Expansion of Von Neumann entropy for small deviations Suppose that your quantum system is described by $\sigma = \rho + \delta\rho$, where both $\sigma$ and $\rho$ are density matrices while $\delta\rho$ is "small".
The Von Neumann entropy of the system is given by:
$$
S(\sigma) = \textrm{Tr}[\sigma \log \sigma]
$$
Does anyone know how to expand the Von Neumann entropy in terms of $\delta\rho$?
I am thinking of something like:
$$
S(\rho+\delta\rho) \simeq S(\rho)+F(\delta\rho)
$$
where I do not know what $F(\delta\rho)$ is.
 A: $\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\Dyad}[1]{\Ket{#1}\!\Bra{#1}}$
I found an answer in this recent paper: https://arxiv.org/abs/1906.08203
We re-define $\sigma$ as $\sigma = \rho + \epsilon \delta \rho$.
Let $\rho = \sum_i p_i \Dyad{i}$ denote the eigendecomposition of $\rho$ while we call $P_i$ the eigenvalues of the matrix $\sigma$.
Since $\sigma$ is Hermitian, standard perturbation theory applies and, assuming the $p_i$ are non-degenerate we may write up to second order in $\epsilon$
$$
P_i \simeq p_i + \epsilon \delta \rho_{ii} + \epsilon^2 \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j}.
$$
Then, we can also expand the logarithm
$$
\log P_i 
= \log \left[p_i \left(1 + \epsilon \frac{\delta \rho_{ii}}{p_i} + \epsilon^2 \frac{1}{p_i} \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j} \right) \right] 
\simeq \log p_i + \epsilon \frac{\delta \rho_{ii}}{p_i} +
\epsilon^2 \left( -\frac{\delta \rho^2_{ii}}{2 p_i^2} + \frac{1}{p_i} \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j} \right)
$$
Lastly, expanding $P_i \log P_i$ in $\epsilon$ up to second order and using $\textrm{Tr} (\delta \rho) = 0$, i.e. $\sum_i \delta \rho_{ii} = 0$ we get
$$
S(\sigma)=-\sum_i P_i \log P_i \simeq S(\rho) - \epsilon\sum_i \delta \rho_{ii} \log p_i
-\epsilon^2 \sum_i \left( \frac{\delta \rho^2_{ii}}{2 p_i} + \sum_{j \neq i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}\log p_i  \right)
$$
Notice that
$$
\sum_i \sum_{j \neq i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
=\sum_i \sum_{j > i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
+\sum_i \sum_{j < i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
=0
$$
because $ \left|\delta \rho_{ij}\right|^2 =  \left|\delta \rho_{ji}\right|^2$.
