# Symmetries in Wilsonian RG

• I wanted to know if there is a theorem that in writing a Lagrangian if one missed out a term which preserves the (Lie?) symmetry of the other terms and is also marginal then that will necessarily be generated by RG flow to the IR.

(..like a massless scalar field Lagrangian in 3+1 will generate the mass term as a counter term..though there are cases like Gross-Neveu (GN) where where the discrete symmetry is broken dynamically and not in perturbative RG and so there are 2 in equivalent QFTs corresponding to the GN Lagrangian but the relationship between the two non-perturbative vacua re-realizes the discrete symmetry..)

• Can one make a statement along these lines in Ward identities? Like what can be in general be said about the fate in the IR of the Ward identities corresponding to the UV symmetries?

• Also if nothing else prohibits is RG flow towards the IR also guaranteed to produce all those terms which are higher dimensional and preserve the symmetries of the UV theory?

Will any such terms necessarily come in the form $O/\Lambda^n$ where $\Lambda$ is the UV cut-off and $O$ a higher dimension operator and "n" is such that the whole term has the marginal dimension. (...and then I can interpret that these effective terms vanish in the UV when I take the cut-off to infinity..)

• And how much of the answers to the above remain if one is doing non-relativistic theory?

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For starters, I don't think this has anything to do with relativity. The same principles should hold in 3d non-reltivistic systems and 4d lorentz invariant theories.

I don't know if my answer will satisfy you and I'm not fully sure about the details, but I'll think out loud.

• Supose the RG flow generated terms which were incompatible with some symmetry. Then, such a term would break the Ward identity (conservation rule for the symm), i.e. it will violate some property of the "more fundamental" theory. So such terms should not be generated by RG flow.

• When you start out without a certain term in your Lagrangian, you are assuming that it is negligible to some approximation. It is not exactly zero, but probably has some small value. In that case, if it a relevant operator in the IR, it's value will increase exponentially (based on it's canonical scaling dimension). By some chance, RG flow might make it zero at that particular energy scale, but that value need not be a fixed point of the RG flow. With the path integral perspective in mind, we can say that any process which is not banned must occur with some small probability. These will add other loop (quantum) corrections to the beta function, which will generically not be propotional to the coupling itself. Such terms will make additive (and not multiplicative) corrections to the oupling, and hence push it off zero even if you happened to start at zero. Then, the exponential increase due to the scaling dimension will take care of increasing it to some significant value.

Your picture of the terms being of the form $O/\Lambda^n$ seems right to me.

Marginal couplings might be a little bit more subtle, since classically they should not be renormalized. But quantum (loop) corrections will make the coupling run logarithmically with the scale. This is called dimensional transmutation.

I would like to mention a related example of custodial symmetry. In condensed matter systems, the UV physics is typically described by a theory on a lattice. Then, the UV theory only has a discrete subgroup of the rotation group as a symmetry. But in the IR, we get a nice effective field theory with full rotational symmetry! To understand this, let's look at the picture in momentum space. For the UV theory, the (pseudo)momentum can take values within, say a cube (assuming our lattice to be cubic). With RG flow to the IR, all operators of dimension $p^4$ or higher are irrelevant and become negligible. So, only the operator with two derivatives ($p^2$) survives in the IR, giving us full rotational symmetry. However, if we didn't have the cubic lattice in the UV, we might have had terms breaking rotational symmetry, which become relevant in the IR. So, a (discrete) symmetry in the UV protects an ehanced symmetry in the IR. This phenomenon is aptly called custodial symmetry. The enhanced IR symmetry may not be exact, but terms breaking that symmetry are suppressed.

Note that the above example is in line with the principles raised in your question and doesn't say anything different. I just found it to be a very cool example, so I hope you indulge me.

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I am a bit wary of thinking that somehow RG flow is guaranteed to produce all terms as you say and as I said earlier. I mean it is true that a free massless scalar field theory will remain so forever. But sometimes the Ward identities can be violated - thats anomalies! – user6818 Apr 1 '13 at 16:31
Also this interpretation about "dimensional transmuation" seems new to me - is it true that it is generic for any marginal coupling? - In QCD you calculate the 1-loop beta function and try to integrate it to get the flow of the coupling and in the answer you see that its possible to tradeoff the bare-coupling and the renormalization scale in terms of a fictitious mass scale - that it requires a logarithm is not obvious a priori- thats lay man's dimensional transmutation. (..though Gross-Nevue does dimensional transmutation non-perturbatively!..) – user6818 Apr 1 '13 at 16:33
If your bare coupling doesn't have a natural scale, then the corresponding operator has the canonical dimension as the spacetime dimension. That would mean that either it doesn't get renormalized (and is a fixed point of the RG flow), or it grows logarithmically in magnitude. That sounds like what one does in QCD. (If the coupling did not get renormalized, then you would be at the fixed point and your theory would be conformal.) I don't know about the Gross-Neveu model. I'll try to have a look at those two calculations and add a comment soon. – Siva Apr 1 '13 at 18:10
In response to your first comment: 1. Free theories are not good prototypes and also quite uninteresting. Interacting theories are very different qualitatively. 2. If you have an anomaly, then you never really had a symmetry, so restrictions on allowed operators might be removed. – Siva Apr 2 '13 at 3:03
@Sia (1) In an anomalous scenario the symmetry exists classically but is broken by QFT. The hard aspect is that if anomaly doesn't exist in one energy scale then it probably doesn't exist anywhere! (apparently some 'tHooft's theorem) - massless interacting theories are the good examples of anomaly - the classical conformal invariance is broken by QFT. – user6818 Apr 2 '13 at 19:04