4
$\begingroup$

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?

$\endgroup$
4
  • 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, 2021 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, 2021 at 23:48
  • 3
    $\begingroup$ 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, 2021 at 23:49
  • $\begingroup$ @JohnWatrous thanks a lot for the comment. Do you know of a reference where this is shown? $\endgroup$
    – user15342
    Feb 28, 2022 at 11:34

2 Answers 2

1
$\begingroup$

No. Consider the trivial case where $\{\Pi_j\} = \{I\} $. Then $p^\Pi = (1,)$ and $ H(p^\Pi) = 0 $. This holds even if $ S(\rho) > 0 $.

$\endgroup$
0
$\begingroup$

As previously pointed out, the correct statement is that the von Neumann entropy of a state $\rho$ equals the smallest (Shannon) entropy that can be obtained measuring the state with a rank-1 POVM. More precisely, given an $n$-dimensional state $\rho\in\mathrm D(\mathbb{C}^n)$, the von Neumann entropy $S(\rho)$ equals $H(\boldsymbol p_\mu)$ minimised over all POVMs $\mu:a\mapsto \mu_a\in\operatorname{Pos}(\mathbb{C}^n)$ such that $\operatorname{rank}(\mu_a)=1$ for all $a$, where $\boldsymbol p_\mu$ is the probability distribution defined as $(\boldsymbol p_\mu)_a\equiv \langle\mu_a,\rho\rangle\equiv \operatorname{Tr}(\mu_a \rho)$, and $H(\boldsymbol p_\mu)$ is its Shannon entropy.

This is shown e.g. in Theorem 11.1.1, page 323 of the current arxiv version, of Wilde's book: https://arxiv.org/abs/1106.1445. I'll give here another line of reasoning to prove this statement.

  1. Let $\mu$ be a POVM with rank-1 elements. We can thus write $\mu_a=u_a u_a^\dagger$ for some $u_a\in\mathbb{C}^n$, and we must have $\sum_a u_a u_a^\dagger=I$. Assume without loss of generality that $\rho$ is diagonal, and write it as $\rho=\sum_j p_j \mathbb{P}_j$, where $\mathbb{P}_j\equiv |j\rangle\!\langle j|$.
  2. Observe that $\langle\mu_a,\rho\rangle=\sum_j p_j M_{aj}=M\boldsymbol p$, where $M_{aj}\equiv |\langle u_a,j\rangle|^2$ and $\boldsymbol p\in\mathbb{R}^n$ is the vector of eigenvalues of $\rho$. Observe that $M$ is "almost bistochastic", meaning $$\sum_a M_{aj}=1, \qquad \sum_j M_{aj}=\| u_a\|^2.$$
  3. Define $\eta:\mathbb{R}^+\to\mathbb{R}$ as $\eta(x)\equiv x\log x$ with $\eta(0)=0$, and observe that $\eta$ is convex and such that $\eta(x)\le0$ for $x\in[0,1]$. Note that the Shannon entropy can be written as $H(p)=-\sum_a \eta(p_a)$.
  4. Observe that $$\eta(\langle \mu_a,\rho\rangle) = \eta\left(\sum_j p_j \frac{M_{aj}}{\|u_a\|^2} \|u_a\|^2\right) \le \sum_j \frac{M_{aj}}{\|u_a\|^2} \eta(\|u_a\|^2 p_j) \\ = \sum_j M_{aj} \eta(p_j) + \sum_j p_j \frac{M_{aj}}{\|u_a\|^2} \eta(\|u_a\|^2) \le \sum_j M_{aj} \eta(p_j),$$ where in the last two steps we used the identity $\eta(ab)=a\eta(b)+b \eta(a)$, and then the fact that $\|u_a\|^2\le1$ and thus $\eta(\|u_a\|^2)\le0$ and $\sum_j p_j \frac{M_{aj}}{\|u_a\|^2} \eta(\|u_a\|^2)\le0$. The bound $\|u_a\|^2\le1$ follows from the normalisation condition $\sum_a u_a u_a^\dagger=I$ and the fact that any rank-1 POVM on an $n$-dimensional space must contain at least $n$ components.
  5. Conclude that $$H(\boldsymbol p_\mu) = -\sum_a \eta(\langle\mu_a,\rho\rangle) \ge \sum_a\sum_j M_{aj}\eta(p_j) = \sum_j \eta(p_j) = S(\rho).$$
$\endgroup$

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.