# Renormalization - basic questions

#### cordovan66

I'm just learning renormalization in QFT and have a few basic questions:

1) It seems to me that renormalization has the status of a *prescription* for extracting a finite number from an infinite one. It cannot be justified except that this prescription leads to agreement with experiment. Is this a correct understanding?

2) It seems to me that we first identify the cutoff dependent terms in loop diagrams and the prescription is to just drop them. We then figure out what counter term to add to the Lagrangian to have the effect of dropping the above cutoff dependent terms. If these counter terms are of a form that existed in the original Lagrangian, we deem the theory renormalizable. Is this right? If so, is the only purpose of finding the required counter terms to figure out if the theory is renormalizable? If we already knew that the theory was renormalizable, could we just drop the cutoff dependent terms and forget about counter terms?

3) In the presentation in our class, the loop amplitude was divided into a cutoff dependent "infinite part" and a finite part. It was stated that this division has some arbitrariness: we can move a finite amount from one part to the other. A couple of examples were given: in one of them (phi ^3 in 3+1 dimensions, 1 loop) the arbitrariness in this division was stated to be up to a constant. In another example (phi^3 in 5+1 dimensions, 1 loop), the arbitrariness was stated to be up to a quadratic in momentum. I don't understand why the arbitrariness in the former (say) is limited to just a constant. Can't we add any momentum dependent term to infinity and still get infinity? Does the problem occur as momentum tends to infinity?

Thanks!

#### JosephButler

Hi there Cordovan -- renormalization can indeed be a slippery subject. Here are my best attempts at (perhaps partial) answers from a fellow student.

cordovan66 said:
1) It seems to me that renormalization has the status of a *prescription* for extracting a finite number from an infinite one. It cannot be justified except that this prescription leads to agreement with experiment. Is this a correct understanding?
To be precise, regularization is the somewhat arbitrary prescription for extracting a finite number from an infinite number. Renormalization, on the other hand, is the description of how quantities evolve with energy scale. The procedure for dealing with infinities in QFT is to first regularize to get a handle on the infinities (i.e. quantitative expressions such as 1/epsilon or cutoff^2, etc.) then renormalize to understand the physics.

The regularization prescription is usually chosen to obey the symmetries of the theory (e.g. gauge invariance, Lorentz invariance, etc.) so that these symmetries are explicitly preserved in the renormalization procedure, though I believe this isn't a strict requirement to get sensible results, it's just the most straightforward way. (To the extent that this whole procedure is `straightforward'!)

What is important to understand is

1. Actual physical observables are independent of the regularization scheme. One regularizes in a way that is convenient. This regularization has nothing to do with experiment.

2. The renormalization procedure is *required* theoretically in quantum field theory (though this took a long time for physicists to understand!) and is not put in "to agree with experiment."

3. On "agreeing with experiment:" When you have a theory (i.e. the Lagrangian for a model) you check consistency with experiment by measuring different quantities that over-constrain your parameters. For each parameter, one needs to make a measurement. Then, on top of this, one needs to make additional independent measurements that check whether these parameters are consistent within the framework of a theory. Renormalization is related to how measurements at one energy scale can be compared to those at another energy scale. Said in another way, experimental input comes into the renormalization conditions, but this is just using experiment to give initial conditions.

cordovan66 said:
2) It seems to me that we first identify the cutoff dependent terms in loop diagrams and the prescription is to just drop them. We then figure out what counter term to add to the Lagrangian to have the effect of dropping the above cutoff dependent terms. If these counter terms are of a form that existed in the original Lagrangian, we deem the theory renormalizable. Is this right? If so, is the only purpose of finding the required counter terms to figure out if the theory is renormalizable? If we already knew that the theory was renormalizable, could we just drop the cutoff dependent terms and forget about counter terms?
Indeed -- if the counter terms are of the form of terms in the original Lagrangian, this means that we can absorb those infinities into redefinitions of the field strengths, coupling constants, and masses.

I don't think that the only purpose of writing out counter terms is to figure out renormalizability. One can usually figure this out by looking at the Lagrangian and using power-counting and symmetry arguments. The counter terms are just the divergent parts of loop diagrams. These end up being reabsorbed ("renormalized") into the Lagrangian parameters if the theory is renormalizable. One needs to the divergence structure to calculate beta functions which tell us how these 'physical' couplings ('renormalized couplings') change with energy scale. I.e calculating loop diagrams in QED will tell you how alpha_em deviates from 1/137 at high energies.

cordovan66 said:
3) In the presentation in our class, the loop amplitude was divided into a cutoff dependent "infinite part" and a finite part. It was stated that this division has some arbitrariness: we can move a finite amount from one part to the other. A couple of examples were given: in one of them (phi ^3 in 3+1 dimensions, 1 loop) the arbitrariness in this division was stated to be up to a constant. In another example (phi^3 in 5+1 dimensions, 1 loop), the arbitrariness was stated to be up to a quadratic in momentum. I don't understand why the arbitrariness in the former (say) is limited to just a constant. Can't we add any momentum dependent term to infinity and still get infinity? Does the problem occur as momentum tends to infinity?
I'm not sure if I understand your question, but I think you might be reading into this too much. The arbitrariness is really a choice of how you want to scale your physical couplings. ("Renormalization scheme.") This is the choice, for example, between MS and MS-bar renormalization.

When calculating loop diagrams in dimensional regularization (which was an 'arbitrary choice' for a regularization scheme!) we end up with factors of (4pi) and the Euler-Mascheroni (is that spelled correctly?) constant. When we subtract infinite terms (absorb them into physical couplings) we have the freedom to subtract (absorb) these annoying finite terms as well.

Cheers,
Joe

#### Haelfix

1) No renormalization is not an arbitrary prescription, although it might seem that way at first. I suggest you wait a few weeks, at which point you will learn about the renormalization group equations. It all starts to make sense then.

"If we already knew that the theory was renormalizable, could we just drop the cutoff dependent terms and forget about counter terms?"

2) No!!! Part of those cutoff dependant terms contains bonafide information about real physical corrections (to the couplings, mass, and various other physical quantities). The renormalization procedure extracts that information order by order. You cannot just drop them until the end in the calculation where you are allowed to do so (depending on the scheme) or alternatively to absorb them into a definition somewhere.

3) Yea so for instance, wiki the difference between say minimal subtraction and minimal subtraction bar. However you have to understand this ambiguity is irrelevant to the actual physical results modulo up to calculations of an extra order. So for instance you *might* (but usually not) have disagreement between two renormalization procedures at say a finite number of loop calculations, but if you go to one higher, everything one procedure misses is then recaptured (+ extra information).

#### blechman

At the risk of repeating some of what's already been said, let me take a stab at answering these questions.

1) It seems to me that renormalization has the status of a *prescription* for extracting a finite number from an infinite one. It cannot be justified except that this prescription leads to agreement with experiment. Is this a correct understanding?
It is true, to some extent, that renormalization is an "arbitrary prescription that works" - but then again, so are Newton's Laws and the Schrodinger Equation! I mean, this is physics, and we are out to describe nature. You can ask "Where do these things come from?" but I'm not sure how to answer that.

But it is also true that the renormalization prescription is actually a well-defined, rigorous mathematical procedure. It didn't start out that way (back in the Feynman/Schwinger days people really thought this was just crap!) but over the last half-century we have really nailed it down through things like the renormalization group mentioned above. But I don't want to get too far ahead so let's just leave it at that.

2) It seems to me that we first identify the cutoff dependent terms in loop diagrams and the prescription is to just drop them. We then figure out what counter term to add to the Lagrangian to have the effect of dropping the above cutoff dependent terms. If these counter terms are of a form that existed in the original Lagrangian, we deem the theory renormalizable. Is this right? If so, is the only purpose of finding the required counter terms to figure out if the theory is renormalizable? If we already knew that the theory was renormalizable, could we just drop the cutoff dependent terms and forget about counter terms?
Hmmm.... the DEFINITION of a "renormalizable" QFT is one that has a FINITE NUMBER OF COUNTERTERMS. For example, in QED, **ALL** infinities can be absorbed into mass charge and field renormalization.

The prescription you give above is misleading. Here's the better way to see it: write down all the operators in your action. Now give EACH of those operators a "counterterm". For example: in scalar $\phi^4$ theory there are 3 operators: a kinetic term, a mass term, and an interaction term. So since there are three operators, there are three counterterms.

QUESTION: when doing calculations in this theory, are those three counterterms all you need to absorb divergences? If yes (in this case, that's right) then the theory is renormalizable. If no (see your next question) then your theory is nonrenormalizable.

3) In the presentation in our class, the loop amplitude was divided into a cutoff dependent "infinite part" and a finite part. It was stated that this division has some arbitrariness: we can move a finite amount from one part to the other. A couple of examples were given: in one of them (phi ^3 in 3+1 dimensions, 1 loop) the arbitrariness in this division was stated to be up to a constant. In another example (phi^3 in 5+1 dimensions, 1 loop), the arbitrariness was stated to be up to a quadratic in momentum. I don't understand why the arbitrariness in the former (say) is limited to just a constant. Can't we add any momentum dependent term to infinity and still get infinity? Does the problem occur as momentum tends to infinity?

Thanks!
So there are a few conceptual faux-pas in this paragraph.

First of all, let me tackle your "momentum-dependent" question. The short answer is NO! But why? Recall that momentum in the amplitude corresponds to a derivative in the operator. Thus, subtracting a momentum-dependent infinity requires a counterterm on an operator with a derivative! Now you see where the nonrenormalization argument from above comes in! Consider the phi^4 theory again. Imagine that you get a divergent amplitude that goes like $p^2\log(\Lambda)$ - this can only be subtracted by a counterterm that has two derivatives in it - in other words, the kinetic term! So this is okay. But now imagine that you have an amplitude that diverges but also goes like $p^4$ - now you're in a lot of trouble, since there is no counterterm in phi^4 theory that has 4 derivatives in it! So IF this was to happen, the theory would be NONrenormalizable. Of course, phi^4 theory is a renormalizable theory, and therefore you know that you can NEVER get this kind of divergence. Phew!

Now the "arbitrariness" in the counterterm just follows from the fact that you can put as much or as little of the finite part of the amplitude in the couterterm as you want. But again: you don't just subtract infinities blindly! You need a counterterm to do it, and therefore you need an operator in the action of the right form (number of derivatives, etc). You can still get a finite result, for example:

$$\mathcal{M}=p^2\left(\log(\Lambda)+\frac{3}{2}\right)$$

or some such thing. You know right away that you can use the kinetic operator's counterterm (2 derivatives) to cancel this infinity, but should you subtract the 3/2? Maybe you should only subtract 1? Or zero!? You're free to do whatever you want! But of course, at the end of the day, you find that the PHYSICAL answer, that is, the cross section, must be independent of your choice of subtraction scheme. And of course, this is always the case. You will see proofs of this as you continue your studies.

Hope this helps!

#### meopemuk

Let me give you a different (not widely accepted) view on renormalization.

In my opinion, the need for renormalization in QFT is an indication of a serious flaw in this theory. A consistent physical theory should not involve renormalization tricks.

There is a way to reformulate QFT in such a way that renormalization is not needed. All loop integrals are automatically convergent. The Hamiltonian does not contain divergent counterterms. This reformulation is based on the "dressed particle" approach

O. W. Greenberg and S. S. Schweber, "Clothed particle
operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

More details can be found in

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139.

E. V. Stefanovich, "Renormalization and dressing in quantum field theory", http://www.arxiv.org/abs/hep-th/0503076v4

E. V. Stefanovich, "Relativistic quantum dynamics",
http://www.arxiv.org/abs/physics/0504062v11

#### cordovan66

Thanks for the replies everyone - especially blechman and Joseph Butler. I've read and reread them over the past few days. While I still have some uneasiness about the subject, my understanding has improved a lot.

We've just started Renormalization Group in class. Maybe I should first just read about it properly but I'd appreciate it if someone can tell me how coupling constants can change with energy scale? I presume it's the renormalized (or physical) couplings that change; the bare ones can't, right? In classical physics, parameters don't change; any dependence on energy scale is described explicitly by the model with parameters being constants - that's why they are called parameters. Why can it not be the same in QFT?

Thanks again

#### RedX

I just wanted to point out that renormalization has been discussed in the quantum physics forum recently, and there were some good conversations there:

As to your quesiton on coupling constants changing with energy, technically, as I understand it, they don't have to change with energy if you keep the same mass scale. however, keeping the same mass scale leads to large logs if the energy of your experiment is much different from the mass scale, meaning that you have to calculate a lot of loops, and not just 1 or 2 loops. So I think there is a terminology difference here. The coupling constants do change with mass scale, but not with energy. But if you don't match mass scale with energy, then your theory looks a lot different at different energies.

#### blechman

We've just started Renormalization Group in class. Maybe I should first just read about it properly but I'd appreciate it if someone can tell me how coupling constants can change with energy scale? I presume it's the renormalized (or physical) couplings that change; the bare ones can't, right? In classical physics, parameters don't change; any dependence on energy scale is described explicitly by the model with parameters being constants - that's why they are called parameters. Why can it not be the same in QFT?

Thanks again
There are two ways to understand what's going on, I think. Let me try to explain:

1. Coupling constants are just constants, as you suspect. After you renormalize, you just have a finite answer, and that's all there is to it. The final answer involves your couplings, and a general function of the energy of the interaction. You're done.

But now you look at your answer and you start to worry. You find that in fact, the answer is not just $\lambda$, but it is actually proportional to $\lambda\log(E/M)$, where E is the energy and M is some mass scale in the problem (if your theory is a massless theory, M is the renormalization scale you needed to introduce when renormalizing). Now imagine, being the ambitious calculator that you are, that you proceed to the two-loop level. You will find you get an answer that goes like $\lambda^2\log^2(E/M)$. In fact: this is generic -- perturbation theory is not just an expansion in the coupling, but the coupling times logs! Well, this is seriously bad news, since it means that sooner or later, E is going to be large (or small!) enough to overcome the small coupling and your perturbation theory will no longer make any sense!

Enter the renormalization group: this is a fancy way of doing a calculation to ALL loop order, but only leading order in the log (so you ignore $\lambda^n\log^{n-1}(E/M)$ for example.

So what does it do? At the end of the day, you can either write:

$$\sigma\sim \sigma_0\left[1+b\lambda\log(E/M)+\mathcal{O}(\lambda^2)\right]$$

Or you can write

$$\sigma\sim\sigma_0\times\overline{\lambda}(E)$$

where

$$\overline{\lambda}(E)=\frac{1}{1-b\lambda\log(E/M)}=1+b\lambda\log(E/M)+\mathcal{O}(\lambda^2)$$

So which is right? Of course, they're exactly the same, as they should be! But notice that while $\lambda\log(E/M)$ is a problem when $E\sim Me^{-1/b\lambda}$ (since now the log term is bigger than the leading term), $\overline{\lambda}(Me^{-1/b\lambda})$ is fine! So perturbation theory works IF you use the running coupling!!

You know, I think I'll stop there. The second way to visualize the running coupling (and the RIGHT way, between ourselves) is in terms of the all-powerful "Effective Field Theory" paradigm. But rather than confuse you further, digest this and wonder at the magnificence of The Theory of Renormalization! As you become more advanced in your studies, you will see how the EFT philosophy works, and how it provides an even more natural feeling for these things. OK, enough divine wondering!

Hope that helps!

#### JosephButler

I figure it might be helpful if I offered some literature which I found very, very helpful when I was learning about renormalization for the first time:

* Zee, Chapter III. At the very least III.1. A very user-friendly explanation of what is meant by bare and physical/renormalized paramters.

* P. M. Stevenson, Dimensional Analysis in field theory, Annals of Physics, Volume 132, Issue 2, 1 April 1981, Pages 383-403, http://dx.doi.org/10.1016/0003-4916(81)90072-5" [Broken].

This is, I think, the *best* introduction to the renormalization group. It won't help you calculate anything, but it gives a somewhat unexpected motivation for why the entire program. It presents renormalization as a necessary consequence of dimensional analysis and removes a lot of the apparent arbitrariness when one first learns the subject. This isn't the way people prefer to think about the RG in textbooks, but renormalization is deeply rooted in the idea of scaling and this is how a lot of the previous generation of theorists thought about RG.

* "Cut-offs and continuum limits: A wilsonian approach to field theory." http://pyweb.swan.ac.uk/~hollowood/" [Broken] [I know, the website says "supersymmetry 2005," but the notes linked are actually about RG.]

This is another very well written work, though it's perhaps at a bit more of an advanced level. It explains a lot of the details of how field/statistical theorists think about RG mathematically while maintaining a close connection to simple toy models. I would probably suggest reading this after your course covers RG as a way to review everything and see how different ideas come together.

* A hint of renormalization, B. Delamotte. http://arxiv.org/abs/hep-th/0212049" [Broken]

This is a very handy introduction to the nuts and bolts of regularization and renormalization. It doesn't actually make use of QFT, it just talks about what is meant by infinite integrals and how we should deal with them when we see them in QFT.

* Regularization, Renormalization, and Dimensional Analysis: Dimensional Regularization meets Freshman E&M, Fredrick Olness, Randall Scalise. http://arxiv.org/abs/0812.3578" [Broken]

This is a really cute paper about applying the high-falutin' techniques of RG to simple electromagnetic problems. It's nice to be able to use the heavy machinery on simple problems so that we can really see what our tools are doing for us.

* A Theoretical Physics FAQ. http://www.mat.univie.ac.at/~neum/physics-faq.txt" [Broken]

The sections on renormalization are rather nice and answer a lot of the usual questions that pop up when one first meets the subject.

Hope those help!
Joeseph

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#### Bob_for_short

I agree with the Eugene Stefanovich opinion. Moreover, I wrote a couple of papers on this subject. You can start from "Reformulation instead of Renormalizations" by Vladimir Kalitvianski available at arxiv:0811.4416. Also it is useful to have a gist of "Atom as a "Dressed" Nucleus" at arxiv:0806.2635 (also available in the Central European Journal of Physics", Volume 7, N 1, pp. 1-11, 2009.

In the first article I show why the fundamental constants obtain perturbative corrections in higher orders, why the renormalizations work and how to reformulate the theory in order to avoid these technical and conceptual complications. It is done on a very simple mechanical example comprehensible for each and everyone.

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