Theorem: Cubic witness away from transfers

For every $m>0$ and $t_0<T$, there are real compact $f$ and a normalized symmetric ${\chi }_3\in {C_c}^{\infty }(({\mathbb{R}}^3{)}^3)$, supported away from all three-creation transfer sets, such that

The normalized states

have equal expectations for every defined polynomial form of degree at most two, but their cubic expectations differ.

Proof. Choose ${\mathbf{p}}_{\ast }\ne 0$, all three signs positive, and all three momenta equal to ${\mathbf{p}}_{\ast }$. Every pair transfer is $2{\mathbf{p}}_{\ast }\ne 0$. Here

At $s=t_0$, the source and spatial initial amplitudes vanish. Their total divergence vanishes as well. Therefore

The ordered response, with the three mode normalizations, has short-time expansion

All cyclic summands agree at this point, so this is already the symmetrized coefficient.

Choose an even nonnegative compact spatial bump $b$ with $\int b(x)e^{-3i{\mathbf{p}}_{\ast }\cdot x}{\mathrm{d}}^{3}x>0$, possible by taking its support sufficiently small. Choose a nonnegative $g_0\in {C_c}^{\infty }((1,2))$ of integral one and let

For sufficiently small $h>0$, the value ${K_{f_h}}^{+++}({\mathbf{p}}_{\ast },{\mathbf{p}}_{\ast },{\mathbf{p}}_{\ast })$ has a nonzero leading term of order $h^2$, hence is nonzero.

Fix such $f_h$. In a small symmetric neighborhood avoiding all transfers the kernel is smooth and nonzero. Take a nonnegative symmetric bump $\beta$ there and set

The vacuum block gives

A polynomial of degree at most two cannot connect vacuum and three particles, so its expectations in ${\Psi }_{\pm }$ agree. The cubic difference is