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Quoting Wikipedia:

renormalization is any of a collection of techniques used to treat infinities arising in calculated quantities.

Is that true? to me, it seems better to define renormalization as a collection of techniques for adjusting the theory to obtain physical results. I'll explain. According to Wilson's renormalization group, a quantum field theory always inherently has a cutoff parameter, so in any case integrals should be done only up to the cutoff, so there are no infinite quantities. Yet the results are still not consistent with observation if you don't renormalize the calculations (e.g. using counterterms).

Am I correct? Is it true that the usual presentation of renormalization as a tool for removing divergences is a misinterpretation of the true purpose of it?

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  • $\begingroup$ Lubos's answer is correct. I want to add that it's sort of unfortunate that we call this collection of techniques 'renormalization'. It makes it sound like the key idea is changing the norm of something. $\endgroup$ – user1504 Jun 28 '13 at 13:47
  • $\begingroup$ Dear @user1504, changing the norm of the fields (by an infinite factor) is a major part of the renormalization although it's not the only part. $\endgroup$ – Luboš Motl Jun 28 '13 at 16:11
  • $\begingroup$ @LubošMotl It's an important detail, but it's still just an implementation detail. (It'd be just as silly to call renormalization "counter-term theory".) The central conceptual ideas have more to do with similarity and scaling. $\endgroup$ – user1504 Jun 28 '13 at 17:39
  • $\begingroup$ I just wish to note an important remark: even physical cut-offs below infinity result in infinities in some cases. The problem is not simply that we integrate to infinity. $\endgroup$ – Y2H Jun 24 '18 at 9:12
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You're totally right. The Wikipedia definition of the renormalization is obsolete i.e. it refers to the interpretation of these techniques that was believed prior to the discovery of the Renormalization Group.

While the computational essence (and results) of the techniques hasn't changed much in some cases, their modern interpretation is very different from the old one. The process of guaranteeing that results are expressed in terms of finite numbers is known as the regularization, not renormalization, and integrating up to a finite cutoff scale only is a simple example of a regularization.

However, the renormalization is an extra step we apply later in which a number of calculated quantities is set equal to their measured (and therefore finite) values. This of course cancels the infinite (calculated) parts of these quantities (I mean parts that were infinite before the regularization) but for renormalizable theories, it cancels the infinite parts of all physically meaningful predictions, too.

However, the renormalization has to be done even in theories where no divergences arise. In that case, it still amounts to a correct (yet nontrivial) mapping between the observed parameters and the "bare" parameters of the theory.

The modern, RG-based interpretation of these issues changes many subtleties. For example, the problem with the non-renormalizable theory is no longer the impossibility to cancel the infinities. The infinities may still be regulated away by a regularization but the real problem is that we introduce an infinite number of undetermined finite parameters during the process. In other words, a non-renormalizable theory becomes unpredictive (infinite input is needed to make it predictive) for all questions near (and above?) its cutoff scale where its generic interactions (higher-order terms) become strongly coupled.

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  • $\begingroup$ Clear, succinct answer. $\endgroup$ – joshphysics Jun 28 '13 at 7:52
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    $\begingroup$ I'd love to upvote a second time! $\endgroup$ – Neuneck Jun 28 '13 at 8:51
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    $\begingroup$ And here is a nicely readable article about how renormalization is currently viewed by the same author in the context of a sad event. $\endgroup$ – Dilaton Jun 28 '13 at 11:27
  • $\begingroup$ Could you provide examples where renormalization is needed even for theories with no divergences $\endgroup$ – Revo Jun 29 '13 at 14:53
  • $\begingroup$ It's a somewhat technical example but it's really strong. The Higgs decay to 2 photons is given by convergent diagrams but they're actually wrong if computed in d=4. One has to compute them with a regularization scheme, ideally in d=4-epsilon, see arxiv.org/abs/1306.5767 and motls.blogspot.cz/2013/06/… - otherwise one gets a wrong result! But even if one gets the "right" result, it's still true that the parameters in the Lagrangian aren't exactly equal to some natural ways to quantify the interaction stength, and... $\endgroup$ – Luboš Motl Jun 30 '13 at 6:50

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