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How it could be proven that a non-trivial theory cannot be both asymptotically free and IR free (g=0 both in the UV and IR with some interpolating function in between)? This is of course contrary to the behaviour of both QED and QCD in which we have monotonically RG flow.

Yair
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If I understand what you're asking, it's false: there are plenty of examples of theories that are asymptotically free and also weakly coupled in the IR. A QCD-like theory with more flavors of quarks would be an example. The phrase to search for is "Banks-Zaks fixed point."

For the revised version of the question: there are certainly RG flows that are free in both the UV and the IR. The simplest is Yang-Mills theory, or QCD with massive quarks: there is a mass gap, so the theory is trivial in the IR (no particles at all). But that seems like a "cheat"; you probably mean a free theory that has actual particles.

In supersymmetric QCD, there are examples of theories in a "free magnetic phase": the UV description is a free QCD-like theory, and so is the IR description, but the gluons in the IR are not the same as the gluons in the UV.

If you want the coupling g to mean the same thing in the UV and in the IR, then I don't know any examples that would do what you want.

Matt Reece
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  • Thanks Matt. Actually I intended to ask a little different question. Can you please comment on the modified question above please? – Yair Jan 16 '14 at 06:32
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    Banks-Zaks is weakly coupled but never free, I believe. You'd need the first coefficient in the beta function to vanish exactly to get that. – Vibert Jan 16 '14 at 15:40
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I think that in some scenarii of the IR limit in Pure Yang-Mills SU(3) (QCD without fermions), the theory is also gaussian (trivial) in the IR (of course, one has to do more than the usual perturbative approach), and thus realize what you are looking for. See for example PhysRevD 84, 045018.

Adam
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  • A similar issue (you might mean the same thing): QCD (with massive quarks) becomes in the IR a theory of pions, and once you flow lower than the lightest pion, you have a empty = free theory. – Vibert Jan 16 '14 at 15:39
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    @Vibert: I don't mean the same thing. I'm talking about QCD without fermions. Furthermore, when there is a mass in the problem, the theory won't be free in the IR, as the mass stop the flow. For instance, QED is not asymptotically free for massive electrons. – Adam Jan 16 '14 at 20:26
  • People usually refer to $SU(N)$ gauge theory without matter as (pure) Yang-Mills, but that's okay. QED isn't empty in the IR because you cannot integrate out the photon, which remains exactly massless. But in Yang-Mills or QCD there are no gluon states left in the IR. I don't understand what you mean with 'mass stopping the RG flow'. – Vibert Jan 16 '14 at 21:14
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    @Vibert: You're right, I changed the terminology. What I mean about QED is that if the electron was massless, the theory would be asymptotically free in the IR ($\alpha(\mu)\to0$ for $\mu\to0$). Because of the mass, the flow stops for $\mu<m$, and the theory is not free in the IR. In a Wilsonian point of view, at low energy the mass of the electron in the loops stops the RG flow. – Adam Jan 16 '14 at 21:21