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I'm trying to understand the proof of the so called subadditivity of the von Neumann entropy, $$S(\rho^{AB}\,)\leq{S}(\rho^A)+S(\rho^B)$$ where $S(\rho)=-\mathrm{tr}\{\rho\log\rho\}$.

In the proof I have, I'm given the definition of relative entropy $$S(\rho\|\sigma)=-S(\rho)-\mathrm{tr}\rho\ln\sigma$$ and the property $S(\rho\|\sigma)\geq0$; then with $\rho=\rho^{AB}$ and $\sigma=\rho^A\otimes\rho^B$ it argues that it follows that $$S(\rho^{AB})\leq-\mathrm{tr}\left[\rho^{AB}(\ln\rho^A+\ln\rho^B)\right]=S(\rho^A)+S(\rho^B)$$ but I can't reproduce this explicitly; as I understand, writing everything out this means \begin{align}S(\rho^{AB})\leq&-\mathrm{tr}\left[\rho^{AB}\ln(\rho^A\otimes\rho^B\right]\\&=-\mathrm{tr}\left[\rho^{AB}\left\{\ln(\rho^A\otimes{I}^B)+\ln(I^A\otimes\rho^B)\right\}\right]\\ &=-\mathrm{tr}\left[\rho^{AB}\left(\ln\rho^A\otimes{I}^B+I^A\otimes\ln\rho^B\right)\right]\\ &=-\mathrm{tr}\left[\rho^{AB}(\ln\rho^A\otimes{I}^B)\right]-\mathrm{tr}\left[\rho^{AB}(I^A\otimes\ln\rho^B)\right]\end{align} so specifically, if this is right, I don't understand how $$S(\rho^{A})\stackrel{?}{=}-\mathrm{tr}\left[\rho^{AB}(\ln\rho^A\otimes{I}^B)\right]$$ and similarly for $S(\rho^B)$.

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  • $\begingroup$ Proof from which reference? $\endgroup$ – Qmechanic Feb 10 '17 at 22:01
  • $\begingroup$ It's from a set of lecture notes (2015) by prof. Joan Simón from the University of Edinburgh. I didn't cite because I don't think they're public. $\endgroup$ – Cal Gibson Feb 10 '17 at 23:11
  • $\begingroup$ Can you use Wikipedia and refer to standard formulas there? It would be a more accessible question. $\endgroup$ – Cosmas Zachos Feb 11 '17 at 1:49
  • $\begingroup$ The notation is pretty much the same; I denote the system $AB$ and the subsystems $A$, $B$ instead of 12 and 1, 2, but that's it, I think. $\endgroup$ – Cal Gibson Feb 11 '17 at 2:59
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The short version of your question is

Why is $\mathrm{tr}_{AB}(X_{AB}(Y_A\otimes I_B)) = \mathrm{tr}_A(X_AY_A)$, where $X_A=\mathrm{tr}_B(X_{AB})$ is the partial trace of $X_{AB}$ over $B$.

It should be immediate that this implies what you ask.

To prove it, you just need the definition of the partial trace: $$ \mathrm{tr}_B(Z_{AB}) = \sum_B \langle i|_B Z_{AB}|i\rangle_B\ , $$ and the fact that $\mathrm{tr}_{AB}(Z)=\mathrm{tr}_A(\mathrm{tr}_B(Z))$.

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