In Weinberg's book "The Quantum theory of fields", Chapter 16 section 1: The Quantum Effective action. There is an equation (16.1.17), and several lines of explanation, please see the Images.

Equation (16.1.17) and explanation The Equation is used to calculate the effective action in section 16.2 I can't understand it. Can someone give an explanation as to why this equation is true?

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    $\begingroup$ Related: physics.stackexchange.com/q/146734/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 5, 2013 at 11:57
  • $\begingroup$ I have seen this formula a number of times in other places (such as Schwartz's QFT book) but no one has ever provided a justification for it. $\endgroup$ Jul 20, 2020 at 14:13

1 Answer 1


The quantum effective action is an object with several interesting properties. By constructing it from a given theory, it is possible to gain the full scattering amplitude already at the tree level of the perturbative expansion. This can be understood as follows: take a field theory with the field content $\phi$, the corresponding quantum action $\Gamma(\phi)$ and sources $J$ and construct the Path integral, which is given by

$$Z_\Gamma(J)=\int\mathcal{D}\phi\exp\left(i\Gamma(\varphi)+i\int dx\,J\phi\right).$$

This expression is now identical to the exponential of the sum of connected diagrams, i.e.


These diagrams have the property that each line corresponds to the exact propagator and each vertex to its exact version given by one-particle irreducible diagrams.

From the stationarity of the action one can derive the so-called quantum equations of motion, given by


Solving this equation for the field and denoting the solution by $\phi_J$ leads us to an intriguing relationship between the quantum action and the sum of connected diagrams, $W$:

$$W(J)=\Gamma(\phi_J)+\int dx J\phi_J.$$

This can be interpreted as a Legendre transform between the sum of connected diagrams and the quantum action.

Another interesting consequence of these relations is the fact that the solution of the quantum equations of motion is also the vacuum expectation value of the field in the presence of a source:


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    $\begingroup$ Your answer is really confusing, because you use the same notation $\Gamma$ for the classical action (in the path integral, noted $I$ in Weinberg) and the effective action (the Legendre transform of $W(J)$). I don't think that will help the OP. $\endgroup$
    – Adam
    Dec 5, 2013 at 15:47
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    $\begingroup$ The background field method maybe useful, see link neurotheory.columbia.edu/~larry/AbbottActPhysPol82.pdf $\endgroup$
    – user35289
    Dec 7, 2013 at 6:42
  • $\begingroup$ "...it is possible to gain the full scattering amplitude already at the tree level of the perturbative expansion." What about the higher order contributions in this case? @Frederic Brunner $\endgroup$
    – SRS
    Jun 7, 2017 at 8:35
  • $\begingroup$ You didn't answer OPs question, how is 16.1.7 derived? $\endgroup$ Jul 20, 2020 at 9:06

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