# What exactly is regularization in QFT?

+ 12 like - 0 dislike
2876 views

The question.

Does there exist a mathematicaly precise, commonly accepted definition of the term "regularization procedure" in perturbative quantum field theory? If so, what is it?

Motivation and background.

As pointed out by user drake in his nice answer to my previous question Regulator-scheme-independence in QFT , it is often said that in a renormalizable quantum field theory, results for physical quantities (such as scattering amplitudes) when written in terms of other physical quantities (like physical masses, physical couplings, etc.) do not depend on the regularization procedure one chooses to use. In fact, user drake takes this desirable property as part of the definition of the term "renormalizable."

In my mind, in order for such a statement or definition to be meaningful and useful, it helps to have a precise, mathematically explicit notion of what constitutes a regularization procedure. That way, when one wants to compute something physical, one can use any procedure one wishes provided it satisfies some general properties.

My current understanding in a (small) nutshell.

When we perturbatively compute, say, correlation functions for some theory pre-regularization/renormalization, we obtain formal power series' in the bare parameters that characterize the theory. Such power series' contain expressions for loop integrals that generically diverge, often due to high-momentum (UV) effects, so we "regularize", namely we implement some procedure by which these integrals are made to depend on some parameter, call it $\Lambda$, and are rendered finite provided $\Lambda$ doesn't take on a certain limiting value $\Lambda_*$ (that could be $\infty$) corresponding to the physical regime (like the UV) that led to the original divergence. We then renormalize and find (in renormalizable theories) that $\Lambda$ drops out of physical results.

What sort of an answer am I looking for?

I am looking for something like this.

A regularization procedure is a prescription by which all divergent integrals encountered in perturbation theory are made to depend on a parameter $\Lambda$ and which satisfies the following properties: (1) All divergent integrals are rendered finite for all but a certain value of $\Lambda$. (2)...

I know that there are other properties, but I don't know what constitutes a sufficiently complete list of such properties such that if you were to show me some procedure, I could say "oh yes, that counts as a valid regularization procedure, good job!" Surely, for example, the putative regularization procedure cannot be too destructive; I cannot, for instance, simply replace every loop integral with $3\Lambda$ and call it a day because that would completely destroy all information about how many loops the corresponding diagrams contained. How much of the "formal structure" of the integrals does the procedure need to preserve?

As far as I can tell, there is no discussion of this in any of the standard QFT texts who simply adopt tried-and-true procedures like a hard UV cutoff, dim reg, Pauli-Villars, etc. without commenting on general conditions sufficient to guarantee that these particular procedures count as "good" ones. There is, of course, a lot of discussion of whether certain regulators preserve certain symmetries, but that's a distinct issue.

Edit. (January 8, 2014)

Discussions with fellow graduate students have led me to believe that the appropriate definition proceeds by appealing to the idea of effective field theory. In particular, if we view our theory as an effective low-energy description of some more complete theory that works at higher energy scales, then imposing a high momentum cutoff has a conceptually privileged position among regulators; it is the natural way of encoding the idea that the putative theory only works below a certain scale.

This can then be used to define a regularization procedure that, in some sense, reproduces the same structure encoded in regularizing with a cutoff. Unfortunately, I'm still not entirely sure if this is the correct way to think about this, and I'm also not sure how to formalize the notion of preserving the structure the comes out of cutoff regularization. My inclination is that the most important structure to preserve is the singular behavior of regularized integrals as the cutoff $\Lambda$ is taken to infinity.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics
retagged May 13, 2015
You might want to take a look at the review of Rosten. Exact RG flow equations are discussed in a very general way.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user Steven Mathey

+ 6 like - 0 dislike

The definite answer to your question is: There is no mathematicaly precise, commonly accepted definition of the term "regularization procedure" in perturbative quantum field theory.

Maybe you'll find Chapter B5: Divergences and renormalization of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html illuminating. There I try to abstract the common features and explain in general terms what is needed to make renormalization work. The general belief is that the details of the regularization scheme don't matter, though in fact it is known that sometimes some regularization schemes give apparently incorrect results.

This is to be expected since the unregularized theory is ill-defined, and can be made well-defined in different ways, just as a divergent infinite series can be given infinitely many different meanings depending how you group the terms to sum them up.

If at any time in the future there will be a positive answer to your question, it will be most likely only when someone found a logically sound nonperturbative definition of the class of renormalizable quantum field theories.

On the other hand, if you want to have a mathematically rigorous treatment of some particular regularization schemes for some particular theories, you should read the books by (i) Salmhofer, Renormalization: an introduction, Springer 1999, and (ii) Scharf, Finite quantum electrodynamics: the causal approach, Springer 1995.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user Arnold Neumaier
answered Mar 21, 2014 by (15,787 points)
+ 6 like - 1 dislike

I'm going to give a silly answer, but I think this is the best we can do. A regulator is any prescription for defining the path integral such that after adding a sum of local counterterms to the action and allowing the physical couplings to depend on the renormalization scale $\mu$, the correlation functions are equal to those obtained by taking a continuum limit of a lattice theory. This is similar in motivation to your most recent edit, except that we really need to use a lattice instead of a Euclidean momentum space cutoff, because the latter breaks gauge invariance.

I'm pessimistic about the existence of a better answer, just because some regulators satisfy some nice properties (i.e. unitarity, symmetries, etc.), while others don't.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user Matthew
answered Feb 4, 2014 by (320 points)
+1: I increasingly feel that this is basically right, and I'm similarly pessimistic.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics
+ 3 like - 0 dislike

As you said regularization is necessary to even begin making sense of the diagrams which appear in perturbation theory. The word "perturbation" already contains a hint towards the answer you may be looking for. If you are trying to make sense of the interacting theory $d\nu$ by perturbation, this implies that there is somebody being perturbed, namely a free theory $d\mu$. I think regularization is a feature of the pair $d\nu,d\mu$ rather than $d\nu$ alone. For instance take the functional measure $d\mu$ corresponding to the usual free field $$d\mu(\phi)=\frac{1}{Z}e^{-\frac{1}{2}\int \{(\nabla\phi)^2+m^2\phi^2\}} D\phi$$ and let $d\nu(\phi)=\frac{1}{Z'}e^{-V(\phi)}d\mu(\phi)$ where $V$ is your favorite interaction potential, e.g., $$V(\phi)=g\int \phi^4\ .$$ Let $\rho(x)$ be a mollifier (not sure how one should spell that), i.e., a smooth function with rapid decay or even compact support around the origin with $\int \rho=1$. Let $\rho_r(x)=2^{-dr}\rho(2^{-r} x)$ where $d$ is the dimension (of Euclidean space-time) and $r\rightarrow -\infty$ is the UV cut-off. Then if one denotes by $d\mu_r$ the law of $\rho_r\ast\phi$ with $\phi$ sampled according to the free probability measure $d\mu$, one has the weak convergence of $d\mu_r$ to $d\mu$ when $r\rightarrow-\infty$. In that sense, the modification we are doing is a sensible regularization of the measure $d\mu$. Now on closer inspection $e^{-V(\phi)}d\mu(\phi)$ does not make any sense. However $e^{-V(\phi)}d\mu_r(\phi)$ is perfectly fine. So the game becomes trying to obtain $d\nu$ as the limit of measures $d\nu_r(\phi)=\frac{1}{Z_r}e^{-V_r(\phi)}d\mu_r(\phi)$ when $r\rightarrow -\infty$. Notice that I changed $V$ into $V_r$ because renormalization theory tells us we have to let couplings like $g$ etc. depend on the cut-off $r$.

If one succeeds in this construction of the interacting theory $d\nu$, independence of the regularization can be understood as showing that the result is independent of the choice of the mollifier $\rho$ (heat kernel reularization or Pauli-Villars are particular cases).

In the related context of singular SPDEs, recent progress was made by Hairer with his theory of regularity structures. An important feature of this theory is that it allows one to prove similar statements about independence with respect to the choice of regularization procedure.

Note that there is no reason to restrict the above considerations to perturbed theories $d\mu$ which are Gaussian. In some sense, conformal perturbation theory is about similar perturbations around other nontrivial QFTs.

Of course, in Fourier space, the propagator for $d\mu_r$ is $$\frac{|\widehat{\rho}(2^r p)|^2}{p^2+m^2}$$ but I prefer the definition in $x$ space as a convolution, since this amounts to taking local averages and is a continuous analogue of Kadanoff's block-spinning procedure.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user Abdelmalek Abdesselam
answered Apr 5, 2015 by (640 points)
+ 0 like - 3 dislike

Regularization is a rewriting of your integral so that you can deal its divergences using other tricks.

For example in QFT you calculate some amplitude to a certain order in perturbation theory. The integrals that represent loop diagrams diverge. The most common regularization procedure is called dimensional regularization where you parametrize the dimension of your loop integral to, for example, d=4-c.

It turns out that your integral was divergent for large momentum, therefore having a branch cut. Now after dimensional regularization the branch cut divergence became a simple pole when c=0. As you know, it is easier to deal with a simple pole than a branch cut.

After making your integral more manageable you can renormalize and then separate the divergent and finite parts of your result. You eventually use other tricks to remove the divergent part, for example add counterterms to your lagrangian.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user Mario
answered Feb 4, 2014 by (-30 points)
I appreciate the response, but I'd ask you to read the question, the other answer, and comments more carefully. I unfortunately am aware of what regularization is at the level you suggest; I'm looking for something more mathematically precise and descriptive.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics
+ 0 like - 4 dislike

You have a theory, then you calculate some physical quantity, then you get some divergence . The idea of "regularization " is to define another theory depend on some parameter, when let the parameter send to some value ( say d->4, Λ->infinity, lattice length to zero...), That another theory will tend to original theory. This procedure will guarantee you get a finite result, yet when the parameter send to some value, recover the original theory.

The following step is "renormalization". We use the regularized theory to calculate some basic physical quantity, such as mass, scattering amplitude. All other physical quantity can be expressed by this basic quantity, the miracle of "renormalizable theory "is these expression is irrelevant of the parameter.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user user35289
answered Jan 5, 2014 by (-40 points)
-1: I'm acutely aware of all of this (which I think is relatively clear from my discussion following the question?). This, unfortunately, doesn't answer the question. In particular, I am aware that once a regularization is chosen, the theory is defined with an extra parameter but that physical results don't depend on it in renormalizable theories. The question is about what sort of such parametrizations are allowed.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics
See My first paragraph, which have answered you question. The key word is "Tend to".

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user user35289
I appreciate your input, but the words "tend to" are not particularly precise or descriptive. Note that I acknowledge the need for the regulator to preserve (at least some) of the formal structure of the "original theory," the main point of the question is how one codifies the "tend to" and makes it mathematically precise and explicit.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics
Condition: 1, Lim S(Λ) -> S, When Λ-> ∞; 2, No divergence with action S(Λ)

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user user35289
-1. Those conditions you've written down are trivial and do not guarantee that you've got a good regularization.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user user1504
@user1504 Care to take a stab at an answer? I'd be curious to hear your thoughts, even if they don't constitute a complete answer by your own standards.

This post imported from StackExchange Physics at 2015-05-13 18:49 (UTC), posted by SE-user joshphysics

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification