What is and is not detected

The contrast of Model 1 changes gravitational preparation at fixed scalar preparation. It vanishes when the mixed term is removed. It is not, by itself, a test excluding every classical stochastic description of that preparation.

It is not a gravitational quadrature alone. By (eq:d1), $u_{e,f_{\ast }}\ne 0$, and on the finite-particle domain

For $\lambda \ne 0$, this is nonzero because $I,Q,{\Pi }_0$ are linearly independent. A gravitational-only operator would commute with every scalar quadrature.