In problem 8.10 of Schaum's Quantum Mechanics they
say:
"We see that under the parity operator $r \rightarrow r$,
$\theta \rightarrow \pi – \theta$ and $\phi \rightarrow
\pi + \phi$
.. since $\frac{d}{d\theta} \rightarrow -\frac{d}{d\theta}$ and
$\frac{d}{d\phi} \rightarrow \frac{d}{d\phi}$,
it follows that the operators $\hat{L}_\pm$
are not affected by the parity operation."
(Here $\theta$ is the $z$-axis spherical angle and $\phi$ is the
azimuthal spherical angle.)
Another source, http://itp.uni-frankfurt.de/~valenti/SS14/QMII_2014_chap3.pdf,
also refers to this idea of an OPERATOR being "odd" under parity.
Do the operators really change? If you represented the $\frac{d}{d\theta}$
operator for example "under the parity operation" (what does that mean?)
as a matrix, wouldn't it be the exact same matrix? It's just that the
input wavefunction that is being input as argument to the $\frac{d}{d\theta}$
operator has had all its $+$ and $-$ signs flipped (from the perspective
of $x y z$ coordinates)
messed around with, so naturally the
$\frac{d}{d\theta}$ outputs $-1$ times its result. Correct?
Best Answer
When we talk about an operator undergoing some unitary transformation, be it spatial inversion, rotation, time reversal, etc., we are saying
$$U^{-1}AU = \,\,?$$
Saying an operator is odd/even means
$$U^{-1}AU = \pm A$$
with $+$ for even and $-$ for odd.
When you look at the matrix elements of a transformed operator,
$$\langle \alpha | U^{-1}AU | \beta \rangle$$
you can see that these are not the same as those of the original operator,
$$\langle \alpha | A | \beta \rangle$$
So the matrix is not the same.
That's a valid way of looking at it as well. Instead of viewing $\langle \alpha | U^{-1}AU | \beta \rangle$ as some new operator $A' = U^{-1}AU$ acting on the old states, you can see it as $A$ acting on the transformed states $\langle \tilde \alpha | = \langle \alpha | U^{-1}$ and $| \tilde \beta \rangle = U | \beta \rangle$. I'll stress that the matrix is still not the same, provided you're using the same basis in either case.
To illustrate this point, let's consider the expectation value of an operator $A$ - which commutes with the position operator - after a parity transformation. We have
$$\langle A' \rangle = \langle \Psi | \Pi^{-1}A\Pi | \Psi \rangle = \int d\mathbf{x'}d\mathbf{x''} \langle \Psi | \Pi^{-1} | \mathbf{x'} \rangle \langle \mathbf{x'} | A | \mathbf{x''} \rangle \langle \mathbf{x''} | \Pi | \Psi\rangle = \int d\mathbf{x'} \Psi^*(-\mathbf{x'})A\Psi(-\mathbf{x'})$$
where in the last step I've made use of the orthogonality of position states and the Hermiticity of the parity operator.
So either way is valid (although the relation $U^{-1}AU = \pm A$ is independent of basis). In the same way that we can speak of time-evolved operators with time-independent kets in the Schrödinger picture and time-evolved kets with time-independent operators in the Heisenberg picture, we can choose to transform the "inputs" or the operator.