Timeline for Equivalence of two ways to recover a map from its Choi state
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2021 at 21:26 | vote | accept | glS♦ | ||
| Jan 31, 2021 at 20:50 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
added 61 characters in body
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| Jan 31, 2021 at 17:06 | 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. | |
| Oct 3, 2020 at 16:08 | 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. | |
| Sep 7, 2020 at 10:12 | comment | added | Markus Heinrich | What I actually meant was that inverting is trivial for $|u\rangle\langle v|$, thus for any matrix $M=\sum_k x_k |u_k\rangle\langle v_k|$ the associated superoperator has to be $\Phi(X) = \sum_k x_k U_k X V_k^\dagger$ (operators not necessarily unitary). The only thing to justify is the inversion formula Eq. (2), which you already did in you first post. | |
| Sep 7, 2020 at 9:46 | comment | added | Markus Heinrich | It is clear how the inverse should look like on rank-one matrices $|u\rangle\langle u|$, or even more generally $|u\rangle\langle v|$. Thus, one only needs to check that Eq. (3) gives the right result for those which is straightforward. The general result follows via a decomposition into those (e.g. SVD for general matrices). But that's basically what you did. | |
| Sep 7, 2020 at 8:42 | comment | added | glS♦ | @MarkusHeinrich can you expand as to which part exactly follows from linearity? | |
| Sep 7, 2020 at 8:06 | comment | added | Markus Heinrich | This follows from linearity of the isomorphism. | |
| S Sep 3, 2020 at 15:19 | answer | added | glS♦ | timeline score: 1 | |
| S Sep 3, 2020 at 15:19 | history | asked | glS♦ | CC BY-SA 4.0 |