= Solution
In <superdense coding>, Alice applies a local <Pauli operator> to her half of the shared <Bell state>. Choose a phase bit $u$ and a parity bit $v$, and encode them using $X^vZ^u$ on Alice's qubit. With the standard definitions of the <Bell states>,
$$
\begin{array}{c|c|c}
uv&\text{Alice's operator}&\text{Bell state, up to global phase}\\\hline
00&I&\Phi^+=(|00\rangle+|11\rangle)/\sqrt2\\
01&X&\Psi^+=(|01\rangle+|10\rangle)/\sqrt2\\
10&Z&\Phi^-=(|00\rangle-|11\rangle)/\sqrt2\\
11&XZ&\Psi^-=(|01\rangle-|10\rangle)/\sqrt2
\end{array}.
$$
Alice sends her <qubit> to Bob. He now holds both halves and performs a <Bell-basis measurement>, recovering $uv$ without error. Equivalently, a <CNOT gate> with Alice's qubit as control, followed by a <Hadamard gate> on that qubit, maps these states to their computational labels up to <global phases>. Four distinguishable alternatives give \b[two classical bits per transmitted qubit], with one shared entangled pair consumed per use.
The initial entanglement distribution is independent of the later message and can be completed beforehand. Bob's local <reduced density matrix> before Alice's transmission is $I/2$ for every encoding, so <quantum no-signalling> prevents it from carrying her choice. This does not make the entangled pair free: if a qubit transmission is used to distribute it, counting that transmission as well gives two qubit transmissions for the two-bit message. The quoted dense-coding rate counts the subsequent message-bearing transmission with entanglement already shared.
Back to article page
