= 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 \b[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 $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$, with Alice holding the first <qubit>. For a two-<bit> message $(a,b)\in\{0,1\}^2$, Alice applies the <local unitary operation> $U_{ab}=X^aZ^b$ to her <qubit>. The four encoded <pure states> are, up to irrelevant <global phases>,
$$
\begin{array}{c|c|c}
(a,b)&U_{ab}&(U_{ab}\otimes I)|\Phi^+\rangle\\\hline
(0,0)&I&|\Phi^+\rangle\\
(0,1)&Z&|\Phi^-\rangle\\
(1,0)&X&|\Psi^+\rangle\\
(1,1)&XZ&-|\Psi^-\rangle
\end{array}
$$
They are <orthogonal>: for any <matrix> $V$ on the first <qubit>, $\langle\Phi^+|(V\otimes I)|\Phi^+\rangle=\operatorname{Tr}V/2$, and the four <Pauli operators> are <orthogonal> in the <trace inner product>. Explicitly,
$$
\langle\Phi^+|(U_{ab}^\dagger U_{a'b'}\otimes I)|\Phi^+\rangle
=\delta_{aa'}\delta_{bb'}.
$$
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, giving
$$
\boxed{\text{one shared Bell pair + one noiseless qubit transmission}
\ \Longrightarrow\ \text{two classical bits}.}
$$
Before 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>.
Back to article page
