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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.

A logical qubit is a two-dimensional encoded quantum subsystem with chosen orthonormal logical basis states $|0_L⟩,|1_L⟩$. An encoding linear isometry sends $\alpha |0⟩+\beta |1⟩$ to $\alpha |0_L⟩+\beta |1_L⟩$. In a concatenated quantum error-correcting code, one level's logical qubit supplies one qubit of the next outer code.

Use a concatenated quantum error-correcting code: each of the five outer-code qubits is encoded in its own five-qubit inner block. Independent physical errors and ideal inner correction make each decoded block act as one outer qubit with error probability $q=f(p)$. Errors in distinct inner blocks remain independent in this model. Outer recovery succeeds if at most one of the five decoded blocks is bad, givingThus the two-level logical failure probability isThis improves the leading error order from $p$ for a bare qubit to $p^2$ for one level and $p^4$ for two. At least two physical errors must occur in each of two different inner blocks to defeat the prescribed two-stage decoder. Many error patterns with more than one error in the full 25-qubit register are therefore corrected: one error in each of all five blocks is removed at the inner level, and a single failed inner block is corrected at the outer level. Concentrating several physical errors in one block can be easier for this decoder than placing two errors in each of two blocks.

More generally, $\ell$ levels use $5^{\ell }$ physical qubits with recursion $p_{\ell +1}=f(p_{\ell })$. For $0<p<p_{\ast }$, one has $f(f(p))<f(p)<p$. Repeated concatenation decreases to zero: its limit must be a fixed point of the continuous function $f$, and below $p_{\ast }$ the only fixed point is zero. A simple quantitative bound is $f(x)\le 10x^2$, obtained by the union bound over the ten pairs of possible error positions. Induction giveswhich proves rapid suppression when $p<1/10$; the exact threshold is the larger value $p_{\ast }$. Above $p_{\ast }$ this recursion amplifies errors instead.

Below threshold, concatenation suppresses logical errors while allowing an entire bad inner block to be corrected at the next level. The cost is additional physical qubits, gates, syndrome extraction and recovery. The calculated threshold concerns the independent transmission-noise model with perfect encoding and recovery; including noisy operations or correlated channel errors requires a different error analysis.