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!


1 Answer 1


I believe the answer can be proven to be yes as follows:

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.


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