Main Theorem of Renormalization, but in physicist-speak

In summary, the mathematical formulation of perturbative QFT, specifically in terms of the Stuckelberg-Petermann RG, the Gell-Mann-Low RG, and their difference, can be difficult to understand without a background in mathematics. However, there is a helpful resource, nLab, which provides explanations and references.
  • #1
AndreasC
Gold Member
547
310
I have been reading a few things about the mathematical formulation of perturbative QFT, specifically in terms of the Stuckelberg-Petermann RG, the Gell-Mann-Low RG, and their difference. Unfortunately I lack the mathematical background to understand these things in depth, and I'm having a little bit of trouble in connecting all that to what we are usually taught in QFT. Just so we can get on the same page, here is a link to nLab that brings up some of these things and relevant references: https://ncatlab.org/nlab/show/renormalization+group

From what I understand so far, we have the Stuckelberg-Petermann RG, which is a true group, and its members are all the vertex redefinitions (which I understand to be reparametrizations of, say, the mass, or the charge, etc) that take you to a different renormalization scheme. This last part I'm not so sure what it means in physicist-speak. Is it related to field reparametrizations in the following familiar form?

$$ \phi_R = Z_R \phi $$

Now about the GLRG. It is claimed that scaling the S matrix gives a renormalization of the theory, so there exists a relevant vertex redefinition as per the Main Theorem. Unfortunately I don't understand how this renormalization corresponds to what I know in the first place, or what exactly their idea of scaling is, so I don't understand what this statement concretely means either. What I know is that Gell-Mann and Low method is about choosing an arbitrary renormalization point, and when you work out the counterterms, you end up with a relation for the renormalized g so that the bare charge does not depend on the parameter μ. It's a bit hard for me to concretely relate it to the mathematical approach. Any help would be greatly appreciated!

Edit: Perhaps even better to get us all on the same page is the PF insight page on the same issue: https://www.physicsforums.com/insights/newideaofquantumfieldtheory.renormalization/

Unfortunately, it also requires a bunch of background that I do not possess to really understand, and it also doesn't connect it to the usual, counterterm manipulation/reparametrization methods we are taught...
 
Last edited:
Physics news on Phys.org
  • #3
  • #4
A. Neumaier said:
Maybe you'll like the explanations in my tutorial

Renormalization without infinities–an elementary tutorial

I skimmed it, and while it is interesting, I didn't find anything that directly clears up what I was trying to understand. What I don't really understand is how the formulation in terms of the Main Theorem corresponds to the typical formulation of Green functions etc. What exactly is meant by "renormalization" of the S matrix in that context? What would it mean expressed in terms of Green's functions etc? I'm trying to figure out first of all if it is all supposed to connect a reparametrization of variables such as mass, charge etc to a normalization choice for the fields.
 
Last edited:
  • #5
AndreasC said:
I'm trying to figure out first of all if it is all supposed to connect a reparametrization of variables such as mass, charge etc to a normalization choice for the fields.
Of course. The details depend on the renormalization scheme (and the level of abstractness of the theoretical framework) but the principle is always the same, as apparent from my renormalization tutorial.

In general, renormalization means to use a singular formal parameterization (the renormalization prescription) of the coupling constants (such as bare mass and bare charge in QED) by new parameters and to find a formally equivalent formulation of the theory to be renormalized in these new parameters, such that one can determine the parameters from suitable conditions (the renormalization conditions) relating the new parameters to physical quantities (such as the physical mass and charge in QED) without encountering UV infinities.

You might also be interested in my Insight article on Causal Perturbation Theory - aka Epstein-Glaser (re)normalization -, where it is shown how to avoid infinities completely from the start, in a much less technical way than in Urs Schreiber's treatment of the same topic.
 
Last edited:
  • Like
Likes vanhees71, dextercioby and AndreasC
  • #6
A. Neumaier said:
Of course. The details depend on the renormalization scheme (and the level of abstractness of the theoretical framework) but the principle is always the same, as apparent from my renormalization tutorial.

In general, renormalization means to use a singular formal parameterization (the renormalization prescription) of the coupling constants (such as bare mass and bare charge in QED) by new parameters and to find a formally equivalent formulation of the theory to be renormalized in these new parameters, such that one can determine the parameters from suitable conditions (the renormalization conditions) relating the new parameters to physical quantities (such as the physical mass and charge in QED) without encountering UV infinities.

You might also be interested in my Insight article on Causal Perturbation Theory - aka Epstein-Glaser (re)normalization -, where it is shown how to avoid infinities completely from the start, in a much less technical way than in Urs Schreiber's treatment.
I see. I have read your insight btw, I think it's very informative and overall I've benefited a lot from your answers in various sites.

So, let's say I wanted to "translate" the theorem:
$$ \hat{S} = S \circ Z $$
To, say, two point Green functions in momentum space (propagators), with the following field renormalization:
$$ φ = Z^{1/2}_{R} φ_R$$
Then I take it to mean that, if m and a are the parameters for the initial theory and m', a' are the renormalized ones, then you have:
$$ D_R(p; m', a') = Z_{R}^{(-1)} D(p; m', a')$$
Which comes from the normalization of the functions, then you can find in a unique way functions that relate m', a' to m, a so that:
$$ Z_{R}^{(-1)} D(p; m', a') = D(p; m(m',a'), a(m',a') $$
And therefore:
$$D_{R}(p; m', a') = D(p; m(m',a'), a(m',a')) $$

Correct me if I'm wrong in this understanding. Now suppose we start with, say, the On Shell scheme, and we try to renormalize the electron propagator, in which case we can find the propagator to have a pole in the physical mass and residue 1. Suppose we want to instead set the self energy equal to zero to some arbitrary point $p^2=μ^2$. Then the residue around the pole is no longer 1, but R. So this is the renormalized propagator, but we can relate it to the on shell propagator if only we change the parameters in the on shell case. I don't know if I've got the right picture...
 
  • Like
Likes vanhees71
  • #7
AndreasC said:
$$D_{R}(p; m', a') = D(p; m(m',a'), a(m',a')) $$
This looks right to me.
AndreasC said:
Now suppose we start with, say, the On Shell scheme, and we try to renormalize the electron propagator, in which case we can find the propagator to have a pole in the physical mass and residue 1.
mass ##m##
AndreasC said:
Suppose we want to instead set the self energy equal to zero to some arbitrary point $p^2=μ^2$. Then the residue around the pole is no longer 1, but R.
... but ##Z_R##, and the pole is now at ##m(m',a')##.
 
  • Like
Likes vanhees71
  • #8
Alright, I see... So in the case of the Gell-Mann-Low "group", the idea is that making the "scale transformation", which could be renormalizing the self energy to an arbitrary μ^2 instead of the physical mass, you get a new renormalization, so as per the Main Theorem, there is a unique way to reparametrize, which of course depends on Μ, hence the sliding coupling constant, right?
 
  • Like
Likes vanhees71
  • #9
One approach, the old-fashioned one, is that the physics (aka the S-matrix) cannot depend on the arbitrarily chosen renormalization scheme, and this independence leads to the various renormalization-group equations, depending on the choice of the renormalization scheme (on-shell scheme, mass-independent schemes, etc.).

You find a discussion from a less mathematically rigorous point of view here:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf (Sects. 5.11-5.12)

The more modern approach is a la Wilson and more in the spirit of many-body theory and coarse-graining/resolution of the ("macroscopic") observables.
 
  • #10
AndreasC said:
Alright, I see... So in the case of the Gell-Mann-Low "group", the idea is that making the "scale transformation", which could be renormalizing the self energy to an arbitrary μ^2 instead of the physical mass, you get a new renormalization, so as per the Main Theorem, there is a unique way to reparametrize, which of course depends on Μ, hence the sliding coupling constant, right?
Yes.
vanhees71 said:
The more modern approach is a la Wilson and more in the spirit of many-body theory and coarse-graining/resolution of the ("macroscopic") observables.
But Wilson's RG is only a semigroup - one cannot undo coarsening, hence there are no inverses.
 
  • Like
Likes vanhees71
  • #11
That's of course true.
 

FAQ: Main Theorem of Renormalization, but in physicist-speak

What is the Main Theorem of Renormalization in simple terms?

The Main Theorem of Renormalization essentially states that, through a process called renormalization, we can remove infinities that arise in quantum field theory calculations and make meaningful, finite predictions about physical phenomena. It involves redefining parameters like mass and charge in such a way that the theory remains consistent and predictive at different energy scales.

Why is renormalization necessary in quantum field theory?

Renormalization is necessary because, in quantum field theory, calculations often lead to infinite results for observable quantities like particle masses and interaction strengths. These infinities arise due to the contributions from all possible energy scales, including very high energies. Renormalization helps to systematically cancel out these infinities, allowing for finite and physically meaningful predictions.

How does the renormalization process work?

The renormalization process involves redefining the parameters of the theory, such as masses and coupling constants, in a way that absorbs the infinities. This is done by introducing counterterms that cancel the infinite parts of the calculations. The remaining finite parts then describe the physical quantities we can measure. This process often involves regularization techniques to handle the infinities and renormalization group equations to study how these parameters change with energy scale.

What is the significance of the renormalization group in this context?

The renormalization group (RG) is a mathematical framework that describes how the parameters of a quantum field theory change with the energy scale. The RG equations help us understand how physical quantities evolve as we move from one energy scale to another. This is crucial for making predictions at different scales, such as from the microscopic world of particles to the macroscopic world of everyday phenomena. The RG provides insights into the behavior of the theory under scale transformations and helps identify fixed points where the theory becomes scale-invariant.

Can you give an example of a physical theory where renormalization is applied?

A classic example of a physical theory where renormalization is applied is Quantum Electrodynamics (QED). In QED, the interactions between electrons and photons lead to infinite contributions in the calculations of observable quantities like the electron's mass and charge. Through the process of renormalization, these infinities are systematically removed, resulting in finite predictions that have been experimentally verified to an extremely high degree of precision. The success of renormalization in QED paved the way for its application in other quantum field theories, such as Quantum Chromodynamics (QCD) and the Standard Model of particle physics.

Similar threads

Replies
5
Views
4K
Replies
4
Views
2K
Replies
7
Views
4K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
64
Views
4K
Back
Top