The Repetition Code Revisited

We met the three-qubit repetition code informally as a first example of protecting a bit. Now we recast it in the stabilizer formalism, which exposes exactly what it does and does not protect and sets the template for every code that follows.

The bit-flip code as a stabilizer code

The bit-flip repetition code encodes

so a general logical state $\alpha|0\rangle_L + \beta|1\rangle_L = \alpha|000\rangle + \beta|111\rangle$. This two-dimensional codespace is the simultaneous $+1$ eigenspace of the two commuting stabilizer generators

Each $g_i$ checks the parity of a pair of qubits without measuring their individual values: $Z_1 Z_2$ returns $+1$ when qubits 1 and 2 agree and $-1$ when they differ. Both codewords $|000\rangle$ and $|111\rangle$ have all pairs agreeing, so both are $+1$ eigenstates — measuring $g_1, g_2$ never disturbs the logical superposition.

Syndrome decoding

A single bit-flip $X_i$ anticommutes with whichever generators touch qubit $i$, flipping those syndrome bits:

| Error | $g_1=Z_1Z_2$ | $g_2=Z_2Z_3$ | |---|---|---| | none | $+1$ | $+1$ | | $X_1$ | $-1$ | $+1$ | | $X_2$ | $-1$ | $-1$ | | $X_3$ | $+1$ | $-1$ |

The three single-qubit flips give three distinct syndromes, so each is identified and undone. This is a $[[3,1,1]]$ code with respect to bit-flips: $n=3$, $k=1$, and a logical operator $\bar X = X_1X_2X_3$ of weight 3 while $\bar Z = Z_1$ has weight 1 — so its distance to phase errors is just 1.

What it cannot do

The repetition code is blind to phase errors. A single $Z_i$ commutes with both $Z$-type stabilizers, produces no syndrome, and acts as $\bar Z$ up to a stabilizer — an undetectable logical phase flip.

Try it

Build the encoded $|+\rangle_L$ of the bit-flip code, the GHZ state $(|000\rangle + |111\rangle)/\sqrt{2}$, using a Hadamard and two CNOTs. The grader derives the target state vector from the reference circuit and checks yours against it.

Sign in on the full site to ask questions and join the discussion.