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Textbooks often defines a one-particle irreducible diagram (1PI diagram) as a connected diagram which does not fall into two pieces if you cut one internal line. Is this internal line the full propagator or the free propagator?

Qmechanic
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amilton moreira
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If there are no tadpoles, it formally amounts to the same notion of one-particle irreducible$^\dagger$ (1PI) since we can rewrite a full (connected) propagator $$G_c~=~G_0\sum_{n=0}^{\infty}(\Sigma G_0)^n$$ as a geometric series of bare/free propagators $G_0$, and vice-versa $$G_0~=~G\sum_{n=0}^{\infty}(-\Sigma G_c)^n.$$ Here $$\Sigma~=~G_0^{-1}-G_c^{-1} $$ is the self-energy, which is 1PI if there are no tadpoles, cf. my Phys.SE answer here.

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$^\dagger$ Note that 1PI is called 2-edge-connected by mathematicians.

Qmechanic
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  • If 1PI means that the deletion of any edge doesn't render the graph disconnected, then the definition is not equivalent to 2-connectedness, which means that the deletion of any vertex and the edges connected to it do not render the graph disconnected. Take two triangles, join them at a vertex (like a sharp "8"), and to each of the other vertices connect two external legs: you got a 1PI graph that is not 2-connected. – Albert May 04 '23 at 12:38
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    Hi @Albert. Thanks for the feedback. It seems you're right, so I plan to do a correction. I'm trying to remember where I read the definition in the first place...Found it: It's in some MIT lecture notes by P. Etingof. – Qmechanic May 04 '23 at 12:41
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    I updated the answer. – Qmechanic May 04 '23 at 13:41
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    @Albert: FYI, arXiv:2202.12296 defines a one-vertex-irreducible (1VI) graph as a connected graph that cannot be disconnected by removal of a single vertex. – Qmechanic Aug 22 '23 at 13:15
  • NB: The geometric series $\sum_{n=0}^{\infty}(\Sigma G_0)^n$ is typically not convergent, however perturbatively it makes sense as a formal power series. – Qmechanic Feb 28 '24 at 07:51