Does the von Neumann entropy equal the smallest accessible Shannon entropy?

Question

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?

Answer 1

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 $.

Answer 2

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 $$H(\boldsymbol p_\mu) = -\sum_a \eta(\langle\mu_a,\rho\rangle) \ge \sum_a\sum_j M_{aj}\eta(p_j) = \sum_j \eta(p_j) = S(\rho).$$