Given a smoothly time-dependent density matrix (positive-semidefinite matrix with trace 1) $\rho(t)$, its von Neumann entropy is defined as $$ S(t)=\mathrm{Tr}(f(\rho(t))),\ f:[0,\infty)\ni x\mapsto -x\ln x\in\mathbb{R}\ (0\ln0=0). $$ I want to know when $S(t)$ is differentiable. If $\rho(t)$ has the same rank, then it is differentiable and we have $$ S'(t)=-\mathrm{Tr}(\rho'(t)\ln\rho(t)). $$ (Related question is here. ) If the rank of $\rho(t)$ varies, I think $S(t)$ is not differentiable in general, because $f$ isn't (one-sided) differentiable at $x=0$.
However, some mathematical physicists have stated that $S(t)$ is differentiable without the rank assumption for the special $\rho(t)$ (Equation (1)&(3) in this paper and Equation (10)&(13), (16)&(26) in this paper).
Specifically, this is the case where $\rho(t)$ is a matrix on the tensor product space $\mathcal{H}_A\otimes\mathcal{H}_B$ and is given by using partial traces as $$ \rho(t)=\mathrm{Tr}_B(e^{-iHt}\rho(0)e^{iHt}), $$ where $H$ is some self-adjoint matrix on $\mathcal{H}_A\otimes\mathcal{H}_B$. They state that the derivative of $S(t)$ (at $t=0$) is expressed as $$ S'(0)=-\mathrm{Tr}(H[\rho(0),\ln\rho(0)_A\otimes I_B]). $$
Is this equation still correct when the rank of $\rho(0)_A$ changes, and if so, what does $\ln \rho(0)_A$ mean? Furthermore, how can this equation be shown? If you have a counterexample, I would like to know. Thanks!
Edit: Kurt.G presented the abstract condition that the eigenvalues should satisfy. Is this condition satisfied in the above case (entropy of the subsystem)? If so, why? I suppose we could use the results on the analyticity of the eigenvalues(e.g. this and this), but I am not sure of the details.
We know that when $\rho$ describes a pure state then $\rho^2=\rho$ and $\rho\log\rho=\rho\log\rho^2=2\rho\log\rho$ so that $S=-\operatorname{tr}(\rho\log\rho)=0\,.$
A typical pure state has density matrix $$ \rho=\pmatrix{1&0\\0&0}\quad\text{ or }\quad\rho=\pmatrix{0&0\\0&1}\,. $$ A typical mixed state is $$ \rho(t)=\pmatrix{a(t)&0\\0&b(t)}\quad\text{ with }\quad a(t)+b(t)=1\,, \;a(t),b(t)\ge 0\,. $$ Then, if $a(t),b(t)>0\,,$ $$ S(t)=a(t)\log a(t)+b(t)\log b(t)\,. $$ l'Hospital allows to extend this continuously to the case $a(t)=0$ or $b(t)=0$ with the convention $0\log 0=0$ (compatible with pure states). This $S(t)$ is obviously differentiable whenever $a$ and $b$ are differentiable regardless if the rank of $\rho(t)$ changes or not.
Edit: When $a$ reaches zero at $t_0$ we need one extra condition to ensure the differentiability of $a(t)\log a(t)$ at this point: $$ \lim_{t\to t_0}\frac{a(t)\log a(t)}{t-t_0}=0\,. $$ In particular $a'(t_0)$ must be zero.
Assuming that an analogous condition holds for $b\,,$ the derivative of $S$ is $$ S'(t)=a'(t)\log a(t)+a'(t)+b'(t)\log b(t)+b'(t)\,. $$ The terms with $a$ are zero at $t_0\,$ and an analogous statement holds for the terms with $b\,.$