Certification cost of quantum models: measurement correlation, not parameter count

Abstract

Reporting the Fisher geometry of a trained variational quantum model is routine; quoting the shot budget that would establish it is not. Certifying an empirical Fisher matrix to relative Frobenius error $\varepsilon$ under coordinate-wise parameter shift costs $\Theta(B p^{2} V/(\varepsilon^{2} G))$ circuit executions, where $V$ is the measured readout variance and $G$ the measured squared gradient norm, with uniform allocation optimal in that class. One constant reproduces the cost of two circuit families whose exponents differ by a full power of $p$. The exponent is an identity in how $nV$ and $nG$ scale with the register, holding family by family to $0.001$ across 624 matrix-product-state cells once the finite-$p$ prefactor is removed. The cubic cost is therefore a finite-size window, set by whether the readout light cone grows with the register. A product family to 256 qubits gives $1.966$ (95% CI $1.934$--$1.997$); a brickwork entangler falls from $2.853$ below ten qubits to $1.715$ beyond sixty-four; a blocked entangler gives $1.984$ at a fixed cone width against $3.034$ at a proportional one. Fixed device connectivity fixes the cone, so a cubic budget from a small simulation overestimates a large machine, on top of hardware multipliers $2.07\times$ ($1.41$--$3.02$), $2.38\times$ and $1.91\times$ on ibm_marrakesh, ibm_fez and ibm_kingston. Cost-optimal readout weights cut the measured shot budget by $2.67\times$ ($1.33$--$4.00$) on hardware, flat from four to twelve qubits. A discrepancy model fitted on cheap circuits transfers its mean inside the calibration grid and, at six larger sizes named before the data, does not: nominal 90% intervals cover 36%, and split conformal is the only rung that stays near nominal.