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 $\delta\rho$? $\endgroup$
    – Dispersion
    Jun 14, 2019 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, 2019 at 8:03

1 Answer 1


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

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=βˆ‘π‘ƒπ‘–βˆ£π‘–βŸ©βŸ¨π‘–βˆ£" -- $\sigma$ generally cannot be expressed in this form. $\endgroup$ Jun 25, 2019 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, 2019 at 7:19

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.