# physical importance of regularization in QFT?

The standard lore in QFT is that one must work with renormalised fields, mass, interaction etc. So we must work with "physical" or renormalised quantities and all our ignorance with respect to its relation with "bare quantities" is taken care by counter terms which (surprisingly!!) cancel infinities. My question is:

1. Why do we care to regulate such field theories, when all physical quantities are rendered finite by the way of renormalisation . Why is there such song and dance about regulator and regularizing scheme(one which break certain symmetries and what not)?

2. The following is not very physical question- Is there a deeper understanding how and why counter terms seem to render all physical quantities finite. I guess a satisfactory answer to this question is a matter of taste, nevertheless what I am looking for is a constructive argument which a priori guarantees finite sensible values of measurable quantities.

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–  Michael Brown May 1 '13 at 0:42
My question is different although I agree its related. I question the need for regularization in the face of convergent combination involving counter terms which never blew up in the first place. The question you have referred to questions the reasoning for the cut off and the subsequent rigor involved once you admit the need to regularize. –  Noob Rev B May 1 '13 at 18:48

Firstly, if I'm correct in understanding your question, you are asking why one needs to regulate field theories since we renormalize them anyway?

Assuming this, here's my understanding:

Quantum field theory in general has several divergences, which we must deal with. We get around such divergences by dealing with renormalized quantities. But how does one renormalize the theory? Essentially (without going into too much details), the statement is that quantities in QFT often have infinities. However, if we calculate the "right" quantity (one has to be more specific about what is right and what is not, that is a slightly harder question to answer), the infinities cancel. But one cannot simply say $\infty - \infty = \text{finite}$. That makes no sense. One has to describe the precise form of the divergence. This implies making the $\infty$ more precise, by writing it as $\infty = \lim\limits_{\Lambda \to \infty} \log \Lambda$ or $\infty = \lim\limits_{\epsilon \to 0} \frac{2}{\epsilon}$, etc. The process of characterizing the infinities in this way is regularization. Once you have done this, the expression $\infty - \infty$ now makes complete sense.

We see that regularization is an essential process in renormalizing a theory. Good!

(Let make another comment about the symmetry statement you made in your question) The problem with introducing regulators is that often the regulator used may not satisfy all the symmetries of your theory, in which case we have quantum anomalies. One has to deal those after regulating the theory.

With regards to your comment on counterterms surprisingly cancelling divergences, I will say this: It is not all that surprising. In fact, counterterms are introduced by hand in the action specifically to cancel all those divergences. However, what may be surprising is that in some theories (called renormalizable theories) a finite number of counterterms are all that you need to cancel out all the divergences and that all the divergences can be swept under the rug by redefining your bare fields and couplings. While this is a little bit non-trivial to see, a simply dimension counting of divergences makes it easy to see (atleast at 1-loop level). For most other theories (non-renormalizable) there aren't a finite number of counterterms and one has to modify the action by introducing an infinite number of counterterms and operators.

However, technical discussion aside, if QFT is to be a good theory of nature, one MUST get finite answers out of it and there MUST be enough counterterms to cancel all the divergences. Either this happens on its own (renormalizable theories) or we have to impose it as a condition and modify our theory to accomodate all the divergences (non-renormalizable theories)

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The $\infty-\infty$ you pointed out never arises if you include all the "right" quantities in the integrand before you integrate over loop momenta. The right quantities always appear as convergent combination, hence it seems redundant to renormalize. What I am saying is that one can always write convergent integrand $I$ as $\int I = \int (I +c -c)$. Where $\int c$ might not converge. But then it seems to me that all of a sudden we are worried that we are getting $\int (I+c)=\infty$ & $-\int c =-\infty$ and hence find the need to regularize. –  Noob Rev B May 1 '13 at 23:38
As far as counter terms are concerned they seem to be just an expression for our ignorance about bare quantities in the following way:- $$L_{Bare,Un-renormalisedField} = L_{FiniteCoupling,RenormalisedField}+ L^{CT}_{UnknownCoupling,RenormalisedField}$$. Thus I would contend that CT(counter-term) is essentially a restatement of our original lagrangian with bare coupling constants in terms of know finite quantity + CT lagrangian in principle having infinite quantities. I could have carried out this separation even in 2 or 3 dimensions where diagrams would,'t have diverged. –  Noob Rev B May 2 '13 at 0:01
It is surprising that when it does diverge however, CT terms are sufficient to cancel it apart from giving a finite correction. I agree with @Prahar that infinities in renormalisable theories can be understood by looking at superficial degree of divergence. Although my 2nd question is vague, I am surprised how you could choose your favourite renormalisation condition and express counter terms in language of feynman diagrams with those condition and you see all infinities vanish. If you look at the algebraic process its obvious, but is there a deeper realization which I am missing? –  Noob Rev B May 2 '13 at 0:18