Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 48 / 1

Use logarithms to base two, so all Von Neumann entropies and Shannon entropies are measured in bits. Write the nonzero eigenvalues of ${\rho }_i$ as ${\lambda }_{ij}$. Orthogonality of the supports makes the mixture a block diagonal matrix, with eigenvalues $p_i{\lambda }_{ij}$. ThereforeEach ${\rho }_i$ has trace one. Terms with zero probability or zero eigenvalue vanish under the convention $0\log 0=0$. This proves the entropy of an orthogonal quantum mixture.

For a bipartite density operator ${\rho }_{AB}$, the Subadditivity of Von Neumann entropy isHere $S(A)=S({\mathrm{Tr}}_B{\rho }_{AB})$ and likewise for $B$, using the partial trace. For a tripartite density operator, the Strong subadditivity of Von Neumann entropy isEquivalently, quantum conditional entropy cannot increase upon adding a conditioning system: $S(A\mid BC)\le S(A\mid B)$.

Introduce an orthonormal classical label register $X$ and the classical-quantum stateThe blocks have orthogonal supports because their label vectors are orthogonal, regardless of whether the original ${\rho }_i$ have orthogonal supports. Part (a) gives $S(XQ)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$. The marginals are ${\omega }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\omega }_Q={\sum }_i{p}_i{\rho }_i$, so $S(X)=H(p)$. Apply the Subadditivity of Von Neumann entropy and cancel $H(p)$:Thus $S({\sum }_i{p}_i{\rho }_i)\ge {\sum }_i{p}_{i}S({\rho }_i)$. This is the Concavity of Von Neumann entropy.