Correcting a Bit Flip

Once syndrome measurement has told us which qubit flipped, the correction step is straightforward: apply the gate that undoes the error. For a bit-flip code this means applying a Pauli $X$ to the identified qubit, because $X^2 = I$.

Recap: the three-qubit bit-flip code

The three-qubit bit-flip code protects one logical qubit by spreading it across three physical qubits:

$|0_L\rangle = |000\rangle, \qquad |1_L\rangle = |111\rangle.$

A general logical state $\alpha|0_L\rangle + \beta|1_L\rangle = \alpha|000\rangle + \beta|111\rangle$ lives in the two-dimensional code space spanned by $|000\rangle$ and $|111\rangle$.

If qubit $k$ suffers a bit flip, the $X_k$ error moves the code space into an error space:

| Error | Logical 0 state | Logical 1 state | |-------|-----------------|-----------------| | none | $|000\rangle$ | $|111\rangle$ | | $X_0$ | $|100\rangle$ | $|011\rangle$ | | $X_1$ | $|010\rangle$ | $|101\rangle$ | | $X_2$ | $|001\rangle$ | $|110\rangle$ |

The four error spaces are mutually orthogonal, so they can be distinguished without any information about $\alpha$ or $\beta$.

The syndrome and what it means

Syndrome measurement computes two parity bits using ancilla qubits and CNOT gates, without ever measuring the data qubits directly:

$s_0 = q_0 \oplus q_1, \qquad s_1 = q_1 \oplus q_2.$

The two bits together identify the error:

| $s_0$ | $s_1$ | Syndrome | Indicated error | |--------|--------|----------|-----------------| | 0 | 0 | 00 | No error | | 1 | 1 | 11 | $X_1$ (qubit 1 flipped) | | 1 | 0 | 10 | $X_0$ (qubit 0 flipped) | | 0 | 1 | 01 | $X_2$ (qubit 2 flipped) |

Applying the correction

Given syndrome $s_0 = s_1 = 1$, we know qubit 1 flipped. The correction is simply

$X_1 \cdot X_1 |\psi_{\text{error}}\rangle = X_1^2 |\psi\rangle = |\psi\rangle,$

because $X^2 = I$. More generally the correction table is:

| Syndrome | Gate to apply | |----------|---------------| | 00 | None ($I$) | | 11 | $X$ on qubit 1 | | 10 | $X$ on qubit 0 | | 01 | $X$ on qubit 2 |

After correction the register is back in the code space, with the logical information intact.

Try it

The starter code injects an $X$ error on qubit 1, placing the register in $|010\rangle$. The syndrome would read $(s_0, s_1) = (1, 1)$. Apply the correction gate to restore the state to $|000\rangle$.

After pressing Check you should see the statevector collapse back to the all-zero state $|000\rangle$, confirming that the logical information has been recovered.

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