I'm revisiting the elementary algorithms of renormalization that are taught in a classroom setting and find that the procedure taught to students is as follows:

  1. Write down the bare Lagrangian: $\mathcal{L}_0$
  2. Renormalize the field strengths: $\phi_0\rightarrow Z^{1/2}\phi_r$
  3. Renormalize coupling constants: e.g. $Z^2\lambda_0\rightarrow Z_\lambda \mu^{2\epsilon}\lambda_R$
  4. For perturbation theory rewrite $Z=1+\delta$, where $\delta$'s are counterterms.

I have thought long and hard about steps 2 and 3 (especially in the context of operator renormalization), and have come to the following conclusions which I would like verified.

In step 3, the renormalization of coupling constants is actually the renormalization of the operator product multiplying the coupling constant. And, in step 2, when they renormalize the field strength, it is not the renormalization of a single operator; they are actually renormalizing another composite operator: the kinetic term $\partial_\mu\phi\,\partial^\mu\phi$. The renormalization of a single operator $\phi$ actually corresponds to tadpoles, the one-point function.

Am I interpreting this correctly?

  • 3
    $\begingroup$ Yes, the source is the kinetic term, so it is the normalization of a composite operator, but it is the same as rescaling $\phi$ by Z, so the use the language of normalizing the field strength. The translation shift of a single operator $\phi$ corresponds to tadpoles, and rescaling $\phi$ is only a translation shift if $\phi$ has a VEV, and then the tadpole formalism is annoying in the region where you need a Maxwell construction. There was a question on tadpoles and effective action which I didn't answer yet, because of the Maxwell construction issue for the effective action. $\endgroup$
    – Ron Maimon
    Commented Nov 4, 2012 at 2:21
  • $\begingroup$ Thanks for the answer. And, so like $\phi$, can I expect any composite operator to have a multiplicative and an additive component to its renormalization? that is, $\mathcal{O}\rightarrow Z\mathcal{O}+Y$ ? $\endgroup$
    – QuantumDot
    Commented Nov 4, 2012 at 2:47
  • 1
    $\begingroup$ Only a multiplicvative one, one constant per monomial. $\endgroup$ Commented Nov 19, 2012 at 9:57

1 Answer 1


In renormalization, one considers a family of Lagrangian densities, with arbitrary factors for each renormalizable monomial in fields and derivatives; in case of symmetries only of the symmetric ones. Thus it is determined by the field and symmetry content only. For perturbation theory, these factors are then written as the renormalized finite term plus the diverging counterterm. The particular way of generating these factors (by scaling fields or coupling constants) is immaterial.

As the regularization scale is sent to infinity, the counterterms are left to diverge in such a way that the renormalization conditions give finite results. The family of renormalized theories is now parameterized by the parameters used in the renormalization prescription.


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