Kraus formula for entanglement fidelity ($F_e(\rho ,\Lambda )={\sum }_a|\mathrm{Tr}(\rho K_a){|}^2$)

If $\{K_a\}$ is any Kraus representation of a quantum channel $\Lambda$, thenFor a purification of a density operator, each branch overlap is $⟨{\Psi }_{\rho }|(I\otimes K_a)|{\Psi }_{\rho }⟩=\mathrm{Tr}(\rho K_a)$. Summing the squared branch overlaps gives entanglement fidelity.