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For $k$ equiprobable quantum code states decoded by a POVM $\{E_m\}$, testing whether the classical message label agrees with the decoded output gives a joint binary POVM with $F_0={\sum }_m|m⟩⟨m|\otimes E_m$. Its acceptance probability is the decoding success $1-ϵ$ on the joint classical-quantum state and $1/k$ on the product of its marginals. Applying quantum relative entropy monotonicity to this test gives the one-shot classical-quantum coding converse.

The classical-quantum state is block diagonal. If ${\lambda }_{xj}$ are the eigenvalues of ${\rho }_x$, its eigenvalues are $p_x{\lambda }_{xj}$. Inserting these into the definition of Von Neumann entropy gives the displayed sum of classical Shannon entropy and mean conditional quantum entropy. Its quantum mutual information is therefore the Holevo quantity $S({\sum }_x{p}_x{\rho }_x)-{\sum }_x{p}_{x}S({\rho }_x)$.

For an ensemble $\{p_x,{\rho }_x\}$ with average $\bar{\rho }$, the Holevo quantity is $\chi =S(\bar{\rho })-{\sum }_x{p}_{x}S({\rho }_x)=I(X:B)$ of the associated classical-quantum state.

Let $A$ be a register with orthonormal basis $\{|x⟩\}$ for Alice's classical symbol, $Q$ the transmitted system, and $B$ a register initially in a fixed blank state. The initial classical-quantum state isA POVM fixes probabilities but not a unique conditional state of $Q$. Choose the Lüders rule instrument $M_y=\sqrt{E_y}$, which satisfies ${\sum }_y{M}_y^{†}{M}_y=I$. Record outcome $y$ in orthogonal states of $B^{\prime }$ and retain the corresponding quantum output $Q^{\prime }$. The final nonselective state isThe positive operators in this sum are unnormalized: their traces already include the outcome probabilities. If $\mathrm{Tr}(E_y{\rho }_x)>0$, the normalized conditional state of $Q^{\prime }$ is $M_y{\rho }_x{M}_y^{†}/\mathrm{Tr}(E_y{\rho }_x)$.

More generally, an instrument may use operators $M_{y\mu }$ with ${\sum }_{\mu }{M}_{y\mu }^{†}{M}_{y\mu }=E_y$, replacing $M_y{\rho }_x{M}_y^{†}$ by ${\sum }_{\mu }{M}_{y\mu }{\rho }_x{M}_{y\mu }^{†}$. All subsequent classical information bounds are unchanged. This explicitly accounts for the fact that a POVM does not determine the post-measurement state of a retained quantum system.

Introduce an orthonormal classical-label basis $|i⟩$ and the classical-quantum stateIts reduced density operators are ${\Gamma }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\Gamma }_A=\overline{\rho }={\sum }_i{p}_i{\rho }_i$. If ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$, the eigenvalues of ${\Gamma }_{XA}$ are $p_i{\lambda }_{ij}$. Thus the entropy of an orthogonal quantum mixture iswhere $H(p)=-{\sum }_i{p}_i{\log }_2{p}_i$ is the Shannon entropy. Apply the proved Subadditivity of Von Neumann entropy to $X$ and $A$ and cancel $H(p)$. This gives Concavity of Von Neumann entropy:Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i{\rho }_i=p_i\overline{\rho }$ follow from ${\Gamma }_{XA}={\Gamma }_X\otimes {\Gamma }_A$.

Write $\overline{\rho }={\sum }_i{p}_i{\rho }_i$. The Holevo quantity iswhere $S$ is Von Neumann entropy, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the classical-quantum stateThe entropy of an orthogonal quantum mixture givesHere $H(p)$ is the Shannon entropy of the classical label. Hence $I(X:B)=\chi (\mathcal{E})$, where $I$ denotes quantum mutual information.

Let $\Phi :B\to C$ be the proposed quantum channel. Using its Kraus representation, define an linear isometry $V={\sum }_{\ell }{A}_{\ell }\otimes |\ell {⟩}_E$, since $V^{†}V=I$. This is a Stinespring representation. Apply it to $B$, obtaining a state ${\eta }_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An linear isometry preserves nonzero eigenvalues and therefore Von Neumann entropy, soTheir difference isThe Strong subadditivity of Von Neumann entropy states that this expression is nonnegative for every tripartite state. ConsequentlyThus the Holevo quantity under a quantum channel decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.

Introduce an orthonormal classical label register $X$ and the classical-quantum stateThe blocks have orthogonal supports because their label vectors are orthogonal, regardless of whether the original ${\rho }_i$ have orthogonal supports. Part (a) gives $S(XQ)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$. The marginals are ${\omega }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\omega }_Q={\sum }_i{p}_i{\rho }_i$, so $S(X)=H(p)$. Apply the Subadditivity of Von Neumann entropy and cancel $H(p)$:Thus $S({\sum }_i{p}_i{\rho }_i)\ge {\sum }_i{p}_{i}S({\rho }_i)$. This is the Concavity of Von Neumann entropy.

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.

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.

Choose $P_X$ uniform on ${\mathcal{A}}_M$ and zero outside it, so the marginal density operator of $Q$ is ${\bar{\rho }}_Q=k^{-1}{\sum }_m\rho (m{)}_Q$. Define the binary test for quantum decoding success byEach $E(m)$ is a positive contraction. The computational basis blocks of $F(0)$ are these effects on ${\mathcal{A}}_M$ and zero elsewhere. Thus $0\le F(0)\le I$, making $F$ a POVM on the whole space, including labels outside ${\mathcal{A}}_M$.

On the correlated classical-quantum state, its acceptance probability isOn the product of the marginal density operators,since ${\sum }_{m}E(m)=I_Q$. The two requested probabilities are therefore $1-ϵ$ and $1/k$. In the notation for binary output density operators, the measurement channel sends the joint input to $\omega [ϵ]$ and the product input to $\omega [1-1/k]$; their parameters describe outcome one, not outcome zero.

By Donald's identity, the objective equalsThe nonnegativity of quantum relative entropy makes the last term at least zero, with equality exactly when $\rho =\overline{\omega }$. Hence the unique minimizing density operator is the ensemble average andFor the classical-quantum state ${\sigma }_{XB}$, its marginal on $X$ has entropy $H(p)$, its marginal on $B$ is $\overline{\omega }$, and its joint entropy is $H(p)+{\sum }_j{p}_{j}S({\omega }_j)$. Substituting in the definition of quantum mutual information givesThis is the relative-entropy barycenter of a quantum ensemble: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the Holevo quantity.

Both states are block diagonal in $X$. On the support of each block,Substitution into quantum relative entropy givesTaking $-\mathrm{Tr}{\rho }_{XB}\log {\rho }_{XB}$ similarly yields the entropy of a classical-quantum state:

For an ensemble $\{p_x,{\rho }_x\}$ and measurement outcome $Y$, the Holevo bound iswhere $\bar{\rho }={\sum }_x{p}_x{\rho }_x$. Form the classical-quantum state ${\omega }_{XB}={\sum }_x{p}_x|x⟩⟨x|\otimes {\rho }_x$. Part a givesThe measurement is a quantum channel from $B$ to a classical register $Y$. Data processing for relative entropy, applied to the mutual-information representation, gives $I(X:Y)\le I(X:B)$ and proves the bound.

The ensemble average $\overline{\omega }$ uniquely minimizes ${\sum }_j{p}_{j}D({\omega }_j\Vert \rho )$ over density operators $\rho$. Donald's identity separates this objective into a constant plus $D(\overline{\omega }\Vert \rho )$, and nonnegativity of quantum relative entropy proves the claim. The minimum is the Holevo quantity $S(\overline{\omega })-{\sum }_j{p}_{j}S({\omega }_j)$, equivalently the quantum mutual information of the associated classical-quantum state.