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Why do we need the large N expansion?

+ 3 like - 0 dislike
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People say that QCD is a zero-parameter theory because the coupling constant g is changing with the RG flow. And to do a first approximation, we should introduce a parameter to expand in. The parameter turns out to be the N in SU(N). But similar arguments can be given for ϕ4 theory where the coupling λ is also a RG flow dependent quantity. Then why do we naturally and comfortably expand in λ there? And even after we introduce the large N expansion, we still do the analysis with the help of feynman diagrams which means that we already use the perturbation theory. So does it mean that the argument 'large N limit is introduced because QCD is a zero-parameter theory' incorrect and actually it is introduced simply for simplifying the calculations?

asked Jun 20, 2016 in Theoretical Physics by Wein Eld (195 points) [ no revision ]

1 Answer

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You can expand confortably in a parameter when this parameter is small. Even if the coupling constant depends on the energy scale, one can expand in terms of the coupling constant at a fixed energy scale. This is a good thing to do only if this number is small. In $\phi^4$-theory (in four dimensions), the coupling is small in the IR but big in the UV: perturbation theory will tell you nothing interesting in the UV (and it is quite likely that $\phi^4$ theory does not exist in the UV). In QCD, the coupling constant is small in the UV and perturbation theory in this regime can be done and is done and is extremely ueful in this regime. It just becomes inadequate to understand the IR physics.  Large $N$ limit is another perturbative scheme which is valid in a different range of parameters of the theory: it is useful for some questions, not for others.

answered Jun 21, 2016 by 40227 (4,660 points) [ no revision ]

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