Checkpoint: Concatenation Math

Putting the recursion to work

This checkpoint ties together the concatenation recursion and the overhead count. Recall the level- $L$ logical error rate for a fault-tolerant concatenated scheme,

p_L = p_{\text{th}}\left(\frac{p}{p_{\text{th}}}\right)^{2^{L}},

and that each level multiplies the physical-qubit count by the base-code size $n$ , so a level- $L$ logical qubit occupies $n^{L}$ physical qubits. The practical question an architect asks is: how many levels do I need to hit a target logical error rate, and what does that cost?

A concrete instance

Take a base code with threshold $p_{\text{th}} = 10^{-2}$ , a physical error rate $p = 10^{-3}$ (so the suppression base is $p/p_{\text{th}} = 0.1$ ), and a target logical error rate $\varepsilon = 10^{-15}$ — a representative requirement for a long algorithm. We need the smallest integer number of levels $L$ such that $p_L \le \varepsilon$ .

Working it out by hand:

p_L = 10^{-2}\,(0.1)^{2^{L}} = 10^{-(2 + 2^{L})} \le 10^{-15} \;\Longleftrightarrow\; 2 + 2^{L} \ge 15 \;\Longleftrightarrow\; 2^{L} \ge 13 .

Since $2^{3} = 8 < 13$ but $2^{4} = 16 \ge 13$ , the smallest level is

\boxed{\,L = 4\,.}

A quick sanity check on the boundary: $L = 3$ gives $p_3 = 10^{-2}(0.1)^{8} = 10^{-10}$ , which is larger than $10^{-15}$ and so insufficient; $L = 4$ gives $p_4 = 10^{-2}(0.1)^{16} = 10^{-18} \le 10^{-15}$ , comfortably below target. With a $7$ -qubit base code that is $7^{4} = 2401$ physical qubits per logical qubit — the price of double-exponential suppression.

Try it

Compute the smallest integer number of concatenation levels $L$ needed so that $p_L = p_{\text{th}}(p/p_{\text{th}})^{2^{L}} \le \varepsilon$ , with $p_{\text{th}} = 10^{-2}$ , $p = 10^{-3}$ , and $\varepsilon = 10^{-15}$ . Loop over $L = 0, 1, 2, \ldots$ , evaluating $p_L$ each time, and return the first $L$ that satisfies the bound. (Derive it in code — do not hardcode $4$ .)

Run your code to see the quantum state.