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The 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 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?

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    $\begingroup$ 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 '16 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 '16 at 12:21
  • $\begingroup$ I've edited the question to make it clearer. $\endgroup$ – SRS Dec 24 '16 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 '16 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 '16 at 12:41
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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).

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