Decoding with a Bell Measurement

In superdense coding, the receiver holds both qubits after the sender transmits one qubit over the quantum channel. That receiver — conventionally Bob — must now extract the two classical bits the sender encoded. The extraction tool is exactly the same Bell measurement used in teleportation: a CNOT followed by a Hadamard, and then ordinary measurement in the computational basis.

From encoded state back to bits

Recall the encoding table. Starting from the shared Bell pair $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ , Alice applies one of four operations to qubit 0 to encode two bits:

| Bits | Gate on qubit 0 | Resulting state | |------|-----------------|-----------------| | 00 | $I$ (none) | $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ | | 01 | $X$ | $|\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$ | | 10 | $Z$ | $|\Phi^-\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$ | | 11 | $ZX$ | $|\Psi^-\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ |

Each encoding leaves the joint system in a different Bell state. Since the four Bell states are orthogonal — they form a basis — a measurement that distinguishes them perfectly is possible. That measurement is the Bell measurement.

How the Bell measurement decodes

The Bell measurement circuit is:

CNOT with control qubit 0 and target qubit 1.
H on qubit 0.
Measure both qubits in the computational basis.

Working through the 10 case step by step illustrates the pattern. After the $Z$ encoding, the state is $|\Phi^-\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$ .

Applying $\mathrm{CNOT}(0 \to 1)$ maps $|00\rangle \to |00\rangle$ and $|11\rangle \to |10\rangle$ (qubit 1 is flipped when qubit 0 is $|1\rangle$ ):

\mathrm{CNOT}\,|\Phi^-\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle - |10\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(|0\rangle - |1\rangle\bigr) \otimes |0\rangle = |{-}\rangle|0\rangle.

Next, $H$ on qubit 0 maps $|{-}\rangle \to |1\rangle$ :

(H \otimes I)\,|{-}\rangle|0\rangle = |1\rangle|0\rangle = |10\rangle.

Measuring now yields the outcome 10 with certainty — exactly the two bits Alice encoded. The same algebra applied to the other three Bell states shows the pattern is universal:

| Encoded state | Bell meas. outcome | |---------------|--------------------| | $|\Phi^+\rangle$ | 00 | | $|\Phi^-\rangle$ | 10 | | $|\Psi^+\rangle$ | 01 | | $|\Psi^-\rangle$ | 11 |

Why this is remarkable

Two classical bits of information were carried by a single qubit. Without the pre-shared entangled pair, sending one qubit can convey at most one classical bit. The Bell pair doubles the capacity — but only because both parties agreed to share it in advance, and the receiver already holds one half. The protocol does not violate any information-theoretic limit: the Bell pair itself must have been distributed before the protocol begins, consuming one shared ebit of entanglement — a quantum resource that requires transmitting one qubit in advance.

Try it

The starter code prepares $|\Phi^+\rangle$ and applies the $Z$ gate to encode the bits 10, leaving the system in $|\Phi^-\rangle$ . Add the Bell measurement (CNOT from qubit 0 to qubit 1, then H on qubit 0, then measure) so that the grader sees the outcome 10 with probability 1.

Run your code to see the quantum state.

The Probabilities panel should show a single bar at 10 at full height, confirming that the Bell measurement correctly decoded the two bits.