Timeline for Find unitary transformation between two sets of matrices that represent group generators
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 3, 2021 at 23:17 | comment | added | Gerson J Ferreira | Actually, $A_i$ and $B_i$ do not need to be hermitian. They are reps of a finite group, so they are unitary. | |
| May 3, 2021 at 16:55 | history | edited | Marcel | CC BY-SA 4.0 |
deleted 25 characters in body
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| May 2, 2021 at 23:10 | comment | added | Gerson J Ferreira | Yes, all comments above are correct. And I'm sorry that I wasn't precise enough in the definitions and that I took some contextualization for granted. But I guess the problem is well established now. | |
| May 2, 2021 at 22:47 | comment | added | LSpice | Sure, it can be arranged that $U$ (and $S$ and $P$) are unitary, but it is not automatic. (But neither is there such a thing as "the matrices that diagonalize"—indeed, the failure for there to be 'the matrix' is the whole point—so maybe it should be read as "some unitary matrices that diagonalize".) | |
| May 2, 2021 at 22:38 | comment | added | Rorsa | Because $S$ and $P$ can always be made unitary (since they are transformations that diagonalize $A$ and $B$). | |
| May 2, 2021 at 22:26 | comment | added | thedude | @LSpice I think he is assuming $A_i$ and $B_i$ to be hermitian | |
| May 2, 2021 at 22:09 | review | Close votes | |||
| May 12, 2021 at 3:02 | |||||
| May 2, 2021 at 21:42 | comment | added | LSpice | Why should $U = S P^{-1}$ be unitary? | |
| May 2, 2021 at 21:03 | answer | added | Gerson J Ferreira | timeline score: 3 | |
| May 1, 2021 at 15:22 | history | asked | Gerson J Ferreira | CC BY-SA 4.0 |