Stabilizer Measurement Cycles

A surface-code computation is a sequence of identical measurement cycles. In each cycle every ancilla is entangled with its data neighbours and then measured, producing one classical bit per check. We never measure the data qubits directly — that would destroy the logical state — so the ancilla acts as a probe that reports a parity while leaving the data's superposition intact.

Extracting a Z-type check with an ancilla

To measure the plaquette operator $Z_0 Z_1$ without learning the individual values of qubits $0$ and $1$ , we use a fresh ancilla $a$ prepared in $|0\rangle$ and apply a CNOT from each data qubit into the ancilla:

\text{CNOT}_{0\to a}\,\text{CNOT}_{1\to a}\;|q_0 q_1\rangle|0\rangle_a = |q_0 q_1\rangle\,|q_0 \oplus q_1\rangle_a.

The ancilla ends up holding the parity $q_0 \oplus q_1$ . Measuring it returns

$0$ when the data parity is even ( $|00\rangle$ or $|11\rangle$ ) — the $+1$ eigenvalue of $Z_0 Z_1$ ;
$1$ when the parity is odd ( $|01\rangle$ or $|10\rangle$ ) — the $-1$ eigenvalue.

Crucially the data register is left in $|q_0 q_1\rangle$ untouched, and for a superposition the ancilla measurement projects only onto a definite parity sector, never onto a single data basis state. That is exactly what lets the same check repeat round after round.

A quiet round

Consider the cleanest possible cycle: both data qubits in $|0\rangle$ , no error present. The parity is even, so the ancilla measures $0$ with certainty. The whole register stays in

|0\rangle_0 |0\rangle_1 |0\rangle_a,

and the syndrome bit reads $0$ — "this check sees nothing wrong." Establishing this error-free baseline is the first thing any decoder needs; the next lesson injects an error and watches the bit flip.

Try it

Build one quiet round of the $Z_0 Z_1$ check. Use qubits $0$ and $1$ as data (both $|0\rangle$ ) and qubit $2$ as the ancilla. Apply c.cx(0, 2) and c.cx(1, 2), then measure the ancilla. The grader checks that the only outcome with non-zero probability is $|000\rangle$ — syndrome $0$ .

Run your code to see the quantum state.

After running, the Probabilities tab should show a single bar at $000$ with probability $1$ .