Timeline for Is this correct? $Tr_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2016 at 16:15 | vote | accept | XXDD | ||
| Feb 25, 2016 at 12:55 | answer | added | Emilio Pisanty | timeline score: 1 | |
| Feb 25, 2016 at 12:05 | history | edited | XXDD | CC BY-SA 3.0 |
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| Feb 25, 2016 at 11:51 | answer | added | XXDD | timeline score: 0 | |
| Feb 25, 2016 at 11:28 | history | edited | XXDD | CC BY-SA 3.0 |
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| Feb 25, 2016 at 11:22 | history | edited | XXDD | CC BY-SA 3.0 |
added 154 characters in body
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| Feb 25, 2016 at 11:10 | comment | added | XXDD | No, I am not using $Tr(XY)=Tr(YX)$ on partial trace. In fact I only used it in computing $Tr(o_{ij})$. | |
| Feb 25, 2016 at 10:34 | answer | added | Martin | timeline score: 2 | |
| Feb 25, 2016 at 10:03 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
LaTeX fixes.
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| Feb 25, 2016 at 9:57 | comment | added | yuggib | It seems to me that you are trying to use the property of the trace $\mathrm{Tr}(XY)=\mathrm{Tr}(YX)$, but you can't do that on the partial trace $\mathrm{Tr}_B(X_{AB}Y_{AB})$ since $Y_{AB}$ (and $X_{AB}$) are still operators with an $A$-dependence (and obviously do not commute) | |
| Feb 25, 2016 at 9:51 | history | asked | XXDD | CC BY-SA 3.0 |