Suppose the logical error rate p_L of a surface code can be approximated as p_L ∝ (p/p_th)^((d+1)/2), using the physical error rate p, the threshold p_th and the code distance d. With p = 0.1% and p_th = 1%, which is closest to the ratio of logical error rates when d is increased from 5 to 9 (the value at d = 5 divided by the value at d = 9)?

Answer: C. About 100

The exponent (d+1)/2 is 3 for d = 5 and 5 for d = 9, so the ratio is (p_th/p) squared = 10² = 100, about 100. If the physical error rate is well below the threshold, each increase of the code distance by 2 reduces the logical error rate by a factor of p_th/p (10 here), so errors fall exponentially as more qubits are used. Conversely, if p is above the threshold, a larger d only adds more chances for errors and makes the logical error rate worse. The surface code threshold depends on the error model but is roughly 1%, and keeping the physical error rate well below the threshold is a prerequisite for error correction.

• A. This uses the ratio of code distances, 9/5, directly. The logical error rate changes exponentially with d.
• B. This is the ratio for an increase of d by 2 (from 5 to 7). From 5 to 9 is two steps, giving 10².
• C. Correct. (0.01/0.001)² = 100.