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Above diagram can be written in terms of series:

$$i\Delta = -\frac{i}{p^2 + m^2} + \Big(-\frac{i}{p^2 + m^2}\Big)(i\Pi)\Big(-\frac{i}{p^2 + m^2}\Big)+ \Big(-\frac{i}{p^2 + m^2}\Big)(i\Pi)\Big(-\frac{i}{p^2 + m^2}\Big)(i\Pi)\Big(-\frac{i}{p^2 + m^2}\Big) + \cdot \cdot \cdot = - \frac{i}{p^2 + m^2 - \Pi}.$$

So to get the exact propagator, I should sum over all the 1PI diagrams. 1PI diagrams are those from which I can't get two diagrams by cutting a line. But then $3$rd diagram and those arising from the higher order terms are really 1PI diagrams? (since I can cut the $3$rd diagram and obtain two diagrams)

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2 Answers 2

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  1. To get the full propagator/exact propagator/connected 2pt function $$\Delta_c~=~\Delta_0\sum_{n=0}^{\infty}(\Pi \Delta_0)^n\tag{A}$$ one should sum over self-energy diagrams $\Pi$ and free propagators $\Delta_0$, cf. e.g. this Phys.SE post.

  2. Since the free propagator $\Delta_0$ is not a 1PI diagram (it can be cut in 2 by a single cut) none of the terms on the RHS of eq. (A) are 1PI diagrams, cf. OP's question.

  3. The 1PI diagrams themselves are by definition amputated, i.e. without external legs.

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The definition of 1PI diagrams may seem quite arbitrary so I would like to start my answer there. Imagine you have an interacting QFT and you want to compute observables using Feynman diagrams. Drawing all possible feynman diagrams, you realise that there are a lot of degeneracies. First of all, a diagram can contribute in 2 ways: either it is a sole diagram or it is multiplied by another diagram. The latter are called disconnected diagrams. However, this doesn't solve all the degeneracies. I can also attach two connected diagrams bridged by a propagator and that would be a valid Feynman Diagram as well. Therefore, you define 1PI diagrams as the diagrams which cannot be further reduced by "cutting up".

This is where 1PI diagrams come in very useful. With a little bit of thought, you realise that any connected 2 point correlation function can be written as a telescope sum of 1PI diagrams. This is what your figure says. By eliminating all the redundancy, you build all feynman diagrams from 1PI diagrams. Furthermore, this means that if you renormalise 1PI diagrams, you are done! They massively simplify calculations.

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