 Solution

Take integer $n\ge 3$ and positive integer $d$, so the interaction is genuinely non-Gaussian. For a connected two-point Feynman diagram with $V$ interaction vertices and $I$ internal lines, counting half-edges and using the loop order givesAt one loop $I=V$, hence $(n-2)V=2$. There are only two possibilities: a two-interaction vertex bubble diagram for $n=3$, or a one-interaction vertex tadpole diagram for $n=4$. The latter is independent of external momentum and changes the mass, not the kinetic term.

For cubic scalar field theory, introduce a positive auxiliary mass $\mu$ if needed to separate infrared divergences from ultraviolet divergences. The one-loop self-energy in the quantum effective action is $-{\lambda }^2{B}_d(p)/2$, withThis Feynman parameter representation follows by shifting the loop momentum; it is directly valid for a translation-invariant regulator, or for the convergent subtracted integral. Differentiating at $p^2=0$ givesThus the coefficient of the kinetic term changes by ${\lambda }^2\int d^{d}q/[(2\pi {)}^d(q^2+{\mu }^2{)}^3]/6$.

Only the cubic interaction has a momentum-dependent one-loop two-point correction.There is a terminology qualification: if “wave-function renormalization” means a required ultraviolet counterterm, rather than a finite correction to the field normalization, the large-$q$ radial integral behaves as $\int dq{q}^{d-7}$, and the answer isIt is logarithmically divergent at $d=6$ and power divergent above $6$ with a momentum cutoff. Among interactions with a relevant coupling or marginal coupling, the divergent case is uniquely $(n,d)=(3,6)$. The massless low-dimensional amplitude needs an infrared divergence prescription; one cannot take a local expansion at zero momentum without one. The $n=1,2$ cases, if admitted, are a source and a Gaussian mass term and have no interaction-induced wave-function renormalization.