Timeline for On commutator of bounded operators
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 6 at 18:33 | history | became hot network question | |||
| May 6 at 14:14 | comment | added | Mikael de la Salle | More generally, the claim is easy if one of $J$ or $K$ has a basis of eigenvectors. Indeed, if $\xi$ is an eigenvector for $J$ or $K$, then $\langle [J,K] \xi,\xi\rangle=0$. The approximation argument does not seem straighforward to me, though. | |
| May 6 at 11:50 | comment | added | Martin Hairer | If one of the operators is trace class, it follows immediately from the fact that the trace of $[J,K]$ (or any commutator) vanishes. It shouldn't be hard to get the general case by approximation. | |
| May 6 at 11:37 | answer | added | Nik Weaver | timeline score: 19 | |
| May 6 at 10:33 | history | asked | Bazin | CC BY-SA 4.0 |