Timeline for Show that $\lambda |\phi^+\rangle \langle\phi^+| + (1-\lambda )|\phi^-\rangle \langle\phi^-|$ is the Choi–Jamiołkowski matrix of a quantum channel
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2021 at 12:22 | vote | accept | mikanim | ||
| Mar 8, 2021 at 12:21 | comment | added | glS♦ | it doesn't matter what $|u\rangle$ is. The operator $|u\rangle\!\langle u|$ is always positive semidefinite. An easy way to see it is that it has a single nonzero eigenvalue which is equal to $\|u\|$. Equivalently, you can verify that $\langle \alpha|u\rangle\!\langle u|\alpha\ge0$ for all $|\alpha\rangle$, which is again true because you can verify picking $|\alpha\rangle$ from an orthonormal basis containing $|u\rangle$. Equivalently, if $\|u\|=1$, just notice that $|u\rangle\!\langle u|$ is the projection onto $|u\rangle$. | |
| Mar 8, 2021 at 12:18 | comment | added | glS♦ | adding questions only makes the post too broad and liable to be closed. You can remove them by editing this post and ask the other questions on a separate thread. | |
| Mar 8, 2021 at 11:54 | comment | added | mikanim | right but $|u \rangle$ in this case is a bipartite state if we look at it in the computational basis. I only see that it is positive semidefinite once I calculate out the matrix into the computational basis, which took 2-3 lines. Could you elaborate a bit on how you found it so quick? | |
| Mar 8, 2021 at 11:45 | comment | added | glS♦ | @mikanim because the sum of positive semidefinite operators is positive semidefinite, and $|u\rangle\!\langle u|$ is always positive semidefinite | |
| Mar 8, 2021 at 11:45 | comment | added | mikanim | Thanks for the comment! How is the positivity immediate? | |
| Mar 8, 2021 at 11:44 | history | edited | glS♦ | CC BY-SA 4.0 |
added 69 characters in body
|
| Mar 8, 2021 at 11:33 | history | answered | glS♦ | CC BY-SA 4.0 |