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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?

SRS
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2 Answers2

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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).

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.

AccidentalFourierTransform
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  • @AFT You said, "in this (Wilsonian) picture one must include all terms that are compatible with some set of symmetries,...". Lets take $\phi^4$ theory. To start with, one keeps only terms whose mass dimensions are $\leq 4$ and doesn't include, for example, $\phi^6$ or $\phi^2(\partial_\mu\phi)^2$ etc to start with. However, these terms are generated after integrating out high-momentum modes. For example, Peskin and Schroeder Eqn. 12.3. Why is that? Why not write all possible terms (allowed by symmetry) in the starting Lagrangian itself? – SRS Mar 01 '17 at 08:45
  • @AFT Do they implicitly assume that the starting Lagrangian holds for an energy scale where higher-dimensional operators are irrelevant? – SRS Mar 01 '17 at 08:45
  • @AFT In what sense $\phi^4$−theory is renormalizable but $\phi^6$ is not (in Wilsonian picture)? Is it that the number of relevant interactions are infinite in this case (unlike $\phi^4$-theory)? – SRS Mar 01 '17 at 10:06
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    @SRS 1. P&S's introduction to the Wilsonian interpretation of QFT is a pedagogical simplification (which only works for perturbatively renormalisable and UV complete theories. The most general case does not fit in their treatment). In reality, one does not starts with only terms whose mass dimensions are $\le 4$. One starts with a finite cutoff, and all possible terms. Integrating out UV modes changes the coefficients of the already existing terms, it does not generates them from nothing. But in order for the whole program to work, you must include all terms from the very beginning. – AccidentalFourierTransform Mar 01 '17 at 14:50
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    @SRS 2. I guess that they start from a Lagrangian where the infinite cutoff limit is already taken (which only works for UV complete theories - and this might not be the case for $\phi^4_4$, I'm not sure off the top of my head). As $\Lambda$ has been removed, all irrelevant operators effectively disappear. But the general case is that the cutoff cannot be removed (the theory is not UV complete), and therefore their treatment doesn't work. Rather, one must include all terms from the very beginning. – AccidentalFourierTransform Mar 01 '17 at 14:56
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    @SRS 3. In the Wilsonian picture, "renormalisable" is not a concept. $\phi$ is not renormalisable and $\phi^6$ is not non-renormalisable. Recall that in the general case the cutoff is finite, and therefore all theories are finite: the interaction $\phi^6$ is just as valid as $\phi^4$. On the other hand, in the Wilsonian picture interactions are classified according to whether they are relevant or irrelevant; and, as I mention in my post, both $\phi^4_4$ and $\phi^6_4$ are irrelevant. – AccidentalFourierTransform Mar 01 '17 at 15:00
  • @SRS See 0909.0859, by Hollowood. – AccidentalFourierTransform Jan 03 '18 at 09:28
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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.

CStarAlgebra
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