Trace of Hermitian Operator and Operator Function

Question

I am having trouble understanding the following step. From:

$$\operatorname{trace}\left(\sum_z |z\rangle\langle z| \rho_A |z\rangle\langle z| \times \log( \sum_z |z\rangle\langle z| \sum_x |\langle x|z \rangle |^2 \langle x | \rho_A | x \rangle)\right) \\ = \sum_z \langle z | \rho_A | z \rangle \times \log(\sum_x |\langle x|z \rangle |^2 \langle x | \rho_A | x \rangle)$$

Where $\rho_A$ is a quantum density operator, $X$ and $Z$ are quantum measurement operators, which of course would have to be hermitian. I think the line of reasoning is that those $|z\rangle$ are orthogonal to each other. So essentially it would be like

$$\operatorname{trace}(\text{diagonal matrix} \times \log(\text{another diagonal matrix}))$$ So the trace would simply be the sum of the diagonal elements. But I don't know how to argue about the orthogonality of them. What would be a good approach to go?

Answer 1

If $|z\rangle$ are orthogonal to each other, then $$ \log(\sum_z |z\rangle\langle z| \cdot b_z) = \sum_z |z\rangle\langle z| \cdot \log(b_z) $$ So $$ \mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime})) $$ $$ =\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^\prime\rangle\langle z^\prime| \cdot a_z \cdot \log( b_{z^\prime})) $$ $$ = \mathrm{trace}(\sum_z |z\rangle\langle z|a_z\cdot \log(b_z)) = \sum_z a_z\cdot \log(b_z) $$