# Do we "get rid of" irrelevant operators in as we consider low-energy effective theory?

On reading the section 12.1 of Peskin and Schroeder's (P&S) QFT on "Wilson's Approach to Renormalization Theory", I gathered the impression that in the Wilsonian approach, one starts by analyzing a Lagrangian containing all possible terms compatible with some given symmetry and having mass dimension $\leq 4$. For example, P&S have considered the $\phi^4-$theory and discarded all other terms such as $\phi^6$, $\phi^2(\partial_\mu\phi)^2$ etc in the starting Lagrangian.

However, according to this answer by AccidentalFourierTransform (AFT) and the following comments, in general, P&S's treatment is a pedagogical simplification. The answer says that one starts with a Lagrangian containing all terms compatible with some given symmetry and not just those with mass dimension $\leq 4$. In this scenario, on integrating out high momentum modes, the irrelevant operators contribute insignificantly and effectively disappear from the low-energy effective Lagrangian.

But as I understand, P&S shows the opposite. They show as we integrate high momentum modes, we generate irrelevant terms. P&S's result is also compatible with the fact that Fermi theory is a low-energy theory containing an irrelevant operator. Therefore, we do not "get rid of" irrelevant operators in the low-energy theory. In some sense, we generate them in the low-energy version of the Standard Model.

*How do I reconcile the result of Peskin and Schroeder with the answer of AFT? Following P&S, if we say non-renormalizable operators are "generated" (was not there to start with) in the low energy theory, aren't we saying that non-renormalizable operators become important in the low-energy approximation? But that's wrong.

The answer here by ACuriousMind also says something similar to P&S. This answer also says that non-renormalizable operators are thought to be absent in the original Lagrangian, and they appear when the cut-off is lowered.

• 'Irrelevant' operators do not cease to exist as energy is lowered. They are 'irrelevant' in the sense that they do not affect the scaling dimension of operators in the ground state. The coarse-graining procedure will generate all possible operators that are consistent with the symmetries of the initial (or 'bare') action (assuming that the coarse graining does not break any of symmetries of the bare action).
– vik
May 19, 2019 at 20:11
• Mar 1 at 22:04

There are three parts to the process of getting an effective Lagrangian which are often conflated in discussion. Hopefully, seeing them separately will help:

1) When considering processes at energy scale $E_{exp}$ we can 'integrate out' modes with $p > E_{exp}$ as such terms will never appear on external legs (by design of the experiment) and so will never need them in the Lagrangian (this discussion applies in momentum space).

2) However, virtual processes with $p > E_{exp}$ can and will still happen 'in loops' and this effect is captured by the extra terms introduced in the lagrangian when integrating out.

3) Nevertheless, if we are only interested in $\tilde E_{exp} \ll E_{exp}$ processes (i.e. ones well below the cutoff used for integrating out) then the 'virtual' and 'irrelevant' processes (mass dimension > 4) have vanishing contribution and it is safe to 'drop' (ignore) them.

A very useful useful toy model to understand this is in David Skinner's lecture notes section 2.5 where he considers a '$0$-dimension QFT'

$$S(\phi,\chi) = \frac{m^2}{2} \phi^2 + \frac{M^2}{2} \chi^2 + \frac{\lambda}{4} \phi^2 \chi^2$$

and integrates out $\chi$.

I explained all of that in my answer to Wilsonian definition of renormalizability so I will not repeat myself here. The Wilsonian RG is a dynamical system. The issue has to do with what question one is asking regarding this dynamical system: is one given a starting point and trying to figure out where one goes from there (statistical mechanics), or is one aiming for a final destination and is trying to figure out the starting point in order to get there (continuum QFT).