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  "finite" QFTs and short-distance singularities and vanishing beta functions

+ 7 like - 0 dislike

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..)

This post has been migrated from (A51.SE)
asked Nov 18, 2011 in Theoretical Physics by Anirbit (585 points) [ no revision ]
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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 $\langle \phi(x)\phi(y)\rangle$.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
@Zohar May be you can elaborate more on the connection between being conformal (non-perturbatively 0 beta function) and the 4 "different" kinds of singularities I listed.

This post has been migrated from (A51.SE)
@Anirbit , even in free field theory there's a loop integral which diverges: i.e. a one-loop diagram without any vertex. This corresponds to the zero-point energy of free oscillators, which makes the partition function diverge in the UV. So, your points 1,2,3 already apply to the theory of one free boson, and are always there.

This post has been migrated from (A51.SE)
Most recent comments show all comments
The partition function would always be volume divergent in non-compact space. In compact space it could still be UV divergent, but sometimes the UV divergence could cancel (for example if there is SUSY). As I tried to suggest, none of this has anything to do with the beta function.

This post has been migrated from (A51.SE)
@Zohar That is quite an exciting statement that you make that the beta function being non-perturbatively 0 has no effect on possible divergences of the partition function or correlation functions. Can you kindly give examples or references for SUSY QFTs where the UV divergence of the partition functions cancel? And are you saying that nothing can save the divergence on non-compact space-times? What do we know about the partition function or 2-point correlations of N=4 SYM which has a non-perturbatively 0 beta function?

This post has been migrated from (A51.SE)

1 Answer

+ 1 like - 0 dislike

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


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.

This post has been migrated from (A51.SE)
answered Nov 20, 2011 by JonLester (345 points) [ no revision ]

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