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The von Neumann entropy is the analogue of the Shannon entropy and is defined for positive semidefinite matrices as

$$S = \text{Tr}(\rho\log\rho)$$

It is computed in terms of the eigenvalues $\lambda_i$ as $S = \sum_i\lambda_i\log\lambda_i$ where we set $0\log(0) = 0$.

  1. Is $S$ differentiable with respect to $\rho$? The answer to this seems to be yes (see for example the question here and links therein) and is given as $S'(\rho) = I + \log\rho$.

  2. If the result above is true, what happens when $\rho$ has at least one eigenvalue of zero? The von Neumann entropy is defined for positive semidefinite matrices while the log is only defined for positive definite matrices.

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    $\begingroup$ The set of all positive semidefinite matrices is not an open set so it is not clear what is meant by the statement "$S$ is differentiable". Consider the case $n = 1$. The set of positive semidefinite matrices is the interval $[0,\infty)$ and $S(\rho) = \rho \log \rho$ is a function of one variable so you can interpret your question as "does $S$ has a one-sided derivative at $\rho = 0$" and the answer is no. $\endgroup$ – levap May 28 at 21:12
  • $\begingroup$ @levap, thank you for the comment and sorry for a belated response. Can you comment on the following - en.wikipedia.org/wiki/Binary_entropy_function#Derivative. It seems like the function is derivative at zero is allowed provided that one admits that it goes to $-\infty$. I'm not sure if this is mathematically sound. $\endgroup$ – user1936752 Jun 2 at 17:02

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