Concavity of Conditional Quantum Entropy

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!

• 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 $\le$ Aug 26 '21 at 12:07

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 $$I(A:X|B)_{\omega}\ge 0$$