Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 5

The Schmidt decomposition is $|\psi ⟩={\sum }_j{s}_j|a_j⟩|b_j⟩$. Its reduced states are ${\rho }_A={\sum }_j{s}_j^2|a_j⟩⟨a_j|$ and ${\rho }_B={\sum }_j{s}_j^2|b_j⟩⟨b_j|$. They have the same nonzero eigenvalues, so

For the pure state ${\sigma }_{RQE}$ induced by a purification and Stinespring dilation, the coherent information is

Choose any pure input $\rho =|\varphi ⟩⟨\varphi |$. Its reference system may be one-dimensional, and the Stinespring output on $QE$ is pure. Its two reduced states have equal entropy by part a, hence

Because ${\sigma }_{RQE}$ is pure, $S(RQ)=S(E)$ and $S(RE)=S(Q)$. ThereforeThus

For an anti-degradable quantum channel, a channel from $E$ simulates $Q$. The data-processing inequality for quantum relative entropy applied to quantum mutual information gives $I(R:Q)\le I(R:E)$. Part d then implies $I_c(\Lambda ,\rho )\le 0$ for every input. Part c supplies a pure input attaining zero, so

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.