# Simple (but wrong) argument for the generality of positive beta-functions

In the introduction (page 5) of Supersymmetry and String Theory: Beyond the Standard Model by Michael Dine (Amazon, Google), he says

(Traditionally it was known that) the interactions of particles typically became stronger as the energies and momentum transfers grew. This is the case, for example, in quantum electrodynamics, and a simple quantum mechanical argument, based on unitarity and relativity, would seem to suggest it is general.

Of course, he then goes on to talk about Yang-Mills theory and the discovery of negative beta-functions and asymptotic freedom. But it is the mention of the simple but wrong argument that caught my attention.

So, does anyone know what this simple argument is? And how is it wrong?

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I think this is very appropriate here, and in fact this is exactly the kind of question I'd like to see more of (advanced but not research; should have a well-defined answer). But... I don't know the answer. –  wsc Jan 13 '12 at 14:42
@Ron: Thanks for the comments Ron. I think I might put down a bounty on this question to try to get a more explicit answer. –  Simon Feb 19 '12 at 9:44
@Simon: I'll do it. –  Ron Maimon Feb 20 '12 at 13:21
I wrote to Michael Dine. It's an argument about the spectral representation but not what anyone has said so far. He says he'll forward the details a few days from now. –  Mitchell Porter Feb 21 '12 at 5:40
Ron: Thanks for pointing your rep points on the line (I was going to wait until the weekend). I hope that @Mitchell gets a reply soon. –  Simon Feb 25 '12 at 5:11

Michael Dine's response, quoted with permission:

I now have to think back, but the argument in QED is based on the spectral representation ("Kallen-Lehman representation"). The argument purports to show that the wave function renormalization for the photon is less than one (this you can find, for example, in the old textbook of Bjorken and Drell, second volume; it also can be inferred from the discussion of the spectral function in Peskin and Schroder). This is enough, in gauge theories, to show that the coupling gets stronger at short distances. The problem is that the spectral function argument assumes unitarity, which is not manifest in a covariant treatment of the gauge theory (and not meaningful for off-shell quantities). In non-covariant gauges, unitarity is manifest, but not Lorentz invariance, so the photon (gluon) renormalization is more complicated. In particular, the Coulomb part of the gluon ($A^0$) is not a normal propagating field.

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Thanks Mitchell, I forgot about this until Manishearth reminded me in the comments above. It would be nice if the answer was more detailed, but I'm happy with Michael Dine's response. (Your bounty will arrive in ~24 hours) –  Simon Apr 11 '12 at 7:04

This is a temporary answer in order to store the generous bounty that Ron offered. When a proper answer to this question is given, I will transfer the 500 rep points (assign an equal bounty) to that answer.

Going by the totalitarian principle of quantum mechanics / quantum field theory, since this move is not explicitly forbidden, it must be compulsory.

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What an answer! Full bounty! –  Ron Maimon Feb 27 '12 at 14:47

I may be wrong, but the following remark at http://alumnus.caltech.edu/~callaway/trivpurs.pdf (PHYSICS REPORTS 167, No. 5 (1988) 241—320) may be relevant: "Then [2.1] it follows that if $[x, y] = \triangle_+(y-x)$ possesses properties implied by the Garding—Wightman axioms, then... the associated field theory is a generalized free field, i.e., it is a trivial theory." Reference [2.1] there is B. Simon, The $P(\varphi)^2$ Euclidean (Quantum) Field Theory (Princeton, 1974).

By the way, you can just ask the author, M. Dine. Sometimes asking the author is the only way to sort out what (s)he wrote:-). I remember I found a book containing a result I had recently obtained myself, but without any proof. I e-mailed one of the two authors of the book and asked for the relevant reference. It took me a couple of months, but eventually he advised me that they meant something different from the result that I obtained, so my result was new:-)

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This is not an accurate statement of the remark. The theorem is that if the commutator is equal to the free commutator than the theory is free. This is both a trivial result (made to sound nontrivial) and unrelated to the beta function properties. The rigorous literature is superficially impressive in this field, but all of it is essentially less than worthless. But no downvote--- I appreciate the sincere effort. –  Ron Maimon Feb 21 '12 at 1:05
@Ron Maimon: Could you please explain that? Do you think the statement in the Callaway's article is wrong? Because I don't see how the assumption in the statement is "if the commutator is equal to the free commutator". As for relevance to the beta function properties, I guess if the beta function is positive, the charge is screened, and the theory is non-interacting. –  akhmeteli Feb 21 '12 at 3:01
Apologies, I wrote the first comment without reading the paper you linked. The paper is not rigorous nonsense, it's a good review of triviality in scalar field theory, with a rigorous subsection. But the result you quote is just saying that if a theory obeys Wicks theorem then it's a free field theory. The result I thought you were talking about is a somewhat deeper: if the two point function is exactly free, than the theory is free. Neither have anything obvious to do with the beta-function, but they are positive spectral weight results, and so they are Kallan-isms. –  Ron Maimon Feb 21 '12 at 15:04
just because the beta-function is positive doesn't mean that the theory is trivial. The beta-function could be positive with a strong coupling fixed point, or even a fixed point at relatively weak coupling. Not every growing coupling has to blow up to infinity. Its a different question. But at least now I know what you were thinking, it makes sense why you would give this answer. –  Ron Maimon Feb 21 '12 at 15:06