Let's consider the theory given by the following lagrangian $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu \phi - \frac{1}{2} M^2 \phi^2 + \bar\psi (i\gamma^\mu\partial_\mu - m)\psi + \frac{1}{4!} \lambda \phi^4 + g \phi \bar{\psi}\psi. $$
Superficial degree of divergence $D$ of 1PI diagram depends as follows on the number of external fermionic $E_f$ and bosonic $E_b$ lines: $$ D = 4 - E_b - 3/2E_f. $$ Thus, there are 7 superficially 1PI diagrams:
- vacuum diagram,
- diagrams with one, two, three or four external bosonic line,
- diagram with two fermionic lines,
- diagram with two fermionic lines and one bosonic.
If I treat the fields and couplings in the original lagrangian as bare, then some counterterms appear. There is, however, no counterterm of the form $\phi^3$ and $\phi$.
Are the 1PI diagrams with one or three external bosinic lines really divergent? If yes, should I add by hand additional counterterms which cancel these divergencies?
In comparison in the pseudoscalar Yukawa theory $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu \phi - \frac{1}{2} M^2 \phi^2 + \bar\psi (i\gamma^\mu\partial_\mu - m)\psi + \frac{1}{4!} \lambda \phi^4 + i g \phi \bar{\psi}\gamma^5\psi. $$ there parity symmetry makes the problematic superficially divergent diagrams with only one or three bosonic lines (for which there are no counterterms in the lagrangian) vanish identically.
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