Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 / 5

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

Subadditivity of Von Neumann entropy applied to $RE$ gives $S(RE)\le S(R)+S(E)$. Together with the environment expression for coherent information,The channel does not act on $R$. Its reduced state has the same nonzero eigenvalues as the original input $\rho$, because $RA$ initially purifies that input. HenceMore precisely, $I_c=S(\rho )-I(R:E)$: the coherent information upper bound by input entropy is a consequence of nonnegative quantum mutual information with the environment. Equality holds when $R$ and $E$ are uncorrelated.

Dilate the second quantum channel by an isometry $W:B_1⟶B_2{E}_2$, and set ${\tau }_{R{B}_2{E}_2}=(I_R\otimes W){\sigma }_{R{B}_1}(I_R\otimes W^{†})$. Its $R{B}_2$ marginal is the final output. Isometry invariance of Von Neumann entropy gives $S(B_1{)}_{\sigma }=S(B_2{E}_2{)}_{\tau }$ and $S(R{B}_1{)}_{\sigma }=S(R{B}_2{E}_2{)}_{\tau }$. The difference between the two values of coherent information is thereforeNonnegativity follows from Strong subadditivity of Von Neumann entropy. This proves the data-processing inequality for coherent information:The dilated state here need not be pure. The proof uses only isometry invariance and strong subadditivity, so it applies even when the first channel has already entangled the input with its own discarded environment.

Both output states have the same reference marginal, and $S(R)=S(\rho )$. Expand quantum mutual information to obtain the mutual information and coherent information identityThe input-entropy term is identical in the two expressions. Subtract them and use the preceding data-processing inequality for coherent information:Thus the final quantum channel cannot increase the reference-output quantum mutual information, as required by data processing for quantum mutual information.