Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 34 / 1

Use base-two logarithms, so Von Neumann entropy is measured in bits, and set $0\log 0=0$. Write $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ for the quantum relative entropy, the quantity denoted $S(\rho \Vert \sigma )$ in the paper. It is $+\infty$ unless the support of a positive operator $\rho$ lies within that of $\sigma$.

We first establish nonnegativity of quantum relative entropy, including its equality condition. Let $\rho ={\sum }_{i:r_i>0}{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and suppose the support condition holds. Put $q_i=⟨u_i|\sigma |u_i⟩>0$. Concavity of the scalar logarithm givesTerms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\ge 1-x$ yieldssince the $q_i$ sum to at most $\mathrm{Tr}\sigma =1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus $D(\rho \Vert \sigma )=0$ exactly when $\rho =\sigma$.

For the bipartite density operator,Indeed, a projector onto the kernel of ${\rho }_A$, tensored with $I_B$, has zero expectation in ${\rho }_{AB}$. Positivity implies that ${\rho }_{AB}$ annihilates its range: $\mathrm{Tr}({\rho }_{AB}P)=\Vert {\rho }_{AB}^{1/2}P{\Vert }_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the quantum relative entropy below is finite. On this support,Using the partial trace to evaluate the two local terms givesThis proves Subadditivity of Von Neumann entropy:The proof also identifies equality exactly with the product state ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

Introduce an orthonormal classical-label basis $|i⟩$ and the classical-quantum stateIts reduced density operators are ${\Gamma }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\Gamma }_A=\overline{\rho }={\sum }_i{p}_i{\rho }_i$. If ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$, the eigenvalues of ${\Gamma }_{XA}$ are $p_i{\lambda }_{ij}$. Thus the entropy of an orthogonal quantum mixture iswhere $H(p)=-{\sum }_i{p}_i{\log }_2{p}_i$ is the Shannon entropy. Apply the proved Subadditivity of Von Neumann entropy to $X$ and $A$ and cancel $H(p)$. This gives Concavity of Von Neumann entropy:Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i{\rho }_i=p_i\overline{\rho }$ follow from ${\Gamma }_{XA}={\Gamma }_X\otimes {\Gamma }_A$.

Omit indices of probability zero. The support of a positive operator of every remaining ${\rho }_i$ lies within that of $\overline{\rho }$, because $\overline{\rho }$ is a positive sum containing $p_i{\rho }_i$. Thus $D({\rho }_i\Vert \overline{\rho })$ is finite.

First suppose $\mathrm{supp}\overline{\rho }\subseteq \mathrm{supp}\sigma$, so all quantities are finite. Expand the trace definitions of quantum relative entropy:Therefore Donald's identity isIf the support condition on $\sigma$ fails, at least one positive-weight ${\rho }_i$ also fails it. The left side and the final term on the right are then $+\infty$, while the first right-hand sum remains finite. The equality is valid in this extended sense, without taking any undefined difference of infinities.