Proof of 1-particle irreducible (1PI) diagrams

Question

If we split the effective action into

$$Γ[Φ] =\frac{1}2ΦiG_0^{-1}Φ + Γ^{int} [Φ]\tag{1}$$

we can show that the full propagator is given by

$$G= i[iG − Σ]^{-1}\tag{2}$$

With

$$Σ=-Γ_{ΦΦ}^{int} [Φ]\tag{3}$$

Here $Γ_{ΦΦ}$ means double functional derivatives in relation to the mean field $Φ$.

How can we show that $Σ$ is made of only 1-particle irreducible diagrams?

Answer 1

Sketched proof:

1. The full (connected) propagator $$G_c~=~-\Gamma_2^{-1}\tag{A}$$ is (minus) the inverse Hessian of the proper/effective action $\Gamma$, cf. e.g. my Phys.SE answer here.

2. OP's effective action (1) apparently has no tadpole term $\Gamma_1=0$. In this case, the self-energy $$ \Sigma~=~G_0^{-1}-G_c^{-1}\tag{B}$$ is an (amputated) 2-pt 1PI vertex, cf. e.g. my Phys.SE answer here. $\Box$