# Symmetry protects against symmetry-breaking counterterms in renormalization

My lecturer said that when we renormalize a theory (any theory, not necessarily a renormalizable one) we can do so by adding counterterms to the original Lagrangian $$\mathscr{L}_B$$, turning it into $$\mathscr{L}_R + \mathscr{L}_C$$. At any order $$n$$, in the counterterms part of the Lagrangian we can expect all possible scalar combinations of fields, with coupling constants with dimensions (in unit of mass, or length) up to some function of $$n$$. The only counterterms that are prohibited "from the start" are the ones that explicitly break the symmetries of $$\mathscr{L}_B$$. For example, if in $$\mathscr{L}_B$$ we have massless fermions, chiral symmetry protects against counterterms such as $$-\delta m \bar{\psi} \psi$$, which would give mass to fermions and break chiral symmetry. I don't have clear why it is so. Why do symmetries protect against explicitly symmetry-breaking counterterms?

• Possible duplicates: physics.stackexchange.com/q/665144/2451 , physics.stackexchange.com/q/210275/2451 Dec 18, 2021 at 12:56
• I did not do Wilsonian renormalization (yet). Is there another way to see this? Because he said it as if it was the most obvious thing in the world, and that made me think there must have been an easy way to prove it Dec 18, 2021 at 13:02
• Morally, it should be equivalent. Dec 18, 2021 at 13:05
• I'm not sure I understand Dec 18, 2021 at 13:07

Sometimes people describe this approach carelessly and claim that the counterterm Lagrangian is a "new piece" that gets added to the original Lagrangian when you want to consider loops. This is not true. It just looks that way because $$\begin{equation} \mathcal{L} = \frac{1}{2} \partial_\mu \phi_0 \partial_\mu \phi_0 + \frac{1}{2}m_0^2 \phi_0^2 + \frac{\lambda_0}{4!}\phi_0^4 \end{equation}$$ takes the same form as the renormalized Lagrangian when you drop the zero subscripts. But $$\begin{equation} \mathcal{L}_{ren} = \frac{1}{2} \partial_\mu \phi \partial_\mu \phi + \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{4!} \phi^4 \end{equation}$$ is a different Lagrangian. Do $$\mathcal{L}$$ and $$\mathcal{L}_{ren}$$ still give the same results at tree level? Not necessarily because that depends on what scheme you use.
In practice, it is common to use minimal subtraction so that the difference between bare and renormalized quantities (though infinite) is higher order in $$\lambda$$. One can therefore calculate poles in $$\mathcal{L}_{ren}$$ loop diagrams to disentangle the counterterms that were already present from the rest of $$\mathcal{L}$$. But it is important to remember that you're not changing the theory. Merely splitting it in a way that allows for better control.
• That's a good point that we also needs to know what the anomalies are. In the case of $m = 0$, someone unfamiliar with this issue can just do one extra step first. Checking that it's inconsistent to only have counterterms with dimensionless couplings. Dec 18, 2021 at 16:15
• So basically you are saying that the counterterm $-\delta m^2 \phi_r^2$ is a symmetry breaking term that can be there only because of scale invariance symmetry breaking. That is, if there was no scale Invariance breaking it would not exist. Right? Dec 21, 2021 at 12:44