Derivative of von Neumann entropy

Similar/related questions to the one I have here that I have looked at are this and this.

Problem description

I have an N-body density matrix (for all intents and purposes, this is just a square matrix), which is given by a product of single-particle density matrices: $$ \rho = \prod_{i,a} \rho_i^a \tag{1} $$

I'm looking at performing the derivation of the von Neumann entropy with respect to the single particle matrix $\rho_i^a$: $$ \frac{\delta}{\delta \rho_i^a} \mathrm{Tr}(\rho\log\rho) ={} ? $$

My attempt

I tried to decompose $\rho = AXB$, where $X = \rho_i^a$ is the parameter I am intending to take the derivative of, and $A$, $B$ are the other single-particle density matrices in the product in eq.(1): $$ \begin{align} \frac{\delta}{\delta X} \mathrm{Tr}(AXB\log AXB) &= \frac{\delta}{\delta X} \mathrm{Tr}(AXD) + \frac{\delta}{\delta X} \mathrm{Tr}(E\log AXB) \\ &= A^\top D^\top + A^\top \frac{E^\top}{AXB} B^\top \\ &= A^\top \biggl( \bigl(\log(AXB) \bigr)^\top + \frac{(AXB)^\top}{AXB} \biggr) B^\top \end{align} $$ where, in the first step, the product rule is applied with $D = B\log AXB$ and $E = AXB$ taken as '$X$-independent', and second step was using results from these notes.

My question

I am actually trying to reproduce / work through the derivation from Appendix A of this paper, page 9. There is a set of four equations (un-numbered) that comes right after eqn A6: the equation in question is the second one.

For those without access to journals, the result they claim to have is:

$$ \frac{\delta}{\delta \rho_i^a} \mathrm{Tr}(\rho\log\rho) = \mathrm{Tr} (\log \rho_i^a) + \mathrm{Tr}(1) $$

What am I missing with my working?

Solution 1:

You're both missing something.

The paper's result is wrong, since the derivative of a scalar with respect to a matrix is a matrix, not a scalar.

The main thing that you seem to be missing is that density matrices for different particles commute, so you don't need all this complicated bookkeeping of the order of products.

The first thing to note is

\begin{eqnarray*} \operatorname{Tr}(\rho\log\rho) &=& \operatorname{Tr}\left(\rho\log\prod_{j,b}\rho_j^b\right) \\ &=& \operatorname{Tr}\left(\rho\sum_{j,b}\log\rho_j^b\right) \\ &=& \sum_{j,b}\operatorname{Tr}\left(\rho\log\rho_j^b\right) \\ &=& \sum_{j,b}\operatorname{Tr}_j^b\left(\rho_j^b\log\rho_j^b\right) \;, \end{eqnarray*}

where $\operatorname{Tr}_j^b$ is the trace over the states of the particle with indices $j,b$ and the trace over the states of the remaining particles is $1$, since for these particles the trace is just over the single-particle density matrices (whose traces are $1$).

So the entropy of a system whose density matrix factorizes into single-particle density matrices is simply the sum of the single-particle entropies.

Now differentiating with respect to $\rho_i^a$ has become rather straightforward, since only the $i,a$ term of the sum contains $\rho_i^a$:

\begin{eqnarray*} \frac\delta{\delta\rho_i^a}\operatorname{Tr}(\rho\log\rho) &=& \frac\delta{\delta\rho_i^a}\operatorname{Tr}_i^a(\rho_i^a\log\rho_i^a) \\&=&\log\rho_i^a+1\;, \end{eqnarray*}

the result in the paper, just without the excess trace.