Why do the counterterms in renormalized $\phi^4$-theory with power two in fields give vertices and not propagators?

I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}z^2\phi^4.$$

It was my understanding that terms of power 2 in the fields always give Feynman rules that are propagators. However, it appears that the counterterms with power 2 in the fields give a Feynman rule looking like $$i(p^2\delta_z-\delta_m),$$ instead of something with a denominator that would be more familiar. Like $\frac{i}{p^2-m^2+i\epsilon}$, from the first terms. Why is this the case? Is the idea that any term with power 2 in the fields gives a propagator wrong?

• Hint: the propagator is an all order resummation of terms related to the terms you write. – innisfree May 26 '17 at 7:50
• – AccidentalFourierTransform May 26 '17 at 10:07

Consider $\phi^4$ theory: $$\mathcal L=\frac12 Z_1(\partial\phi)^2-\frac12 Z_m m^2\phi^2-\frac{1}{4!}\lambda_0\phi^4$$

There are two approaches to perturbation theory:

First

The propagator is given by $$\Delta=\frac{1}{Z_1p^2-Z_m m^2}$$ and there is one type of vertex, with value $$-i\lambda_0$$

Second

The propagator is given by $$\Delta=\frac{1}{p^2-m^2}$$ and there are two types of vertices, with value $$-i((Z_1-1)p^2-(Z_m-1)m^2),\qquad -i\lambda_0$$

The two approaches are completely equivalent, and they give rise to the exact same expression for a given scattering process.

Note that the coefficients $Z_1,Z_m$ depend on the expansion parameter $\lambda$. This means that the first approach is more cumbersome because it is in general not clear which diagrams contribute to a given order in perturbation theory, inasmuch as both the vertices and the propagators contain powers of $\lambda$. On the other hand, the second approach leads to more diagrams (because there is one more vertex) but it is more convenient (because the propagators are independent of $\lambda$).

I want to add to AccidentalFourierTransform's answer:

Assuming the $\delta$'s are small, then we can expand the renormalized term in powers of $(\delta_2p^2-\delta_m)$: $$\frac{i}{Z_2p^2-Z_mm^2}=\frac{i}{p^2-m^2 }\left(1+\frac{\delta_2p^2-\delta_m}{p^2-m^2}\right)^{-1}=\frac{i}{p^2-m^2 }\left(1-\frac{\delta_2p^2-\delta_m}{p^2-m^2}+\dots\right)=\frac{i}{p^2-m^2 }+\frac{i}{p^2-m^2 }\left(i\delta_2p^2-i\delta_m\right)\frac{i}{p^2-m^2 }+\dots$$

Which is the sum of all the diagrams consisting of the original term + the counter-term, so by identifying $\frac{i}{p^2-m^2}$ as the momentum term, we identify $i(\delta_2p^2-\delta_m)$ as the momentum counter-term.

• but some of the $\delta$ terms are also divergent, so how do we make sense out of the expansion? – M. Zeng Jan 1 '18 at 2:31
• @M.Zeng The assumption that the $\delta$'s are small is used many times in RG computations. e.g: To one loop order $\frac{\partial\log{(Z)}}{\partial \mu}=\frac{\partial\log{(1+\delta)}}{\partial \mu} \approx \frac{\partial\delta}{\partial \mu}$. However, to be frank, I cannot argue to the validity of this assumption. – EZLearner Feb 5 '18 at 20:40