Why does the state fidelity satisfy $\operatorname{tr}|\sqrt{\rho}\sqrt{\sigma}|=\operatorname{tr}\sqrt{\sigma^{1/2}\rho\sigma^{1/2}}$?

Question

Given two quantum states $\rho$ and $\sigma$, the fidelity is defined as: $$F(\rho,\sigma)=\max_{|\psi\rangle,|\varphi\rangle}|\langle\psi|\varphi\rangle|$$ with $|\psi\rangle$ and $|\varphi\rangle$ respective purifications. During the derivation of the expresssion, $|\langle\psi|\varphi\rangle|$ satisfies this inequality: $$ |\langle\psi|\varphi\rangle|\leq {\rm tr}|\sqrt{\rho}\sqrt{\sigma}|={\rm tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}}$$ This follows from the following computation: \begin{align*} {\rm tr}|\sqrt{\rho}\sqrt{\sigma}|&={\rm tr}\sqrt{(\sqrt{\rho}\sqrt{\sigma})^\dagger(\sqrt{\rho}\sqrt{\sigma})}\\&={\rm tr}\sqrt{(\sqrt{\sigma}\sqrt{\rho})(\sqrt{\rho}\sqrt{\sigma})}\\&={\rm tr}\sqrt{\sigma^{1/2}\rho\sigma^{1/2}}\end{align*} Why do I get an expression that is different from the definition?

Answer 1

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.