Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 3 / ii

A real function $f$ on $I=[0,1]$ is an operator convex function when, for all Hermitian operators $X,Y$ whose spectra lie in $I$ and every $0\le \lambda \le 1$,where $⪯$ is the Loewner order. Reversing the inequality defines an operator concave function, equivalently $-f$ is operator convex.

Because the two flags $E_C$ and $F_C$ are orthogonal projections, the flagged state is block diagonal. If $h(\lambda )$ denotes the binary entropy, thenandIts unflagged marginals are ${\rho }_{AB}=\lambda {\rho }_{AB}^{\prime }+(1-\lambda ){\rho }_{AB}^{\prime \prime }$ and ${\rho }_B=\lambda {\rho }_B^{\prime }+(1-\lambda ){\rho }_B^{\prime \prime }$. Substitution in Strong subadditivity of Von Neumann entropy cancels the two binary-entropy terms and givesThis is exactly the concavity of quantum conditional entropy.

For $t>0$, positive homogeneity and concavity giveAfter subtracting $f(X)$, dividing by $t$, and taking the one-sided directional derivative at zero,

Extend the quantum conditional entropy from normalized states to positive operators bySince $\mathrm{Tr}{X}_{AB}=\mathrm{Tr}{X}_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is positively homogeneous, and part (b) extends its concavity from states to the positive cone.

Apply part (c) with $X={\sigma }_{AB}$ and $Y={\rho }_{AB}$. Differentiating the matrix logarithm under the trace givesThe inequality from part (c), after moving $F(\rho )$ to the left, becomesThis is the data-processing inequality for quantum relative entropy under partial trace. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.