# If renormalization scale is arbitrary, why do we care about running couplings?

For the bounty please verify the following reasoning

[copied from comment below]

Ah right, so the idea is that overall observable quantities must be independent of the renormalization scale. But at each order in perturbation theory the result can depend on renormalization scale, right? And they do so in exactly the right way to invalidate the use of a perturbation series when external momenta are of the wrong order. This is because the external momenta get caught up in loop integrals due to momenta conservation at vertices and the renormalization group equation comes along and means the right things cancel.

[original question]

In Brief

People say that perturbation theory breaks down when the couplings run to high values. But given that this running depends on an arbitrary mass scale, how is this argument logical?!

A Longer Exposition (TL;DR)

It's obvious that Feynman diagram techniques work best when the coupling constant is small, since then you can neglect higher order terms. But various sources (e.g. Peskin and Schroeder) make claims about running couplings that seem incomplete to me.

I often read sentences like

• if the renormalized coupling is small at low energy, then Feynman diagrams are good in that region
• for asymptotically free theories, Feynman diagrams are good for high energy calculations

I understand exactly why these are true. But surely something much more general is true, namely

• if the coupling constant is small at any renormalization scale the Feynman diagrams are good for calculations

My reasoning is as follows. Observable quantities are completely independent of the running of the coupling, so you can just choose your scale appropriately so that the coupling is small for the expansion. Hey presto, you can use Feynman diagrams.

Is this true? I strongly expect the answer to be no but I can't convince myself why. Here's my attempt at a self-rebuttal.

• my argument above is incorrect, because I've assumed there's only a single coupling. In reality there's contributions from "irrelevant" terms at higher energies whose couplings can't be fixed from low energy observations.

This leads me to hypothesise that the following is true (would you agree?)

• if the coupling constant is small at any renormalization scale above the scale of your observations then Feynman diagrams are good

This seems more plausible to me, but it would mean that Feynman diagrams would be good for low energy strong interaction processes for example. This feels wrong is a sense because the renormalized coupling is large there. But then again the renormalization scale is arbitrary, so maybe that doesn't matter.

Many thanks in advance!

• Short answer (due to time): the running of the coupling is a way of summing the leading logarithmic corrections. So if you choose the "wrong" scale you end up with large logarithms in your Feynman diagrams. So in the series you trade a large running coupling and $\mathcal{O}(1)$ coefficients for a small coupling and large coefficients involving the log of the ratio of scales between the process of interest and the renormalization point. Either way the series is no good in that regime. Oct 16 '13 at 16:43
• Ah right, so the idea is that overall observable quantities must be independent of the renormalization scale. But at each order in perturbation theory the result can depend on renormalization scale, right? And they do so in exactly the right way to invalidate the use of a perturbation series when external momenta are of the wrong order. This is because the external momenta get caught up in loop integrals due to momenta conservation at vertices and the renormalization group equation comes along and means the right things cancel. Oct 16 '13 at 17:07
• Do you know a reference which goes through this problem in detail? I've tried to construct an example myself, but it's not much use because I can't check my workings against an example. Most textbooks just gloss over it (at least Weinberg and P&S do...) Oct 16 '13 at 17:08
• I am afraid I don't understand correctly your question. However, here is a practical remark. If, in QED, we take in account the first loop correction, we get, for the coupling $\alpha$, the following relation, where $\mu$ is the energy scale : $\alpha(\mu) = \dfrac{\alpha(\mu_0)}{1 - \frac{2}{3\pi} \alpha(\mu_0)log \large \frac{\mu}{\mu_0}}$. (This relation is only valid if the second term in the denominator is $<<1$). From this, we see, that, for increasing momenta (decreasing distances), the coupling $\alpha$ is increasing. Oct 26 '13 at 11:44
• Did you take a look at Weinberg Volume II page 111-130? I think your answer lies there Sep 10 '15 at 23:38

## 1 Answer

The main problem lies in the "large logarithms". Indeed, suppose you want to calculate some quantity in Quantum Field Theory, for instance a Green Function. In perturbation theory this is something like:

$$\tilde{G}(p_1,...,p_n)=\sum_k g^k F_k(p_1,...,p_n)$$

for some generic functions $F$ and $g$ is the coupling constant. It's not enough to require a small $g$. You need small $g$ AND small $F$, for every value of the momenta $p$ (so for every value of the energy scale of your system).

A nice little calculation to understand this point. It's obvious that:

$$\int_0^\infty \frac{dx}{x+a}=[log(x+a)]_0^\infty=\infty$$

Let's use a cutoff: $$\int_0^\Lambda \frac{dx}{x+a}=log\frac{(\Lambda+a)}{a}$$

This is still infinite if the (unphysical) cutoff is removed. The whole point of renormalization is to show that a finite limit exist (this is "Fourier-dual" to send the discretization interval of the theory to zero). This quantity is finite:

$$\int_0^\Lambda \frac{dx}{x+a}-\int_0^\Lambda \frac{dx}{x+b} \rightarrow log\frac{b}{a}$$

But if $a \rightarrow \infty$ the infinite strikes back! So for a generic quantity F(p) regularized to F(p)-F(0) we want at least two things: that the coupling is small at that momentum $p$ and that $p$ is not far away from zero. But zero is arbitrary, we can choose an arbitrary (subtraction) scale. So we can vary this arbitrary scale $\mu$ in such a way that it is always near the energy scale we are probing.

Is convenient to take this scale $\mu$ at the same value of the renormalization scale. This is the energy at which you take some finiteness conditions (usually two conditions on the two point Green function and one condition on the 4 point one). The finiteness conditions are real physical measures at an arbitrary energy scale, so they fix the universe in which you live. If you change $\mu$ and you don't change mass, charge, ecc. you are changing universe. The meaning of renormalization group equations is to span the different subtraction points of the theory, remaining in your universe. And of course every physical quantity is independent of these arbitrary scale.

EDIT: Some extra motivations for the running couplings and renormalization group equations, directly for Schwartz:

The continuum RG is an extremely practical tool for getting partial results for high- order loops from low-order loops. [...]

Recall [...] that the difference between the momentum-space Coulomb potential V (t) at two scales, t1 and t2 , was proportional to [...] ln t1 for t1 ≪ t2. The RG is able to reproduce this logarithm, and similar logarithms of physical quantities. Moreover, the solution to the RG equation is equivalent to summing series of logarithms to all orders in perturbation theory. With these all-orders results, qualitatively important aspects of field theory can be understood quantitatively. Two of the most important examples are the asymptotic behavior of gauge theories, and critical exponents near second-order phase transitions.

[...] $$e^2_{eff}(p^2)=\frac{e^2_R}{1-\frac{e^2_R}{12 \pi^2}ln\frac{p^2}{\mu^2}}$$ $$e_R=e_{eff}(\mu)$$ ￼￼￼ This is the effective coupling including the 1-loop 1PI graphs, This is called leading- logarithmic resummation. Once all of these 1PI 1-loop contributions are included, the next terms we are missing should be subleading in some expansion. [...] However, it is not obvious at this point that there cannot be a contribution of the form $ln^2\frac{p^2}{\mu^2}$ from a 2-loop 1PI graph. To check, we would need to perform the full zero order calculation, including graphs with loops and counterterms. As you might imagine, trying to resum large logarithms beyond the leading- logarithmic level diagrammatically is extremely impractical. The RG provides a shortcut to systematic resummation beyond the leading-logarithmic level.

Another example: In supersymmetry you usually have nice (theoretically predicted) renormalization conditions at very high energy for your couplings (this is because you expect some ordering principle from the underlying fundamental theory, string theory for instance). To get predictions for the couplings you must RG evolve all the couplings down to electroweak scale or scales where human perform experiments. Using RG equations ensures that the loop expansions for calculations of observables will not suffer from very large logarithms.

A suggested reference: Schwartz, Quantum Field Theory and the Standard model. See for instance pag. 422 and pag.313.

• When you say "a quantity regularized to $F(p)-F(0)$", do you mean that $0$ is an arbitrary momentum scale at which we measure the renormalized coupling? And I'm afraid I still don't understand. I know that you can take $\mu$ to be whatever scale you want, but I don't see how this answers the original question: Why can we extract physical information from the dependence of $\mu$ on renormalization scale, if the latter is arbitrary? Sep 14 '15 at 18:24
• Yes, for the first question. For the rest, I will clarify in the main answer. Sep 14 '15 at 22:32
• The page 313 reference is actually really interesting. I don't think I have seen an explicit distinction between renormalized charge and effective charge anywhere else. It makes sense, though, because the behavior of the effective charge with energy is physically relevant. I appreciate your answer; I will take a little time to think about it more deeply. Sep 15 '15 at 1:22