Comprehensive Study Notes

Certain quantum-mechanical operators have no classical analog. An example is the parity operator. Recall that the harmonic-oscillator wave functions are either even or odd. We shall show how this property is related to the parity operator.

The parity operator $\hat{\Pi}$ is defined in terms of its effect on an arbitrary function $f$ :

$
\begin{equation}
\hat{\Pi} f(x, y, z)=f(-x,-y,-z) \tag{7.52}
\end{equation}
$

The parity operator replaces each Cartesian coordinate with its negative. For example, $\hat{\Pi}\left(x^{2}-z e^{a y}\right)=x^{2}+z e^{-a y}$.

As with any quantum-mechanical operator, we are interested in the eigenvalues $c{i}$ and the eigenfunctions $g{i}$ of the parity operator:

$
\begin{equation}
\hat{\Pi} g{i}=c{i} g_{i} \tag{7.53}
\end{equation}
$

The key to the problem is to calculate the square of $\hat{\Pi}$ :

$
\hat{\Pi}^{2} f(x, y, z)=\hat{\Pi}[\hat{\Pi} f(x, y, z)]=\hat{\Pi}[f(-x,-y,-z)]=f(x, y, z)
$

Since $f$ is arbitrary, we conclude that $\hat{\Pi}^{2}$ equals the unit operator:

$
\begin{equation}
\hat{\Pi}^{2}=\hat{1} \tag{7.54}
\end{equation}
$

We now operate on (7.53) with $\hat{\Pi}$ to get $\hat{\Pi} \hat{\Pi} g{i}=\hat{\Pi} c{i} g{i}$. Since $\hat{\Pi}$ is linear (Prob. 7.26), we have $\hat{\Pi}^{2} g{i}=c{i} \hat{\Pi} g{i}$, which becomes

$
\begin{equation}
\hat{\Pi}^{2} g{i}=c{i}^{2} g_{i} \tag{7.55}
\end{equation}
$

where the eigenvalue equation (7.53) was used. Since $\hat{\Pi}^{2}$ is the unit operator, the left side of (7.55) is simply $g_{i}$, and

$
g{i}=c{i}^{2} g_{i}
$

The function $g{i}$ cannot be zero everywhere (zero is always rejected as an eigenfunction on physical grounds). We can therefore divide by $g{i}$ to get $c_{i}^{2}=1$ and

$
\begin{equation}
c_{i}= \pm 1 \tag{7.56}
\end{equation}
$

The eigenvalues of $\hat{\Pi}$ are +1 and -1 . Note that this derivation applies to any operator whose square is the unit operator.

What are the eigenfunctions $g_{i}$ ? The eigenvalue equation (7.53) reads

$
\begin{gathered}
\hat{\Pi} g{i}(x, y, z)= \pm g{i}(x, y, z) \
g{i}(-x,-y,-z)= \pm g{i}(x, y, z)
\end{gathered}
$

If the eigenvalue is +1 , then $g{i}(-x,-y,-z)=g{i}(x, y, z)$ and $g{i}$ is an even function. If the eigenvalue is -1 , then $g{i}$ is odd: $g{i}(-x,-y,-z)=-g{i}(x, y, z)$. Hence, the eigenfunctions of the parity operator $\hat{\Pi}$ are all possible well-behaved even and odd functions.

When the parity operator commutes with the Hamiltonian operator $\hat{H}$, we can select a common set of eigenfunctions for these operators, as proved in Section 7.4. The eigenfunctions of $\hat{H}$ are the stationary-state wave functions $\psi_{i}$. Hence when

$
\begin{equation}
[\hat{\Pi}, \hat{H}]=0 \tag{7.57}
\end{equation}
$

the wave functions $\psi_{i}$ can be chosen to be eigenfunctions of $\hat{\Pi}$. We just proved that the eigenfunctions of $\hat{\Pi}$ are either even or odd. Hence, when (7.57) holds, each wave function can be chosen to be either even or odd. Let us find out when the parity and Hamiltonian operators commute.

We have, for a one-particle system,

$
[\hat{H}, \hat{\Pi}]=[\hat{T}, \hat{\Pi}]+[\hat{V}, \hat{\Pi}]=-\frac{\hbar^{2}}{2 m}\left(\left[\frac{\partial^{2}}{\partial x^{2}}, \hat{\Pi}\right]+\left[\frac{\partial^{2}}{\partial y^{2}}, \hat{\Pi}\right]+\left[\frac{\partial^{2}}{\partial z^{2}}, \hat{\Pi}\right]\right)+[\hat{V}, \hat{\Pi}]
$

Since

$
\begin{aligned}
\hat{\Pi}\left[\frac{\partial^{2}}{\partial x^{2}} f(x, y, z)\right]=\frac{\partial}{\partial(-x)} \frac{\partial}{\partial(-x)} f(-x,-y,-z) & =\frac{\partial^{2}}{\partial x^{2}} f(-x,-y,-z) \
& =\frac{\partial^{2}}{\partial x^{2}} \hat{\Pi} f(x, y, z)
\end{aligned}
$

where $f$ is any function, we conclude that

$
\left[\frac{\partial^{2}}{\partial x^{2}}, \hat{\Pi}\right]=0
$

Similar equations hold for the $y$ and $z$ coordinates, and $[\hat{H}, \hat{\Pi}]$ becomes

$
\begin{equation}
[\hat{H}, \hat{\Pi}]=[\hat{V}, \hat{\Pi}] \tag{7.58}
\end{equation}
$

Now

$
\begin{equation}
\hat{\Pi}[V(x, y, z) f(x, y, z)]=V(-x,-y,-z) f(-x,-y,-z) \tag{7.59}
\end{equation}
$

If the potential energy is an even function, that is, if $V(-x,-y,-z)=V(x, y, z)$, then (7.59) becomes

$
\hat{\Pi}[V(x, y, z) f(x, y, z)]=V(x, y, z) f(-x,-y,-z)=V(x, y, z) \hat{\Pi} f(x, y, z)
$

so $[\hat{V}, \hat{\Pi}]=0$. Hence, when the potential energy is an even function, the parity operator commutes with the Hamiltonian:

$
\begin{equation}
[\hat{H}, \hat{\Pi}]=0 \quad \text { if } V \text { is even } \tag{7.60}
\end{equation}
$

These results are easily extended to the $n$-particle case. For an $n$-particle system, the parity operator is defined by

$
\begin{equation}
\hat{\Pi} f\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)=f\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right) \tag{7.61}
\end{equation}
$

It is easy to see that (7.57) holds when

$
\begin{equation}
V\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)=V\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right) \tag{7.62}
\end{equation}
$

If $V$ satisfies this equation, $V$ is said to be an even function of the $3 n$ coordinates. In summary, we have:

THEOREM 7. When the potential energy $V$ is an even function, we can choose the stationary-state wave functions so that each $\psi_{i}$ is either an even function or an odd function.

A function that is either even or odd is said to be of definite parity.
If the energy levels are all nondegenerate (as is usually true in one-dimensional problems), then only one independent wave function corresponds to each energy eigenvalue and there is no element of choice (apart from an arbitrary multiplicative constant) in the wave functions. Thus, for the nondegenerate case, the stationary-state wave functions must be of definite parity when $V$ is an even function. For example, the one-dimensional harmonic oscillator has $V=\frac{1}{2} k x^{2}$, which is an even function, and the wave functions have definite parity.

The hydrogen-atom potential-energy function is even, and the hydrogenlike orbitals can be chosen to have definite parity (Probs. 7.22 and 7.29).

For the degenerate case, we have an element of choice in the wave functions, since an arbitrary linear combination of the functions corresponding to the degenerate level is an eigenfunction of $\hat{H}$. For a degenerate energy level, by taking appropriate linear combinations we can choose wave functions that are of definite parity, but there is no necessity that they be of definite parity.

Parity aids in evaluating integrals. We showed that $\int_{-\infty}^{\infty} f(x) d x=0$ when $f(x)$ is an odd function [Eq. (4.51)]. Let us extend this result to the $3 n$-dimensional case. An odd function of $3 n$ variables satisfies

$
\begin{equation}
g\left(-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}\right)=-g\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right) \tag{7.63}
\end{equation}
$

If $g$ is an odd function of the $3 n$ variables, then

$
\begin{equation}
\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} g\left(x{1}, \ldots, z{n}\right) d x{1} \cdots d z{n}=0 \tag{7.64}
\end{equation}
$

where the integration is over the $3 n$ coordinates. This equation holds because the contribution to the integral from the value of $g$ at $\left(x{1}, y{1}, z{1}, \ldots, x{n}, y{n}, z{n}\right)$ is canceled by the contribution from ( $-x{1},-y{1},-z{1}, \ldots,-x{n},-y{n},-z{n}$ ). Equation (7.64) also holds when the integrand is an odd function of some (but not necessarily all) of the variables. See Prob. 7.30.