Comprehensive Study Notes

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

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

\(
\begin{equation<em>}
\left|\left\langle g_{b}(q) \mid \Psi(q, t)\right\rangle\right|^{2} d b \tag{7.76}
\end{equation</em>}
\)