# Prove that $Tr(\chi(\rho_A) \log(\chi(\rho_A)) = Tr(\rho_A \log(\chi(\rho_A))$

I am trying to see how the following statement about trace $$Tr$$ is true.

$$Tr(\chi(\rho_A) \log(\chi(\rho_A)) = Tr(\rho_A \log(\chi(\rho_A)),$$ for some quantum state $$\rho_A$$, Where,

$$\chi(.) = \sum_x |X^x\rangle\langle X^x| (.) |X^x\rangle\langle X^x|$$ is a measurement map. I have tried to write out a spectral decomposition of the state $$\rho_A$$,

$$\rho_A = \sum_i \lambda_i |\lambda_i \rangle \langle \lambda_i|,$$ and then insert it into the equation. However, I don't see how to work with s sum inside a matrix logarithm. Any help would be highly appreciated.

• You can refer to operator functions in the book of Nielsen and Chuang. We have $\log \sum_i{\lambda _i|i\rangle \langle i|}=\sum_i{\left( \log \lambda _i \right) |i\rangle \langle i|}$. Commented Jul 10, 2023 at 1:29

## 1 Answer

Let $$\rho$$,$$\sigma$$ be arbitrary states, and let $$f$$ be some "well-behaved" function defined on the eigenvalues of both operators. Then $$\operatorname{tr}(\rho f(\sigma)) = \sum_{ij} f(\lambda_j(\sigma)) |\langle \lambda_i(\rho)|\lambda_j(\sigma)\rangle|^2,$$ where $$\rho=\sum_i \lambda_i(\rho)|\lambda_i(\rho)\rangle\!\langle\lambda_i(\rho)|$$ and $$\rho=\sum_i \lambda_i(\sigma)|\lambda_i(\sigma)\rangle\!\langle\lambda_i(\sigma)|$$ are the eigendecompositions of the operators.

It follows that the only thing about $$\rho$$ that enters this expression are the projections of its eigenstates onto those of $$\sigma$$. Therefore if we were to replace $$\rho$$ with $$\chi(\rho)$$ defined as $$\chi(\rho) = \sum_i \langle \lambda_i(\sigma)|\rho|\lambda_i(\sigma)\rangle \, |\lambda_i(\sigma)\rangle\!\langle \lambda_i(\sigma)|$$ nothing would change.