Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 4 / c

If $\epsilon =0$, then ${\rho }_{AB}={\sigma }_{AB}$ and the desired continuity bound is immediate, so assume $\epsilon >0$. Apply the positive-negative decomposition of a Hermitian operator toBecause $\mathrm{Tr}X=0$ and $\Vert X{\Vert }_1=2\epsilon$, one hasThus ${\Delta }_{AB}=X_{+}/\epsilon$ is positive with trace one, hence is a density operator. DefineIt is a convex combination of states. The equation $\epsilon {\Delta }_{AB}^{\prime }=(1+\epsilon ){\omega }_{AB}-{\rho }_{AB}$ giveswhich is likewise positive and has trace one.

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The two forms of ${\omega }_{AB}$ found in part (i) giveTherefore, in the Löwner order,

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Set $q=(1+\epsilon {)}^{-1}$, so $1-q=\epsilon /(1+\epsilon )$ and $\omega =q\rho +(1-q){\Delta }^{\prime }$. For every state ${\xi }_B$, the entropy bound for a binary mixture givesTaking the minimum over ${\xi }_B$ and using the variational characterization of quantum conditional entropy on each term givesSince binary entropy satisfies $H(q)=H(1-q)$, this is

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The other convex decomposition is $\omega =q\sigma +(1-q)\Delta$. Applying the concavity of quantum conditional entropy, which follows from the Strong subadditivity of Von Neumann entropy, yields

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Combining parts (iii) and (iv), then multiplying by $1+\epsilon$, givesThe dimension bound for quantum conditional entropy isIndeed, Subadditivity of Von Neumann entropy gives $H(A|B{)}_{\tau }\le S({\tau }_A)\le \log d_A$, while the Araki–Lieb inequality gives $H(A|B{)}_{\tau }\ge -S({\tau }_A)\ge -\log d_A$. HenceInterchanging $\rho$ and $\sigma$ proves the continuity bound for quantum conditional entropy:

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