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.

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

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

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 $$\left|\delta \rho_{ij}\right|^2 = \left|\delta \rho_{ji}\right|^2$$.

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