Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 65 / 3

Use base-two logarithms throughout, so information and entropy are in bits. The relative entropy of classical probability distributions isSet $0{\log }_2(0/q)=0$; if $p(x)>0$ and $q(x)=0$, define the value to be $+\infty$. In the finite case, $\ln u\le u-1$ givesHence $D(p\Vert q)\ge 0$, with equality exactly when $p=q$: equality in the logarithm inequality requires $q(x)=p(x)$ on the support, and equality in the last step leaves no mass outside it.

The quantum relative entropy is $S(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ when the support of a positive operator $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Write spectral decompositions $\rho ={\sum }_i{r}_i|i⟩⟨i|$ and $\sigma ={\sum }_j{t}_j|u_j⟩⟨u_j|$, and set $q_i=⟨i|\sigma |i⟩$.

The weights $w_{ij}=|⟨i|u_j⟩{|}^2$ sum to one. Concavity of the scalar logarithm gives the diagonal logarithm concavity boundFor $r_i>0$, support inclusion removes overlaps with zero eigenvalues of $\sigma$; alternatively take a positive regularization and pass to the limit. The two diagonal vectors $r$ and $q$ are probability distributions. ThereforeThis proves nonnegativity of quantum relative entropy without assuming the states commute. Equality in the strict logarithmic concavity condition makes each occupied $|i⟩$ an eigenvector of $\sigma$; classical equality gives $q_i=r_i$ and no weight outside the support, hence equality occurs exactly for $\rho =\sigma$.

A deterministic quantum operation here is a CPTP map. By Stinespring dilation, there is an isometry $V:B\to B^{\prime }E$ with $\Lambda (\tau )={\mathrm{Tr}}_E(V\tau V^{†})$. Let ${\omega }_{A{B}^{\prime }E}=(I\otimes V){\rho }_{AB}(I\otimes V^{†})$. An isometry preserves all nonzero eigenvalues, soThe quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$. The difference after discarding the environment iswhere the 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 data processing for quantum mutual information givesSubsystem $A$ is unchanged; its prime simply labels the output. Keeping the environment would preserve the mutual information, while tracing it out cannot increase it.

Put $u=\epsilon /2$. The state is well defined for $0\le \epsilon \le 2$, with eigenvalues $1-u,u$. The first state is pure and has zero Von Neumann entropy, so the exact difference isThis is the entropy of a binary diagonal qubit mixture. Its trace distance from the pure state is $u$, not $\epsilon$. An explicit elementary upper bound follows from $-(1-u)\ln (1-u)\le u$:At zero the limit is zero. For small $\epsilon$, the sharp exact answer behaves as $(\epsilon /2){\log }_2(2/\epsilon )+\epsilon /(2\ln 2)+O({\epsilon }^2)$, displaying the logarithmic continuity of entropy near a pure state.