# Trace of Hermitian Operator and Operator Function

I am having trouble understanding the following step. From:

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

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

$$\operatorname{trace}(\text{diagonal matrix} * \log(\text{another diagonal matrix}))$$ 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?

• it should have previously been defined that $|z\rangle$ is an orthonormal basis. It's not something that you can just pull out of the air at this point. – DaftWullie Oct 2 '19 at 12:13
• Could you provide a reference for this equation? – Jonathan Trousdale Oct 2 '19 at 15:44
• i was reading this paper: arxiv.org/pdf/1511.04857.pdf. at page 58, second column top, this reduction is present. – Hasan Iqbal Oct 2 '19 at 15:49
• I don't understand what is the question. Are you trying to prove the identity in the first equation, or something else? – glS Oct 4 '19 at 18:00

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 $$\mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime}))$$ $$=\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^\prime\rangle\langle z^\prime| \cdot a_z \cdot \log( b_{z^\prime}))$$ $$= \mathrm{trace}(\sum_z |z\rangle\langle z|a_z\cdot \log(b_z)) = \sum_z a_z\cdot \log(b_z)$$