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Jul 28, 2022 at 3:52 history edited Qmechanic
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Jul 28, 2022 at 3:07 answer added hft timeline score: 0
Jul 28, 2022 at 3:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Mar 3, 2022 at 20:45 history edited Cosmas Zachos CC BY-SA 4.0
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Mar 3, 2022 at 20:04 answer added Cosmas Zachos timeline score: 2
Mar 3, 2022 at 12:59 history edited Qmechanic
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Mar 3, 2022 at 9:32 comment added electrovolt Oh, I didn't think of commutation. I think it must include some commutator term as well. Thanks for pointing that out.
Mar 3, 2022 at 9:31 comment added AfterShave I think that identity only works if all the operators commute, the first term looks suspicious.
Mar 3, 2022 at 9:31 comment added ACuriousMind The "vector triple product relation" is presumably something derived for vectors with numbers as entries, i.e. vectors whose components commute. Why do you think you can blindly apply it to a vector of operators?
Mar 3, 2022 at 9:30 comment added electrovolt But if I write the vector product $p \times (r \times p)$, I would get $(p \cdot p)r - (p \cdot r)p$. If I take the hermitian conjugate, it appears to be Hermitian. Where am I possibly going wrong?
Mar 3, 2022 at 9:27 comment added electrovolt Yeah, it doesn't appear to be Hermitian. Also, I wanted to state that $p^2$ is not Hermitian in general, and $rp^2$ won't commute as well.
Mar 3, 2022 at 9:26 comment added AfterShave I was wrong sorry, the operator appears not to be hermitian.
Mar 3, 2022 at 9:15 vote accept electrovolt
Mar 3, 2022 at 9:15
Mar 3, 2022 at 8:47 history edited electrovolt CC BY-SA 4.0
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S Mar 3, 2022 at 8:18 review First questions
Mar 3, 2022 at 12:25
S Mar 3, 2022 at 8:18 history asked electrovolt CC BY-SA 4.0