Wilsonian picture of renormalization and $\phi^4-$theory

Question

What is the criterion of renormalizability in the Wilsonian picture? In this picture, why is a theory with $\phi^4$ interactions renormalizable and a theory with $\phi^6$ interaction is non-renormalizable?

Answer 1

There is no renormalisable vs. non-renormalisable dichotomy in the Wilsonian interpretation of QFT. In fact, in this picture one must include all terms that are compatible with some set of symmetries, and all coupling constants typically run. All interactions are valid, and there is no fundamental criterion to reject some over the rest.

On the other hand, in the Wilsonian picture interactions are classified according to their scaling dimension, which is given by their mass-dimension plus a certain correction, the anomalous dimension (which is sometimes misleadingly called the "quantum" contribution; more properly, it should be called the "non-linear" contribution). As you already know, the coupling constants of the operators with positive scaling dimension decrease as you integrate out UV degrees of freedom, and as such, they "disappear" from the Lagrangian if you look at your theory in the IR. They are irrelevant, because their contribution to low-energy experiments is highly suppressed. By contrast, the coupling constants with negative (or zero) scaling dimension do not flow towards the fixed-points (if any) of the theory, and so they are relevant: they do affect low-energy physics.

If it were not for of the anomalous contribution, the scaling dimension would agree with the mass dimension of the operators, and therefore the irrelevant operators would agree with the non-renormalisable interactions. Nevertheless, if the theory is perturbative then the anomalous dimension is very small and it does not change the sign of the scaling dimension. In this sense, for perturbative theories irrelevant operators are non-renormalisable, and relevant operators are renormalisable. The Wilsonian classification agrees with the power-counting renormalisability classification (for non-perturbative theories, the power-counting renormalisability classification is meaningless anyway).

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Note that the operator $\phi^6_4$ is irrelevant in the Wilsonian sense, while $\phi^4_4$ is naïvely marginal (due to its mass-dimension). Taking into account the anomalous contribution, this operator turns out to be irrelevant too. The theory is, in a sense, free, even though in perturbation theory it seems to make sense as an interacting theory. See 0806.0789 for more details.

Answer 2

Let's say your lagrangian has the interaction term $g\phi^6$. In d = 4 dimensions the lagrangian needs to have a mass dimension of 4 i.e. $[L] = m^4$. Now $\phi^6$ has mass dimension $m^6$, so the coupling constant g must have mass dimension $m^{-2}$. Now if you put a cutoff of $\Lambda$ in your theory, its not hard to convince yourself that scattering amplitudes, which are dimensionless, must go like $\frac{\Lambda^2}{g}$ such that the amplitude is dimensionless. As you take the cutoff $\Lambda$ to infinity, this amplitude diverges quadratically.

Contrast this to $\lambda \phi^4$ theory, where it's easy to see the coupling consant is dimensionless, so no powers of the cutoff appear in your scattering amplitudes. This theory is renormalizable.