I am learning renormalisation in QFT and I had a question regarding the counter-terms we put in the Lagrangian. For the purpose of this question I shall consider a $\phi^4$ theory in 4D. The "bare" Lagrangian is given by:

$$\mathcal{L} = \frac{1}{2} \partial_{\mu}\phi_0\partial^{\mu}\phi_0 - \frac{1}{2}m_0^2\phi_0^2 - \frac{\lambda_0}{4!}\phi_0^4.\tag{1}$$

However, due to loop diagrams, the field and the parameters receive some "quantum" corrections and so it is more appropriate to expand the lagrangian in terms of the parameters we physically observe. Thus, the Lagrangian with the counter-terms is given by

$$\mathcal{L} = \frac{1}{2} \partial_{\mu}\phi\partial^{\mu}\phi - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4 + \frac{\delta Z_{\phi}}{2} \partial_{\mu}\phi\partial^{\mu}\phi - \frac{\delta m^2}{2}m^2\phi^2 - \frac{\delta \lambda}{4!}\phi^4. \tag{2}$$

I understand the reasons why we add the counter-terms and how we use them to eliminate infinities that arise from loop diagrams that contribute to $\langle\phi(x)\phi(y)\rangle$ and $\langle\phi(x)\phi(y)\phi(z)\phi(w)\rangle$. We effectively treat these counter-terms as perturbations to our "physical" lagrangian and absorb the infinities that arise from loop integrals into the counter-term coefficients.

My question is concerned about the calculation of higher order correlators (e.g. 6-point correlator). In this theory, specifically, 6-point correlators do not suffer from loop divergences. However, now that we have added the counter-terms to our lagrangian, we should also include their contribution to these higher order correlators. My issue is that since these counterterm parameters are basically infinite, wouldn't the higher order correlators also suffer from infinities coming from the parameters? What am I missing?

I have tried searching online for a calculation of this type for $\phi^4$ theory; however, I haven't managed to find one. I would appreciate if someone could refer me to a textbook or online source where this calculation is done explicitly.


1 Answer 1


TL;DR: Although far from trivial, in a renormalizable theory, the diagrams with divergent subdiagrams and the diagrams with counterterms together yield finite correlator functions.

Renormalization is a huge topic, but here are a few comments:

  1. It should perhaps be stressed that OP's eqs. (1) and (2) both represent the same bare Lagrangian.

  2. It is enough to consider connected propagators $G_c$ and amputated 1PI correlator functions $\Gamma_n$, because all correlator functions can be built as trees of vertices $\Gamma_n$ and lines $G_c$.

  3. Renormalization of $\phi^4$ theory in 4D is discussed in many textbooks, see e.g. Refs. 1 & 2. In particular, the 6-point correlator function is mentioned in Fig. 9.1 of Ref. 2.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT. 1995; sections 10.2 + 10.5.

  2. L.H. Ryder, QFT, 2nd eds., 1996; sections 9.1 + 9.2 + 9.3.

  • $\begingroup$ Thank you very much for the response and the references! $\endgroup$
    – emir sezik
    Commented Feb 21, 2023 at 10:58

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