Timeline for Is the trace distance upper bounded by the Euclidean distance?
Current License: CC BY-SA 4.0
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| Feb 29 at 16:09 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 2, 2023 at 7:17 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 2, 2023 at 6:55 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 2, 2023 at 6:41 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 2, 2023 at 6:40 | comment | added | glS♦ | @ZehongFan well, I'm not sure, the comment is hard parse. But the correct calculation is at the end of the post. You should be careful how you define things though. What exactly is "euclidean distance" for you in this context? Are you asking about $\|\rho-\sigma\|_2$ or are you asking about the Euclidean distance between the ket vectors, that is, $\| |\psi\rangle-|\phi\rangle\|_2$? | |
| Aug 2, 2023 at 4:35 | comment | added | Zehong Fan | In the case of pure state, my understanding is as following: $|| |\psi\rangle, |\phi\rangle ||_{tr} = \sqrt{1 - |<\phi|\psi>|^2}$, and the Euclidean distance is $|| |\psi\rangle - |\phi\rangle || = \sqrt{2 - <\phi|\psi> - <\psi|\phi>}$ and the latter is large and equal to the former because $(<\phi|\psi> - 1)(<\psi|\phi> - 1)\ge 0$. Does that have some problem? | |
| Aug 1, 2023 at 20:05 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 1, 2023 at 19:47 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 1, 2023 at 19:34 | history | answered | glS♦ | CC BY-SA 4.0 |