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Is every QFT non-local in the U.V.?

+ 1 like - 0 dislike
41 views

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...)?

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot
asked May 29, 2012 in Theoretical Physics by toot (445 points) [ no revision ]
retagged May 1, 2014

1 Answer

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user Luboš Motl
answered May 29, 2012 by Luboš Motl (10,248 points) [ no revision ]
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").

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot
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?

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot
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.

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user Luboš Motl

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