It is known that the following result holds for quantum relative entropy.

$S(\rho_{AB} ||\sigma_{AB}) \geq S(\rho_A ||\sigma_{A})$,

where we have traced out system B on the right hand side. The proof of this is quite complex and proved by Lieb and Ruskai.

The usual interpretation of this is that given a true state $\rho$ and a hypothesis state $\sigma$, we need at least $N$ copies where $NS\gg 1$ before we can know our hypothesis was wrong.

Let us now interpret the quantum relative entropy (after symmetrizing it) as our ability to distinguish between two density matrices $\rho$ and $\sigma$. It is well known and easily proved that taking the partial trace (or indeed doing any trace preserving operation) always decreases the trace distance between two density matrices. In other words, taking the partial trace makes states less distinguishable. But if this connection between distinguishability and relative entropy is proven rigorously, then the monotonicity of relative entropy becomes trivial.

My questions are:

1) Is my interpretation of quantum relative entropy as a measure of distinguishability correct?

2) If not (as I guess it must be), can one provide a counter example of two pairs of states that show why this interpretation is incorrect?

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    $\begingroup$ It is a measure of distinguishability, but different from the trace distance. I am not sure I understand the problem. $\endgroup$ – Norbert Schuch Dec 13 '18 at 13:07

I believe that your question is fully answered by the Wikipedia article. Indeed, Lindblad (1975) showed that Lieb and Ruskai's result implies that the relative entropy satisfies the data-processing inequality, that is,

$$ S(\Lambda(\rho) \| \Lambda(\sigma)) \geq S(\rho \| \sigma) $$

for any completely positive trace-preserving map $\Lambda$. Therefore, it can be regarded as a measure of distinguishability in the way that you describe.


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