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](http://userpage.fu-berlin.de/%7Emarekgluza/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?](https://mathoverflow.net/questions/68249/bound-on-the-von-neumann-entropy-of-a-positive-positive-semi-definite-and-unit) also talks about Schur-concave functions. But how can we know density matrix with off-diagonal elements equal to zero has maximum entropy?



Dose density matrix with off-diagonal elements equal to zero has maximum von-Neumann entropy?
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](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiG5ODx8ef7AhXlzIUKHbxGCK0QFnoECA4QAQ&url=https%3A%2F%2Fwww-m5.ma.tum.de%2Ffoswiki%2Fpub%2FM5%2FAllgemeines%2FMichaelWolf%2FQChannelLecture.pdf&usg=AOvVaw0ifu39ST08JdtnUx7QZI5s)).


