delta(x-a)$ can have any real value, the eigenvalues of $\hat{x}$ form a continuum: $-\infty<a<\infty$. As usual for continuum eigenfunctions, $\delta(x-a)$ is not quadratically integrable (Prob. 7.43).

Summarizing, the eigenfunctions and eigenvalues of position are

\(
\begin{equation}
\hat{x} \delta(x-a)=a \delta(x-a) \tag{7.94}
\end{equation}
\)

where $a$ is any real number.
The delta function is badly behaved, and consequently the manipulations we performed are lacking in rigor and would make a mathematician shudder. However, one can formulate things rigorously by considering the delta function to be the limiting case of a function that becomes successively more peaked at the origin (Fig. 7.5). The delta function is not really a function but is what mathematicians call a distribution (see en.wikipedia .org/wiki/Dirac_delta_function).

FIGURE 7.5 Functions that approximate $\delta(x)$ with successively increasing accuracy. The area under each curve is 1 .

This section summarizes the postulates of quantum mechanics introduced in previous chapters.

POSTULATE 1. The state of a system is described by a function $\Psi$ of the coordinates of the particles and the time. This function, called the state function or wave function, contains all the information that can be determined about the system. We further postulate that $\Psi$ is single-valued, continuous, and quadratically integrable. For continuum states, the quadratic integrability requirement is omitted.

The designation "wave function" for $\Psi$ is perhaps not the best choice. A physical wave moving in three-dimensional space is a function of the three spatial coordinates and the time. However, for an $n$-particle system, the function $\Psi$ is a function of $3 n$ spatial coordinates and the time. Hence, for a many-particle system, we cannot interpret $\Psi$ as any sort of physical wave. The state function is best thought of as a function from which we can calculate various properties of the system. The nature of the information that $\Psi$ contains is the subject of Postulate 5 and its consequences.

POSTULATE 2. To every physically observable property there corresponds a linear Hermitian operator. To find this operator, write down the classical-mechanical expression for the observable in terms of Cartesian coordinates and corresponding linearmomentum components, and then replace each coordinate $x$ by the operator $x$. and each momentum component $p_{x}$ by the operator $-i \hbar \partial / \partial x$.

We saw in Section 7.2 that the restriction to Hermitian operators arises from the requirement that average values of physical quantities be real numbers. The requirement of linearity is closely connected to the superposition of states discussed in Section 7.6. In our derivation of (7.70) for the average value of a property $B$ for a state that was expanded as a superposition of the eigenfunctions of $\hat{B}$, the linearity of $\hat{B}$ played a key role.

When the classical quantity contains a product of a Cartesian coordinate and its conjugate momentum, we run into the problem of noncommutativity in constructing the correct quantum-mechanical operator. Several different rules have been proposed to handle this case. See J. R. Shewell, Am. J. Phys., 27, 16 (1959); E. H. Kerner and W. G. Sutcliffe, J. Math. Phys., 11, 391 (1970); A. de Souza Dutra, J. Phys. A: Math. Gen., 39, 203 (2006) (arxiv.org/abs/0705.3247).

The process of finding quantum-mechanical operators in non-Cartesian coordinates is complicated. See K. Simon, Am. J. Phys., 33, 60 (1965); G. R. Gruber, Found. Phys., 1, 227 (1971).

POSTULATE 3. The only possible values that can result from measurements of the physically observable property $B$ are the eigenvalues $b{i}$ in the equation $\hat{B} g{i}=b{i} g{i}$, where $\hat{B}$ is the operator corresponding to the property $B$. The eigenfunctions $g_{i}$ are required to be well behaved.

Our main concern is with the energy levels of atoms and molecules. These are given by the eigenvalues of the energy operator, the Hamiltonian $\hat{H}$. The eigenvalue equation for $\hat{H}, \hat{H} \psi=E \psi$, is the time-independent Schrödinger equation. However, finding the possible values of any property involves solving an eigenvalue equation.

POSTULATE 4. If $\hat{B}$ is a linear Hermitian operator that represents a physically observable property, then the eigenfunctions $g_{i}$ of $\hat{B}$ form a complete set.

There are Hermitian operators whose eigenfunctions do not form a complete set (see Griffiths, pp. 99, 106; Messiah, p. 188; Ballentine, Sec. 1.3). The completeness requirement is essential to developing the theory of quantum mechanics, so it is necessary to postulate that all Hermitian operators that correspond to observable properties have a complete set of eigenfunctions. Postulate 4 allows us to expand the wave function for any state as a superposition of the orthonormal eigenfunctions of any quantum-mechanical operator:

\(
\begin{equation}
\Psi=\sum{i} c{i} g{i}=\sum{i}\left|g{i}\right\rangle\left\langle g{i} \mid \Psi\right\rangle \tag{7.95}
\end{equation}
\)

POSTULATE 5. If $\Psi(q, t)$ is the normalized state function of a system at time $t$, then the average value of a physical observable $B$ at time $t$ is

\(
\begin{equation}
\langle B\rangle=\int \Psi^{} \hat{B} \Psi d \tau \tag{7.96}
\end{}
\)

The definition of the quantum-mechanical average value is given in Section 3.7 and should not be confused with the time average used in classical mechanics.

From Postulates 4 and 5, we showed in Section 7.6 that the probability of observing the nondegenerate eigenvalue $b{i}$ in a measurement of $B$ is $P\left(b{i}\right)=\left|\int g{i}^{*} \Psi d \tau\right|^{2}=$ $\left|\left\langle g{i} \mid \Psi\right\rangle\right|^{2}$, where $\hat{B} g{i}=b{i} g{i}$. If $\Psi$ happens to be one of the eigenfunctions of $\hat{B}$, that is, if $\Psi=g{k}$, then $P\left(b{i}\right)$ becomes $P\left(b{i}\right)=\left|\int g{i}^{*} g{k} d \tau\right|^{2}=\left|\delta{i k}\right|^{2}=\delta{i k}$, where the orthonormality of the eigenfunctions of the Hermitian operator $\hat{B}$ was used. We are certain to observe the value $b{k}$ when $\Psi=g{k}$.

POSTULATE 6. The time development of the state of an undisturbed quantum-mechanical system is given by the Schrödinger time-dependent equation

\(
\begin{equation}
-\frac{\hbar}{i} \frac{\partial \Psi}{\partial t}=\hat{H} \Psi \tag{7.97}
\end{equation}
\)

where $\hat{H}$ is the Hamiltonian (that is, energy) operator of the system.

The time-dependent Schrödinger equation is a first-order differential equation in the time, so that, just as in classical mechanics, the present state of an undisturbed system
determines the future state. However, unlike knowledge of the state in classical mechanics, knowledge of the state in quantum mechanics involves knowledge of only the probabilities for various possible outcomes of a measurement. Thus, suppose we have several identical noninteracting systems, each having the same state function $\Psi\left(t{0}\right)$ at time $t{0}$. If we leave each system undisturbed, then the state function for each system will change in accord with (7.97). Since each system has the same Hamiltonian, each system will have the same state function $\Psi\left(t{1}\right)$ at any future time $t{1}$. However, suppose that at time $t{2}$ we measure property $B$ in each system. Although each system has the same state function $\Psi\left(t{2}\right)$ at the instant the measurement begins, we will not get the same result for each system. Rather, we will get a spread of possible values $b{i}$, where $b{i}$ are the eigenvalues of $\hat{B}$. The relative number of times we get each $b{i}$ can be calculated from the quantities $\left|c{i}\right|^{2}$, where $\Psi\left(t{2}\right)=\sum{i} c{i} g{i}$ with the $g_{i}$ 's being the eigenfunctions of $\hat{B}$.

If the Hamiltonian is independent of time, we have the possibility of states of definite energy $E$. For such states the state function must satisfy

\(
\begin{equation}
\hat{H} \Psi=E \Psi \tag{7.98}
\end{equation}
\)

and the time-dependent Schrödinger equation becomes

\(
-\frac{\hbar}{i} \frac{\partial \Psi}{\partial t}=E \Psi
\)

which integrates to $\Psi=A e^{-i E t / \hbar}$, where $A$, the integration "constant," is independent of time. The function $\Psi$ depends on the coordinates and the time, so $A$ is some function of the coordinates, which we designate as $\psi(q)$. We have

\(
\begin{equation}
\Psi(q, t)=e^{-i E t / \hbar} \psi(q) \tag{7.99}
\end{equation}
\)

for a state of constant energy. The function $\psi(q)$ satisfies the time-independent Schrödinger equation

\(
\hat{H} \psi(q)=E \psi(q)
\)

which follows from (7.98) and (7.99). The factor $e^{-i E t / \hbar}$ simply indicates a change in the phase of the wave function $\Psi(q, t)$ with time and has no direct physical significance. Hence we generally refer to $\psi(q)$ as "the wave function." The Hamiltonian operator plays a unique role in quantum mechanics in that it occurs in the fundamental dynamical equation, the time-dependent Schrödinger equation. The eigenstates of $\hat{H}$ (known as stationary states) have the special property that the probability density $|\Psi|^{2}$ is independent of time.

The time-dependent Schrödinger equation (7.97) is $(i \hbar \partial / \partial t-\hat{H}) \Psi=0$. Because the operator $i \hbar \partial / \partial t-\hat{H}$ is linear, any linear combination of solutions of the time-dependent Schrödinger equation (7.97) is a solution of (7.97). For example, if the Hamiltonian $\hat{H}$ is independent of time, then there exist stationary-state solutions $\Psi{n}=e^{-i E{n} t / \hbar} \psi_{n}(q)$ [Eq. (7.99)] of the time-dependent Schrödinger equation. Any linear combination

\(
\begin{equation}
\Psi=\sum{n} c{n} \Psi{n}=\sum{n} c{n} e^{-i E{n} t \hbar} \psi_{n}(q) \tag{7.100}
\end{equation}
\)

where the $c{n}$ 's are time-independent constants is a solution of the time-dependent Schrödinger equation, although it is not an eigenfunction of $\hat{H}$. Because of the completeness of the eigenfunctions $\psi{n}$, any state function can be written in the form (7.100) if $\hat{H}$ is independent of time. (See also Section 9.8.) The state function (7.100) represents a state that does not have a definite energy. Rather, when we measure the energy, the probability of getting $E{n}$ is $\left|c{n} e^{-i E{n} t / \hbar}\right|^{2}=\left|c{n}\right|^{2}$.

To find the constants $c{n}$ in (7.100), we write (7.100) at time $t{0}$ as $\Psi\left(q, t{0}\right)=$ $\sum{n} c{n} e^{-i E{n} t{0} / \hbar} \psi{n}(q)$. Multiplication by $\psi_{j}^{*}(q)$ followed by integration over all space gives

\(
\left\langle\psi{j}(q) \mid \Psi\left(q, t{0}\right)\right\rangle=\sum{n} c{n} e^{-i E{n} t{0} / \hbar}\left\langle\psi{j} \mid \psi{n}\right\rangle=\sum{n} c{n} e^{-i E{n} t{0} / \hbar} \delta{j n}=c{j} e^{-i E{j} t{0} / \hbar}
\)

so $c{j}=\left\langle\psi{j} \mid \Psi\left(q, t{0}\right)\right\rangle e^{i E{j} t_{0} / \hbar}$ and (7.100) becomes

\(
\begin{equation}
\Psi(q, t)=\sum{j}\left\langle\psi{j}(q) \mid \Psi\left(q, t{0}\right)\right\rangle e^{-i E{j}\left(t-t{0}\right) / \hbar} \psi{j}(q) \quad \text { if } \hat{H} \text { ind. of } t \tag{7.101}
\end{equation}
\)

where $\hat{H} \psi{j}=E{j} \psi{j}$ and the $\psi{j}$ 's have been chosen to be orthonormal. Equation (7.101) tells us how to find $\Psi$ at time $t$ from $\Psi$ at an initial time $t_{0}$ and is the general solution of the time-dependent Schrödinger equation when $\hat{H}$ is independent of $t$. [Equation (7.101) can also be derived directly from the time-dependent Schrödinger equation; see Prob. 7.46.]

EXAMPLE

A particle in a one-dimensional box of length $l$ has a time-independent Hamiltonian and has the state function $\Psi=2^{-1 / 2} \psi{1}+2^{-1 / 2} \psi{2}$ at time $t=0$, where $\psi{1}$ and $\psi{2}$ are particle-in-a-box time-independent energy eigenfunctions [Eq. (2.23)] with $n=1$ and $n=2$, respectively. (a) Find the probability density as a function of time. (b) Show that $|\Psi|^{2}$ oscillates with a period $T=8 m l^{2} / 3 h$. (c) Use a spreadsheet or Mathcad to plot $l|\Psi|^{2}$ versus $x / l$ at each of the times $j T / 8$, where $j=0,1,2, \ldots, 8$.
(a) Since $\hat{H}$ is independent of time, $\Psi$ at any future time will be given by (7.100) with $c{1}=2^{-1 / 2}, c{2}=2^{-1 / 2}$, and all other c's equal to zero. Therefore,

\(
\begin{aligned}
\Psi & =\frac{1}{\sqrt{2}} e^{-i E{1} t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \frac{\pi x}{l}+\frac{1}{\sqrt{2}} e^{-i E{2} t / \hbar}\left(\frac{2}{l}\right)^{1 / 2} \sin \frac{2 \pi x}{l} \
& =\frac{1}{\sqrt{2}} e^{-i E{1} t / \hbar} \psi{1}+\frac{1}{\sqrt{2}} e^{-i E{2} t / \hbar} \psi{2}
\end{aligned}
\)

We find for the probability density (Prob. 7.47)

\(
\begin{equation}
\Psi^{} \Psi=\frac{1}{2} \psi{1}^{2}+\frac{1}{2} \psi{2}^{2}+\psi{1} \psi{2} \cos \left[\left(E{2}-E{1}\right) t / \hbar\right] \tag{7.102}
\end{}
\)

(b) The time-dependent part of $|\Psi|^{2}$ is the cosine factor in (7.102). The period $T$ is the time it takes for the cosine to increase by $2 \pi$, so $\left(E{2}-E{1}\right) T / \hbar=2 \pi$ and $T=2 \pi \hbar /\left(E{2}-E{1}\right)=8 m l^{2} / 3 h$, since $E{n}=n^{2} h^{2} / 8 m l^{2}$. (c) Using (7.102), the expressions for $\psi{1}$ and $\psi{2}$, and $T=2 \pi \hbar /\left(E{2}-E{1}\right)$, we have
$l|\Psi|^{2}=\sin ^{2}\left(\pi x{r}\right)+\sin ^{2}\left(2 \pi x{r}\right)+2 \sin \left(\pi x{r}\right) \sin \left(2 \pi x{r}\right) \cos (2 \pi t / T)$
where $x{r} \equiv x / l$. With $t=j T / 8$, the graphs are easily plotted for each $j$ value. The plots show that the probability-density maximum oscillates between the left and right sides of the box. Using Mathcad, one can produce a movie of $|\Psi|^{2}$ as time changes (Prob. 7.47). (An online resource that allows one to follow $|\Psi|^{2}$ as a function of time for systems such as the particle in a box or the harmonic oscillator for any chosen initial mixture of stationary states is at www.falstad.com/qm1d; one chooses the mixture by clicking on the small circles at the bottom and dragging on each rotating arrow within a circle.)

Equation (7.100) with the $c{n}$ 's being constant is the general solution of the timedependent Schrödinger equation when $\hat{H}$ is independent of time. For a system acted on by an external time-dependent force, the Hamiltonian contains a time-dependent part: $\hat{H}=\hat{H}^{0}+\hat{H}^{\prime}(t)$, where $\hat{H}^{0}$ is the Hamiltonian of the system in the absence of the external force and $\hat{H}^{\prime}(t)$ is the time-dependent potential energy of interaction of the system with the external force. In this case, we can use the stationary-state time-independent eigenfunctions of $\hat{H}^{0}$ to expand $\Psi$ in an equation like (7.100), except that now the $c{n}$ 's depend on $t$. An example is an atom or molecule exposed to the time-dependent electric field of electromagnetic radiation (light); see Section 9.8.

What determines whether a system is in a stationary state such as (7.99) or a nonstationary state such as (7.100)? The answer is that the history of the system determines its present state. For example, if we take a system that is in a stationary state and expose it to radiation, the time-dependent Schrödinger equation shows that the radiation causes the state to change to a nonstationary state; see Section 9.8.

You might be wondering about the absence from the list of postulates of the Born postulate that $|\Psi(x, t)|^{2} d x$ is the probability of finding the particle between $x$ and $x+d x$. This postulate is a consequence of Postulate 5, as we now show. Equation (3.81) is $\langle B\rangle=\sum{b} P{b} b$, where $P{b}$ is the probability of observing the value $b$ in a measurement of the property $B$ that takes on discrete values. The corresponding equation for the continuous variable $x$ is $\langle x\rangle=\int{-\infty}^{\infty} P(x) x d x$, where $P(x)$ is the probability density for observing various values of $x$. According to Postulate 5, we have $\langle x\rangle=\int{-\infty}^{\infty} \Psi^{*} \hat{x} \Psi d x=\int{-\infty}^{\infty}|\Psi|^{2} x d x$. Comparison of these two expressions for $\langle x\rangle$ shows that $|\Psi|^{2}$ is the probability density $P(x)$.

Chapter 10 gives two further quantum-mechanical postulates that deal with spin and the spin-statistics theorem.

In quantum mechanics, the state function of a system changes in two ways. [See E. P. Wigner, Am. J. Phys., 31, 6 (1963).] First, there is the continuous, causal change with time given by the time-dependent Schrödinger equation (7.97). Second, there is the sudden, discontinuous, probabilistic change that occurs when a measurement is made on the system. This kind of change cannot be predicted with certainty, since the result of a measurement cannot be predicted with certainty; only the probabilities (7.73) are predictable. The sudden change in $\Psi$ caused by a measurement is called the reduction (or collapse) of the wave function. A measurement of the property $B$ that yields the result $b{k}$ changes the state function to $g{k}$, the eigenfunction of $\hat{B}$ whose eigenvalue is $b{k}$. (If $b{k}$ is degenerate, $\Psi$ is changed to a linear combination of the eigenfunctions corresponding to $b{k}$.) The probability of finding the nondegenerate eigenvalue $b{k}$ is given by Eq. (7.73) and Theorem 9 as $\left|\left\langle g{k} \mid \Psi\right\rangle\right|^{2}$, so the quantity $\left|\left\langle g{k} \mid \Psi\right\rangle\right|^{2}$ is the probability the system will make a transition from the state $\Psi$ to the state $g_{k}$ when $B$ is measured.

Consider an example. Suppose that at time $t$ we measure a particle's position. Let $\Psi\left(x, t_{-}\right)$be the state function of the particle the instant before the measurement is made (Fig. 7.6a). We further suppose that the result of the measurement is that the particle is found to be in the small region of space

\(
\begin{equation}
a<x<a+d a \tag{7.104}
\end{equation}
\)

We ask: What is the state function $\Psi\left(x, t{+}\right)$the instant after the measurement? To answer this question, suppose we were to make a second measurement of position at time $t{+}$.

FIGURE 7.6 Reduction of the wave function caused by a measurement of position.

Since $t{+}$differs from the time $t$ of the first measurement by an infinitesimal amount, we must still find that the particle is confined to the region (7.104). If the particle moved a finite distance in an infinitesimal amount of time, it would have infinite velocity, which is unacceptable. Since $\left|\Psi\left(x, t{+}\right)\right|^{2}$ is the probability density for finding various values of $x$, we conclude that $\Psi\left(x, t{+}\right)$must be zero outside the region (7.104) and must look something like Fig. 7.6b. Thus the position measurement at time $t$ has reduced $\Psi$ from a function that is spread out over all space to one that is localized in the region (7.104). The change from $\Psi\left(x, t{-}\right)$to $\Psi\left(x, t_{+}\right)$is a probabilistic change.

The measurement process is one of the most controversial areas in quantum mechanics. Just how and at what stage in the measurement process reduction occurs is unclear. Some physicists take the reduction of $\Psi$ as an additional quantum-mechanical postulate, while others claim it is a theorem derivable from the other postulates. Some physicists reject the idea of reduction [see M. Jammer, The Philosophy of Quantum Mechanics, Wiley, 1974, Section 11.4; L. E. Ballentine, Am. J. Phys., 55, 785 (1987)]. Ballentine advocates Einstein's statistical-ensemble interpretation of quantum mechanics, in which the wave function does not describe the state of a single system (as in the orthodox interpretation) but gives a statistical description of a collection of a large number of systems each prepared in the same way (an ensemble). In this interpretation, the need for reduction of the wave function does not occur. [See L. E. Ballentine, Am. J. Phys., 40, 1763 (1972); Rev. Mod. Phys., 42, 358 (1970).] There are many serious problems with the statistical-ensemble interpretation [see Whitaker, pp. 213-217; D. Home and M. A. B. Whitaker, Phys. Rep., 210, 223 (1992); Prob. 10.4], and this interpretation has been largely rejected.
"For the majority of physicists the problem of finding a consistent and plausible quantum theory of measurement is still unsolved. . . The immense diversity of opinion . . . concerning quantum measurements . . . [is] a reflection of the fundamental disagreement as to the interpretation of quantum mechanics as a whole" (M. Jammer, The Philosophy of Quantum Mechanics, pp. 519, 521).

The probabilistic nature of quantum mechanics has disturbed many physicists, including Einstein, de Broglie, and Schrödinger. These physicists and others have suggested that quantum mechanics may not furnish a complete description of physical reality. Rather, the probabilistic laws of quantum mechanics might be simply a reflection of deterministic laws that operate at a subquantum-mechanical level and that involve "hidden variables." An analogy given by the physicist Bohm is the Brownian motion of a dust particle in air. The particle undergoes random fluctuations of position, and its motion is not completely determined by its position and velocity. Of course, Brownian motion is a result of collisions with the gas molecules and is determined by variables existing on the level of
molecular motion. Analogously, the motions of electrons might be determined by hidden variables existing on a subquantum-mechanical level. The orthodox interpretation (often called the Copenhagen interpretation) of quantum mechanics, which was developed by Heisenberg and Bohr, denies the existence of hidden variables and asserts that the laws of quantum mechanics provide a complete description of physical reality. (Hidden-variables theories are discussed in F. J. Belinfante, A Survey of Hidden-Variables Theories, Pergamon, 1973.)

In 1964, J. S. Bell proved that, in certain experiments involving measurements on two widely separated particles that originally were in the same region of space, any possible local hidden-variable theory must make predictions that differ from those that quantum mechanics makes (see Ballentine, Chapter 20). In a local theory, two systems very far from each other act independently of each other. The results of such experiments agree with quantum-mechanical predictions, thus providing very strong evidence against all deterministic, local hidden-variable theories but do not rule out nonlocal hidden-variable theories. These experiments are described in A. Aspect, in The Wave-Particle Dualism, S. Diner et al. (eds.), Reidel, 1984, pp. 377-390; A. Shimony, Scientific American, Jan. 1988, p. 46; A. Zeilinger, Rev. Mod. Phys, 71, S288 (1999).

Further analysis by Bell and others shows that the results of these experiments and the predictions of quantum mechanics are incompatible with a view of the world in which both realism and locality hold. Realism (also called objectivity) is the doctrine that external reality exists and has definite properties independent of whether or not we observe this reality. Locality excludes instantaneous action-at-a-distance and asserts that any influence from one system to another must travel at a speed that does not exceed the speed of light. Clauser and Shimony stated that quantum mechanics leads to the "philosophically startling" conclusion that we must either "totally abandon the realistic philosophy of most working scientists, or dramatically revise our concept of space-time" to permit "some kind of action-at-a-distance" [J. F. Clauser and A. Shimony, Rep. Prog. Phys., 41, 1881 (1978); see also B. d’Espagnat, Scientific American, Nov. 1979, p. 158; A. Aspect, Nature, 446, 866 (2007); S. Gröblacher et al., Nature, 446, 871 (2007)].

Quantum theory predicts and experiments confirm that when measurements are made on two particles that once interacted but now are separated by an unlimited distance the results obtained in the measurement on one particle depend on the results obtained from the measurement on the second particle and also depend on which property of the second particle is measured. (Such particles are said to be entangled. For more on entanglement, see en.wikipedia.org/wiki/Quantum_entanglement; chaps. 7-10 of J. Baggott, Beyond Measure, Oxford, 2004; L. Gilder, The Age of Entanglement, Vintage, 2009; Part II of A. Whitaker, The New Quantum Age, Oxford, 2012.) Such instantaneous "spooky actions at a distance" (Einstein's phrase) have led one physicist to remark that "quantum mechanics is magic" (D. Greenberger, quoted in N. D. Mermin, Physics Today, April 1985, p. 38).

The relation between quantum mechanics and the mind has been the subject of much speculation. Wigner argued that the reduction of the wave function occurs when the result of a measurement enters the consciousness of an observer and thus "the being with consciousness must have a different role in quantum mechanics than the inanimate measuring device." He believed it likely that conscious beings obey different laws of nature than inanimate objects and proposed that scientists look for unusual effects of consciousness acting on matter. [E. P. Wigner, "Remarks on the Mind-Body Question," in The Scientist Speculates, I. J. Good, ed., Capricorn, 1965, p. 284; Proc. Amer. Phil. Soc., 113, 95 (1969); Found. Phys., 1, 35 (1970).]

In 1952, David Bohm (following a suggestion made by de Broglie in 1927 that the wave function might act as a pilot wave guiding the motion of the particle) devised a nonlocal deterministic hidden-variable theory that predicts the same experimental results as quantum mechanics [D. Bohm, Phys. Rev., 85, 166, 180 (1952)]. In Bohm's theory, a
particle at any instant of time possesses both a definite position and a definite momentum (although these quantities are not observable), and it travels on a definite path. The particle also possesses a wave function $\Psi$ whose time development obeys the time-dependent Schrödinger equation. In Bohm's theory, the wave function is a real physical entity that determines the motion of the particle. If we are given a particle at a particular position with a particular wave function at a particular time $t$, Bohm's theory postulates a certain equation that allows us to calculate the velocity of the particle at that time from its wave function and position; knowing the position and velocity at $t$, we can find the position at time $t+d t$ and can use the time-dependent Schrödinger equation to find the wave function at $t+d t$; then we calculate the velocity at $t+d t$ from the position and wave function at $t+d t$; and so on. Hence the path can be calculated from the initial position and wave function (assuming we know the potential energy). In Bohm's theory, the particle's position turns out to obey an equation like Newton's second law $m d^{2} x / d t^{2}=-\partial V / \partial x$ [Eqs. (1.8) and (1.12)], except that the potential energy $V$ is replaced by $V+Q$, where $Q$ is a quantum potential that is calculated in a certain way from the wave function. In Bohm's theory, collapse of the wave function does not occur. Rather, the interaction of the system with the measuring apparatus follows the equations of Bohm's theory, but this interaction leads to the system evolving after the measurement in the manner that would occur if the wave function had been collapsed.

Bohm's work was largely ignored for many years, but interest in his theory has increased. For more on Bohm's theory, see Whitaker, Chapter 7; D. Bohm and B. J. Hiley, The Undivided Universe, Routledge, 1992; D. Z Albert, Scientific American, May 1994, p. 58; S. Goldstein, "Bohmian Mechanics," plato.stanford.edu/entries/qm-bohm; en.wikipedia .org/wiki/De_Broglie-Bohm_theory. One of the main characters in Rebecca Goldstein's novel Properties of Light (Houghton Mifflin, 2000) is modeled in part on David Bohm.

Some physicists argue that the wave function represents merely our knowledge about the state of the system (this epistemic interpretation is used in the Copenhagen viewpoint), whereas others argue that the wave function corresponds directly to an element of physical reality (the ontic interpretation). A paper published in 2012 used certain mild assumptions to prove a result that the authors argued strongly favored the ontic interpretation; M. F. Pusey, J. Barrett, and T. Rudolph, Nature Phys. 8, 476 (2012) (arxiv.org/abs/1111.3328). For discussion of this result (called the PBR theorem), see www.aps.org/units/gqi/newsletters/upload/vol6num3.pdf; mattleifer.info/2011/11/20/ can-the-quantum-state-be-interpreted-statistically.

Although the experimental predictions of quantum mechanics are not arguable, its conceptual interpretation is still the subject of heated debate. Excellent bibliographies with commentary on this subject are B. S. DeWitt and R. N. Graham, Am. J. Phys., 39, 724 (1971); L. E. Ballentine, Am. J. Phys., 55, 785 (1987). See also B. d'Espagnat, Conceptual Foundations of Quantum Mechanics, 2nd ed., Benjamin, 1976; M. Jammer, The Philosophy of Quantum Mechanics, Wiley, 1974; Whitaker, Chapter 8; P. Yam, Scientific American, June 1997, p. 124. An online bibliography by A. Cabello on the foundations of quantum mechanics lists 12 different interpretations of quantum mechanics (arxiv.org/ abs/quant-ph/0012089); a Wikipedia article lists 14 interpretations of quantum mechanics (en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics).

Matrix algebra is a key mathematical tool in doing modern-day quantum-mechanical calculations on molecules. Matrices also furnish a convenient way to formulate much of the theory of quantum mechanics. Matrix methods will be used in some later chapters, but this book is written so that the material on matrices can be omitted if time does not allow this material to be covered.

A matrix is a rectangular array of numbers. The numbers that compose a matrix are called the matrix elements. Let the matrix A have $m$ rows and $n$ columns, and let $a_{i j}$ $(i=1,2, \ldots, m$ and $j=1,2, \ldots, n)$ denote the element in row $i$ and column $j$. Then

\(
\mathbf{A}=\left(\begin{array}{cccc}
a{11} & a{12} & \cdots & a{1 n} \
a{21} & a{22} & \cdots & a{2 n} \
\cdot & \cdot & \cdots & \cdot \
a{m 1} & a{m 2} & \cdots & a_{m n}
\end{array}\right)
\)

$\mathbf{A}$ is said to be an $m$ by $n$ matrix. Do not confuse $\mathbf{A}$ with a determinant (Section 8.3); a matrix need not be square and is not equal to a single number.

A row matrix (also called a row vector) is a matrix having only one row. A column matrix or column vector has only one column.

Two matrices $\mathbf{R}$ and $\mathbf{S}$ are equal if they have the same number of rows, and the same number of columns, and have corresponding elements equal. If $\mathbf{R}=\mathbf{S}$, then $r{j k}=s{j k}$ for $j=1, \ldots, m$ and $k=1, \ldots, n$, where $m$ and $n$ are the dimensions of $\mathbf{R}$ and $\mathbf{S}$. A matrix equation is thus equivalent to $m n$ scalar equations.

The sum of two matrices $\mathbf{A}$ and $\mathbf{B}$ is defined as the matrix formed by adding corresponding elements of $\mathbf{A}$ and $\mathbf{B}$; the sum is defined only if $\mathbf{A}$ and $\mathbf{B}$ have the same dimensions. If $\mathbf{P}=\mathbf{A}+\mathbf{B}$, then we have the $m n$ scalar equations $p{j k}=a{j k}+b_{j k}$ for $j=1, \ldots, m$ and $k=1, \ldots, n$.

\(
\begin{equation}
\text { If } \mathbf{P}=\mathbf{A}+\mathbf{B}, \text { then } p{j k}=a{j k}+b_{j k} \tag{7.105}
\end{equation}
\)

The product of the scalar $c$ and the matrix $\mathbf{A}$ is defined as the matrix formed by multiplying every element of $\mathbf{A}$ by $c$.

\(
\begin{equation}
\text { If } \mathbf{D}=c \mathbf{A}, \quad \text { then } \quad d{j k}=c a{j k} \tag{7.106}
\end{equation}
\)

If $\mathbf{A}$ is an $m$ by $n$ matrix and $\mathbf{B}$ is an $n$ by $p$ matrix, the matrix product $\mathbf{R}=\mathbf{A B}$ is defined to be the $m$ by $p$ matrix whose elements are

\(
\begin{equation}
r{j k} \equiv a{j 1} b{1 k}+a{j 2} b{2 k}+\cdots+a{j n} b{n k}=\sum{i=1}^{n} a{j i} b{i k} \tag{7.107}
\end{equation}
\)

To calculate $r{j k}$ we take row $j$ of $\mathbf{A}$ (this row's elements are $a{j 1}, a{j 2}, \ldots, a{j n}$ ), multiply each element of this row by the corresponding element in column $k$ of $\mathbf{B}$ (this column's elements are $b{1 k}, b{2 k}, \ldots, b_{n k}$ ), and add the $n$ products. For example, suppose

\(
\mathbf{A}=\left(\begin{array}{rrr}
-1 & 3 & \frac{1}{2} \
0 & 4 & 1
\end{array}\right) \quad \text { and } \quad \mathbf{B}=\left(\begin{array}{rrr}
1 & 0 & -2 \
2 & 5 & 6 \
-8 & 3 & 10
\end{array}\right)
\)

The number of columns of $\mathbf{A}$ equals the number of rows of $\mathbf{B}$, so the matrix product $\mathbf{A B}$ is defined. $\mathbf{A B}$ is the product of the 2 by 3 matrix $\mathbf{A}$ and the 3 by 3 matrix $\mathbf{B}$, so $\mathbf{R} \equiv \mathbf{A B}$ is a 2 by 3 matrix. The element $r{21}$ is found from the second row of $\mathbf{A}$ and the first column of $\mathbf{B}$ as follows: $r{21}=0(1)+4(2)+1(-8)=0$. Calculation of the remaining elements gives

\(
\mathbf{R}=\left(\begin{array}{rrr}
1 & 16 \frac{1}{2} & 25 \
0 & 23 & 34
\end{array}\right)
\)

Matrix multiplication is not commutative; the products $\mathbf{A B}$ and $\mathbf{B A}$ need not be equal. (In the preceding example, the product $\mathbf{B A}$ happens to be undefined.) Matrix multiplication can be shown to be associative, meaning that $\mathbf{A}(\mathbf{B C})=(\mathbf{A B}) \mathbf{C}$ and can be shown to be distributive, meaning that $\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A B}+\mathbf{A C}$ and $(\mathbf{B}+\mathbf{C}) \mathbf{D}=\mathbf{B D}+\mathbf{C D}$.

A matrix with equal numbers of rows and columns is a square matrix. The order of a square matrix equals the number of rows.

If $\mathbf{A}$ is a square matrix, its square, cube, $\ldots$ are defined by $\mathbf{A}^{2} \equiv \mathbf{A A}$, $\mathbf{A}^{3} \equiv \mathbf{A A A}, \ldots$

The elements $a{11}, a{22}, \ldots, a_{n n}$ of a square matrix of order $n$ lie on its principal diagonal. A diagonal matrix is a square matrix having zero as the value of each element not on the principal diagonal.

The trace of a square matrix is the sum of the elements on the principal diagonal. If $\mathbf{A}$ is a square matrix of order $n$, its trace is $\operatorname{Tr} \mathbf{A}=\sum{i=1}^{n} a{i i}$.

A diagonal matrix whose diagonal elements are each equal to 1 is called a unit matrix or an identity matrix. The $(j, k)$ th element of a unit matrix is the Kronecker delta $\delta{j k}$; $(\mathbf{I}){j k}=\delta_{j k}$, where $\mathbf{I}$ is a unit matrix. For example, the unit matrix of order 3 is

\(
\left(\begin{array}{lll}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
\end{array}\right)
\)

Let $\mathbf{B}$ be a square matrix of the same order as a unit matrix $\mathbf{I}$. The $(j, k)$ th element of the product IB is given by (7.107) as (IB $){j k}=\sum{i}(\mathbf{I}){j i} b{i k}=\sum{i} \delta{j i} b{i k}=b{j k}$. Since the $(j, k)$ th elements of $\mathbf{I B}$ and $\mathbf{B}$ are equal for all $j$ and $k$, we have $\mathbf{I B}=\mathbf{B}$. Similarly, we find $\mathbf{B I}=\mathbf{B}$. Multiplication by a unit matrix has no effect.

A matrix all of whose elements are zero is called a zero matrix, symbolized by $\mathbf{0}$. A nonzero matrix has at least one element not equal to zero. These definitions apply to row vectors and column vectors.

Most matrices in quantum chemistry are either square matrices or row or column matrices.

Matrices and Quantum Mechanics

In Section 7.1 , the integral $\int f{m}^{*} \hat{A} f{n} d \tau$ was called a matrix element of $\hat{A}$. We now justify this name by showing that such integrals obey the rules of matrix algebra.

Let the functions $f{1}, f{2}, \ldots$ be a complete, orthonormal set and let the symbol $\left{f{i}\right}$ denote this complete set. The numbers $A{m n} \equiv\left\langle f{m}\right| \hat{A}\left|f{n}\right\rangle \equiv \int f{m}^{*} \hat{A} f{n} d \tau$ are called matrix elements of the linear operator $\hat{A}$ in the basis $\left{f{i}\right}$. The square matrix
is called the matrix representative of the linear operator $\hat{A}$ in the $\left{f{i}\right}$ basis. Since $\left{f_{i}\right}$ usually consists of an infinite number of functions, $\mathbf{A}$ is an infinite-order matrix.

Consider the addition of matrix-element integrals. Suppose $\hat{C}=\hat{A}+\hat{B}$. A typical matrix element of $\hat{C}$ in the $\left{f_{i}\right}$ basis is

\(
\begin{aligned}
C{m n} & =\left\langle f{m}\right| \hat{C}\left|f{n}\right\rangle=\left\langle f{m}\right| \hat{A}+\hat{B}\left|f{n}\right\rangle=\int f{m}^{}(\hat{A}+\hat{B}) f{n} d \tau \
& =\int f{m}^{} \hat{A} f{n} d \tau+\int f{m}^{*} \hat{B} f{n} d \tau=A{m n}+B_{m n}
\end{aligned}
\)

Thus, if $\hat{C}=\hat{A}+\hat{B}$, then $C{m n}=A{m n}+B_{m n}$, which is the rule (7.105) for matrix addition. Hence, if $\hat{C}=\hat{A}+\hat{B}$, then $\mathbf{C}=\mathbf{A}+\mathbf{B}$, where $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ are the matrix representatives of the operators $\hat{A}, \hat{B}, \hat{C}$.

Similarly, if $\hat{P}=c \hat{S}$, where $c$ is a constant, then we find (Prob. 7.52) $P{j k}=c S{j k}$, which is the rule for multiplication of a matrix by a scalar.

Finally, suppose that $\hat{R}=\hat{S} \hat{T}$. We have

\(
\begin{equation}
R{m n}=\int f{m}^{} \hat{R} f{n} d \tau=\int f{m}^{} \hat{S} \hat{T} f_{n} d \tau \tag{7.109}
\end{equation}
\)

The function $\hat{T} f{n}$ can be expanded in terms of the complete orthonormal set $\left{f{i}\right}$ as [Eq. (7.41)]:

\(
\hat{T} f{n}=\sum{i} c{i} f{i}=\sum{i}\left\langle f{i} \mid \hat{T} f{n}\right\rangle f{i}=\sum{i}\left\langle f{i}\right| \hat{T}\left|f{n}\right\rangle f{i}=\sum{i} T{i n} f_{i}
\)

and $R_{m n}$ becomes

\(
\begin{equation}
R{m n}=\int f{m}^{} \hat{S} \sum{i} T{i n} f{i} d \tau=\sum{i} \int f{m}^{*} \hat{S} f{i} d \tau T{i n}=\sum{i} S{m i} T{i n} \tag{7.110}
\end{}
\)

The equation $R{m n}=\sum{i} S{m i} T{i n}$ is the rule (7.107) for matrix multiplication. Hence, if $\hat{R}=\hat{S} \hat{T}$, then $\mathbf{R}=\mathbf{S T}$.

We have proved that the matrix representatives of linear operators in a complete orthonormal basis set obey the same equations that the operators obey. Combining Eqs. (7.109) and (7.110), we have the useful sum rule

\(
\begin{equation}
\sum_{i}\langle m| \hat{S}|i\rangle\langle i| \hat{T}|n\rangle=\langle m| \hat{S} \hat{T}|n\rangle \tag{7.111}
\end{equation}
\)

Suppose the basis set $\left{f{i}\right}$ is chosen to be the complete, orthonormal set of eigenfunctions $g{i}$ of $\hat{A}$, where $\hat{A} g{i}=a{i} g{i}$. Then the matrix element $A{m n}$ is

\(
A{m n}=\left\langle g{m}\right| \hat{A}\left|g{n}\right\rangle=\left\langle g{m} \mid \hat{A} g{n}\right\rangle=\left\langle g{m} \mid a{n} g{n}\right\rangle=a{n}\left\langle g{m} \mid g{n}\right\rangle=a{n} \delta_{m n}
\)

The matrix that represents $\hat{A}$ in the basis of orthonormal $\hat{A}$ eigenfunctions is thus a diagonal matrix whose diagonal elements are the eigenvalues of $\hat{A}$. Conversely, one can prove (Prob. 7.53) that, when the matrix representative of $\hat{A}$ using a complete orthonormal set is a diagonal matrix, then the basis functions are the eigenfunctions of $\hat{A}$ and the diagonal matrix elements are the eigenvalues of $\hat{A}$.

We have used the complete, orthonormal basis $\left{f{i}\right}$ to represent the operator $\hat{A}$ by the matrix $\mathbf{A}$ of (7.108). The basis $\left{f{i}\right}$ can also be used to represent an arbitrary function $u$, as follows. We expand $u$ in terms of the complete set $\left{f{i}\right}$, according to $u=\sum{i} u{i} f{i}$, where the expansion coefficients $u{i}$ are numbers (not functions) given by Eq. (7.40) as $u{i}=\left\langle f{i} \mid u\right\rangle$. The set of expansion coefficients $u{1}, u{2}, \ldots$ is formed into a column matrix (column vector), which we call $\mathbf{u}$, and $\mathbf{u}$ is said to be the representative of the function $u$ in the $\left{f{i}\right}$ basis. If $\hat{A} u=w$, where $w$ is another function, then we can show (Prob. 7.54) that $\mathbf{A u}=\mathbf{w}$, where $\mathbf{A}, \mathbf{u}$, and $\mathbf{w}$ are the matrix representatives of $\hat{A}, u$, and $w$ in the $\left{f_{i}\right}$ basis. Thus, the effect of the linear operator $\hat{A}$ on an arbitrary function $u$ can be found if the matrix representative $\mathbf{A}$ of $\hat{A}$ is known. Hence, knowing the matrix representative $\mathbf{A}$ is equivalent to knowing what the operator $\hat{A}$ is.

To deal with the time-independent Schrödinger equation for systems (such as atoms or molecules) that contain interacting particles, we must use approximation methods. This chapter discusses the variation method, which allows us to approximate the ground-state energy of a system without solving the Schrödinger equation. The variation method is based on the following theorem:

THE VARIATION THEOREM

Given a system whose Hamiltonian operator $\hat{H}$ is time independent and whose lowest-energy eigenvalue is $E_{1}$, if $\phi$ is any normalized, well-behaved function of the coordinates of the system's particles that satisfies the boundary conditions of the problem, then

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E_{1}, \quad \phi \text { normalized } \tag{8.1}
\end{}
\)

The variation theorem allows us to calculate an upper bound for the system's groundstate energy.

To prove (8.1), we expand $\phi$ in terms of the complete, orthonormal set of eigenfunctions of $\hat{H}$, the stationary-state eigenfunctions $\psi_{k}$ :

\(
\begin{equation}
\phi=\sum{k} a{k} \psi_{k} \tag{8.2}
\end{equation}
\)

where

\(
\begin{equation}
\hat{H} \psi{k}=E{k} \psi_{k} \tag{8.3}
\end{equation}
\)

Note that the expansion (8.2) requires that $\phi$ obey the same boundary conditions as the $\psi_{k}$ 's. Substitution of (8.2) into the left side of (8.1) gives

\(
\int \phi^{} \hat{H} \phi d \tau=\int \sum{k} a{k}^{} \psi{k}^{*} \hat{H} \sum{j} a{j} \psi{j} d \tau=\int \sum{k} a{k}^{} \psi_{k}^{} \sum{j} a{j} \hat{H} \psi_{j} d \tau
\)

Using the eigenvalue equation (8.3) and assuming the validity of interchanging the integration and the infinite summations, we get

\(
\begin{aligned}
\int \phi^{} \hat{H} \phi d \tau & =\int \sum{k} a{k}^{} \psi{k}^{*} \sum{j} a{j} E{j} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} E{j} \int \psi_{k}^{} \psi{j} d \tau \
& =\sum{k} \sum{j} a{k}^{*} a{j} E{j} \delta_{k j}
\end{aligned}
\)

where the orthonormality of the eigenfunctions $\psi_{k}$ was used. We perform the sum over $j$, and, as usual, the Kronecker delta makes all terms zero except the one with $j=k$, giving

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k} a{k}^{} a{k} E{k}=\sum{k}\left|a{k}\right|^{2} E_{k} \tag{8.4}
\end{equation}
\)

Since $E{1}$ is the lowest-energy eigenvalue of $\hat{H}$, we have $E{k} \geq E{1}$. Since $\left|a{k}\right|^{2}$ is never negative, we can multiply the inequality $E{k} \geq E{1}$ by $\left|a{k}\right|^{2}$ without changing the direction of the inequality sign to get $\left|a{k}\right|^{2} E{k} \geq\left|a{k}\right|^{2} E{1}$. Therefore, $\sum{k}\left|a{k}\right|^{2} E{k} \geq \sum{k}\left|a{k}\right|^{2} E_{1}$, and use of (8.4) gives

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k}\left|a{k}\right|^{2} E{k} \geq \sum{k}\left|a{k}\right|^{2} E{1}=E{1} \sum{k}\left|a_{k}\right|^{2} \tag{8.5}
\end{}
\)

Because $\phi$ is normalized, we have $\int \phi^{*} \phi d \tau=1$. Substitution of the expansion (8.2) into the normalization condition gives

\(
\begin{gather}
1=\int \phi^{} \phi d \tau=\int \sum{k} a{k}^{} \psi_{k}^{} \sum{j} a{j} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} \int \psi{k}^{} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} \delta{k j} \
1=\sum{k}\left|a{k}\right|^{2} \tag{8.6}
\end{gather}
\)

[Note that in deriving Eqs. (8.4) and (8.6) we essentially repeated the derivations of Eqs. (7.70) and (7.69), respectively.]

Use of (8.6) in (8.5) gives the variation theorem (8.1):

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E_{1}, \quad \phi \text { normalized } \tag{8.7}
\end{}
\)

Suppose we have a function $\phi$ that is not normalized. To apply the variation theorem, we multiply $\phi$ by a normalization constant $N$ so that $N \phi$ is normalized. Replacing $\phi$ by $N \phi$ in (8.7), we have

\(
\begin{equation}
|N|^{2} \int \phi^{} \hat{H} \phi d \tau \geq E_{1} \tag{8.8}
\end{}
\)

$N$ is determined by $\int(N \phi)^{} N \phi d \tau=|N|^{2} \int \phi^{} \phi d \tau=1$; so $|N|^{2}=1 / \int \phi^{*} \phi d \tau$ and (8.8) becomes

\(
\begin{equation}
\frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau} \geq E_{1} \tag{8.9}
\end{equation}
\)

where $\phi$ is any well-behaved function (not necessarily normalized) that satisfies the boundary conditions of the problem.

The function $\phi$ is called a trial variation function, and the integral in (8.1) [or the ratio of integrals in (8.9)] is called the variational integral. To arrive at a good approximation to the ground-state energy $E{1}$, we try many trial variation functions and look for the one that gives the lowest value of the variational integral. From (8.1), the lower the value of the variational integral, the better the approximation we have to $E{1}$. One way to disprove quantum mechanics would be to find a trial variation function that made the variational integral less than $E{1}$ for some system where $E{1}$ is known.

Let $\psi_{1}$ be the true ground-state wave function:

\(
\begin{equation}
\hat{H} \psi{1}=E{1} \psi_{1} \tag{8.10}
\end{equation}
\)

If we happened to be lucky enough to hit upon a variation function that was equal to $\psi{1}$, then, using (8.10) in (8.1), we see that the variational integral will be equal to $E{1}$. Thus the ground-state wave function gives the minimum value of the variational integral. We therefore expect that the lower the value of the variational integral, the closer the trial variational function will approach the true ground-state wave function. However, it turns out that the variational integral approaches $E{1}$ a lot faster than the trial variation function approaches $\psi{1}$, and it is possible to get a rather good approximation to $E_{1}$ using a rather poor $\phi$.

In practice, one usually puts several parameters into the trial function $\phi$ and then varies these parameters so as to minimize the variational integral. Successful use of the variation method depends on the ability to make a good choice for the trial function.

Let us look at some examples of the variation method. Although the real utility of the method is for problems to which we do not know the true solutions, we will consider problems that are exactly solvable so that we can judge the accuracy of our results.

EXAMPLE

Devise a trial variation function for the particle in a one-dimensional box of length $l$.
The wave function is zero outside the box and the boundary conditions require that $\psi=0$ at $x=0$ and at $x=l$. The variation function $\phi$ must meet these boundary conditions of being zero at the ends of the box. As noted after Eq. (4.57), the ground-state $\psi$ has no nodes interior to the boundary points, so it is desirable that $\phi$ have no interior nodes. A simple function that has these properties is the parabolic function

\(
\begin{equation}
\phi=x(l-x) \quad \text { for } 0 \leq x \leq l \tag{8.11}
\end{equation}
\)

and $\phi=0$ outside the box. Since we have not normalized $\phi$, we use Eq. (8.9). Inside the box the Hamiltonian is $-\left(\hbar^{2} / 2 m\right) d^{2} / d x^{2}$. For the numerator and denominator of (8.9), we have

\(
\begin{gather}
\int \phi^{} \hat{H} \phi d \tau=-\frac{\hbar^{2}}{2 m} \int{0}^{l}\left(l x-x^{2}\right) \frac{d^{2}}{d x^{2}}\left(l x-x^{2}\right) d x=\frac{\hbar^{2}}{m} \int{0}^{l}\left(l x-x^{2}\right) d x=\frac{\hbar^{2} l^{3}}{6 m} \tag{8.12}\
\int \phi^{} \phi d \tau=\int_{0}^{l} x^{2}(l-x)^{2} d x=\frac{l^{5}}{30} \tag{8.13}
\end{gather}
\)

Substituting in the variation theorem (8.9), we get

\(
E_{1} \leq \frac{5 h^{2}}{4 \pi^{2} m l^{2}}=0.1266515 \frac{h^{2}}{m l^{2}}
\)

From Eq. (2.20), $E_{1}=h^{2} / 8 m l^{2}=0.125 h^{2} / m l^{2}$, and the energy error is $1.3 \%$.
Since $\int|\phi|^{2} d \tau=l^{5} / 30$, the normalized form of (8.11) is $\left(30 / l^{5}\right)^{1 / 2} x(l-x)$.
Figure 7.3 shows that this function rather closely resembles the true ground-state particle-in-a-box wave function.

EXERCISE A one-particle, one-dimensional system has the potential energy function $V=V{0}$ for $0 \leq x \leq l$ and $V=\infty$ elsewhere (where $V{0}$ is a constant). (a) Use the variation function $\phi=\sin (\pi x / l)$ for $0 \leq x \leq l$ and $\phi=0$ elsewhere to estimate the ground-state energy of this system. (b) Explain why the result of (a) is the exact groundstate energy. Hint: See one of the Chapter 4 problems. (Answer: (a) $V_{0}+h^{2} / 8 m l^{2}$.)

The preceding example did not have a parameter in the trial function. The next example does.

EXAMPLE

For the one-dimensional harmonic oscillator, devise a variation function with a parameter and find the optimum value of the parameter. Estimate the ground-state energy.
The variation function $\phi$ must be quadratically integrable and so must go to zero as $x$ goes to $\pm \infty$. The function $e^{-x}$ has the proper behavior at $+\infty$ but becomes infinite at $-\infty$. The function $e^{-x^{2}}$ has the proper behavior at $\pm \infty$. However, it is dimensionally unsatisfactory, since the power to which we raise $e$ must be dimensionless. This can be seen from the Taylor series $e^{z}=1+z+z^{2} / 2!+\cdots$ [Eq. (4.44)]. Since all the terms in this series must have the same dimensions, $z$ must have the same dimensions as 1 ; that is, $z$ in $e^{z}$ must be dimensionless. Hence we modify $e^{-x^{2}}$ to $e^{-c x^{2}}$, where $c$ has units of length ${ }^{-2}$. We shall take $c$ as a variational parameter. The true ground-state $\psi$ must have no nodes. Also, since $V=\frac{1}{2} k x^{2}$ is an even function, the ground-state $\psi$ must have definite parity and must be an even function, since an odd function has a node at the origin. The trial function $e^{-c x^{2}}$ has the desired properties of having no nodes and of being an even function.
Use of (4.30) for $\hat{H}$ and Appendix integrals gives (Prob. 8.3)

\(
\begin{aligned}
\int \phi^{} \hat{H} \phi d \tau & =-\frac{\hbar^{2}}{2 m} \int{-\infty}^{\infty} e^{-c x^{2}} \frac{d^{2} e^{-c x^{2}}}{d x^{2}} d x+2 \pi^{2} v^{2} m \int{-\infty}^{\infty} x^{2} e^{-2 c x^{2}} d x \
& =\frac{\hbar^{2}}{m}\left(\frac{\pi c}{8}\right)^{1 / 2}+v^{2} m\left(\frac{\pi^{5}}{8}\right)^{1 / 2} c^{-3 / 2} \
\int \phi^{} \phi d \tau & =\int{-\infty}^{\infty} e^{-2 c x^{2}} d x=2 \int{0}^{\infty} e^{-2 c x^{2}} d x=\left(\frac{\pi}{2 c}\right)^{1 / 2}
\end{aligned}
\)

The variational integral $W$ is

\(
\begin{equation}
W \equiv \frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau}=\frac{\hbar^{2} c}{2 m}+\frac{\pi^{2} \nu^{2} m}{2 c} \tag{8.14}
\end{equation}
\)

We now vary $c$ to minimize the variational integral (8.14). A necessary condition that $W$ be minimized is that

\(
\begin{gather}
\frac{d W}{d c}=0=\frac{\hbar^{2}}{2 m}-\frac{\pi^{2} \nu^{2} m}{2 c^{2}} \
c= \pm \pi \nu m / \hbar \tag{8.15}
\end{gather}
\)

The negative root $c=-\pi \nu m / \hbar$ is rejected, since it would make $\phi=e^{-c x^{2}}$ not quadratically integrable. Substitution of $c=\pi \nu m / \hbar$ into (8.14) gives $W=\frac{1}{2} h \nu$. This is the exact ground-state harmonic-oscillator energy. With $c=\pi \nu m / \hbar$ the variation function $\phi$ is the same (except for being unnormalized) as the harmonic-oscillator ground-state wave function (4.53) and (4.31).
For the normalized harmonic-oscillator variation function $\phi=(2 c / \pi)^{1 / 4} e^{-c x^{2}}$, a large value of $c$ makes $\phi$ fall off very rapidly from its maximum value at $x=0$. This makes the probability density large only near $x=0$. The potential energy $V=\frac{1}{2} k x^{2}$ is low near $x=0$, so a large $c$ means a low $\langle V\rangle=\langle\phi| V|\phi\rangle$. [Note also that $\langle V\rangle$ equals the second term on the right side of (8.14).] However, because a large $c$ makes $\phi$ fall off very rapidly from its maximum, it makes $|d \phi / d x|$ large in the region near
$x=0$. From Prob. 7.7b, a large $|d \phi / d x|$ means a large value of $\langle T\rangle$ [which equals the first term on the right side of (8.14)]. The optimum value of $c$ minimizes the sum $\langle T\rangle+\langle V\rangle=W$. In atoms and molecules, the true wave function is a compromise between the tendency to minimize $\langle V\rangle$ by confining the electrons to regions of low $V$ (near the nuclei) and the tendency to minimize $\langle T\rangle$ by allowing the electron probability density to spread out over a large region.
EXERCISE Consider a one-particle, one-dimensional system with $V=0$ for $-\frac{1}{2} l \leq x \leq \frac{1}{2} l$ and $V=b \hbar^{2} / m l^{2}$ elsewhere (where $b$ is a positive constant) (Fig. 2.5 with $V_{0}=b \hbar^{2} / m l^{2}$ and the origin shifted). (a) For the variation function $\phi=(x-c)(x+c)=x^{2}-c^{2}$ for $-c \leq x \leq c$ and $\phi=0$ elsewhere, where the variational parameter $c$ satisfies $c>\frac{1}{2} l$, one finds that the variational integral $W$ is given by

\(
W=\frac{\hbar^{2}}{m l^{2}}\left[\frac{5 l^{2}}{4 c^{2}}+b\left(1-\frac{15 l}{16 c}+\frac{5 l^{3}}{32 c^{3}}-\frac{3 l^{5}}{256 c^{5}}\right)\right]
\)

Sketch $\phi$ and $V$ on the same plot. Find the equation satisfied by the value of $c$ that minimizes $W$. (b) Find the optimum $c$ and $W$ for $V_{0}=20 \hbar^{2} / \mathrm{ml}^{2}$ and compare with the true ground-state energy $2.814 \hbar^{2} / m l^{2}$ (Prob. 4.31c). (Hint: You may want to use the Solver in a spreadsheet or a programmable calculator to find $c / l$.) (Answer: (a) $48 t^{4}-24 t^{2}-128 t^{3} / b+3=0$, where $t \equiv c / l$. (b) $c=0.6715 l, W=3.454 \hbar^{2} / m l^{2}$.)

The variation method as presented in the last section gives information about only the ground-state energy and wave function. We now discuss extension of the variation method to excited states. (See also Section 8.5.)

Consider how we might extend the variation method to estimate the energy of the first excited state. We number the stationary states of the system $1,2,3, \ldots$ in order of increasing energy:

\(
E{1} \leq E{2} \leq E_{3} \leq \cdots
\)

We showed that for a normalized variational function $\phi$ [Eqs. (8.4) and (8.6)]

\(
\int \phi^{} \hat{H} \phi d \tau=\sum{k=1}^{\infty}\left|a{k}\right|^{2} E_{k} \quad \text { and } \quad \int \phi^{} \phi d \tau=\sum{k=1}^{\infty}\left|a{k}\right|^{2}=1
\)

where the $a{k}$ 's are the expansion coefficients in $\phi=\sum{k} a{k} \psi{k}$ [Eq. (8.2)]. We have $a{k}=\left\langle\psi{k} \mid \phi\right\rangle$ [Eq. (7.40)]. Let us restrict ourselves to normalized functions $\phi$ that are orthogonal to the true ground-state wave function $\psi{1}$. Then we have $a{1}=\left\langle\psi_{1} \mid \phi\right\rangle=0$ and

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k=2}^{\infty}\left|a{k}\right|^{2} E{k} \quad \text { and } \quad \int \phi^{*} \phi d \tau=\sum{k=2}^{\infty}\left|a_{k}\right|^{2}=1 \tag{8.16}
\end{}
\)

For $k \geq 2$, we have $E{k} \geq E{2}$ and $\left|a{k}\right|^{2} E{k} \geq\left|a{k}\right|^{2} E{2}$. Hence

\(
\begin{equation}
\sum{k=2}^{\infty}\left|a{k}\right|^{2} E{k} \geq \sum{k=2}^{\infty}\left|a{k}\right|^{2} E{2}=E{2} \sum{k=2}^{\infty}\left|a{k}\right|^{2}=E{2} \tag{8.17}
\end{equation}
\)

Combining (8.16) and (8.17), we have the desired result:

\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E{2} \quad \text { if } \quad \int \psi{1}^{} \phi d \tau=0 \quad \text { and } \quad \int \phi^{} \phi d \tau=1 \tag{8.18}
\end{}
\)

The inequality (8.18) allows us to get an upper bound to the energy $E{2}$ of the first excited state. However, the restriction $\left\langle\psi{1} \mid \phi\right\rangle=0$ makes this method troublesome to apply.

For certain systems, it is possible to be sure that $\left\langle\psi{1} \mid \phi\right\rangle=0$ even though we do not know the true ground-state wave function. An example is a one-dimensional problem for which $V$ is an even function of $x$. In this case the ground-state wave function is always an even function, while the first excited-state wave function is odd. (All the wave functions must be of definite parity. The ground-state wave function is nodeless, and, since an odd function vanishes at the origin, the ground-state wave function must be even. The first excited-state wave function has one node and must be odd.) Therefore, for odd trial functions, it must be true that $\left\langle\psi{1} \mid \phi\right\rangle=0$; the even function $\psi_{1}$ times the odd function $\phi$ gives an odd integrand whose integral from $-\infty$ to $\infty$ is zero.

Another example is a particle moving in a central field (Section 6.1). The form of the potential energy might be such that we could not solve for the radial factor $R(r)$ in the eigenfunction. However, the angular factor in $\psi$ is a spherical harmonic [Eq. (6.16)], and spherical harmonics with different values of $l$ are orthogonal. Thus we can get an upper bound to the energy of the lowest state with any given angular momentum $l$ by using the factor $Y_{l}^{m}$ in the trial function. This result depends on the extension of (8.18) to higher excited states:

\(
\begin{equation}
\frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau} \geq E{k+1} \quad \text { if } \quad \int \psi{1}^{} \phi d \tau=\int \psi{2}^{*} \phi d \tau=\cdots=\int \psi{k}^{} \phi d \tau=0 \tag{8.19}
\end{equation}
\)

Section 8.5 discusses a kind of variation function that gives rise to an equation involving a determinant. Therefore, we now discuss determinants.

A determinant is a square array of $n^{2}$ quantities (called elements); the value of the determinant is calculated from its elements in a manner to be given shortly. The number $n$ is the order of the determinant. Using $a_{i j}$ to represent a typical element, we write the $n$ th-order determinant as

\(
\operatorname{det}\left(a{i j}\right)=\left|\begin{array}{ccccc}
a{11} & a{12} & a{13} & \cdots & a{1 n} \tag{8.20}\
a{21} & a{22} & a{23} & \cdots & a{2 n} \
\cdot & \cdot & \cdot & \cdots & \cdot \
\cdot & \cdot & \cdot & \cdots & \cdot \
\cdot & \cdot & \cdot & \cdots & \cdot \
a{n 1} & a{n 2} & a{n 3} & \cdots & a_{n n}
\end{array}\right|
\)

The vertical lines in (8.20) have nothing to do with absolute value. Before considering how the value of the $n$ th-order determinant is defined, we consider determinants of first, second, and third orders.

A first-order determinant has one element, and its value is simply the value of that element. Thus

\(
\begin{equation}
\left|a{11}\right|=a{11} \tag{8.21}
\end{equation}
\)

where the vertical lines indicate a determinant and not an absolute value.
A second-order determinant has four elements, and its value is defined by

\(
\left|\begin{array}{ll}
a{11} & a{12} \tag{8.22}\
a{21} & a{22}
\end{array}\right|=a{11} a{22}-a{12} a{21}
\)

The value of a third-order determinant is defined by

\(
\left|\begin{array}{lll}
a{11} & a{12} & a{13} \tag{8.23}\
a{21} & a{22} & a{23} \
a{31} & a{32} & a{33}
\end{array}\right|=a{11}\left|\begin{array}{ll}
a{22} & a{23} \
a{32} & a{33}
\end{array}\right|-a{12}\left|\begin{array}{ll}
a{21} & a{23} \
a{31} & a{33}
\end{array}\right|+a{13}\left|\begin{array}{ll}
a{21} & a{22} \
a{31} & a{32}
\end{array}\right|
\)

\(
\begin{align}
= & a{11} a{22} a{33}-a{11} a{32} a{23}-a{12} a{21} a{33}+a{12} a{31} a{23} \
& +a{13} a{21} a{32}-a{13} a{31} a{22} \tag{8.24}
\end{align}
\)

A third-order determinant is evaluated by writing down the elements of the top row with alternating plus and minus signs and then multiplying each element by a certain secondorder determinant; the second-order determinant that multiplies a given element is found by crossing out the row and column of the third-order determinant in which that element appears. The $(n-1)$-order determinant obtained by striking out the $i$ th row and the $j$ th column of the $n$ th-order determinant is called the minor of the element $a{i j}$. We define the cofactor of $a{i j}$ as the minor of $a{i j}$ times the factor $(-1)^{i+j}$. Thus (8.23) states that a thirdorder determinant is evaluated by multiplying each element of the top row by its cofactor and adding up the three products. [Note that (8.22) conforms to this evaluation by means of cofactors, since the cofactor of $a{11}$ in (8.22) is $a{22}$, and the cofactor of $a{12}$ is $-a_{21}$.] A numerical example is

\(
\begin{aligned}
\left|\begin{array}{rrr}
5 & 10 & 2 \
0.1 & 3 & 1 \
0 & 4 & 4
\end{array}\right| & =5\left|\begin{array}{ll}
3 & 1 \
4 & 4
\end{array}\right|-10\left|\begin{array}{rr}
0.1 & 1 \
0 & 4
\end{array}\right|+2\left|\begin{array}{rr}
0.1 & 3 \
0 & 4
\end{array}\right| \
& =5(8)-10(0.4)+2(0.4)=36.8
\end{aligned}
\)

Denoting the minor of $a{i j}$ by $M{i j}$ and the cofactor of $a{i j}$ by $C{i j}$, we have

\(
\begin{equation}
C{i j}=(-1)^{i+j} M{i j} \tag{8.25}
\end{equation}
\)

The expansion (8.23) of the third-order determinant can be written as

\(
\operatorname{det}\left(a{i j}\right)=\left|\begin{array}{lll}
a{11} & a{12} & a{13} \tag{8.26}\
a{21} & a{22} & a{23} \
a{31} & a{32} & a{33}
\end{array}\right|=a{11} C{11}+a{12} C{12}+a{13} C{13}
\)

A third-order determinant can be expanded using the elements of any row and the corresponding cofactors. For example, using the second row to expand the third-order determinant, we have

\(
\begin{gather}
\operatorname{det}\left(a{i j}\right)=a{21} C{21}+a{22} C{22}+a{23} C{23} \tag{8.27}\
\operatorname{det}\left(a{i j}\right)=-a{21}\left|\begin{array}{ll}
a{12} & a{13} \
a{32} & a{33}
\end{array}\right|+a{22}\left|\begin{array}{ll}
a{11} & a{13} \
a{31} & a{33}
\end{array}\right|-a{23}\left|\begin{array}{ll}
a{11} & a{12} \
a{31} & a_{32}
\end{array}\right| \tag{8.28}
\end{gather}
\)

and expansion of the second-order determinants shows that (8.28) is equal to (8.24). We may also use the elements of any column and the corresponding cofactors to expand the determinant, as can be readily verified. Thus for the third-order determinant, we can write

\(
\begin{array}{ll}
\operatorname{det}\left(a{i j}\right)=a{k 1} C{k 1}+a{k 2} C{k 2}+a{k 3} C{k 3}=\sum{l=1}^{3} a{k l} C{k l}, & k=1 \text { or } 2 \text { or } 3 \
\operatorname{det}\left(a{i j}\right)=a{1 k} C{1 k}+a{2 k} C{2 k}+a{3 k} C{3 k}=\sum{l=1}^{3} a{l k} C{l k}, & k=1 \text { or } 2 \text { or } 3
\end{array}
\)

The first expansion uses one of the rows; the second uses one of the columns.

We define determinants of higher order by an analogous row (or column) expansion. For an $n$ th-order determinant,

\(
\begin{equation}
\operatorname{det}\left(a{i j}\right)=\sum{l=1}^{n} a{k l} C{k l}=\sum{l=1}^{n} a{l k} C_{l k}, \quad k=1 \text { or } 2 \text { or } \ldots \text { or } n \tag{8.29}
\end{equation}
\)

Some theorems on determinants are as follows (for proofs, see Sokolnikoff and Redheffer, pp. 702-707):
I. If every element of a row (or column) of a determinant is zero, the value of the determinant is zero.
II. Interchanging any two rows (or columns) multiplies the value of a determinant by -1 .
III. If any two rows (or columns) of a determinant are identical, the determinant has the value zero.
IV. Multiplication of each element of any one row (or any one column) by some constant $k$ multiplies the value of the determinant by $k$.
V. Addition to each element of one row of the same constant multiple of the corresponding element of another row leaves the value of the determinant unchanged. This theorem also applies to the addition of a multiple of one column to another column.
VI. The interchange of all corresponding rows and columns leaves the value of the determinant unchanged. (This interchange means that column one becomes row one, column two becomes row two, etc.)

EXAMPLE

Use Theorem V to evaluate

\(
B=\left|\begin{array}{llll}
1 & 2 & 3 & 4 \tag{8.30}\
4 & 1 & 2 & 3 \
3 & 4 & 1 & 2 \
2 & 3 & 4 & 1
\end{array}\right|
\)

Addition of -2 times the elements of row one to the corresponding elements of row four changes row four to $2+(-2) 1=0,3+(-2)(2)=-1,4+(-2) 3=-2$, and $1+(-2) 4=-7$. Then, addition of -3 times row one to row three and -4 times row one to row two gives

\(
B=\left|\begin{array}{rrrr}
1 & 2 & 3 & 4 \tag{8.31}\
0 & -7 & -10 & -13 \
0 & -2 & -8 & -10 \
0 & -1 & -2 & -7
\end{array}\right|=1\left|\begin{array}{rrr}
-7 & -10 & -13 \
-2 & -8 & -10 \
-1 & -2 & -7
\end{array}\right|
\)

where we expanded $B$ in terms of elements of the first column. Subtracting twice row three from row two and seven times row three from row one, we have

\(
B=\left|\begin{array}{rrr}
0 & 4 & 36 \tag{8.32}\
0 & -4 & 4 \
-1 & -2 & -7
\end{array}\right|=(-1)\left|\begin{array}{rr}
4 & 36 \
-4 & 4
\end{array}\right|=-(16+144)=-160
\)

The diagonal of a determinant that runs from the top left to the lower right is the principal diagonal. A diagonal determinant is a determinant all of whose elements are zero except those on the principal diagonal. For a diagonal determinant,

\(
\begin{align}
\left|\begin{array}{ccccc}
a{11} & 0 & 0 & \cdots & 0 \
0 & a{22} & 0 & \cdots & 0 \
0 & 0 & a{33} & \cdots & 0 \
\cdot & \cdot & \cdot & \cdots & \cdot \
0 & 0 & 0 & \cdots & a{n n}
\end{array}\right| & =a{11}\left|\begin{array}{cccc}
a{22} & 0 & \cdots & 0 \
0 & a{33} & \cdots & 0 \
\cdot & \cdot & \cdots & \cdot \
0 & 0 & \cdots & a{n n}
\end{array}\right|=a{11} a{22}\left|\begin{array}{cccc}
a{33} & 0 & \cdots & 0 \
0 & a{44} & \cdots & 0 \
\cdot & \cdot & \cdots & \cdot \
0 & 0 & \cdots & a{n n}
\end{array}\right| \
& =\cdots=a{11} a{22} a{33} \ldots a_{n n} \tag{8.33}
\end{align}
\)

A diagonal determinant is equal to the product of its diagonal elements.
A determinant whose only nonzero elements occur in square blocks centered about the principal diagonal is in block-diagonal form. If we regard each square block as a determinant, then a block-diagonal determinant is equal to the product of the blocks. For example,

The dashed lines outline the blocks. Equation (8.34) is readily proved by expanding the left side in terms of elements of the top row and expanding several subsequent determinants using their top rows (Prob. 8.21).