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 $9 h^{2} / 8 m l^{2}$.