Consider a classical scalar field theory for a real scalar field $\phi$ given by $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-V(\phi)$$ where $V(\phi)$ is the classical potential. In quantum field theory, one defines an effective potential $V_{eff}(\phi)$. And unlike classical field theory where spontaneous symmetry breaking (SSB) is analyzed by minimizing $V(\phi)$, SSB in quantum field theory is analyzed by minimizing $V_{eff}(\phi)$.

For this purpose, one defines a new functional $\Gamma[\phi]$, called the effective action. Intuitively, the name suggests that $\Gamma[\phi]$ must be a modification to the classical action $S[\phi]$ when one takes quantum corrections into account. Indeed when one calculates $\Gamma[\phi]$, one obtains $$\Gamma[\phi]=S[\phi]+\text{quantum corrections of O($\hbar$)}.$$ But that may or may not contain all possible corrections.

However, $\Gamma[\phi]$ is not defined as $$\Gamma[\phi]=S[\phi]+\text{all possible quantum loop corrections}\tag{1}$$ but as $$\Gamma[\phi]=W[J]-\int d^4x j(x)\phi(x).$$

From this definition, how can one be so sure, in general, that evaluation of $\Gamma[\phi]$ gives all possible quantum corrections to $S[\phi]$ in powers of $\hbar$ and nothing is left out? In other words, is there a way to show/see that $(1)$ holds for a generic potential $V(\phi)$?

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    $\begingroup$ OP's question (v2) seems unclear: On one hand, the effective action is defined as the Legendre transform of the generating functional for connected Feynman diagrams. On the other hand, OP's phrase all possible quantum loop corrections is not a well-defined notion. E.g. the normalization of each loop correction term is unspecified. $\endgroup$
    – Qmechanic
    Dec 24, 2016 at 9:41
  • $\begingroup$ @SRS the perturbative series of QFT are asymptotic expansions. Renormalizability only guarantees finiteness up to any a-priori given order, but the series itself doesn't converge. It is why we say that perturbative QFTs aren't be well-defined mathematically. $\endgroup$ Dec 27, 2016 at 2:02

1 Answer 1


The effective action is defined so that $\frac{\delta \Gamma \left[ \phi \right]}{\delta \phi^i \left( x \right)}=J_{\phi}^i \left( x \right)$, where $J$ is a classical current, so we see that this definition coincides with the usual definition of the action in classical field theory.

Moreover, the procedure of evaluation of this action involves evaluation of all the appropriate loop-diagrams. Is it enough to convince you that all the necessary corrections are taken into account and action is well-defined?

There is detailed discussion of the subject in Ch. 16 of Weinberg.

  • $\begingroup$ I think, unless I can show $\Gamma[\phi]=S[\phi]+\text{all corrections of order} \hbar+\text{all corrections of order} \hbar^2+...$, I'm not convinced. $\endgroup$
    – SRS
    Dec 24, 2016 at 7:54
  • $\begingroup$ The definition $\frac{\delta \Gamma \left[ \phi \right]}{\delta \phi^i \left( x \right)}=J_{\phi}^i \left( x \right)$ is analogous to the definition of classical action. True. But this in no way makes it clear what I asked. $\endgroup$
    – SRS
    Dec 24, 2016 at 7:58
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    $\begingroup$ @SRS Again, one sums over all the loops for evaluating the effective action. What are the problems? $\endgroup$ Dec 24, 2016 at 9:32

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