Note: I'm reposting this question as it was deleted by the original author, so that we do not lose out on the existing answer there, by Prof. Watrous. Further answers are obviously welcome.

I have two questions:

  1. What are degradable channels?

  2. Given the dephasing channel $$\Phi^D\begin{pmatrix} \rho_{11} & \rho_{12}\\ \rho_{21} & \rho_{22} \end{pmatrix}$$ $$=\begin{pmatrix} \rho_{11} & \rho_{12} e^{-\Gamma(t)}\\ \rho_{21} e^{-\Gamma(t)} & \rho_{22} \end{pmatrix},$$ the complementary map is given by $$\Phi^D\begin{pmatrix} \rho_{11} & \rho_{12}\\ \rho_{21} & \rho_{22} \end{pmatrix}$$ $$= \begin{pmatrix} \frac{1+e^{-\Gamma(t)}}{2} & \frac{\sqrt{1-e^{-2\Gamma(t)}}}{2} (\rho_{11}-\rho_{22})\\ \frac{\sqrt{1-e^{-2\Gamma(t)}}}{2} (\rho_{11}-\rho_{22}) & \frac{1-e^{-\Gamma(t)}}{2} \end{pmatrix}.$$

    How can one prove that the quantum channel capacity is given by $Q_D = 1 - H_2(\frac{1+e^{-\Gamma(t)}}{2} )$, where $H_2(\cdot)$ is the binary Shanon entropy.

Reference: Eq. 13 of this article.

†: Bylicka, B., D. Chruściński, and Sci Maniscalco. "Non-Markovianity and reservoir memory of quantum channels: a quantum information theory perspective." Scientific reports 4 (2014): 5720.

  • $\begingroup$ I thought an answer with a positive score was sufficient to block deletion of a question? $\endgroup$ – Mark S May 9 at 15:44
  • $\begingroup$ @MarkS True. But unfortunately, I could upvote it only after it was deleted (mods can do that by undeleting only the answer and not the question). At the time of deletion, Prof. Watrous' answer had a score of 0. :( $\endgroup$ – Sanchayan Dutta May 9 at 15:45

A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$.

The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is complementary to $\Phi$. If $\Phi$ is applied to a state $\rho$, then the output of the channel is $\Phi(\rho)$ (of course), while $\Psi(\rho)$ represents whatever the environment effectively receives. A channel being degradable therefore means that given the output of the channel, one could reconstruct whatever the environment receives.

On its own, the property of being degradable may be difficult to motivate, except for the fact that we can actually calculate the quantum capacity for degradable channels (while we often cannot for other channels). This is based on two facts:

  1. If $\Phi$ is degradable, then its quantum capacity is equal to its maximum coherent information.

  2. The coherent information is a concave function for degradable channels.

Now, the channel in part 2 of the question is degradable. (In fact, dephasing channels are always degradable, in every dimension.) Perhaps the easiest way to see this for this particular channel is to observe that we can take $\Xi = \Psi$, (i.e., $\Psi \Phi = \Psi$), where I am using $\Phi$ as the name of the original channel and $\Psi$ for the complementary channel given in the question.

It remains to compute the maximum coherent information of this channel. For a given state $\rho$, the coherent information of $\rho$ through $\Phi$ is defined as $$ \text{I}_{\text{C}}(\rho; \Phi) = \text{H}(\Phi(\rho)) - \text{H}(\Psi(\rho)). $$ It is perhaps not obvious, but the maximum value over all states $\rho$ is obtained by the completely mixed state $\rho = \mathbb{1}/2$, and therefore $$ Q(\Phi) = \max_{\rho} \text{I}_{\text{C}}(\rho; \Phi) = \text{I}_{\text{C}}(\mathbb{1}/2; \Phi) = \text{H}(\mathbb{1}/2) - \text{H}(\Psi(\mathbb{1}/2)) = 1 - \text{H}_2\Bigl(\frac{1 + e^{-\Gamma(t)}}{2}\Bigr). $$

One way to argue that the maximum coherent information is obtained by the completely mixed state is to use the fact that $\Phi$ is a Pauli channel, together with the concavity of coherent information for degradable channels mentioned above.


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