I have am action $S_0$ valid up to energy scale $\Lambda_0$ with renormalisable terms.

I want to study the EFT at a lower scale $\Lambda \ll \Lambda_0$, by using the Wilsonian RG. It will give me an effective action $S^\Lambda$ with an infinite number of terms, renormalisable and not.

In which terms is this "Effective Action" different from the general effective action I can write down just by adding terms preserving the symmetry of the starting action?

[edit:] and in particular

  1. are terms generated by RG flow ALL the ones possible respecting the symmetry of the action $S_0$, or just a subset of all the (theoretically) symmetry-respecting terms? Can WRG "miss" some terms?

  2. Apart from anomalies, can one get a Wilsonian effective action $S^\Lambda$ with a different (enhanced or reduced) symmetry wrt the starting $S_0$?

With respect to these questions:

  1. How is the Coleman-Weinberg potential understood in the Wilsonian RG flow?

1 Answer 1


The terms generated by the RG will all respect the symmetries of the microscopic action (though one have to be careful with anomalies). That's why people tend to directly write a low energy effective action, and do not bother calculating the RG flow of the parameters.

However, this implies that you do not know how the effective parameters relates to the microscopic ones (this information can be interesting depending on the problem). For example, if one studies the Ising model (say, on a cubic lattice), and is just interested in the long wavelength physics, one can do calculation starting with a low energy effective action, that is, a scalar $\phi^4$ field theory. But then, one cannot know how to related the real correlation length, or the real magnetization, to the effective ones computed from the effective theory. Neither can one compute the critical temperature of the model. (The only information that one can infer from the effective theory, if we don't know how to relate the effective parameters to the microscopic ones, is the universal features close to a 2nd order phase transition.)

But, at least in principle, one could do a RG calculation to relate the microscopic physics to the macroscale. In fact, in the example above, the perturbative Wilsonian RG fails (because the problem is strongly coupled), but one can use non-perturbative approximations to do exactly that.

EDIT Concerning the additional questions.

1) Generically$^*$, all terms are generated (this is easy to see, if one draws the diagrams that can contribute for a given term). However, their value will depend on where the flow is going. If it goes toward the Gaussian fixed point, all terms but the quadratic ones will be small, of order $(\Lambda/\Lambda_0)^\Delta$, where $\Delta$ is the canonical dimension of the term. If it flows somewhere else, all terms can be large.

2) The low energy action can have additional (emergent) symmetries, but not less symmetry$^{**}$. For instance, at a critical point, a lattice model will have a conformal invariance at long wavelength. Or in the context of the Mott transition in the Bose-Hubbard model, the special particle-hole symmetric point have an emergent relativistic symmetry (whereas the microscopic model only has a Galilean invariance).

3) To me, the best way to understand the flow of the effective potential is to learn about the non-perturbative/functional version of the RG, where one write down a flow equation for the full potential, and one is able to treat all couplings (even the ``non-renormalizable'' ones). For an introduction, see arXiv:0702.365.

$^*$ However, some models with specific symmetries (for example SUSY) can be protected by some non-renormalization theorem, implying that the effective potential won't be renormalized. Hence, the generic terms one would expect are not present in the low-energy effective action.

$^{**}$ This is in fact true in most cases, but not always. In some specific model, it might happen that a symmetry is broken dynamically during the flow, for example if a singularity of the effective potential is generated along the RG trajectory (this implies doing some functional RG). In these rare cases, while the effective potential had a symmetry at high energy, it loses at low energy. For an example of this, in the context of the random field Ising model (where a (non-dynamical) supersymmetry is lost during the flow, see arXiv:1103.4812.

  • $\begingroup$ @Adam Actually, not every term allowed by the symmetries is necessarily generated by the RG. It happens in special cases but it does happen. For example, in Susy if you don't write terms in the superpotential they won't be generated by the RG flow even if they were allowed by the symmetries. $\endgroup$
    – TwoBs
    Oct 3, 2015 at 11:21
  • $\begingroup$ @TwoBs: do you have a specific model in mind (with a reference), I would love to see how it works in this case. How come there is no diagram that can generate these terms ? As always, when trying to answer general questions, there are usually no general answer that cover all cases. That's why I used "Generically" ;-) But when I understand how it works in your example, I'll add another asterisk... $\endgroup$
    – Adam
    Oct 3, 2015 at 11:26
  • $\begingroup$ I don't have a specific example, since every susy N=1 model has this property. See e.g. Chapter 3 of these lectures arxiv.org/abs/0907.0039v1 (there there is a nice simple example, too). In the case of susy, the non-generation of allowed terms boils down to the holomorphy of the potential. But there exist other non-susy theories where the non- renormalization takes place (the Galileon models are one such example). It's certainly true that it is not a generic feature, the theory must be somehow special, I just wanted to remark this because the OP was emphasizing the 'all' in the question $\endgroup$
    – TwoBs
    Oct 3, 2015 at 12:22
  • $\begingroup$ @TwoBs: if I understand correctly, it is because one starts with an holomorphic potential that this is the case (question : can one start with a SUSY non-holomorphic potential ?). But this can be seen as a property of the potential (some kind of symmetry) that is not broken by the RG, right ? In the same sense that if I only have O(N) invariant terms in the initial Lagrangian, the flow will stay in this subspace. $\endgroup$
    – Adam
    Oct 3, 2015 at 12:36
  • $\begingroup$ @Adam It's not a symmetry that protects it, it is the analytic structure associated with susy (super)potentials (and yes, they must be holomorphic). It's definitely not like the case of an O(N) symmetry. To see the difference starts with a superpotential $W=\phi^3$. It's holomorphic in $\phi$, and yet $\phi^2$, that is also holomorphic, is not going to be generated! For the O(n) instead, there is nothing that would prevents the generation of other O(n) symmetric terms in the lagrangian (from $\mathcal{L}=\lambda(\phi^a\phi^a)^2+..$, the RG flow generate a term $\phi^a \phi^a$ that is allowed). $\endgroup$
    – TwoBs
    Oct 3, 2015 at 12:58

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