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.

  • $\begingroup$ Have you tried $S(\rho+\delta\rho)\approx S(\rho)+\delta\rho \cdot S'(\rho)+$ for small $\delta\rho$? $\endgroup$
    – Zachary
    Jun 14 '19 at 20:54
  • 1
    $\begingroup$ Could you comment more on what you think about when you write $S' (\rho)$? Thanks for the help! $\endgroup$
    – Knomes
    Jun 17 '19 at 8:03

$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\Dyad}[1]{\Ket{#1}\!\Bra{#1}}$

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$.

  • $\begingroup$ "while 𝜎=∑𝑃𝑖∣𝑖⟩⟨𝑖∣" -- $\sigma$ generally cannot be expressed in this form. $\endgroup$ Jun 25 '19 at 17:21
  • $\begingroup$ Sorry, of course, they are not on the same basis in general. I changed the text, thanks. $\endgroup$
    – Knomes
    Jun 26 '19 at 7:19

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