Superdense Coding: The Idea

Quantum teleportation and superdense coding are, in a precise sense, inverses of each other. Teleportation uses two classical bits and one Bell pair to send one qubit. Superdense coding uses one qubit and one Bell pair to send two classical bits. Both protocols showcase entanglement as a communication resource, not just a curiosity.

The setup

Alice wants to send one of four possible messages to Bob: $00$ , $01$ , $10$ , or $11$ — two classical bits. The catch: she is only allowed to send Bob a single qubit over the quantum channel. If qubits were classical two-level systems this would obviously be impossible; a single classical bit can carry at most one bit of information. Superdense coding shows the restriction can be beaten, but only if Alice and Bob share an entangled pair prepared beforehand.

The shared resource is the Bell state

$|\Phi^+\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr),$

with Alice holding qubit $A$ and Bob holding qubit $B$ .

Alice's four operations

Alice encodes her two-bit message by applying a single-qubit gate to her qubit only — she never touches Bob's qubit. There are exactly four operations, one per message:

| Message | Gate Alice applies | Resulting joint state | |---|---|---| | $00$ | $I$ (do nothing) | $\tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = |\Phi^+\rangle$ | | $01$ | $X$ (bit-flip) | $\tfrac{1}{\sqrt{2}}(|10\rangle + |01\rangle) = |\Psi^+\rangle$ | | $10$ | $Z$ (phase-flip) | $\tfrac{1}{\sqrt{2}}(|00\rangle - |11\rangle) = |\Phi^-\rangle$ | | $11$ | $ZX$ (both) | $\tfrac{1}{\sqrt{2}}(|10\rangle - |01\rangle) = |\Psi^-\rangle$ |

Each operation steers the joint state into a different one of the four Bell states $\{|\Phi^\pm\rangle, |\Psi^\pm\rangle\}$ , which form an orthonormal basis for the two-qubit space. Because orthonormal states are perfectly distinguishable by a single measurement, four distinct messages can be recovered with certainty.

Bob's decoding

After receiving Alice's qubit, Bob holds both qubits. He performs a Bell measurement: a CNOT with Alice's qubit as control (Bob's qubit as target), followed by a Hadamard on Alice's qubit, and then a standard computational-basis measurement on both qubits. This exactly inverts the Bell-state preparation circuit, mapping each Bell state back to a unique computational-basis state:

Bob reads off Alice's two-bit message directly from the two measurement outcomes.

Why this does not violate information theory

Sending two classical bits using one physical qubit sounds like it breaks Shannon's channel-capacity theorem. The resolution is that the Bell pair is a pre-shared resource: its creation and distribution cost two qubits worth of quantum communication before the protocol even starts. The total quantum communication is one qubit sent (the encoding step) plus the two qubits originally distributed to set up the entanglement — three qubits in total, which is more than two classical bits. No law of physics is broken; entanglement is simply being traded for channel capacity.

Resources at a glance

| Resource | Quantity | |---|---| | Shared entangled pairs (Bell pairs) | 1 (consumed) | | Qubits sent Alice → Bob | 1 | | Classical bits communicated | 2 (the decoded message) |

The comparison with teleportation is vivid: teleportation sends 1 qubit of quantum information by spending 1 Bell pair and 2 classical bits; superdense coding sends 2 bits of classical information by spending 1 Bell pair and 1 qubit. Together they reveal that quantum entanglement is an independent resource — distinct from classical bits and from bare qubits — that can be exchanged for either kind of communication capacity.