# Renormalization group and summation of diagrams

Currently I'm studying renormalization group, and I'm having trouble understanding the following statement which I see almost everywhere in books on QFT: renormalization group sums a series of divergent diagrams.

In particular examples, like 1-loop corrections to photon propagator, it is clear: we consider a following series It sums as a geometric progression and gives the desirable answer, same as RG equation provides if we consider a contribution to beta-function from the first diagram. But is there any way to construct and look at concrete series of diagrams, which we have resummed via the use of RG equation in a given order, in an arbitrary case, in order to find what contributions we missed?

Say, we have a $$\phi^4$$ theory. The beta-function in one-loop is $$\beta(\lambda) = \frac{3\lambda^2}{16\pi^2}$$ is given by a first divergent contribution to 4-point function - diagram with one "bubble"; the factor 3 comes from crossing symmetry. By solving the RG equations, we get for a running coupling constant on scale $$p$$

$$\lambda(p) = \frac{\lambda(\mu)}{1 - \frac{3\lambda}{16\pi^2} \log p/\mu}$$

where $$\mu$$ is the reference scale. If we expand the denominator, we see a series which looks like it's given by some set of perturbative expansion terms; the first one is just the "one-bubble-diagrams". But I wasn't able to find what diagrams correspond to different terms even in the next order, especially to reproduce the strange factor of 9.

• What do you call "the strange factor of 9"? The diagrammatic expansion for the renormalization of the $\phi^4$ vertex looks like >< + >O< + >OO< + ... (sorry for using such a basic drawing for Feynman diagrams, but I think it should be clear, no?) – M.Jo Sep 23 at 13:12
• Yeah, pics are understandable :) the factor 9 appears when you expand the answer that RG gives for running coupling to the 3rd order in lambda. If we interpret that answer as if we summed some series of diagrams, which lots of authors propose to do, that suggests that the diagrams which contribute something of order lambda^2 log^2 (Lambda/p) should give in total a factor of 9/(16 pi^2)^2. The >00< diagram gives only 1/9 of that, with crossing symmetry it becomes 3 times larger, but it's still not enough; in higher orders of expansion the situation becomes even worse. – Aleksandr Artemev Sep 23 at 19:58

There isn't a sense in which the extra terms obtained by using the renormalization group correspond to any specific subset of Feynman diagrams. As you have already pointed out, for $$\phi^4$$ theory, it is not true that you just sum the "bubble" diagrams; you need to calculate all of the corrections, and then these corrections will contain the correct $$\log^2(p/\mu)$$ dependence predicted by expanding your effective coupling, but will also contain other terms.
The argument that you can predict the form of these higher-order terms can go as follows, using dimensional regularization. At first-order in $$\phi^4$$ theory, you obtain $$\Gamma^{(4)}(p) = \mu^{\epsilon} u_0 \left\{ 1 - \frac{3 u_0}{16 \pi^2 \epsilon} \left[ 1 + \epsilon \log(p/\mu) \right] + \cdots \right\}.$$ Here, I am taking $$\Gamma^{(4)}(k_i)$$ to be the four-point function, defined with total momentum $$p$$ flowing through it. The omitted terms in the ellipsis are momentum-independent and finite for $$\epsilon \rightarrow 0$$.
At this point, one introduces a renormalized coupling to subtract the divergent term, $$u_0 = u \left( 1 + \frac{3 u}{16 \pi^2 \epsilon} \right),$$ and this is sufficient to renormalize the correlation function to $$O(u^{2})$$.
How do we use this result to obtain information about higher-order contributions? Well we can already read off a particular $$O(u^3)$$ contribution just from noticing that we will have a term $$\Gamma^{(4)} \supset \frac{18 u^3}{(16 \pi^2) \epsilon} \log(p/\mu)$$ coming from the counter-term for $$u_0$$ defined above. Such a term is initially very worrying, because it is a momentum-dependent divergence - we cannot subtract this using counter-terms! Therefore, for the theory to make sense, it must be that a corresponding divergence with identical momentum-dependence will arise at two-loop to cancel this off. Of course, within dimensional regularization, the $$log(p/\mu)$$ dependence always occurs due to expanding a function like $$(p/\mu)^{\epsilon}$$. In particular, the above divergence would need to come from a term like $$-\frac{18 u^3}{(16 \pi^2) \epsilon^2} (p/\mu)^{\epsilon} = -\frac{18 u^3}{(16 \pi^2) \epsilon^2} - \frac{18 u^3}{(16 \pi^2) \epsilon} \log(p/\mu) - \frac{9 u^3}{(16 \pi^2)} \log^2(p/\mu)$$ Therefore, if this term shows up at two-loop (and it needs to in order for this renormalization scheme to make sense), it follows that one also needs the $$- \frac{9 u^3}{(16 \pi^2)} \log^2(p/\mu)$$ term. But in this particular case, the term is generated by several (all?) two-loop diagrams, which in turn also contribute other terms which one-loop RG doesn't know about.
• I would definitely be interested if there were an easy correspondence to be made between specific diagrams and the terms summed by RG. In $\phi^4$ theory at two-loop, I believe every diagram is needed, but it would be cool if that dramatically simplified at higher loops. – Seth Whitsitt Oct 19 at 17:02