Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 66 / 1

In the state-picture convention, a Stinespring representation of a completely positive map $\mathcal{N}:\mathcal{L}({\mathcal{H}}_Q)\to \mathcal{L}({\mathcal{H}}_{Q^{\prime }})$ consists of an auxiliary Hilbert space ${\mathcal{H}}_E$ and a linear map $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q^{\prime }}\otimes {\mathcal{H}}_E$ such thatThe partial trace discards the environment. For example, from a Kraus representation $\mathcal{N}(X)={\sum }_j{K}_{j}X{K}_j^{†}$, take $V={\sum }_j{K}_j\otimes |j{⟩}_E$. The adjoint, or observable-picture, form is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$.

A general completely positive map does not require $V$ to be an linear isometry of Hilbert spaces. If $\mathcal{N}$ is trace preserving, then $V^{†}V={\sum }_j{K}_j^{†}{K}_j=I$, so $V$ is an linear isometry of Hilbert spaces. This is the Stinespring dilation of a quantum channel, which will be used in the data-processing proof.

By spectral decomposition, the Von Neumann entropy is the Shannon entropy of the eigenvalues:Throughout, logarithms have base two, so the Von Neumann entropy and Shannon entropy are measured in bits. The continuous convention $0{\log }_{2}0=0$ includes zero eigenvalues.

For a bipartite density operator ${\rho }_{AB}$, let $S(A{)}_{\rho }=S({\rho }_A)$, with ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$, and similarly for other systems. The quantum conditional entropy and coherent information areUnlike classical conditional entropy, quantum conditional entropy can be negative. For a Bell state, $S(AB)=0$ and $S(B)=1$, giving $H(A|B)=-1$ and coherent information equal to one bit.

Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.

Take a purification of a density operator $|\Psi {⟩}_{ABCD}$ of ${\rho }_{ABC}$. Apply Strong subadditivity of Von Neumann entropy to the reduced state on $ACD$, with $A$ the conditioning system:Complementary subsystems of a pure state have the same Von Neumann entropy, by the Schmidt decomposition. Therefore $S(AD)=S(BC)$ and $S(ACD)=S(B)$. Substitution givesThis is weak monotonicity of quantum entropy. The purification of a density operator is only a proof device; no purity assumption is imposed on ${\rho }_{ABC}$.