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Oct 14, 2022 at 16:41 vote accept peter mafai
Sep 29, 2022 at 13:11 answer added peter mafai timeline score: 0
Sep 29, 2022 at 7:37 history edited Tobias Fünke CC BY-SA 4.0
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Sep 28, 2022 at 22:01 comment added Tobias Fünke Note that $\rho_{12}\,\rho_{21}=|\rho_{12}|^2 \leq \rho_{11} \,\rho_{22}$, cf. here, and thus, regarding your $2\times 2$ example (where I assume you mean an ONB here): $\mathrm{Tr}\rho^2 \leq \rho_{11}^2 +2\rho_{11}\,\rho_{22}+\rho_{22}^2=(\rho_{11}+\rho_{22})^2 = (\mathrm{Tr}\rho)^2=1$. As shown in the linked answer, the equality holds if and only if $\rho$ is pure.
Sep 28, 2022 at 21:39 answer added hft timeline score: 2
Sep 28, 2022 at 21:26 history edited peter mafai CC BY-SA 4.0
expand question with an example
Sep 28, 2022 at 10:27 history edited Qmechanic CC BY-SA 4.0
edited tags; edited title
S Sep 28, 2022 at 10:26 history suggested lcv CC BY-SA 4.0
Fixed title and a few typesetting errors
Sep 28, 2022 at 9:23 review Suggested edits
S Sep 28, 2022 at 10:26
Sep 28, 2022 at 9:18 answer added lcv timeline score: 5
Sep 28, 2022 at 8:59 comment added flippiefanus It is a simple property of any set of positive values that add up to one that their squares would add up to a value smaller than one.
Sep 28, 2022 at 8:53 history edited peter mafai CC BY-SA 4.0
added 6 characters in body
Sep 28, 2022 at 8:45 comment added flippiefanus No, you should have learned that tr$\{\rho\}=1$ for all states due to normalization. For pure states tr$\{\rho^2\}=1$ and for mixed states tr$\{\rho^2\}<1$. The situation tr$\{\rho^2\}>1$ is then not possible.
Sep 28, 2022 at 7:48 history asked peter mafai CC BY-SA 4.0