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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?

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If you represented the ddθ operator for example "under the parity operation" (what does that mean?) as a matrix, wouldn't it be the exact same matrix?

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.

It's just that the input wavefunction that is being input as argument to the ddθ operator has had all its + and − signs flipped

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.

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  • $\begingroup$ Thanks for pointing out that you can think of a unitary operator transforming another operator, with the expression $U^{-1}AU$. Will think more on your other points.. $\endgroup$ – a00 Sep 7 '15 at 19:08
  • $\begingroup$ It makes more sense now. If we consider $U^{-1}AU$ to be "moving into the frame created by the unitary transformation, for example the 'mirror image of reality' created by the parity operator, then performing the original operation $A$ in that 'alternate reality', then switching back to the original 'home coordinates' frame", then it makes sense to call this transforming the operator itself. Thank you for clearing that up. $\endgroup$ – a00 Sep 8 '15 at 12:58
  • $\begingroup$ I believe that in going from $\langle \Psi | \Pi^{-1} A \Pi | \Psi \rangle$ to $\int d\mathbf{x'} d\mathbf{x''} \langle \Psi | \Pi^{-1} | \mathbf{x'} \rangle \langle \mathbf{x'} | A | \mathbf{x''} \rangle \langle \mathbf{x''} | \Pi | \Psi \rangle$ you are just saying that to evaluate expectation value in the position basis you have to stick those $| \mathbf{x'} \rangle \langle \mathbf{x'} |$ $| \mathbf{x''} \rangle \langle \mathbf{x''} |$ in those places. $\endgroup$ – a00 Sep 10 '15 at 13:52
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    $\begingroup$ Yes, that's correct $\endgroup$ – Kyle Arean-Raines Sep 10 '15 at 13:54
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    $\begingroup$ Well, let me clarify that the parity operator, because it's Hermitian, can work on the $\langle \mathbf{x''} \mid$ bra to give $\langle -\mathbf{x''} \mid$, and when you take its inner product with $\mid \Psi \rangle$ you get $\Psi(-\mathbf{x''})$ $\endgroup$ – Kyle Arean-Raines Sep 10 '15 at 14:00

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