# How to compute derivatives of partial traces of the form $\frac{\partial \operatorname{Tr}_B(F(\mathbf{X}))}{\partial \mathbf{X}}$?

The Matrix Cookbook says that for any differentiable matrix function $$F(\cdot)$$, it holds that

$$\frac{\partial \operatorname{Tr}(F(\mathbf{X}))}{\partial \mathbf{X}}=f(\mathbf{X})^{T},$$

where $$f(\cdot)$$ is the scalar derivative of $$F(\cdot)$$. Is there a relation like this for the case where one has the partial trace i.e., for some bipartite operator $$X_{AB}$$

$$\frac{\partial \operatorname{Tr}_B(F(\mathbf{X}))}{\partial \mathbf{X}}= \ ?$$

I will eventually apply this to optimization problems where $$X_{AB}$$ will be a quantum state so assumptions like positive semidefiniteness and unit trace are okay.

• Note that this formula is given for real matrices $X$ Dec 13, 2021 at 18:16
• Although I do not know the answer of this question, Appendix A of this paper [Computer Physics Communications 183 (2012) 155–165] may be helpful to you. In this paper, the authors provide the calculations for the derivatives of entropy of entanglement, which has partial trace. Dec 14, 2021 at 2:30

The question amounts to whether there are nice enough expressions for the partial traces of powers of an operator in a bipartite (finite-dimensional) space. That's where the result with the standard trace comes from: just observe that $$\frac{\partial}{\partial X_{ij}}\operatorname{Tr}(X^n)=n (X^{n-1})_{ji}$$ and apply it to the series expansions of the given expressions. When we have the partial trace things seem to not be that nice.
The main objects of interest is then a four-indices object $$X_{12,34}$$, and the operators of the form $$\operatorname{Tr}_B(X^n)$$. To refer to the operator itself, rather than its components, I will write $$X=X_{12,34}(E_{13}\otimes E_{24})$$, where $$E_{ij}\equiv |i\rangle\!\langle j|\equiv e_i e_j^\dagger$$ are the canonical operatorial basis in the relevant space. As a couple of examples to set the notation, note that: $$\operatorname{Tr}_B(X) = X_{12,32} E_{13} \equiv \sum_{123} X_{12,32}E_{13}, \\ \operatorname{Tr}_B(X^2) = X_{12,34}X_{34,56} \delta_{26} E_{15} = X_{12,34}X_{34,52} E_{15}.$$ Now, about the partial derivatives. Observe that these essentially amount to setting specific indices equal to those in the partial derivative, and thus: $$\frac{\partial}{\partial X_{ij,k\ell}}\operatorname{Tr}_B(X) \equiv \partial_{ij,k\ell} \operatorname{Tr}_B(X) = \delta_{j\ell} E_{ik}, \\ \partial_{ij,k\ell} \operatorname{Tr}_B(X^2) = X_{k\ell,5j} E_{i5} + X_{1\ell,ij} E_{1k} = [(E\otimes I)X^T+X^T(E\otimes I)]_{ij,k\ell}, \\ \partial_{ij,k\ell} \operatorname{Tr}_B(X^3) = [(E\otimes I) (X^T)^2 + X^T (E\otimes I)X^T + (X^T)^2(E\otimes I)]_{ij,k\ell}.$$ An important note: I wrote "matrix products" of the form $$(E\otimes I)X^T$$ for notational brevity, but these expressions should be probably be understood as mostly just formal ones: the elements of $$X$$ are numbers, whereas the elements of $$E$$ are operators.
I haven't checked explicitly, but I'm going to assume the above pattern holds in general. Nevertheless, from expressions like these, I'm not confident that there is a nice enough expression to be found for $$\partial\operatorname{Tr}_B[F(X)]$$. This is not a full answer, but I'll leave it here in case the calculations and methodology might be of use.