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.

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

1 Answer 1


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.


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