Consider the quantum state fidelity $F(\rho,\sigma)$ defined as (I will use the notation used in Nielsen & Chuang here): $$ F(\rho,\sigma) \equiv \operatorname{Tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}} = \operatorname{Tr}\lvert\sqrt\sigma\sqrt\rho\rvert. $$ It is a standard result that this function is symmetric: $F(\rho,\sigma)=F(\sigma,\rho)$.

This is straightforwardly seen through Uhlmann's theorem or the representation of the fidelity as maximum over all possible POVMS of the fidelity of the associated probability distributions: $$F(\rho,\sigma) = \max\lvert\langle\psi_\rho\rvert\psi_\sigma\rangle\rvert $$ when maximising over all possible purifications of $\rho$ and $\sigma$, and $$ F(\rho,\sigma) = \min F(p_m, q_m) $$ when minimising over all possible POVMs $\{E_m\}_m$ and $p_k=\operatorname{Tr}(E_k \rho)$ and $q_k=\operatorname{Tr}(E_k \sigma)$.

However, the way these results are obtained does not make it very clear to me why $F(\rho,\sigma)$ should be symmetric. It seems weird that one would need to study an enlarged space, like is the case when passing through Uhlmann's theorem, to only show the symmetry of $F$. On the other hand, to show $F(\rho,\sigma) = \min F(p_m, q_m)$ one requires a few manipulations that make it a bit obscure to me how similar ideas could be used to prove the simpler result of the symmetry of $F$.

Is there a simpler, more direct way to prove that the fidelity is symmetric?


This follows directly from the fact that the trace norm is symmetric, $\mathrm{tr}\,|X|=\mathrm{tr}\,|X^\dagger|$.


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