Dose density matrix with off-diagonal elements equal to zero has maximum von-Neumann entropy?

Question

von-Neumann entropy

I know von-Neumann entropy on density matrix $S=-{\rm Tr}(\rho \ln\rho)$ is similar to Shannon entropy $S=-\sum_i p_i\ln p_i$ in classical mechanics. And I want to get Bose-Einstein distribution and Fermi-Dirac distribution with the principle of maximum entropy. I work with some key points shown in this document http://userpage.fu-berlin.de/~marekgluza/ASM2_16/sheet06.pdf.

Statement of Problem

What I was confused is "Among all density matrices with the same diagonal (in some basis), the matrix with all entries outside the diagonal equal to zeros has the largest von-Neumann entropy". How can we get this conclusion?

Schur's theorem

The document also says it's a consequence of Schur's theorem. Does Schur's theorem say von-Neumann entropy is concave? And this problem Bound on the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix? also talks about Schur-concave functions. But how can we know density matrix with off-diagonal elements equal to zero has maximum entropy?

Answer 1

The following observation allows to reduce the problem to the classical, commuting case.

I will assume finite dimensionality although generalizations are possible. Let the Hamiltonian have the following spectral decomposition, $H = \sum_k E_k \Pi_k $, where $E_k, \Pi_k$ are respectively the eigenvalues, spectral projectors.

Define the following dephasing map:

\begin{align} D_H(X) &:= \lim_{T\to \infty} \frac{1}{T} \int_0^T \!\! dt \, e^{-it H} X e^{itH} \\ & = \sum_k \Pi_k X \Pi_k . \end{align}

Note that the first line is also valid in infinite dimensions. It is easy to see that $D$ is positive, trace preserving and unital (in fact it is completely positive as one has a Kraus representation).

Let the entropy of a state $\rho$ be given by

$$S(\rho) = -\operatorname{Tr} [\rho \ln \rho]. $$

The observation is the following. It turns out that, for any quantum state $\rho$

$$S (D_H (\rho)) \ge S(\rho) . $$

Since we want to maximize the entropy we can do so among the fixed points of $D_H$ which are states that commutes with $H$. From here on you can follow the classical proof with Lagrange multipliers.

The observation is part of the characterization of entropy increasing channels. In particolar, if $T$ is positive, trace preserving and unital, and $h: \mathbb{R} \rightarrow \mathbb{R}$ is a concave function, then

$$ \operatorname{Tr} [h(T(\rho))] \ge \operatorname{Tr}[h(\rho)]$$ for all density operators $\rho$.

For a proof, see Theorem 8.8 and corollary 8.1 of Quantum channels & operations guided tour by M. Wolf (freely available here).