Theorem: Model 3 record: fixed scalar cutoff and finite M

Fix $J_{\psi },z_{\ast },\Lambda$ as above, and take

Choose $\lambda >0$, let $\nu ={\Vert z_{\ast }\Vert }^2$, and put

For the common instrument (eq:Kraus), choose

Then for every finite $M$,

where

The detector, states, $\Lambda ,\lambda ,\tau ,\theta$ are independent of $M$. In particular the contrast is positive for every sufficiently large finite $M$.

At $\lambda =0$, both probabilities are exactly $1/2$. Replacing the stress source by zero also makes both probabilities exactly $1/2$. No graviton factor is present, and neither $\Lambda$ nor the measured number cutoff is removed in this theorem.

Proof. The vacuum mean vanishes. Equations (eq:dLambda) and (eq:tail3) give a mean difference at least $\lambda (d_{\Lambda }-{E_M}^{(3)})$. The number bound gives

since the vacuum value of ${\Vert (\mathsf{N}+2)\Omega \Vert }^2$ is four. At $\theta =\pi /2$, Lemma lem:Ramsey gives

The tail tends to zero.

At $\lambda =0$, the generator is zero and $U=I$; the chosen phase gives $p_{+}=1/2$ in every state. If $T$ is replaced by zero, linearity of the response makes every kernel and generator zero, proving the same control. Setting the scalar mass to zero is not this source-off operation.