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If we split the effective action into

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

we can show that the full propagator is given by

$$G= i[iG − Σ]^{-1}$$

With

$$Σ=-Γ_{ΦΦ}^{int} [Φ]$$

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?

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Sketched proof:

  1. The full (connected) propagator $$G~=~-\Gamma_2^{-1}$$ is (minus) the inverse Hessian of the proper/effective action $\Gamma$, cf. e.g. eq. (6) in my Phys.SE answer here.

  2. The self-energy $$ \Sigma~=~G_0^{-1}-G^{-1}$$ is an (amputated) 2-pt 1PI vertex, cf. e.g. my Phys.SE answer here. $\Box$

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