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Every entanglement-breaking channel $\mathcal{E}$ has additive Holevo capacity with every finite-dimensional quantum channel $\mathcal{N}$. A measure-and-prepare channel factorization exposes a classical measurement outcome. The quantum mutual information balance identity and data processing for quantum mutual information then bound the joint output Holevo quantity by the sum of the individual capacities, using the conditional input ensemble after a local measurement for the other channel. Independent product ensembles attain the reverse inequality.

Attach an orthogonal classical label register to an ensemble. Its quantum mutual information with the quantum system equals the ensemble Holevo quantity. A local quantum channel changes the ensemble states but preserves their labels and probabilities. Data processing for quantum mutual information then proves that the output Holevo quantity cannot exceed the input value.

For a purified input with reference $R$ and channel output $B$, $I(R:B)=S(R)+S(B)-S(RB)=S(\rho )+I_c(\Lambda ,\rho )$. The reference entropy is fixed through subsequent channels. Thus data-processing inequality for coherent information immediately implies data processing for quantum mutual information between that reference and successive outputs.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.

Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.

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.

Encode the ensemble label in an orthogonal classical register:The classical-quantum state has $S(XB)=H(p)+{\sum }_x{p}_{x}S({\rho }_x)$, $S(X)=H(p)$ and $S(B)=S(\bar{\rho })$, where $\bar{\rho }={\sum }_x{p}_x{\rho }_x$. HenceApplying ${\mathrm{id}}_X\otimes \Lambda$ changes the conditional states to the output ensemble. The same identity gives $I(X:B^{\prime })=\chi ({\mathcal{E}}^{\prime })$. Data processing for quantum mutual information, proved in Question 3, therefore yields the Holevo quantity under a quantum channel:This concerns the information available in the whole ensemble, not the change in the entropy of each individual state.

Both output states have the same reference marginal, and $S(R)=S(\rho )$. Expand quantum mutual information to obtain the mutual information and coherent information identityThe input-entropy term is identical in the two expressions. Subtract them and use the preceding data-processing inequality for coherent information:Thus the final quantum channel cannot increase the reference-output quantum mutual information, as required by data processing for quantum mutual information.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

Let a uniformly random bit $A$ be copied into a three-state classical register $B$ with probability $\epsilon$, and otherwise replaced in $B$ by a flagged erasure independent of $A$. Then $I(A:B)=\epsilon$ bits. Projecting $B$ onto its two nonerased states and conditioning on success gives a perfectly correlated bit pair with mutual information one. The postselection branch is completely positive and trace decreasing, but renormalization after conditioning is not a deterministic quantum channel. This does not contradict data processing for quantum mutual information, which applies to trace-preserving operations or an unconditioned instrument including its outcome flag.