Theorem: No positive- correction

For the prescribed scalar-only pair, the output test in (eq:Rone) has no $\hslash$-dependence:

It is also independent of the choice of Hadamard contraction kernel with the stated commutator. This is tree-exactness of this one map, not an all-orders interacting identity.

Proof. For the verification, introduce only the auxiliary kernel

and the expression

Here $V=-\frac{1}{2}⟨\psi ,L\psi ⟩$, with $L$ formally self-adjoint and compactly supported. Both ${\mathsf{E}}_{\mathrm{A}}c$ and ${\Delta }_{+}c$ are smooth: the Hadamard kernel has no wavefront covector with a zero second slot. Thus the expression is well defined.

Since ${F_c}^{(n)}=0$ for $n\ge 2$,

Consequently

The numerator is exactly divisible by $\hslash$. The zero-contraction term and the Hadamard kernel cancel; a second cross-contraction is absent. Taking $L=L_{1,g}[h]$ proves the assertion.

Changing a quadratic vertex's free Wick ordering adds only a scalar-independent term. Its first functional derivative is zero and does not change this calculation.