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  Why is this an absolutely necessary condition for power-counting nonrenormalizable theory to make sense perturbatively?

+ 5 like - 0 dislike

We learn from Wilson and Weinberg etc. that there's no holy reason to exclude interactions with negative-mass-dimension coupling while probing physics. However, according to Weinberg, this doesn't mean possibilities can go completely wild, in page 2 of this article(and many other places including his biblicalized textbooks) he stressed:

The second, ‘modern,’ sense in which a theory may be said to be renormalizable is that the infinities from loop graphs are constrained by the symmetries of the bare action in such a way that there is a counterterm available to absorb every infinity. Unlike the Dyson criterion, this condition is absolutely necessary for a theory to make sense perturbatively. 

(I should make it clear that here Weinberg is allowing effective theories with infinitely many counterterms)

I simply don't get why this is absolutely necessary. Isn't it quite conceivable that maybe a UV completed theory simply doesn't possess as many symmetries as its low-energy incarnation? That the symmetries of the effective theory are only emergent? With this understanding, even if there are infinities not constrained by the symmetry of the original action, can't we just use counterterms that do not have the symmetry either? 

asked Nov 26, 2014 in Theoretical Physics by Jia Yiyang (2,640 points) [ revision history ]
edited Nov 26, 2014 by Jia Yiyang

As I understand it, even in the Wilsonian picture a theory should not have infinitely many operators with negative mass dimension, that would need to be fixed by counter terms to be renormalizable. If for example in the UV infinitely many counterterms are needed to fix the infinities, the theory is no longer predictive because by putting in an infinite amount of information you can learn nothing new from it. If this is the case, string theory for example can save the day because it has infinite towers of excitations.

The concept of non-renormalizability of a theory somehow somehow reminds me of the concept of an "essential singularity" in complex analysis, which is a singularity that would produce a Laurent series with an infinite number of terms in the principle (negative exponent) part. So maybe one could say a theory should have no "essential singularity" to be renormalizable?

@Dilaton, The UV complete theory should have only finite number of coupling constants, that there should be no question. That's why I take Weinberg's word as referring to a low energy effective theory with infinitely counterterms. I just don't get why we are not allowed to sacrifice some symmetries for the sake of canceling infinities.

1 Answer

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If a divergent term cannot be absorbed by a corresponding divergent counterterm, the renormalization prescription doesn't make sense, as the results will not be finite. Therefore it is absolutely necessary for a theory to make sense perturbatively that the required counterterms exist. This is the content of Weinberg's statement.

The symmetries play only a subsidiary role as they affect the possibilities - depending on which symmetry group one assumes, a different set of loop graphs and interaction terms must be considered. Thus ''that the infinities from loop graphs are constrained by the symmetries of the bare action in such a way that there is a counterterm available to absorb every infinity'' just means that at least for the loop graphs allowed by some assumed symmetries, each divergent term can be absorbed by a corresponding divergent counterterm. This doesn't say which symmetries one is imposing; Weinberg's wording allows the symmetry group used to be smaller than the full symmetry group. But this would allow more loop graphs, hence possibly more infinities, and hence possibly infinities that cannot be cancelled. (And conversely, imposing more symmetry allows fewer possible counterterms, hence possibly infinites that can no longer be cancelled.)

The number of counterterms needed (finite or infinite) is independent of this absolute necessity. It just means that while any in the old sense renormalizable theory describes a family of QFTs parameterized by a few numbers (renormalized coupling constants), any in the old sense non-renormalizable theory describes a family of QFTs parameterized by infinitely many numbers. Thus the non-renormalizable case defines a far bigger, and therefore less predictive, family of theories.

But this is not really a handicap. We all know that power series (which need infinitely many coefficients to be well-defined) define a much larger class of functions than low degree polynomials (where a few parameters suffice), but no physicist takes this as a reason for restricting attention to functions that are low degree polynomials.
answered Nov 26, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
edited Nov 26, 2014 by Arnold Neumaier

Thank you! I actually briefly thought about the possibility that, quoting you

Weinberg's wording allows the symmetry group used to be smaller than the full symmetry group. But this would allow more loop graphs, hence possibly more infinities, and hence possibly infinities that cannot be cancelled.

But I couldn't quite conceive what kind of divergence can't be cancelled by some counterterms if we don't impose symmetry condition on the counterterms at all. Do you have a concrete example?

There is no example, since Weinberg says on the same page that

It is automatically satisfied if the only limitations imposed on the terms in the bare action arise from global, linearly realized symmetries. 

 The point is that Weinberg wants to impose (nonlinearly realized or gauge) symmetries. But then one has to be careful to make sure that it is still possible to cancel every divergence. And in the last sentence of p.3 he states that there is indeed a cohomological condition that must be satisfied to achieve the wanted cancellation.

Ok. Indeed the context can reasonably be interpreted as "Given a prescribed (possibly infinite) list of interactions, does the list all by itself define a perturbative theory.", if I'm not misinterpreting you. Then indeed everything makes sense. I'll take it this way until something doesn't make sense shows up:-).

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