= Solution
Assume an anti-degradable channel transmits perfectly with encoder $\mathcal E$ and decoder $\mathcal D$. Apply its Stinespring isometry to $\mathcal E(\rho)$, producing receiver system $Q$ and environment $E$. The receiver obtains $\rho$ by $\mathcal D$; anti-degradability lets the environment simulate the receiver output using $\mathcal N$, and then $\mathcal D\circ\mathcal N$ also produces $\rho$.
Apply these two local decoding channels simultaneously to $Q$ and $E$. Each marginal of the resulting bipartite state is the original pure state $\rho$. A bipartite state with a pure marginal is a product, so the joint output is $\rho\otimes\rho$. We have therefore constructed one quantum channel mapping every pure $\rho$ to $\rho\otimes\rho$, contradicting the <no-cloning theorem for two pure states>. Hence no anti-degradable channel can transmit arbitrary quantum information perfectly in one use.
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