Timeline for Eigenvalues of Exchange Operator
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 23, 2022 at 22:57 | history | edited | ZeroTheHero | CC BY-SA 4.0 |
added 1042 characters in body
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| Jun 23, 2022 at 22:40 | comment | added | SimoBartz | so maybe these are both true.. interesting | |
| Jun 23, 2022 at 22:38 | comment | added | SimoBartz | Sorry i deleted the comment before you posted. I agree, but consider this $\hat P \hat P\psi(1,2)=\psi(1,2)$ then since $\psi(1,2)=e^{i \lambda} \psi(1,2)$ we have $\hat P \hat P\psi(1,2)=\psi(1,2)=e^{i \lambda} \psi(1,2)$ namely $\hat P \hat P\psi(1,2)=e^{i \lambda} \psi(1,2)$ | |
| Jun 23, 2022 at 22:35 | comment | added | ZeroTheHero | you can’t, which is why you need properly symmetrized states. This way none of the measurable quantities depend on the assignment of labels. | |
| Jun 23, 2022 at 22:03 | comment | added | ZeroTheHero | Particles are labelled by their coordinate (ignoring spin as per your example) so how do you differentiate between exchanging coordinates and exchanging particles? | |
| Jun 23, 2022 at 21:53 | comment | added | SimoBartz | I don't understand why this should answer to the question, if $\hat P$ is seen as coordinate exchange operator i can see $\hat P \hat P=I$ but if you interpret it as operator that exchanges particles then it's not ture | |
| Jun 23, 2022 at 14:45 | history | answered | ZeroTheHero | CC BY-SA 4.0 |