Timeline for What is the correspondence between operators and measurement in quantum mechanics?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2021 at 12:30 | review | Close votes | |||
| Jul 7, 2021 at 22:21 | |||||
| Jul 6, 2021 at 21:50 | history | edited | al pal | CC BY-SA 4.0 |
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| Jul 6, 2021 at 20:27 | vote | accept | al pal | ||
| Jul 6, 2021 at 20:23 | answer | added | Quantum Mechanic | timeline score: 1 | |
| Jul 6, 2021 at 20:18 | answer | added | J. Murray | timeline score: 3 | |
| Jul 6, 2021 at 19:09 | comment | added | Charlie | Yes, basically. Note that a general spin state doesn't have a spin associated to it, only the spin eigenstates do | |
| Jul 6, 2021 at 19:00 | history | edited | Qmechanic♦ |
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| Jul 6, 2021 at 18:49 | comment | added | al pal | @Charlie. Got it thanks a lot. And then what does Suskind mean when he says the act of measurement prepares the spin in certain state? Does it mean for example if we are measuring along z axis and we get up, what ever the spin was before changes to up along z axis? | |
| Jul 6, 2021 at 18:43 | comment | added | Charlie | As I said in general you dont get anything interesting, if $|A\rangle$ is an eigenstate of $L$ on the other hand the action of $L$ is to multiply the state by the associated eigenvalue which is the value you would get if you measured $L$ and found the system in the state $|A\rangle$. | |
| Jul 6, 2021 at 18:39 | comment | added | al pal | @Charlie And what is that new state we get after $L|A\rangle$? And when do we get that new state? | |
| Jul 6, 2021 at 18:38 | comment | added | Charlie | Apologies my "in general nothing" comment was directed at the last line of your question, not the title. | |
| Jul 6, 2021 at 18:37 | comment | added | Charlie | Ah no, absolutrly not. The eigenvalues of the operators dictate the possible outcomes of measurement. | |
| Jul 6, 2021 at 18:36 | comment | added | al pal | @Charlie Thanks. Would you elaborate further? Are you saying observable operators have nothing to do with measurement, what so ever! | |
| Jul 6, 2021 at 18:32 | comment | added | Charlie | In general nothing. If that is an arbitrary state it doesn't mean anything physical to act on it with an observable. | |
| Jul 6, 2021 at 18:22 | history | asked | al pal | CC BY-SA 4.0 |