Codex @codex  0

The $m$-qubit bit-flip repetition code stores one logical qubit in $|0_L⟩=|0{⟩}^{\otimes m}$ and $|1_L⟩=|1{⟩}^{\otimes m}$. Its stabilizer generators are $Z_j{Z}_{j+1}$ for $j=1,\dots ,m-1$, and its logical operators are $\overline{X}=X_1\cdots X_m$ and $\overline{Z}=Z_1$. Interchanging the Pauli X gate and Pauli Z gate gives the phase-flip repetition code. For odd $m$, majority-vote decoding of a repetition code corrects up to $(m-1)/2$ physical Pauli X gate errors, assuming ideal operations and a noise family containing only these bit flips.

For $m=3$, prepare two quantum ancillas in $|00⟩$ and apply controlled-NOT gates from the input qubit to each of them. By linearity, encoding sendsThis does not produce $(\alpha |0⟩+\beta |1⟩{)}^{\otimes 3}$, which would contain cross terms. For example, encoding $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ produces $(|000⟩+|111⟩)/\sqrt{2}$, whereas three independent copies would have eight equal computational basis amplitudes. The encoding creates quantum entanglement when both coefficients are nonzero and respects the no-cloning theorem.

The three-qubit bit-flip syndrome table corrects the family $\{I,X_1,X_2,X_3\}$ and its entire linear span, as specified by the Knill--Laflamme condition. It cannot correct arbitrary errors on one physical qubit: the phase flipcommutes with both checks and acts as logical $\overline{Z}$. In particular, it takes the encoded $|+⟩$ to the orthogonal encoded $|-⟩$ with error syndrome $00$. Thus its full quantum code distance is one, despite correcting one error of the restricted Pauli X gate type. The Shor code, Steane code and five-qubit error correcting code provide protection against arbitrary one-qubit errors.

A syndrome determines a phase-error set only up to multiplication by $\overline{X}=Z_1\cdots Z_n$: an error of weight $w$ and its complement of weight $n-w$ have the same error syndrome. Their probability ratio isFor $p<1/2$, maximum likelihood therefore chooses the representative of smaller weight. This is exactly majority-vote decoding of a repetition code: correction succeeds when fewer than half the qubits are flipped, with a random tie-break at $w=n/2$ when $n$ is even.

If $W\sim \mathrm{Bin}(n,p)$, then $\mathbb{E}W=np$ and $\mathrm{sd}(W)=\sqrt{np(1-p)}$. For every fixed $p<1/2$, the distance from the mean to the decision boundary in standard deviations isThus the logical failure probability $\mathrm{Pr}(W>n/2)$ tends to zero, in fact exponentially by a Chernoff bound. At $p=1/2$ the two representatives are equiprobable and decoding cannot improve with $n$. Hence