Comprehensive Study Notes

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: $h^{2} / 4 l^{2}$.)