Concatenated Codes

Codes inside codes

Concatenation is the simplest route to arbitrarily strong protection: take a code, then re-encode each of its physical qubits using the same code again, and repeat. A level-1 qubit is one logical qubit of the base code. A level-2 qubit replaces every physical qubit of that level-1 block by another full block, and so on. With a base code on $n$ qubits, a level- $L$ logical qubit occupies

n^{L}

physical qubits. The Steane $[[7,1,3]]$ code, for example, uses $7$ physical qubits at level 1, $49$ at level 2, $343$ at level 3 — the cost grows geometrically, but as we will see the protection grows much faster.

The recursion for the logical error rate

Concatenation is powerful because of how the failure probability recurses. Suppose the base code together with its fault-tolerant gadgets corrects any single fault, so a level-1 logical error requires at least two independent faults among the $A$ fault locations in the gadget. Then

p_1 \;\le\; A\,p^2 ,

where $p$ is the physical error rate. Crucially, a level-2 block is built from level-1 components that behave exactly like physical components but with error rate $p_1$ . The identical argument applies one level up:

p_2 \;\le\; A\,p_1^2 = A\,(A p^2)^2 .

Iterating and writing $p_{\text{th}} \equiv 1/A$ , the level- $L$ logical error rate satisfies

\boxed{\,p_L \;\le\; p_{\text{th}}\left(\frac{p}{p_{\text{th}}}\right)^{2^{L}}\,.}

Double-exponential suppression

The exponent $2^L$ is the whole story. Below threshold, $p < p_{\text{th}}$ , so $p/p_{\text{th}} < 1$ and each added level squares the suppression. Concretely, take $p_{\text{th}} = 10^{-2}$ and a physical rate $p = 10^{-3}$ , so the ratio is $p/p_{\text{th}} = 0.1$ :

p_1 \le 10^{-2}\,(0.1)^{2} = 10^{-4}, \quad p_2 \le 10^{-2}\,(0.1)^{4} = 10^{-6}, \quad p_3 \le 10^{-2}\,(0.1)^{8} = 10^{-10}.

Three levels of a small code turn a $10^{-3}$ physical error into a $10^{-10}$ logical error. Because the suppression squares while the qubit count merely multiplies by $n$ , reaching a target $\varepsilon$ needs only $L \sim \log\log(1/\varepsilon)$ levels — the polylogarithmic overhead of the threshold theorem.

Try it

You will compute the level- $L$ logical error rate from the recursion. Use the boxed formula $p_L = p_{\text{th}}\,(p/p_{\text{th}})^{2^{L}}$ with a threshold $p_{\text{th}} = 10^{-2}$ , a physical error rate $p = 10^{-3}$ , and $L = 3$ levels. Return $p_L$ as a plain number. (The grader checks it against the value derived from the same formula, so compute it from first principles — don't paste a literal.)

Run your code to see the quantum state.