Timeline for Does approximate equality of quantum states imply operator inequality in a large subspace?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 29, 2023 at 16:25 | comment | added | Iosif Pinelis | @GuillaumeAubrun : Right. Somehow I perceived $H$ as $\max(0,\cdot)$. | |
| Jun 29, 2023 at 16:18 | comment | added | Guillaume Aubrun | @IosifPinelis applying a $\{0,1\}$-valued function to a self-adjoint operator always produces a projector. | |
| Jun 29, 2023 at 16:17 | history | edited | Guillaume Aubrun | CC BY-SA 4.0 |
added 37 characters in body
|
| Jun 29, 2023 at 16:14 | comment | added | Guillaume Aubrun | @Noel oops, you are right (and I should have read your notes better) | |
| Jun 29, 2023 at 15:56 | comment | added | Noel | You get $ \Pi (1+\sqrt{\epsilon})\sigma \Pi \geq \Pi \rho \Pi$, how do you transform this to $(1+\sqrt{\epsilon})\sigma \geq \Pi \rho \Pi $? It doesn't seem to me that $\Pi$ and $\sigma$ commute. This seems to me to be the same argument as Theorem 2.2 in my notes (github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/…) | |
| Jun 29, 2023 at 15:25 | comment | added | Iosif Pinelis | I think your $\Pi$ is not in general a projector, even when $\rho$ and $\sigma$ are diagonal. | |
| Jun 29, 2023 at 7:04 | history | answered | Guillaume Aubrun | CC BY-SA 4.0 |