Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As much as I understand the renormalization group transformation and the concept of relevant/irrelevant operators, I'd say that if we push the reasoning of only looking at relevant operators when we follow a "renormalization group orbit" (a curve in the space of the parameters) to the infra-red regime the other way, so going to the U.V., we should keep all the operators this time (as they are no more suppressed by huge factors), or there is a infinite number of them, depending on an infinite number of the derivatives, and we end up with a non-local theory. As this reasoning can be applied to every QFT, does this mean that they are all non-local in the U.V.? or the reasoning fooled somewhere (breakdown of the QFT framework somewhere for example...)?

share|cite|improve this question
up vote 3 down vote accepted

Quite on the contrary, every genuine and genuinely consistent QFT is exactly local in the UV; it must converge to a scale-invariant theory, a fixed point.

One may obtain nonlocalities in the IR for most theories (through the higher-derivative corrections with arbitrarily many derivatives) – if we compute the effective field theory description of the dynamics, either directly from the exact UV starting point, or otherwise.

share|cite|improve this answer
Yes but if the theory is not renormalizable, it shouldn't converge to any fixed point? (maybe that's what you ruled out by saying "genuinely consistent"). – toot May 29 '12 at 13:14
Ok I think I just got that you may mean asymptotically free as genuinely consistent, in that case I get that you should have a vanishing $\beta$ function and conformal invariance. Or am I wrong again? – toot May 29 '12 at 13:53
Dear toot, if a theory is nonrenormalizable, then it doesn't exist as a theory at very short distances. In fact, it doesn't exist at any distances shorter than the cutoff distance - which is usually of the same order as the characteristic length scales in this theory. ... I didn't say it has to be asymptotically free; it must be a fixed point but the fixed point may be interacting, not necessarily free. Yes, it should be conformal, that's what "fixed point" means, and the vanishing of all beta functions follows from the conformality. – Luboš Motl May 29 '12 at 16:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.