Yes. <Uhlmann's theorem> says that, for the fixed purification $|\Psi^\rho\rangle_{AR}$,
$$
F(\rho_A,\sigma_A)
=\max_{|\Phi^\sigma\rangle}
|\langle\Psi^\rho|\Phi^\sigma\rangle|.
$$
Choose a maximizing purification of $\sigma_A$ on the same reference space. In the unsquared fidelity convention,
$$
\boxed{F(|\Psi^\rho\rangle,|\Phi^\sigma\rangle)
=F(\rho_A,\sigma_A)\geq1-\epsilon}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 2 / b / ii / Solution

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Uhlmann's theorem states that the fidelity of two density operators is the maximum absolute overlap between their purifications on a common reference space.


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