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Assume that you have two bipartite systems $\rho_1^{AB},\rho_2^{AB}$ then I would like to prove the following:

$$S(\frac{1}{2}( \rho_1^{AB}+I^A\otimes\rho_2^B))+S(\frac{1}{2}(\rho_2^{AB}+I^A\otimes\rho_1^B)) \geq S(\frac{1}{2}(\rho_1^{AB}+I^A\otimes\rho_1^B))+S(\frac{1}{2}(\rho_2^{AB}+I^A\otimes\rho_2^B))$$

where $S$ is the von Neumann entropy, $\rho_1^B=tr_A(\rho_1^{AB}),\rho_2^B=tr_A(\rho_2^{AB})$ and $I^A$ is the maximally mixed state on $A$. It looks like it should pass with some monotony property, any hints or counterexample are welcome.

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    $\begingroup$ Are you sure you mean $\rho_1^A=tr(\rho_1^{AB})$, etc.? Because that's just a number... One. You probably wanted to take the partial trace over $B$? $\endgroup$
    – Bubble
    Commented Sep 30, 2013 at 21:33

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(very) partial answer:

In a very particular case, this is true.

Let's have $\rho_1^{AB} = I^A \otimes \rho_1^{B}$ and $\rho_2^{AB} = I^A \otimes \rho_2^{B}$

The left hand side is $L = 2 S( \frac{1}{2}(\rho_1^{AB}+\rho_2^{AB})) $

The right hand side is $R = S(\rho_1^{AB}) + S(\rho_2^{AB})$

By the concavity of the Von Neumann entropy, we have $L \ge R$

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  • $\begingroup$ Well, it sheds some light on what arguments can be used. I'm thinking of how it could be generalized... $\endgroup$
    – Ando Masahashi
    Commented Oct 1, 2013 at 10:38
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    $\begingroup$ In some sense, this very particular case is very "trivial". For instance, a next step of difficulty would be something like $\rho_1^{AB} = \rho^A \otimes \rho_1^{B}, \rho_2^{AB} = \rho^A \otimes \rho_2^{B}$, but I have absolute no idea to prove it ... $\endgroup$
    – Trimok
    Commented Oct 1, 2013 at 10:52

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