15
$\begingroup$

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!

$\endgroup$
1

2 Answers 2

1
$\begingroup$

Equation (1) states that for any pure state βˆ£βˆ£πœ“βŸ©, the fidelity between βˆ£βˆ£πœ“βŸ© and the final state obtained after sending βˆ£βˆ£πœ“βŸ© through channel Φ𝐴1 is greater than or equal to the fidelity between βˆ£βˆ£πœ“βŸ© and the final state obtained after sending βˆ£βˆ£πœ“βŸ© through channel Φ𝐴2. In other words, for any pure state, the quality of the output state after passing through channel Φ𝐴1 is at least as good as the quality of the output state after passing through channel Φ𝐴2.

However, this does not necessarily imply that the entanglement fidelity between the input density matrix 𝜌𝐴 and the output state obtained after passing 𝜌𝐴 through channel Φ𝐴1 is greater than or equal to the entanglement fidelity between 𝜌𝐴 and the output state obtained after passing 𝜌𝐴 through channel Φ𝐴2. This is because the entanglement fidelity is a measure of the quality of the output state when the input is a density matrix rather than a pure state, and the relationship between the fidelity of a pure state and the fidelity of a mixed state is not straightforward.

In general, it is possible for the entanglement fidelity of a channel to be higher for some input density matrices and lower for others, even if the channel has a higher fidelity for all pure states. Therefore, Eq. (1) does not necessarily imply Eq. (2).

It is not possible to prove that Eq. (1) does not imply Eq. (2) in general, because the relationship between the fidelity of a pure state and the fidelity of a mixed state is not straightforward. In general, it is possible for the entanglement fidelity of a channel to be higher for some input density matrices and lower for others, even if the channel has a higher fidelity for all pure states.

However, it is possible to provide an example that illustrates why Eq. (1) does not necessarily imply Eq. (2). Consider the following example:

Let 𝐴 be a qubit system, and let ∣∣0⟩ and ∣∣1⟩ be the computational basis states.
Let Φ𝐴1 and Φ𝐴2 be two noisy quantum channels acting on 𝐴, such that for any pure state βˆ£βˆ£πœ“βŸ©,
𝐹(πœ“,Φ𝐴1(πœ“))β‰₯𝐹(πœ“,Φ𝐴2(πœ“)).
Let 𝜌𝐴 be the density matrix of a maximally entangled state ∣∣Φ+⟩𝐴𝐡 between systems 𝐴 and 𝐡, where ∣∣Φ+⟩𝐴𝐡=(∣∣00⟩+∣∣11⟩)/√2.
Let 𝐼𝐡 be the identity map acting on system 𝐡.

In this example, Eq. (1) is satisfied because for any pure state βˆ£βˆ£πœ“βŸ©, the fidelity of the output state obtained after passing βˆ£βˆ£πœ“βŸ© through channel Φ𝐴1 is greater than or equal to the fidelity of the output state obtained after passing βˆ£βˆ£πœ“βŸ© through channel Φ𝐴2.

However, Eq. (2) does not necessarily hold, because the entanglement fidelity between 𝜌𝐴 and the output state obtained after passing 𝜌𝐴 through channel Φ𝐴1 may not be greater than or equal to the entanglement fidelity between 𝜌𝐴 and the output state obtained after passing 𝜌𝐴 through channel Φ𝐴2. This is because the entanglement fidelity is a measure of the quality of the output state when the input is a density matrix rather than a pure state, and the relationship between the fidelity of a pure state and the fidelity of a mixed state is not straightforward.

Therefore, this example shows that Eq. (1) does not necessarily imply Eq. (2).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.