# Should $\delta_1\phi$, $\delta_3\phi^3$, and $\delta_4\phi^4$ be included in the 4D Yukawa Lagrangian with counterterms? [duplicate]

Given the 4D Yukawa Lagrangian

$$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi)(\partial^\mu \phi) - \frac{1}{2} m^2 \phi^2 + \bar\psi(i\not\partial-m)\psi+ g \bar{\psi} \phi \psi \tag{1}$$

The theory with counterterms is given as

$$\mathcal{L}^c = \frac{1}{2} (\partial_\mu \phi)(\partial^\mu \phi)(1+\delta_z) - \frac{1}{2} \phi^2 (m^2+\delta_m) + \bar\psi(i\not\partial(1+\delta_z')-(m+\delta_m'))\psi- (g+\delta_g) \bar{\psi} \phi \psi \tag{2}$$

For the term $$g\phi\bar\psi\psi$$, we can find the superficial degree of divergence as $$\omega = 4-(3/2)f-b, \tag{3}$$ where $$f$$ and $$b$$ are the numbers of external fermion and scalar legs of Feynman diagrams. Then, I can draw all the possible divergent diagrams as follows:

My question is about the three boxed cases: I don't think I can find the associated terms in $$\mathcal{L}^c$$. Why they are not included? In this case ($$\phi^3$$ theory), we included $$-\delta_1\phi$$ in the Lagrangian, so I wonder if we should do something similar in this case.

• In order to formulate a consistent renormalizable model, you must include all possible terms (respecting space-time symmetries and possibly further symmetries) up to dimension 4 in the Lagrangian you are starting with. Otherwise the diagrams you have drawn will generate divergences, which cannot be absorbed in the parameters of your tree Lagrangian. So your question "Why are they not included?" has a simple answer: Because you did not write them down in $\mathcal{L}$. Commented Dec 22, 2022 at 18:05
• Essentially a duplicate of physics.stackexchange.com/q/741879/2451 Commented Dec 22, 2022 at 19:41
• @Hyperon Thanks so much! Do you mean if we add those counterterms (like $\delta_4\phi^4$), we will need to have more terms in the original Lagrangian, like $\phi^4$?
– IGY
Commented Dec 22, 2022 at 22:26
• @Hyperon In the last question I posted, we have a $\phi^3$ theory, there is no $\phi^1$ term, but we still included $\delta_1\phi$ in the Lagrangian with counterterm, that's where I found confusing.
– IGY
Commented Dec 22, 2022 at 22:30
• You have to include all the terms I had mentioned in my comment above in the Lagrangian. So you need also a $\phi^3$ term (dimension 3) and a $\phi^4$ term (dimension 4) with associated coupling constants. These coupling constants are related to certain measurable cross sections. The only purpose of the (more or less trivial) $\delta_1 \phi$ counterterm is to absorb the divergent part of the vacuum expectation value of the scalar field . If you wish, you can put this vacuum expectation value to zero (order by order) by imposing this as one of your renormalization conditions. Commented Dec 22, 2022 at 22:59