The Three-Qubit Phase-Flip Code

The three-qubit bit-flip code (module 14, lessons 3–5) protected against $X$ errors by encoding each logical qubit as three physical qubits in the $Z$ basis. A phase-flip error is the $Z$ gate, which does nothing to $|0\rangle$ but flips the sign of $|1\rangle$:

$Z|0\rangle = |0\rangle, \qquad Z|1\rangle = -|1\rangle.$

A $Z$ error on the logical $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ state turns it into $|-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, flipping the relative phase. Since $Z$ commutes with the computational basis it is invisible to the bit-flip code. We need a different approach.

Basis change: from Z errors to X errors

The key insight is that $Z$ and $X$ are related by the Hadamard:

$H Z H = X.$

This means a $Z$ error in the $Z$ basis looks like an $X$ error in the $X$ basis. If we first rotate every physical qubit from the $Z$ basis into the $X$ basis by applying $H$, a phase-flip $Z$ becomes a bit-flip $X$ — and the bit-flip code already handles those.

Encoding procedure

Starting from a logical qubit in state $|\psi_L\rangle = \alpha|0\rangle + \beta|1\rangle$:

1. Fan out using two CNOTs so qubits 1 and 2 mirror qubit 0: $(\alpha|0\rangle + \beta|1\rangle)|00\rangle \xrightarrow{CNOT_{01},\,CNOT_{02}} \alpha|000\rangle + \beta|111\rangle.$

2. Rotate into the X basis by applying $H$ to every qubit: $\alpha|000\rangle + \beta|111\rangle \xrightarrow{H^{\otimes 3}} \alpha|{+}{+}{+}\rangle + \beta|{-}{-}{-}\rangle.$

Here $|{+}\rangle = H|0\rangle$ and $|{-}\rangle = H|1\rangle$. The logical codewords are therefore

$|0_L\rangle = |{+}{+}{+}\rangle, \qquad |1_L\rangle = |{-}{-}{-}\rangle.$

A phase-flip $Z$ on any single physical qubit maps $|{+}\rangle \leftrightarrow |{-}\rangle$ on that qubit — exactly a bit flip in the $X$ basis. Majority-vote decoding (in the $X$ basis) then corrects any single-qubit $Z$ error.

Why this works

After encoding, a $Z$ error on physical qubit $k$ sends $|{+}\rangle \to Z|{+}\rangle = |{-}\rangle$ on that qubit. Applying $H^{\otimes 3}$ before measuring converts this to a plain $X$ error in the computational basis, and the two-ancilla parity checks $Z_0 Z_1$ and $Z_1 Z_2$ (syndrome measurement, lesson 4) locate and correct it. The original logical state is recovered by the inverse process.

The three-qubit phase-flip code corrects any single-qubit phase-flip error with only three physical qubits and two ancilla parity checks, at the cost of being unable to correct bit-flip errors — a trade-off that motivates the nine-qubit Shor code, which combines both codes.

Try it

Build the encoding circuit for the logical $|+\rangle = H|0\rangle$ state. The grader checks the full three-qubit statevector, which must equal $\alpha|{+}{+}{+}\rangle + \beta|{-}{-}{-}\rangle$ for $\alpha = \beta = \tfrac{1}{\sqrt{2}}$.

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