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?

  • 1
    $\begingroup$ 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. $\endgroup$ – DaftWullie Oct 2 '19 at 12:13
  • 1
    $\begingroup$ Could you provide a reference for this equation? $\endgroup$ – Jonathan Trousdale Oct 2 '19 at 15:44
  • $\begingroup$ i was reading this paper: arxiv.org/pdf/1511.04857.pdf. at page 58, second column top, this reduction is present. $\endgroup$ – Hasan Iqbal Oct 2 '19 at 15:49
  • $\begingroup$ I don't understand what is the question. Are you trying to prove the identity in the first equation, or something else? $\endgroup$ – 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 $$ \mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime})) $$ $$ =\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^\prime\rangle\langle z^\prime| \cdot a_z \cdot \log( b_{z^\prime})) $$ $$ = \mathrm{trace}(\sum_z |z\rangle\langle z|a_z\cdot \log(b_z)) = \sum_z a_z\cdot \log(b_z) $$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.