Timeline for Showing that measurement of spin parity does not conserve total angular momentum
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2023 at 3:22 | vote | accept | DisposableGuy | ||
| Aug 16, 2023 at 14:17 | comment | added | Buzz♦ | @DisposableGuy Yes, that is correct. | |
| Aug 16, 2023 at 4:01 | comment | added | DisposableGuy | Could you confirm that this is right what I said? | |
| Aug 15, 2023 at 2:21 | vote | accept | DisposableGuy | ||
| Aug 20, 2023 at 3:22 | |||||
| Aug 14, 2023 at 2:22 | comment | added | DisposableGuy | So I found out that $S^2$ and $S_z$ do commute with $P$, but $S_x$ and $S_y$ do not commute with $P$, which should mean that the operator $P$ does not conserve total angular momentum. So it is sufficient for only one of the components of $S$ to not commute with an operator $O$ to say that the total angular momentum is not conserved? | |
| Aug 13, 2023 at 22:07 | comment | added | Buzz♦ | @DisposableGuy See my edit. | |
| Aug 13, 2023 at 22:07 | history | edited | Buzz♦ | CC BY-SA 4.0 |
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| Aug 13, 2023 at 8:58 | comment | added | DisposableGuy | Thanks for the answer. I have two thoughts on this: 1. I think I got it initially wrong because I found out that $S^2$ actually only covers the magnitude of the spin operator, I think I was really looking for $\hat{S}$ but I don't really know how to get this for two spin half particles. 2. Regarding your answer: This means I need to show that every component $S_x$, $S_y$ and $S_z$ does not commute with $P$? Does this need to be true for all of them or is it enough that only one component does not commute with $P$? | |
| Aug 13, 2023 at 7:17 | history | answered | Buzz♦ | CC BY-SA 4.0 |