Timeline for Does separability of a matrix implies the matrix is a density matrix?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2023 at 7:24 | answer | added | DaftWullie | timeline score: 2 | |
| Oct 24, 2023 at 15:53 | answer | added | forky40 | timeline score: 2 | |
| Oct 24, 2023 at 13:21 | history | edited | Physkid | CC BY-SA 4.0 |
added 27 characters in body
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| Oct 24, 2023 at 13:21 | comment | added | Physkid | @Callum I meant the latter. To your response, it wouldn't even be a matrix if the former. Edited the OP to remove ambiguity. | |
| Oct 24, 2023 at 13:20 | comment | added | Callum | When you say the matrix can be shown to be a tensor product of two states. Do you mean $|\psi\rangle \otimes |\phi\rangle$ or $|\psi\rangle\langle \psi | \otimes |\phi\rangle\langle \phi |$? The former wouldn't be a density matrix as its the wrong shape. | |
| Oct 24, 2023 at 13:15 | history | asked | Physkid | CC BY-SA 4.0 |