1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Not sure the second term in the final equation is correct, shouldn't it be the Shannon entropy of the probability distribution of the measurement? How did you get rid of the logarithm? $\endgroup$
    – Rammus
    Jul 19, 2023 at 11:06
  • $\begingroup$ @Rammus you are right, thank you! I have edited the equation. $\endgroup$ Jul 19, 2023 at 11:48

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Why do we have $D(\rho||\rho_{\textrm{coarse}})= -S(\rho)+S(\rho_{\textrm{coarse}})$? The first term is fine but I'm unable to derive $-\textrm{tr}\rho \log{\rho_{\textrm{coarse}}}= S(\rho_{\textrm{coarse}})$ (whereas in the case of $\rho_{\textrm{rank-one}}$ this works out easily because we know the eigenbasis of $\rho_{\textrm{rank-one}}$). $\endgroup$ Jul 20, 2023 at 9:26
  • $\begingroup$ $\rho_{\mathrm{coarse}}$ is block diagonal with blocks defined by the orthogonal projectors. So $\log(\rho)$ is also block diagonal and so the trace with $\rho$ is the same if we apply the same pinching to $\rho$. Effectively, $$ \mathrm{tr}[ \rho \log (\sum_i \Pi_i \rho \Pi_i)] = \mathrm{tr}[\rho\sum_i \log \Pi_i \rho \Pi_i ] = \mathrm{tr}[\rho \sum_i \Pi_i \log(\Pi_i \rho \Pi_i) \Pi_i] = S(\rho_{\mathrm{coarse}}) $$ $\endgroup$
    – Rammus
    Jul 20, 2023 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.