Let a quantum channel $\Phi(\cdot)$ between two Hilbert spaces $\mathcal{H}_{in}$ and $\mathcal{H}_{out}$. What is the quantum channel $\Phi_{inv}(\cdot)$ that best reverses $\Phi(\cdot)$ ?

$\forall $ states $\rho$, the state $\widetilde{\rho} \equiv \Phi_{inv}(\Phi(\rho))$ should be the best approximation of $\rho$ (according to some appropriate distance measure).

If we assume that the input state comes from a known probability distribution, according to the quantum data processing inequality there are cases, depending on the source entropy and the channel properties that allow for perfect decoding.

However, how do you perform decoding in cases where perfect decoding is not possible, and in particular if no source distribution is known (so you can't assume that the state has been encoded with an error-correcting code)? I presume somebody has already thought about this, but I can't find it in the literature.


1 Answer 1


First, your two criteria (that it is a quantum channel and that for all input states, the inverse channel gives the best approximation of the input state) aren't compatible. If they were, then the optimal inverse channel would be independent of whether you choose to optimize the worst-case fidelity or the average-case fidelity. It's not.

There are quite a few papers on this question. Here's one Andrew Fletcher, Moe Win, and I wrote. It shows you can use a semidefinite program to optimize the recovery channel when the criterion to be optimized is the entanglement fidelity, and you assume a specific entangled input state and a specific channel $\Phi$. You should be able to find other papers by looking at our bibliography and using some forward citation tool.

  • 1
    $\begingroup$ Thanks for the answer. Just a few questions: - The definition of entanglement fidelity refers to a specific input state. I suppose this means that if you have a distribution $\left \{ (p_i,\ \rho_i) \right \}$ you will optimize respect to the state $\rho \equiv \sum_i p_i \rho_i$, right? - Does the entanglement fidelity estimates the performance of the channel also with respect to entangled mixed states $\rho_{All} \to (I_{Rest} \otimes \Phi)(\rho_{All})$ ? I suppose it does. $\endgroup$ May 19, 2011 at 11:15
  • $\begingroup$ (I rewrote the comment) $\endgroup$ May 19, 2011 at 11:16

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