# Does higher channel fidelity imply higher entanglement fidelity?

Consider two noisy quantum channels (CPTP maps), $$\Phi_1^A$$ and $$\Phi_2^A$$, acting on a system $$A$$. Suppose that for any pure state $$\left|\psi\right>\in \mathcal H_A$$, $$F\big(\psi, \Phi_1^A(\psi)\big) \geq F\big(\psi, \Phi_2^A(\psi)\big) \tag{1}$$ where $$F(\psi, \Phi(\psi))= \big<\psi\big|\Phi\big(\left| \psi \right>\left< \psi \right|\big) \big|\psi \big>$$ is the fidelity between the initial state $$\left|\psi \right>$$ and the final state obtained after sending $$\left|\psi \right>$$ through the channel $$\Phi$$.

My question is: does Eq. (1) imply that $$F_e\big(\rho^A,\Phi^A_1\big) \geq F_e\big(\rho^A,\Phi^A_2\big) \tag{2}$$ for any density matrix $$\rho^A$$? Here, $$F_e$$ denotes the entanglement fidelity, defined as $$F_e(\rho^A, \Phi^A) = \big<\phi^{AB}\big|\big(\Phi^A\otimes I^B\big)\big(\left| \phi^{AB} \right>\left< \phi^{AB} \right|\big) \big|\phi^{AB}\big>,$$ where $$\left|\phi\right>^{AB} \in \mathcal H_A \otimes \mathcal H_B$$ is a purification of the state $$\rho^A$$ to a system $$B$$. In other words, for any pure state $$\left|\phi\right>^{AB} \in \mathcal H_A \otimes\mathcal H_B$$, does Eq. (1) imply that $$F\big(\phi^{AB}, (\Phi_1^A\otimes I^B)(\phi^{AB})\big) \geq F\big(\phi^{AB}, (\Phi_2^A\otimes I^B)(\phi^{AB})\big)?$$

I believe that the answer to the question is yes, but I haven't found a way to prove it. The answer is certainly yes when $$\rho_A$$ is a pure state (i.e. when $$\left|\phi\right>^{AB}$$ is a product state) because $$F_e(\left|\psi\right>\left<\psi\right|,\Phi) = F(\psi,\Phi(\psi))$$. I've tried numerically searching for mixed states that violate the inequality (2) for a few simple choices of $$\Phi_{1,2}$$. For instance, taking $$\Phi_1$$ as the amplitude-damping channel and $$\Phi_2$$ as the depolarizing channel (with error rates chosen so that (1) is satisfied), I was not able to find a $$\rho^A$$ that violates (2). This suggests that (1) implies (2) at least for these particular channels.

To summarize, I'm looking for a proof that (1) implies (2), or else a counterexample showing that (1) does not imply (2). Thanks!

Suppose $$|\phi^{AB}\rangle$$ is a 2-qubit state $$a_0|00\rangle + a_1|11\rangle$$.
Then $$F(\phi^{AB},(\Phi^A\otimes I_B)(\phi^{AB}))\\ = |a_0|^2F(|0\rangle\langle 0|,\Phi^A(|0\rangle\langle 0|))+|a_1|^2F(|1\rangle\langle 1|, \Phi^A(|1\rangle\langle1|))$$
Thus if $$F(\psi, \Phi_1^A(\psi)) \ge F(\psi, \phi_2^A(\psi))$$ for all pure $$\psi$$ then $$F(\phi^{AB},(\Phi_1^A\otimes I_B)(\phi^{AB})) \ge F(\phi_2^{AB},(\Phi_2^A\otimes I_B)(\phi^{AB}))$$.
But, via the Schmidt decomposition, the form of $$|\phi^{AB}\rangle$$ is general for pure 2-qubit states (and the analogous result true for higher dimensions), hence the result holds for all pure $$\phi^{AB}$$ and hence for the entanglement fidelity.