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I am not sure that I can frame this question coherently enough - it springs from various things in QFT that I have recently been thinking and reading about. May be these thoughts are mis-directed but it would still help to know why they are if they are so! I would like to hear some discussions about these.

  • I guess there are QFTs which are "exact" or "finite" as in they don't require a regulator or cut-off for their partition function to be defined (..and I guess one can evaluate the partition function exactly..) I guess the really non-trivial "finite" QFTs would be ones which are so even on non-compact spacetimes. Are there such?

  • I guess there exists QFTs which have a non-perturbatively 0 beta function. (like $\cal {N} =4$ SYM?)

Are the above two properties related?

Like does being "finite" imply it has an exactly $0$ beta function or conversely? (..seems no..see below..)

  • I guess CFTs (..or any QFT sitting at a zero of its beta-function -"critical" ?..) sort of by definition have a 0 beta function but they do have non-trivial OPEs coming from short-distance singularities. This is somehow not intuitive because one would have naively thought that large momentum is like short distance and hence if the theory requires no regulator and hence has no large momentum divergence then it should also correspondingly not have any short-distance singularity. But this seems to be wrong - hence I guess one is led to thinking that having an exactly 0 beta function has nothing to do with the theory being finite.

Its not clear to me that there is any direct relationship between having short-distance singularities and whether or not momentum space integrals diverge (..which should possibly imply that the partition function also diverges..)(..Like Yuji Tachikawa in the comments points out the simple case that even free boson theory has short-distance singularity but since there are no loop processes in it I guess it doesn't make sense to ask whether momentum space integrals converge but I guess its partition function is not always well-defined..)

Like on page 441, Weinberg in his volume 1 of his QFT books, says in italics that "renormalization of masses and fields has nothing directly to do with the presence of infinities and would be necessary even in a theory in which all momentum space integrals were convergent"

To may be summarize my query - is one saying that there are conceptually different multiple sources of infinity in a QFT like,

  1. divergence of the momentum space integrals
  2. short distance singularity
  3. divergent partition functions
  4. coupling constants being sent to infinity by the beta function

(.I thought of also adding the phenomenon of Landau pole in the above list but I guess that is not so fundamental a property and is only an indication of the failure of the perturbation technique..thought I may be wrong..)

So is there a way to think of these "different" infinities as cause and effect of one another?

Or is it possible that any combination of these can show up in some QFT?

And/How are these related to the property of the beta-function being non-perturbatively 0 or not? (..except for the "by definition" case that for non-perturbatively 0 beta-function (4) can't happen..)

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Please rephrase your question. Even a theory of free bosons $\phi$ has a short distance singularity in its two-point function $\langle \phi(x)\phi(y)\rangle$. –  Yuji Nov 18 '11 at 12:09
    
@Yuji: Is this tow-point function involved into calculations of a free boson field evolution? –  Vladimir Kalitvianski Nov 18 '11 at 21:14
    
@Yuji Thanks for the comment - I reworded some of the sentences. Yes - free boson theory is an example of a QFT where the two-point function diverges and that theory does not need a regulator but I guess its not "finite" since its partition function most probably is not convergent - does that functional determinant always have a finite meaning? Thats the point of my confusion - what or is there a relationship between these properties of (not)having short-distance singularity, (not)having a non-trivial beta-function and (not)having a finite partition function. Would be delighted if you explain –  user6818 Nov 19 '11 at 1:29
    
Agree with Yuji, question too vague. Clearly finiteness does not imply conformal invariance. For instance take a massive free field. All the correlation functions at separated points are finite but the theory is not conformal. Alternatively, take N=4 deformed by a mass term. The UV does not know about the mass and so all correlation functions of all fundamental fields are finite. The theory is clearly not conformal. Also a conformal theory does not have to be finite. One only needs to require that the divergences conspire to be such that they can be eliminated by a field redefinition. –  Zohar Ko Nov 19 '11 at 2:25
    
@Zohar Komargodski Thanks for your reply. I don't know how to make the question precise but I guess experts like you would know as to what is the concept that explains what I am being fuzzy about. Also I don't know what is the correct terminology - BY "finite" I did not mean that correlation functions are finite at finite non-zero separation - I was using that term to mean that the partition function is well-defined - that is it doesn't need a regulator etc. –  user6818 Nov 19 '11 at 3:08
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1 Answer 1

This is a rather subtle question as also at classical level the parameters of the theory could turn out to be redefined in a finite way. But, as far as you may know, we are very good with free theories and these mostly happen as fixed points for known theories. If you want an example you can take a look to the scalar field theory. You can consider a standard action like

$$S=\int d^4x\left[\frac{1}{2}(\partial\phi)^2-\frac{\lambda}{4}\phi^4\right]$$

This theory is trivial and this means that it reaches a trivial fixed point both at ultraviolet and infrared that makes it useless to describe physics unless some cut-off is introduced explicitly somewhere. But in the infrared you will get a beta function going like

$$\beta(\lambda)=4\lambda+\frac{c_1}{\sqrt{\lambda}}+O(1/\lambda)$$

and so, if you have a starting coupling $\lambda=\lambda_0$ you will get a running coupling going to zero like $p^4$ lowering momenta. The theory becomes free but these free excitations are all massive with a mass proportional to $\lambda_0^\frac{1}{4}$ as can also be seen from lattice computations. You can see from this that, notwithstanding we are coping with a trivial fixed point, all the parameters of the theory turn out to be properly redefined and in a finite way!

The meaning of this is that renormalization just expresses a physical property of a quantum theory: The simple fact that interaction changes all the parameters of a given theory when the couplings are turned on. But a trace of this can be found in the fixed points of the theory itself.

Now, if you look at the classical theory, you will be able to solve it exactly but the solutions have not finite energy unless you work with a finite volume or redefine the coupling $\lambda$, exactly as happens to the quantum theory. Also in this case you will get a mass even if you started from a massless theory and your coupling will run.

Again, we see that the effect of the interaction, the fact that the coupling $\lambda$ is not zero, is exactly to modify all the parameters of a theory.

As you see, this is true independently from the fact that you are coping with infinities or not.

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