How to compute the quantum effective action from 1PI Feynman diagrams?

Question

On page 33 of these notes by David Skinner, it is claimed that

[starting from a connected graph and removing the bridges] tells us how to compute $\Gamma(\Phi)$ perturbatively from the original action: $\Gamma(\Phi)$ consists of all possible 1PI Feynman graphs that may be constructed using the propagators and vertices in $S(\phi)$.

However, I cannot decipher exactly what this means. How does one go about computing $\Gamma(\Phi)$ using Feynman diagrams as described? By writing out the 1PI Feynman diagrams, should I not just get a number, rather than the effective action with an explicit $\Phi$ dependence?

EDIT: I have read Proof that the effective/proper action is the generating functional of one-particle-irreducible (1PI) correlation functions, but I do not understand how this allows us to directly calculate $\Gamma$?

Answer 1

The explicit calculation of the full effective potential in terms of Feynman diagrams is first laid out in “Functional evaluation of the effective potential,” R. Jackiw, Phys. Rev. D 9, 1686 (1974). The results are nontrivial, in several different ways. For one thing, the structure of the one-loop contribution to the effective action is fundamentally different from the higher-loop terms. The one-loop term is a functional determinant, and it was already known how to calculate it prior to the paper in question. [For example, this kind of calculation is carried out more awkwardly in in “Radiative corrections as the origin of spontaneous symmetry breaking.” S. Coleman, E. Weinberg, Phys. Rev. D 7, 1888 (1973).]

However, the higher-loop terms involve a sum over one-particle irreducible vacuum bubble diagrams, and moreover, the Feynman rules for those diagrams are not the Feynman rules for the original theory. For example, in $\phi^{4}$ theory, the Feynman rules for the vacuum bubbles actually involve both 3-$\phi$ and 4-$\phi$ vertices, even when the underlying action has no $\phi^{3}$ term. And the “coupling constants” for the new Feynman rules depend on the “classical” field $\Phi$, which explains how the final result retains a dependence on $\Phi$.

It is, frankly, a lot of work to follow through the calculations in the Jackiw paper. Even elements that might seem relatively simple, like the function Legendre transformation that eliminates the diagrams that are not 1PI, is tricky to evaluate explicitly. A great deal of familiarity with radiative corrections is needed to parse and understand the whole analysis.

Answer 2

Well, the proof in Ref. 1 does strictly speaking not compute the quantum effective action $\Gamma[\Phi_{\rm cl}]$ directly, but rather the generating functional $W_c[J]$ of connected diagrams in 2 different ways:

1. As trees constructed from full propagators, 1PI vertices, and sources $J$, via a combinatoric argument.

2. As trees constructed from $\Gamma$-propagators and $\Gamma$-vertices of the $\Gamma$-action, and sources $J$, due to the WKB approximation.

However, because of the bijective nature of the Legendre transformation, we conclude that the $\Gamma$-propagators are full propagators and the $\Gamma$-vertices are 1PI vertices. For more details, see this related Phys.SE post.

References:

1. D. Skinner, QFT in 0D; p. 32-33.