Timeline for Cross Product of two Hermitian Operators
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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 | ||
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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 | |||
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S Mar 3, 2022 at 8:18 | history | asked | electrovolt | CC BY-SA 4.0 |