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.