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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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2 Answers2

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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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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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