Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 48 / 2

The channel acts only on the input system $Q$; the reference $R$ is untouched. WriteThe coherent information isThe reference marginal remains ${\rho }_R$, whose nonzero eigenvalues agree with those of $\rho$. Unlike a classical information quantity, coherent information can be negative.

The lower diagram uses a Stinespring dilation: append an environment in a fixed pure state, apply a joint unitary operator, and discard the environment by partial trace. Before discarding it, the global $RBE$ state is pure.

Implement the quantum channel by a unitary dilation with a pure initial environment. The final state on $RBE$ is pure. The Schmidt decomposition across either bipartition gives equal complementary Von Neumann entropies, so $S(RB)=S(E)$ and $S(B)=S(RE)$. By the Subadditivity of Von Neumann entropy,The reference is untouched and purifies the input, hence $S(R)=S(\rho )$. Therefore $\boxed{I_c(\Phi ,\rho )\le S(\rho )}$.

After the first quantum channel, denote the joint reference-output state by ${\omega }_{R{B}_1}$. Dilate the second CPTP map as an linear isometry $B_1\to B_2{E}_2$, with the resulting state denoted $\tau$. Linear isometries preserve nonzero eigenvalues, both globally and on the transformed output, givingTracing out $E_2$ implements the second channel, soThe Strong subadditivity of Von Neumann entropy for $R,B_2,E_2$ gives $S(R{B}_2)+S(B_2{E}_2)\ge S(B_2)+S(R{B}_2{E}_2)$. Rearranging proves the data-processing inequality for coherent information: