I've been reading about the von Neumann entropy of a state, as defined via $S(\rho)=-\operatorname{tr}(\rho\ln \rho)$. This equals the Shannon entropy of the probability distribution corresponding to the basis of eigenvectors of $\rho$.

Given a POVM measurement $\{\Pi_j\}$, this maps a state $\rho$ into a vector of probabilities $p_j^\Pi=\operatorname{tr}(\Pi_j \rho)$, and we can associate a Shannon entropy $H(p^\Pi)$ to any such probability distribution $p^\Pi$.

Is it always the case that $H(p^\Pi) \ge S(\rho)$? Or more precisely, can we say that, for any state $\rho$, the von Neumann entropy $S(\rho)$ equals the minimal Shannon entropy, minimized over the set of possible measurements performed on $\rho$?

This is clear for pure states, as for those we have $S(\rho)=0$, but I'm not sure how to see it in the general case. I'm not even sure this is actually true, as for a maximally mixed state I think this would mean that the Shannon entropy is equal regardless of the measurement basis. Maybe it holds if the minimization is restricted to measurements with a number of components equal to the dimension of the state?

  • 1
    $\begingroup$ If the space is finite-dimensional: since $\rho$ diagonalizes, the set of projectors onto the eigenspaces of $\rho$ will be orthogonal and give a POVM $\Pi$ where $H(p^\Pi)=S(\rho)$. Maybe you can make a convexity argument from there to express any other POVM as a convex combination of $\Pi$, and use the convexity of Shannon entropy to prove your conjecture. $\endgroup$
    – Sam Jaques
    May 31 at 17:25
  • 2
    $\begingroup$ For general measurements, the answer is no. Take the trivial measurement consisting of just the identity operator, for instance. There is only one possible measurement outcome, so the output distribution always has zero entropy, for every $\rho$. $\endgroup$ May 31 at 23:48
  • 2
    $\begingroup$ However, if you restrict your attention to complete projective measurements (i.e., measuring with respect to an orthonormal basis) then the answer is yes. This can be proved using some facts about majorization for Hermitian operators, the concavity and unitary invariance of the von Neumann entropy, and Schur's half of the Schur-Horn theorem (which implies that the vector of measurement probabilities for a complete projective measurement will always be majorized by the vector of eigenvalues of $\rho$). $\endgroup$ May 31 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.