# Relation between perturbative renormalization and RG group flow

I have been trying to understand renormalization for a few days. I have consulted many sources and physics.stackexhange posts but I have unresolved issues, so I am hoping someone can help me!

I think I properly understand RG group flow, and how integrating out higher modes whilst keeping low energy physics constant leads to flows in coupling constants depending on the momentum cut-off. What I am struggling to understand, is exactly how this relates to the perturbative algorithm employed in getting rid of infinities in QFT.

It appears that the modern understanding of effective field theories is that a momentum cut-off is perfectly valid. For example, in the article by G. Peter Lepage at arXiv: hep-ph/0506330, he notes that one can simply leave the cut-off as an explicit parameter and compute physical quantities using this cut-off. Once the coupling constants have been tuned so that the theory agrees with some experimental measurements one obtains a predictive theory with an explicit cut-off.

However, in Chapter 5 of my lecture notes, Prof. Skinner employs a perturbative calculation where he redefines couplings constants by shifting them by an amount dependent on the cut-off in order to get rid of the cut-off dependence in an amplitude calculation. This idea and process seems at odds with the understanding outline in my previous paragraph. What is the resolution?

Let's say you do "normal" perturbation theory with cutoff regularization with cutoff $$\Lambda$$ in a theory with bare parameters $$a_i$$. The bare parameters will be functions of $$\Lambda$$, $$a_i(\Lambda)$$. You can use words like "we choose counter terms to cancel the divergences in loop integrals". This amounts to choosing the $$\Lambda$$ dependence in $$a_i(\Lambda)$$ order by order to cancel the $$\Lambda$$ dependence of loop integrals in correlation functions. The theory explicitly depends on $$\Lambda$$, because the $$a_i$$ are functions of $$\Lambda$$. But, observable quantities do not depend on $$\Lambda$$.

We don't have to change the words that much to connect with a renormalization group picture. The main difference is that rather than thinking of $$\Lambda$$ as a fixed cutoff parameter, we think of it as a parameter that can vary. The $$\Lambda$$ dependence of observable quantities must cancel out. We can express condition as saying that the $$\Lambda$$ derivative of a correlation function must vanish. This condition implies that the $$a_i(\Lambda)$$ will obey a differential equation (in fact there are a tower of differential equations coming from different correlation functions). What this equation expresses is how we must update the value of the $$a_i$$ parameters as we vary $$\Lambda$$.

• Thank you for the reply! So what you're saying is that solving the Callan-Symanzik equations vs adding counter terms ultimately gives the same $\Lambda$ dependence in the coupling constants? Is the finite part of the coupling constant that we are free to fix in the counter term approach then equivalent to the integration constant from the differential equations? And this is ultimately fixed by experimental data, right? Commented Mar 30, 2022 at 15:09
• @BrettOertel At least at a high level, that's right. In practice there are differences between different regularization schemes and so on. But, conceptually, yes -- whether you use counterterms or the RG equations, you fix the cutoff dependence of the coupling constants so that physical quantities do not depend on the cutoff. And you fix the finite part to match experiments. Commented Mar 30, 2022 at 15:39

he notes that one can simply leave the cut-off as an explicit parameter and compute physical quantities using this cut-off. Once the coupling constants have been tuned so that the theory agrees with some experimental measurements one obtains a predictive theory with an explicit cut-off.

Sure.

he redefines couplings constants by shifting them by an amount dependent on the cut-off in order to get rid of the cut-off dependence in an amplitude calculation.

Indeed. And he does this for each $$\Lambda$$. And by solving the so-called beta function equation for the RG flow, what you are doing is constructing "the optimal Gaussian theory that can compute the given observable" for each cutoff, such that the desired observable is cutoff independent. This means finding the renormalized couplings $$\alpha(\Lambda)$$ for all $$\Lambda$$ given the particular initial conditions. This construction can be done for a given observable. If you try to compute a different observable, or even go to higher orders in perturbation theory, you might regain cutoff-dependence for that observable even if using the same level of perturbation theory as before. Introducing more renormalized coupling constants can net you more observables that become cutoff-independent in this procedure, at this level of perturbation theory.

The two authors don't really conflict in their approaches. There are multiple ways to do/interpret RG, and as I said in general different observables will not necessarily be cutoff-independent, and will change with the order of perturbation theory done.

A geometric interpretation of RG as a sequence of best-fitting curves dependent on a parameter can be found in this YouTube video: https://www.youtube.com/watch?v=6JN-DJuSFUY as well as this paper: Kunihiro, Teiji. "A geometrical formulation of the renormalization group method for global analysis." Progress of theoretical physics 94.4 (1995): 503-514. This formulation of RG is used as a tool for global analysis of ODEs/PDEs, and technically one can write down a system of ODEs/PDEs for observables in a theory (the Dyson-Schwinger equations).

• Thank you for your answer! Perhaps you can comment on my reply to Andrew's answer above? This would be appreciated! Commented Mar 30, 2022 at 15:13
• I don't think I can add anything to Andrew's answer/reply. Our answers agree from what I can tell. Commented Apr 4, 2022 at 1:16