Quantum relative entropy between pre- and post-measurement states

Question

The quantum relative entropy between the states $\rho$ and $\sigma$ is defined by $$D(\rho||\sigma)= \textrm{tr}\Big(\rho \big(\log\rho - \log \sigma \big) \Big)\,,$$ as long as the support of $\rho$ is contained within the support of $\sigma$. I want to show that the relative entropy between a state $\rho$ and the post-measurement state after performing a projective measurement on $\rho$ is greater if that measurement is rank-one than if it is coarse-grained. More particularly, denote a coarse-grained projective measurement by $\{\Pi_i\}_{i}=\sum_j | f_{j}^{i} \rangle \langle f_{j}^{i}|$, where $\{|f_{j}^{i}\rangle\}_{i,j}$ is an orthonormal basis, and the post-measurement states $\rho'_{\textrm{coarse}}$ and $\rho'_{\textrm{rank-one}}$ by $$\rho'_{\textrm{coarse}}= \sum_{i}\Pi_i \rho \Pi_i\,, \qquad \rho'_{\textrm{rank-one}}= \sum_{i, j}| f_{j}^{i} \rangle \langle f_{j}^{i}| \rho | f_{j}^{i} \rangle \langle f_{j}^{i}|\,. $$ I want to show that $$ D(\rho || \rho'_{\textrm{rank-one}}) \geq D(\rho || \rho'_{\textrm{coarse}})\,. $$ Intuitively this makes sense to me as with a fine-grained measurement, we are measuring the state 'more' than with a coarse-grained measurement, so the final state should be 'more different' from the initial one, but I don't know how to prove this. $D(\rho || \rho'_{\textrm{rank-one}})$ has the form $$D(\rho || \rho'_{\textrm{rank-one}})= -S(\rho) -\sum_{i, j} \langle f_{j}^{i}| \rho | f_{j}^{i} \rangle \log \langle f_{j}^{i}| \rho | f_{j}^{i} \rangle\,, $$ where $S(\rho)$ is the von Neumann entropy $S(\rho)=-\textrm{tr} \rho \log \rho$, but I can't see any simple form for the second term in $D(\rho || \rho'_{\textrm{coarse}})$. Any help appreciated.

Answer 1

Firstly notice that the divergences are $$ D(\rho\|\rho_{\mathrm{rank-one}}) = - S(\rho) + S(\rho_{\mathrm{rank-one}}) $$ and $$ D(\rho\|\rho_{\mathrm{coarse}}) = - S(\rho) + S(\rho_{\mathrm{coarse}}) $$ and so the difference in the two divergences only boils down to the difference $$ S(\rho_{\mathrm{rank-one}}) - S(\rho_{\mathrm{coarse}}). $$

And it turns out that we will always have $$ S(\rho_{\mathrm{rank-one}}) \geq S(\rho_{\mathrm{coarse}}). $$ This follows from a known result about the von Neumann entropy of a state being equivalent to the smallest Shannon entropy of rank-one measurements on the state (see Theorem 11.1.1 or this answer). In the language of the question we would have that $$ S(\sigma) = \min_{\text{ONBs }\{|v_i\rangle\}} S\left(\sum_i |v_i\rangle \langle v_i| \sigma |v_i\rangle \langle v_i |\right)\,. $$ Now because $\rho_{\mathrm{rank-one}} = \sum_{ij}|f_{ij}\rangle \langle f_{ij}| \rho_{\mathrm{coarse}}|f_{ij}\rangle \langle f_{ij}|$ for some fixed ONB $\{|f_{ij}\rangle\}$ we immediately have the inequality we want.