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Let's say I have a bipartite density operator $\gamma_{12} = (1 - \epsilon) \rho_{12} + \epsilon\sigma_{12}$, for $0 \le \epsilon \le 1$, i.e., a convex combination of $\rho_{12}$ and $\sigma_{12}$. I want to show that ($S$ represents Von Neumann entropy):

$$ S(\gamma_{12} | \gamma_2) \ge (1 - \epsilon) S(\rho_{12} | \rho_2) + \epsilon S( \sigma_{12} | \sigma_2). $$ The note that I am following says that this is due to the concavity of conditional entropy, which is not immediately obvious to me. I tried to derive it in the following way:

$$ \begin{align} S(\gamma_{12} | \gamma_2) &= S(\gamma_{12}) - S(\gamma_2) \\ &= S((1 - \epsilon) \rho_{12} + \epsilon\sigma_{12}) - S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \;\; \text{[definition of $\gamma_{12}$ and using partial trace] } \\ &\ge (1 - \epsilon) S(\rho_{12}) + \epsilon S(\sigma_{12}) - S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \;\; \text{[using concavity in the first S] } \\ &\stackrel{?}{\ge} (1 - \epsilon) S(\rho_{12}) + \epsilon S(\sigma_{12}) - (1 - \epsilon)S(\rho_2) - \epsilon S(\sigma_2). \end{align} $$ This of course, gives me the desired inequality. But how come the last inequality is true? Isn't $S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \ge (1 - \epsilon)S(\rho_2) - \epsilon S(\sigma_2)$ due to concavity? Thanks!

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    $\begingroup$ The last inequality isn't true. If $S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \ge (1 - \epsilon)S(\rho_2) + \epsilon S(\sigma_2)$, then the inequality on the last line should be $\le$ $\endgroup$ Aug 26 at 12:07
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First, encode the bipartite ensemble $\gamma_{12}$ into a CQ state $\omega$ $$\omega_{XAB}=\sum_{x}p_{x}(x)|x\rangle\langle x|\otimes\rho_{AB}^{x}$$ now we can take the difference between $$H(A|B)\gamma_{12}-\sum_{x}p_{x}(x)H(A|B)_{\rho^{x}}$$ where $\rho_{x}$ will be one of the density operators you have in your ensemble. Using the CQ state to rewrite the above difference: $$H(A|B)_{\omega}-H(A|BX)_{\omega}=I(A:X|B)_{\omega}=H(X|B)_{\omega}-H(X|AB)_{\omega}$$ $$=D(\omega_{XAB}||I_{X}\otimes\omega_{AB})-D(\omega_{AB}||I_{X}\otimes\omega_{B})$$ now notice that I can derive the second relative entropy by the first via the action of a partial trace map in subsystem A. Due to the monotonicity of the relative entropy, I then get $$D(\omega_{XAB}||I_{X}\otimes\omega_{AB})- D(\omega_{AB}||I_{X}\otimes\omega_{B}) \ge0$$ subbing this back in $$H(A|B)\gamma_{12}-\sum_{x}p_{x}(x)H(A|B)_{\rho^{x}}=D(\omega_{XAB}||I_{X}\otimes\omega_{AB})- D(\omega_{AB}||I_{X}\otimes\omega_{B}) \ge0$$ so $$H(A|B)\gamma_{12}-\sum_{x}p_{x}(x)H(A|B)_{\rho^{x}}\ge 0$$

Alternatively, after getting to $I(A:X|B)_{\omega}$ you could just use strong subadditivity to show $I(A:X|B)_{\omega}\ge 0$

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