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Averaging conjugation by all Heisenberg-Weyl operators completely depolarizes a system:Summing over phases $Z^m$ kills off-diagonal matrix elements, and averaging shifts $X^k$ makes all diagonal elements equal. Acting on subsystem $B$ of a bipartite operator gives $({\mathrm{Tr}}_{B}A)\otimes I_B/d$. This identity turns joint convexity of quantum relative entropy into monotonicity under partial trace.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

First prove monotonicity for a partial trace. Let $r=\dim {\mathcal{H}}_B$ and define the Heisenberg-Weyl twirling channelThe supplied identity on individual operators extends to bipartite operators by expanding them in a tensor-product basis. The joint convexity of quantum relative entropy and its unitary invariance implyThe additivity of quantum relative entropy makes the left side $D({\rho }_A\Vert {\sigma }_A)$, proving the partial-trace case.

For a general deterministic quantum operation, meaning a quantum channel, use a Stinespring dilation $\Lambda (X)={\mathrm{Tr}}_E(VX{V}^{†})$ with $V^{†}V=I$. An isometric embedding preserves quantum relative entropy, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result givesThe other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the Lindblad-Uhlmann monotonicity theorem. The proof applies to deterministic channels; normalized postselection is not such a linear channel.

Data processing states $D(\Phi (\rho )\Vert \Phi (\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $\Phi$. Joint convexity statesStrong subadditivity states $S(ABC)+S(B)\le S(AB)+S(BC)$.

For strong subadditivity, apply data processing under ${\mathrm{Tr}}_C$ to ${\rho }_{ABC}$ and $I_A/d_A\otimes {\rho }_{BC}$. Expanding both relative entropies cancels $\log d_A$ and gives precisely the stated inequality. For joint convexity, use flagged states $\hat{\rho }={\sum }_i{p}_i|i⟩⟨i|\otimes {\rho }_i$ and $\hat{\sigma }={\sum }_i{p}_i|i⟩⟨i|\otimes {\sigma }_i$. Part a gives $D(\hat{\rho }\Vert \hat{\sigma })={\sum }_i{p}_{i}D({\rho }_i\Vert {\sigma }_i)$; discarding the flag and applying data processing gives joint convexity of quantum relative entropy.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.