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?


1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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