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.

An error syndrome is the list of outcomes obtained by measuring a chosen ordered family of stabilizer generators. Write $s_j=0$ for eigenvalue $+1$ and $s_j=1$ for eigenvalue $-1$. For a definite Pauli error $E$ on a code state, the rule isThe commuting checks admit joint orthogonal projectorsFor density operator $\rho$, the Born rule gives outcome probability $\mathrm{tr}({\Pi }_s\rho )$ and conditional state ${\Pi }_s\rho {\Pi }_s/\mathrm{tr}({\Pi }_s\rho )$ when that probability is nonzero. A definite Pauli error has one deterministic error syndrome independent of the encoded amplitudes. A coherent linear combination of Pauli errors may occupy several sectors, so measuring the error syndrome selects a sector; recovery preserves the logical qubit when the entire error family is jointly correctable. ^[ref].

An error syndrome locates an error only relative to an assumed family and a recovery rule. Two Pauli errors differing by a logical operator can share an error syndrome; see stabilizer-syndrome coset. In the three-qubit bit-flip syndrome table, $X_3$ and $X_1{X}_2$ both give $01$. Recovering with $X_3$ corrects the first but converts the second into the logical operator $X_1{X}_2{X}_3$. Likewise, $Z_1$ has error syndrome $00$ yet changes the encoded relative phase.

Assume that each physical qubit undergoes an independent quantum bit-flip channel with probability $p$ during one storage interval, and that encoding, error syndrome extraction, recovery and decoding are all ideal. Use the three-qubit bit-flip syndrome table to apply the indicated minimum-weight recovery. Zero or one bit flips are corrected. Two bit flips have the same error syndrome as the complementary one-site error: for example, $X_1{X}_2$ gives $01$, so recovery $X_3$ leaves $\overline{X}=X_1{X}_2{X}_3$. Three bit flips give $00$ and also leave $\overline{X}$. This exchanges $|000⟩$ and $|111⟩$, and therefore acts as a logical Pauli X gate after decoding.

The number of physical bit flips has a binomial distribution with three trials, hence the logical-error probability isFor a logical density operator $\rho$, the decoded quantum channel is exactly $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$. Here logical failure means a nontrivial logical Pauli X gate error; a particular input such as $|+⟩$ can be invariant under that error. At $p=0.1$, exactly two flips have probability $3(0.1{)}^2(0.9)=0.027$ and three flips have probability $(0.1{)}^3=0.001$, giving $p_L=0.028$. The block succeeds with probability $0.972$ compared with $0.9$ for an unencoded qubit. At $p=0.01$, $p_L=0.000298$, illustrating the quadratic suppression of rare independent bit flips.

Factoring the improvement givesThus one level reduces logical-error probability for $0<p<1/2$, is equal at $p=0,1/2,1$, and makes it worse for $1/2<p<1$ under this same decoder. The suppression arises because a logical error requires at least two physical bit flips. With independent disjoint inner blocks, the same noise model and ideal correction at every level, a concatenated quantum error-correcting code has $p_{\ell +1}=3p_{\ell }^2-2p_{\ell }^3$, with $p_0=p$. At $p_0=0.1$, two levels use nine data qubits and give $p_2=3(0.028{)}^2-2(0.028{)}^3=0.002308096$.

This calculation describes ideal correction of stored information with the specified quantum bit-flip channel. Repeated correction in time additionally requires a fresh specified noise model for each interval; imperfect quantum gates, faulty measurements and correlated errors change the calculation. The timed bit flip during repetition-code syndrome extraction and exact encoded-state failure with a faulty parity check show concretely why this storage calculation alone is not a fault-tolerance result.