ϕ \mapsto - ϕ

excludes odd

b

ψ

and

\overset{ˉ}{ψ}

, so

f

is even. From b, the complete list of potentially overall divergent vertices with external legs is:

$Γ_{ϕϕ}$ : $(b, f) = (2, 0)$ , $ω = 2 - n_{2}$ , potentially divergent for $n_{2} \leq 2$ .
$Γ_{ϕ^{4}}$ : $(b, f) = (4, 0)$ , $ω = 1 - n_{2}$ , potentially divergent for $n_{2} \leq 1$ .
$Γ_{ϕ^{6}}$ : $(b, f) = (6, 0)$ , $ω = - n_{2}$ , potentially divergent only for $n_{2} = 0$ .
$Γ_{\overset{ˉ}{ψ} ψ}$ : $(b, f) = (0, 2)$ , $ω = 1 - n_{2}$ , potentially divergent for $n_{2} \leq 1$ .
$Γ_{ϕ^{2} \overset{ˉ}{ψ} ψ}$ : $(b, f) = (2, 2)$ , $ω = - n_{2}$ , potentially divergent only for $n_{2} = 0$ .

The zero-leg vacuum diagrams, with

ω = 3 - n_{2}

ω \leq - 1

and do not require a new four-fermion counterterm. These are superficial bounds: a particular allowed diagram can vanish or be convergent.

Locality and Lorentz invariance restrict the divergent parts to a finite basis. The scalar two-point vertex can contain a constant and

p^{2}

; there is no Lorentz scalar linear in a single momentum. The fermion two-point vertex can contain

I

and

\neq p

L_{ct} = - δ Λ + \frac{δ Z _{ϕ}}{2} (\partial_{μ} ϕ) (\partial^{μ} ϕ) - \frac{δ M ^{2}}{2} ϕ^{2} + δ Z_{ψ} \overset{ˉ}{ψ} i γ^{μ} \partial_{μ} ψ - δ m \overset{ˉ}{ψ} ψ - \frac{δ g _{1}}{2} ϕ^{2} \overset{ˉ}{ψ} ψ - \frac{δ g _{2}}{4 !} ϕ^{4} - \frac{δ g _{3}}{6 !} ϕ^{6} .

The coefficients supply wavefunction renormalization, mass counterterms and renormalizations of the three interactions already present. Recursive subtraction of ultraviolet subdivergences then leaves only these local overall divergences. This finite counterterm closure is why the complete theory is expected to have perturbative renormalizability. It does not mean every listed coefficient first becomes nonzero at one loop, or that setting an allowed coupling to zero is stable under renormalization.

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