= Solution
Use <one-particle-irreducible correlation functions> to identify independent local <counterterms>. The exact internal symmetry $\phi\mapsto-\phi$ excludes odd $b$, and <Dirac fermion number conservation> requires equal numbers of $\psi$ and $\bar\psi$, so $f$ is even. From <3/b>, the complete list of potentially overall divergent vertices with external legs is:
* $\Gamma_{\phi\phi}$: $(b,f)=(2,0)$, $\omega=2-n_2$, potentially divergent for $n_2\leq2$.
* $\Gamma_{\phi^4}$: $(b,f)=(4,0)$, $\omega=1-n_2$, potentially divergent for $n_2\leq1$.
* $\Gamma_{\phi^6}$: $(b,f)=(6,0)$, $\omega=-n_2$, potentially divergent only for $n_2=0$.
* $\Gamma_{\bar\psi\psi}$: $(b,f)=(0,2)$, $\omega=1-n_2$, potentially divergent for $n_2\leq1$.
* $\Gamma_{\phi^2\bar\psi\psi}$: $(b,f)=(2,2)$, $\omega=-n_2$, potentially divergent only for $n_2=0$.
The zero-leg <vacuum diagrams>, with $\omega=3-n_2$, may also require a <vacuum energy density> subtraction. Four-fermion vertices have $\omega\leq-1$ 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 $\not p$. The remaining listed vertices have no allowed derivative divergence. A complete <counterterm Lagrangian>, up to <total derivatives>, is therefore
$$
\boxed{\begin{aligned}
\mathcal L_{\rm ct}={}&-\delta\Lambda+\frac{\delta Z_\phi}{2}(\partial_\mu\phi)(\partial^\mu\phi)-\frac{\delta M^2}{2}\phi^2
+\delta Z_\psi\bar\psi i\gamma^\mu\partial_\mu\psi-\delta m\bar\psi\psi\\
&-\frac{\delta g_1}{2}\phi^2\bar\psi\psi-\frac{\delta g_2}{4!}\phi^4-\frac{\delta g_3}{6!}\phi^6.
\end{aligned}}
$$
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>.
Back to article page
