Timeline for Reasoning Check: Trace of squared mixed-state density matrix
Current License: CC BY-SA 3.0
16 events
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| Jul 31, 2019 at 19:03 | 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. | |
| Mar 30, 2019 at 17:02 | 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. | |
| Feb 22, 2019 at 15:00 | history | tweeted | twitter.com/StackPhysics/status/1098960646446739456 | ||
| Feb 22, 2019 at 3:15 | answer | added | Nguyen D. H. Minh | timeline score: 6 | |
| Aug 24, 2018 at 11:27 | history | edited | Qmechanic♦ |
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| Aug 24, 2018 at 10:40 | answer | added | sslucifer | timeline score: 2 | |
| Nov 11, 2016 at 23:31 | comment | added | miggle | @CraigGidney I understand that, that's not the issue. It's OK anyway, I realized that my proof posted above only needs the Cauchy-Schwarz inequality to be rigorous anyway. | |
| Nov 11, 2016 at 11:42 | comment | added | Craig Gidney | @mflynn You'll always be able to rewrite it in terms of orthogonal projectors. Those projectors correspond to the eigenvectors. | |
| Nov 11, 2016 at 7:03 | comment | added | miggle | @CraigGidney right, so I understand the pure case - that makes sense. However the mixed case will in general be written in terms of non-orthogonal projectors. I can change my basis and the trace will be invariant, but then my expansion of $\rho^{2}$ will not be in terms of $p_{i}$, the original expansion coefficients. So I'm not sure how to show the result in the diagonal basis, since the expansion coefficients no longer satisfy any normalization condition. | |
| Nov 11, 2016 at 6:21 | comment | added | pppqqq | @mflynn sorry, I think I don't understand your question. You say that you want to prove $\text {tr}(\rho ^2) <1 \implies \text{state is mixed}$. But this is clearly implied by $\text {state is pure} \implies \text {tr} (\rho ^2) =1$. It is totally irrelevant whether $\rho$ is written as a sum of orthogonal or nonorthogonal projectors, all that matters here is its trace. Perhaps you want to know if it's possible that $\text {tr} (\rho ^2)=1$ for a mixed state? | |
| Nov 11, 2016 at 5:58 | comment | added | miggle | @CraigGidney, could you write something a little more detailed? I understand your comment, but I can't see how I can relate the expansion coefficients in the basis where $\rho^{2}$ is diagonal to the original $p_{i}$ which obey the nice normalization condition. | |
| Nov 11, 2016 at 4:34 | comment | added | miggle | Umm OK, I'll need to think about that a little. I can hit $\rho^{2}$ with unitaries to make it diagonal but I need to think about what the coefficients in the expansion of $\rho^{2}$ will look like. | |
| Nov 11, 2016 at 2:29 | comment | added | Craig Gidney | A Hermitian matrix is guaranteed to have a decomposition into orthogonal eigenvectors. Just use that decomposition instead of the non-orthogonal one. | |
| Nov 10, 2016 at 22:56 | comment | added | miggle | The point is to check that trace($\rho^{2}$) < 1 for mixed states that sum over a set of non-orthonormal quantum states. I understand the pure state case. | |
| Nov 10, 2016 at 22:07 | comment | added | pppqqq | A pure state has a density matrix $\rho = \vert \rangle \langle \vert$, which has $\rho ^2=\rho$ and trace $=1$. Am I missing something? | |
| Nov 10, 2016 at 19:16 | history | asked | miggle | CC BY-SA 3.0 |