 Solution

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.