Detecting Errors as Syndromes

A single stabilizer measurement returns one bit. The full set of those bits, read together, is the syndrome — the code's report on what kind of error (if any) has occurred. The decoder never sees the data qubits; it sees only this stream of parity bits.

What a flipped check means

In the previous lesson a quiet $Z_0 Z_1$ check returned $0$ . Now suppose a bit-flip error $X_0$ strikes data qubit $0$ between rounds. The data parity becomes odd, so when the ancilla copies it,

\text{CNOT}_{0\to a}\,\text{CNOT}_{1\to a}\;X_0|00\rangle|0\rangle_a = |10\rangle\,|1\rangle_a,

and the syndrome bit reads $1$ . The check has fired. An $X$ error on either data qubit of a $Z$ -type check flips that check; an $X$ error on a qubit shared by two $Z$ -checks flips both. This is the whole detection mechanism: errors live on qubits, but they announce themselves on the checks that touch them.

Syndromes are differences in time

A subtle but essential point: a decoder does not act on the raw value of a check, but on whether it changed from the previous round. A check that reads $1$ in two consecutive rounds usually signals a stable measurement artefact, not a freshly arrived error. So the detection event a decoder actually consumes is

\text{detector} = s_t \oplus s_{t-1},

the XOR of a check's outcome in successive rounds. A new error produces a pair of detection events in spacetime — one when the error appears and one when it is later corrected or leaves — and matching those pairs is the job of the decoder we meet next.

Try it

Reproduce a single fired check. Inject a bit-flip on data qubit $0$ with c.x(0), then run the $Z_0 Z_1$ extraction (c.cx(0, 2), c.cx(1, 2)) into ancilla qubit $2$ and measure it. The data is now $|10\rangle$ and the ancilla reports parity $1$ , so the only outcome is $|101\rangle$ — syndrome bit $1$ , the check has fired.

Run your code to see the quantum state.

Compare with the quiet round of the previous lesson: there the bar sat at $000$ ; here it moves to $101$ , and the ancilla bit has flipped from $0$ to $1$ .