Syndrome Extraction Circuits

Syndrome extraction is the heart of the QEC cycle. It answers the question "did an error occur, and where?" without revealing the encoded data — no individual data qubit is ever measured. The tool is the stabilizer: a multi-qubit Pauli operator whose eigenvalue we read out through an ancilla.

Stabilizers of the bit-flip code

The codewords $|000\rangle$ and $|111\rangle$ are both $+1$ eigenstates of the two operators

S_1 = Z_0 Z_1, \qquad S_2 = Z_1 Z_2.

These are the stabilizer generators. Any superposition of the codewords is also stabilized: $S_1|\psi_L\rangle = S_2|\psi_L\rangle = +|\psi_L\rangle$ . A $Z_i Z_j$ operator measures the parity of qubits $i$ and $j$ — it returns $+1$ when they agree and $-1$ when they differ — while saying nothing about whether they are both $0$ or both $1$ . That is exactly the data-blind information we need.

What an error does to the syndrome

An $X$ error on a single data qubit flips that qubit, changing one or two parities:

| Error | $Z_0Z_1$ | $Z_1Z_2$ | Syndrome $(s_1,s_2)$ | |-------|----------|----------|----------------------| | none | $+1$ | $+1$ | $(0,0)$ | | $X_0$ | $-1$ | $+1$ | $(1,0)$ | | $X_1$ | $-1$ | $-1$ | $(1,1)$ | | $X_2$ | $+1$ | $-1$ | $(0,1)$ |

Here we write a $-1$ eigenvalue as the bit $1$ . Each single-qubit error has its own unique two-bit syndrome — the property that makes the code correctable.

Measuring a parity with an ancilla

We cannot apply $Z_0 Z_1$ and "look at" its eigenvalue directly. Instead we copy the parity onto a fresh ancilla. Starting the ancilla in $|0\rangle$ , two CNOTs do the job:

\texttt{cx(0,3)};\ \texttt{cx(1,3)}.

The first writes qubit 0 into the ancilla; the second XORs qubit 1 on top. The ancilla ends in $|q_0 \oplus q_1\rangle$ — precisely the $Z_0 Z_1$ parity bit. A second ancilla collects $q_1 \oplus q_2$ the same way.

Try it

A bit-flip error has been injected on data qubit 0. Wire up the two parity measurements onto the ancillas and measure the register. The grader checks the measurement distribution: the syndrome must come out as $(s_1, s_2) = (1, 0)$ with certainty.

Run your code to see the quantum state.

Open the Probabilities tab after running: a single outcome should carry all the weight — the bitstring 10010, whose last two bits 10 are the syndrome pointing at qubit 0.