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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.