Timeline for If all the eigenvalues of an operator are real, is the operator Hermitian?
Current License: CC BY-SA 3.0
11 events
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| S Apr 25, 2018 at 21:44 | history | suggested | user74560 | CC BY-SA 3.0 |
Title changed to reflect the actual question
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| Apr 25, 2018 at 21:15 | review | Suggested edits | |||
| S Apr 25, 2018 at 21:44 | |||||
| Nov 8, 2011 at 18:31 | comment | added | Vladimir Kalitvianski | @RonMaimon: Yes, I have published it. First, it was a preprint in Russian and then, two articles in Russian journals. I submitted an English version of it on arXiv: arxiv.org/abs/0906.3504 | |
| Nov 8, 2011 at 18:02 | comment | added | Ron Maimon | @Vladimir: this is very fasionable nowadays--- you should publish your analysis (if you haven't already)--- this is PT quantum mechanics. Do you have a reference or a better description of V? | |
| Nov 8, 2011 at 17:35 | vote | accept | a06e | ||
| Nov 8, 2011 at 14:15 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
hermitic-> hermitian; removed greeting; retagged;
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| Nov 8, 2011 at 11:20 | history | tweeted | twitter.com/#!/StackPhysics/status/133866592479088641 | ||
| Nov 8, 2011 at 10:45 | comment | added | Vladimir Kalitvianski | In my practice I encountered an differential operator (a Hamiltonian), which was Hermitian in a space of functions with a non trivial scalar product, like $(\psi,\phi) = \int \psi(x)\phi (x)\rho (x) dx$. This Hamiltonian could be split into two parts: $\hat H = \hat H_0 + \hat V$, where $\hat H_0$ was hermitian in another space - with a different scalar product ($\rho =1$). Naturally the "perturbation" operator $\hat V$ was not Hermitian in this new basis. Yet, the perturbation theory worked. | |
| Nov 8, 2011 at 10:29 | answer | added | Qmechanic♦ | timeline score: 16 | |
| Nov 8, 2011 at 5:15 | answer | added | Ron Maimon | timeline score: 18 | |
| Nov 8, 2011 at 4:42 | history | asked | a06e | CC BY-SA 3.0 |