1
$\begingroup$

Given the the two states $\rho$ and $\sigma$ of a quantum system, with $|\psi\rangle$ and $|\varphi\rangle$ as their purification respectively, the fidelity is defined as:

$$F(\rho,\sigma)=\max_{|\psi\rangle,|\varphi\rangle}|\langle\psi|\varphi\rangle|$$

During the derivation of the expresssion, $|\langle\psi|\varphi\rangle|$ follows this inequality:

$$ |\langle\psi|\varphi\rangle|\leq tr|\sqrt{\rho}\sqrt{\sigma}|=tr\sqrt{\rho^{\frac{1}{2}}\sigma\rho^{\frac{1}{2}}}$$

My workings for $tr|\sqrt{\rho}\sqrt{\sigma}|$ is as follows:

$$tr|\sqrt{\rho}\sqrt{\sigma}|=tr\sqrt{(\sqrt{\rho}\sqrt{\sigma})^\dagger(\sqrt{\rho}\sqrt{\sigma})}$$ $$=tr\sqrt{(\sqrt{\sigma}\sqrt{\rho})(\sqrt{\rho}\sqrt{\sigma})}$$ $$=tr\sqrt{\sigma^{\frac{1}{2}}\rho\sigma^{\frac{1}{2}}} $$

Why do I get an expression that is different from the definition?

$\endgroup$
3
$\begingroup$

The expressions are equivalent:

$$ F(\rho,\sigma) =\operatorname{tr}\sqrt{\sigma^{1/2}\rho\sigma^{1/2}}= \operatorname{tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}} \\ = \operatorname{tr}|\sqrt\rho\sqrt\sigma| = \operatorname{tr}|\sqrt\sigma\sqrt\rho| = \max_{\psi_\rho,\psi_\sigma}|\langle\psi_\rho|\psi_\sigma\rangle|.$$

See also this question about the symmetry of the fidelity, and the relevant Wikipedia page, as well as this question and links therein.

| improve this answer | |
$\endgroup$

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.