Entanglement fidelity ($F_e(\rho ,\Lambda )$)

For a purification $|{\Psi }_{\rho }{⟩}_{RA}$, the entanglement fidelity of a channel $\Lambda$ isIt is independent of the chosen purification.

In the unsquared quantum fidelity convention, the fidelity between a purification and its channel output is the square root of entanglement fidelity. Tracing out the purifying reference leaves the input and output marginals. Monotonicity of quantum fidelity under partial trace therefore gives the bound. Preserving an input marginal is a weaker requirement than preserving its correlations with a reference.

For a quantum channel on a $d\ge 2$ dimensional system, purify the input to $|\Psi {⟩}_{QR}$ and set $\sigma =(\mathcal{N}\otimes \mathrm{id})(|\Psi ⟩⟨\Psi |)$. Its entanglement fidelity $F_e=⟨\Psi |\sigma |\Psi ⟩$ satisfies $S(\sigma )\le h(F_e)+(1-F_e){\log }_2(d^2-1)$. This is the entropy bound from overlap with a pure state in dimension $d^2$. For a trivial one-dimensional system the output entropy is zero.

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.

In the unsquared quantum fidelity convention, operation fidelity compares a purification of a density operator with its image under a quantum channel acting on the source subsystem:Its square is entanglement fidelity. Thus this convention is distinct from the squared quantity $F_e$; the two should not be installed as synonyms.