The bare Lagrangian of the $\phi^4$-theory can be written in terms of bare parameters as $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_0)^2-\frac{1}{2}m_0^2\phi_0^2+\frac{\lambda_0}{4!}\phi_0^4\tag{1}.$$ The same bare Lagrangian, in terms of renormalized parameters and counterterms as $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2+\frac{\lambda}{4!}\phi_r^4+\frac{\delta_Z}{2}(\partial_\mu\phi_r)^2-\frac{\delta_m}{2}\phi_r^2+\frac{\delta_\lambda}{4!}\phi_r^4 $$ $$=\mathcal{L}_{renorm}+\mathcal{L}_{counterterm}\tag{2}. $$

I'm interested in evaluating the 1PI effective action for this theory. For this, I have to compute the integral $$Z[j]=e^{iW[j]}=\int D\phi \exp[{i\int d^4x( \mathcal{L} +j\phi)}]$$

I think, it is the total Lagrangian $\mathcal{L}$, in either forms (1) or (2), can be used to evaluate $Z$.

But if I understand it correct, A. Zee's book on Quantum Field Theory in Nutshell, calculates Z using $\mathcal{L}_{renorm}$ part of $\mathcal{L}$ and not using full $\mathcal{L}$. See Chapter IV.3 Eqn. (1), (11). He uses, $A,B,C$ for counterterms.

Why is it that only the renormalized part of the Lagrangian used for the calculation of effective action? Am I missing something?

  • 1
    $\begingroup$ Realise that when we perform field redefinitions and introduce these counter-terms, it's just a way re-expressing the Lagrangian to perform renormalisation. In the end, it's the exact same Lagrangian we started with, so partition functions remain the same. $\endgroup$
    – JamalS
    Dec 24, 2016 at 12:07
  • $\begingroup$ @JamalS -I think, you got me wrong. I agree that it is the same Lagrangian in (1) and (2). Now while calculating $Z$, in place of $\mathcal{L}$, I can substitute either (1) or (2) because they are same. But in Zee's calculation, it seems that he used only $\mathcal{L}_{renorm}$ part of $\mathcal{L}$, and rejected $\mathcal{L}_{counterterm}$ part of $\mathcal{L}$. $\endgroup$
    – SRS
    Dec 24, 2016 at 12:21
  • $\begingroup$ I've edited the question to make it clearer. $\endgroup$
    – SRS
    Dec 24, 2016 at 12:29
  • $\begingroup$ Zee, keeping in tone with the "in a nutshell" theme, is often a bit sloppy. Have you looked at other references? $\endgroup$
    – ACuriousMind
    Dec 24, 2016 at 12:32
  • $\begingroup$ @ACuriousMind I have looked at Ryder. Probably he uses the bare Lagrangian $\mathcal{L}$. But he didn't use $0$ subscript for the parameters and hence I'm confused again. $\endgroup$
    – SRS
    Dec 24, 2016 at 12:41

1 Answer 1


A.Z. uses the full Lagrangian $\mathcal L$, not only $\mathcal L_\mathrm{renorm}$. He omits the counter-terms at first to keep the notation as simple as possible, but he includes them back later on: see equation $(15)$ (it seems odd to me that you decided to stop reading at equation $(11)$). You can repeat the calculations that led to equation $(11)$ but including the counter-terms as well, and you will end up with equation $(15)$ (but note that this is not really necessary: after all, the counter-terms have the same structure as the renormalised Lagrangian, and so it suffices to redefine $\mu^2\to\mu^2+B$ and $\lambda\to\lambda+C$ to get the correct expression).

As ACM mentions in the comments, Zee is not particularly rigorous in his book. For an alternative derivation of the Coleman-Weinberg effective potential, see Itzykson & Zuber's Quantum field theory, section 9-2-2 (in particular, page 454). See also Coleman's Aspects of symmetry, chapter 5, section 3.3 (in particular, page 138).

  • $\begingroup$ According to Zee, the difference between including and not including the counterterms is going from equation (14) to equation (15). Why doesn't the counterterm Lagrangian show up then in the second-order Gaussian approximation? Why don't the counterterms show up in the logarithm in equation (15)? $\endgroup$ Apr 18, 2020 at 16:35

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