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.