Timeline for "This operator is odd under parity"
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2015 at 19:38 | vote | accept | a00 | ||
| Sep 11, 2015 at 19:37 | comment | added | a00 | I now understand that, as well as what you meant by using the orthogonality of position states: $\int d\mathbf{x'} d\mathbf{x''} \langle\Psi|\Pi^{-1}|\mathbf{x'}\rangle \langle|\mathbf{x'}|A|\mathbf{x''}\rangle \langle\mathbf{x''}| \Pi|\Psi\rangle$ being a double integral, there are tons of terms of the form $c_0\langle\mathbf{x'}=(777.88,9,-10)|A (c_1|\mathbf{x''}=(-56.5,-1,22.22)\rangle)$ for example, but they all evaluate to 0 except when $\mathbf{x'} = \mathbf{x''}$. Thanks a lot for the educational exposition! | |
| Sep 10, 2015 at 14:00 | comment | added | Kyle Arean-Raines | 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 $\Psi(-\mathbf{x''})$ | |
| Sep 10, 2015 at 13:54 | comment | added | Kyle Arean-Raines | Yes, that's correct | |
| Sep 10, 2015 at 13:52 | comment | added | a00 | Once you've done that, then resolving the right hand side you are just noting that "Clearly, applying $| \mathbf{x''} \rangle \langle \mathbf{x''} |$ to the ket $(\Pi | \Psi \rangle)$ is clearly just going to be the position representation of $|\Psi\rangle$ but with its input position mirrored" (that's what the parity operator does). Hence, $\Psi(-\mathbf{x''})$. | |
| Sep 10, 2015 at 13:52 | comment | added | a00 | 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. | |
| Sep 8, 2015 at 12:58 | comment | added | a00 | 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. | |
| Sep 7, 2015 at 19:08 | comment | added | a00 | 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.. | |
| Sep 7, 2015 at 16:19 | history | answered | Kyle Arean-Raines | CC BY-SA 3.0 |