# Criterion for a Feynman loop diagram to give a finite value

The contribution of loop diagrams in QFT are often divergent and sometimes convergent as well. For example, the self-energy corrections in QED are divergent. On the other hand, the Zee model of radiative neutrino mass (for example) induces a finite neutrino mass at one-loop.

1. Is there a necessary and/or sufficient condition to tell apriori i.e., without computing the loop explicitly that the loop effect is certainly convergent (or certainly divergent)?

2. Are there cases where no such conclusion can be drawn?

If one is dealing with renormalizable theories a rule to quickly determine whether a diagram is UV divergent or not is to ask the following question: does the vertex realized by all its external legs belong to the Lagrangian already?

Let me make this clearer: in $$\lambda \phi^4$$ theory in $$d=4$$ we only have vertices with two or four fields. So if I compute the diagram with, say, $$8$$ external $$\phi$$ legs, I am guaranteed that it will be convergent. If I instead compute a diagram with two legs that will diverge to contribute to the mass renormalization.

This is of course a bit of a silly answer because I am assuming that you renormalized the theory already, which requires the knowledge of the divergent diagrams in the first place. So how do I actually know whether a diagram converges or not?

@CStarAlgebra's answer is not wrong. You need to compute the so-called ''superficial degree of divergence.'' This is obtained by rescaling the loop momenta by a factor $$\lambda$$, send $$\lambda \to \infty$$ and look at the behavior of the integral. More precisely $$I = \int d^Dp \,f(p,k) \;\;\mbox{has degree A}\;\;\Leftrightarrow\;\; \int d^Dp\,\lambda^D \,f(\lambda p,k) \underset{\lambda \to \infty}{\sim} \lambda^A I$$ When $$A\geq 0$$ it might signal a divergence ($$A=0$$ is a $$\log$$ divergence). But it is called superficial for a reason. There are some exceptions which sadly would require an explicit computation.

1. Gauge symmetry: Let me just make an example for this one. Take in QED a diagram with a loop of electrons and four external photons 1. This has a superficial degree that suggests it might diverge, but if it did we would need to renormalize the Lagrangian by introducing a term $$(A_\mu A^\mu)^2$$ which violates gauge invariance. Indeed by explicitly computing it we see that it does not diverge. And also it follows the rule of thumb I stated at the beginning.
2. Subdivergences: This applies to $$>1$$ loop. If we rescale all loop momenta at the same rate we are sensitive to what are usually called ''overall'' divergences. Some diagrams might instead diverge when we send to infinity only a subset of the momenta. Thus the integral might be infinite even if $$A<0$$. Diagrammatically this means that a subdiagram in my multi loop diagram is itself divergent. These subdivergences are totally fine because, if we renormalized correctly at the previous loop order, they should cancel identically in the final sum of all diagrams in the process. So at every loop level the new information is coming by the overall divergences.
3. IR divergences: this entire story applies to UV divergences. IR divergences appear when you have some massless degrees of freedom and the propagator diverges for soft momenta. The resolution of these kinds of divergences has nothing to do with renormalization and it is a somewhat orthogonal matter.

One can calculate the superficial degree of divergence of any diagram. This is covered in every standard introductory QFT text.

It's no different than looking at the divergence of a regular integral, and it really doesn't have to do with "loops" per-se. It's just dressed up in a lot of terminology which I honestly think just complicates something that is actually quite simple.

(1) If you integrate over a pole, then your integral will diverge.

(2) If your integral looks like $$\int d^4k\ \frac{k^n}{f(k)}$$ then you can see that unless $$f(k) \sim k^{n+2}$$ as $$k$$ gets large, then the integral will diverge at infinity. The reason it's not convergent for $$f(k)\sim k^{n+1}$$ is because the integral will go as $$\log(k)$$, which also diverges. For the cases of $$f(k)\sim k^{m\leq n}$$ convergence is more than hopeless, since you can see that the integrand itself doesn't even go to zero at infinity (for $$m), and the case of $$m=n$$ you are integrating a constant from $$-\infty$$ to $$\infty$$ which also clearly diverges.

When you want to analyze convergence or divergence at large $$k$$ you can ignore all factors except $$k$$, which you can gather easily by just looking at the loop diagram.

• It's true that the $i\epsilon$ makes the integral around $1/(p^2-m^2)$ well defined. But physically that's not there to cure a divergence. The $i\epsilon$ prescription serves the purpose of making the correct choice for the propagator (i.e. the Green's function) because it is not unique in Lorentzian signature. – MannyC Dec 20 '18 at 20:52
• Okay, fair I'll clear that up. – InertialObserver Dec 20 '18 at 20:53