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For input dimension $D$, an input of entropy $S(\rho )$ gives output entropy $h_2(p)+(1-p)S(\rho )$ and reference-output entropy $h_2(p)+pS(\rho )$. These follow from entropy of an orthogonal quantum mixture and purity of the input purification. Thus the quantum mutual information is $2(1-p)S(\rho )$, maximized at $\rho =I_D/D$. The entanglement-assisted classical capacity is consequently $2(1-p){\log }_{2}D$ bits per use; for a binary input it is $2(1-p)$.

For a memoryless quantum channel $\mathcal{N}$, successive uses are described by ${\mathcal{N}}^{\otimes n}$. Its entanglement-assisted classical capacity is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared quantum entanglement, independent of the message. Only the channel is used to transmit the message.

For input state ${\rho }_A$, choose a purification $|{\psi }_{\rho }{⟩}_{RA}$ and set ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The capacity formula isPurifications differ only by an isometry on the reference, so this expression depends only on the input density matrix. The quantity optimized is quantum mutual information.

The binary quantum erasure channel sends a qubit to a qubit subspace plus an orthogonal erasure flag:On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho )$. The entropy of an orthogonal quantum mixture givesThe reference-output state has two orthogonal branches:The first branch is pure; the second has entropy $S({\rho }_R)=s$. Thus $S({\omega }_{RB})=h_2(p)+ps$ andA qubit has entropy at most one bit, with equality at $\rho =I_2/2$. We concludeThe endpoint capacities are two and zero. As an operational check, superdense coding encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased qubit yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p){\log }_{2}4=2(1-p)$, agreeing with the entropy calculation.

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.

Use logarithms to base two, so all Von Neumann entropies and Shannon entropies are measured in bits. Write the nonzero eigenvalues of ${\rho }_i$ as ${\lambda }_{ij}$. Orthogonality of the supports makes the mixture a block diagonal matrix, with eigenvalues $p_i{\lambda }_{ij}$. ThereforeEach ${\rho }_i$ has trace one. Terms with zero probability or zero eigenvalue vanish under the convention $0\log 0=0$. This proves the entropy of an orthogonal quantum mixture.