Superficial Degree of Divergence in Renormalization of Phi^3 Theory

[The black vegetable]

TL;DR Summary

When calculating the superficial degree of divergence do you only do it for diagrams with closed loops?

Hi

When calculating the superficial degree of divergence do you only do it for diagrams with closed loops?

Many thanks :)

 

[vanhees71]

Since tree-level diagrams are finite, it doesn't make sense to think about their superficial degree of divergence.

The superficial degree of divergence rests on Weinberg's theorem about the asymptotic behavior of loop diagrams, and the nice thing is that it's entirely expressed in terms of topologies of diagrams. Take $\phi^3$ theory (which is a toy model only, because it has no stable ground state, but it's used in the famous book by Collins, Renormalization, to explain perturbative renormalization techniques, because of its simplicity).

In $d$ dimensions the any integral counts as $d$, i.e., you have a term $d L$, where $L$ is the number of loops of a diagram. Each propagator has dimension $-2$, so you have a term $-2 I$ for $I$ internal lines in a diagram. So the superficial degree of divergence is
$$D_s=d L-2I.$$

To finally count the degree of divergence in terms of the number external lines $E$ and the number of vertices $V$. In $\phi^3$ theory at each vertex 3 lines meet, which means you have $3V$ lines all together. Now an internal line belongs to two verices and and external line only to 1. Thus $3V=2I+E$. Finally we have from momentum conservation at each vertex $I-V$ independent loop momenta, but one conservation law is trivial due to overall momentum conservation of the external momenta, i.e., $L=I-(V-1)$. Thus we finally have
$$D_s=d(I-V+1)-2I = d+(d-2)I -d V = d+(d-2) (3V-E)/2 - d V=d+\left (1-\frac{d}{2} \right) E + \frac{d-6}{2} V.$$
This shows that for $d=4$ you have
$$D_s|_{d=4}=4-E-V.$$
This means that for $d=4$ the theory is superficially super-renormalizable. This means that first of all only diagrams with $E \leq 3$ can be divergent to begin with, i.e., only for those diagrams $D_s \geq 0$, but it's even only a finite number of diagrams for each $E$ that is divergent, because there's the term $-V$ in $D_s$. It's renormalizable even for $d=6$, leading to
$$D_s|_{d=6}=2(3-E).$$
Again a diagram is only divergent for $E \leq 3$, and $D_s$ doesn't depend on $V$, i.e., there's to be expected divergences at any order of perturbation theory.

 

[The black vegetable]

That's great thanks for this :)