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?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/665144/2451 , physics.stackexchange.com/q/210275/2451 $\endgroup$
    – Qmechanic
    Commented Dec 18, 2021 at 12:56
  • $\begingroup$ 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 $\endgroup$ Commented Dec 18, 2021 at 13:02
  • $\begingroup$ Morally, it should be equivalent. $\endgroup$
    – Qmechanic
    Commented Dec 18, 2021 at 13:05
  • $\begingroup$ I'm not sure I understand $\endgroup$ Commented Dec 18, 2021 at 13:07

1 Answer 1


The "renormalized plus counterterm" decomposition is a rewriting of the original Lagrangian. Any symmetry present in one notation needs to be present in the other.

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.

  • $\begingroup$ Sure, but this rewriting can add some terms that were not present at first. For example if you put m=0 in scalar theory renormalization might force you to add a mass term because no symmetry is protecting it. For fermions on the other hand chiral symmetry does the job $\endgroup$ Commented Dec 18, 2021 at 15:06
  • $\begingroup$ 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. $\endgroup$ Commented Dec 18, 2021 at 16:15
  • $\begingroup$ 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? $\endgroup$ Commented Dec 21, 2021 at 12:44
  • $\begingroup$ Yes... exactly. $\endgroup$ Commented Dec 21, 2021 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.