Timeline for How are the eigenvalues of $\rho=\frac12(|a\rangle\!\langle a| +|b\rangle\!\langle b|)$ derived?
Current License: CC BY-SA 4.0
7 events
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| Feb 10, 2021 at 20:58 | comment | added | Adam Zalcman | I intentionally tried to avoid expanding $|a\rangle$ and $|b\rangle$ in a basis since both the density matrix and the formula for eigenvalues were given using Dirac notation in terms of kets and bras like $|a\rangle$. I thought to myself that we should be able to go from one to the other entirely using Dirac notation and the two kets and bras. Happily, it turned out that it works out :-) | |
| Feb 10, 2021 at 20:50 | comment | added | Adam Zalcman | No shortcut. I computed $\rho^2$ under the trace getting $\frac{1}{4}(|a\rangle\langle a|a\rangle\langle a| + |a\rangle\langle a|b\rangle\langle b| + |b\rangle\langle b|a\rangle\langle a| + |b\rangle\langle b|b\rangle\langle b|)$. Then, I computed the inner products, e.g. $\langle a|a\rangle=1$ and $\langle a|b\rangle$ (which I left unchanged). Next, I applied trace to each term, e.g. $\mathrm{tr}(|a\rangle\langle a|) = \langle a|a\rangle = 1$ and $\mathrm{tr}(|a\rangle\langle b|) = \langle b|a\rangle$. Finally, I collected all terms into $\frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2}$. | |
| Feb 10, 2021 at 20:19 | comment | added | forky40 | is there a shortcut that you used to evaluate $\text{Tr} \rho^2 = \frac{1}{4} \text{Tr} (|a\rangle\langle a| + \dots )$ instead of substituting $|a\rangle,|b\rangle$ written in terms of orthonormal basis like $|a\rangle = a_0 |0\rangle + a_1 |1\rangle$ ? | |
| Feb 10, 2021 at 7:46 | comment | added | KAJ226 | Nice. I like this! | |
| Feb 7, 2021 at 0:23 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Feb 7, 2021 at 0:04 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Feb 6, 2021 at 23:56 | history | answered | Adam Zalcman | CC BY-SA 4.0 |