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From the LSZ reduction formula, it is clear that only the connected Feynman diagrams that contribute to a scattering amplitude. However, connected diagrams are of two types: 1PR and 1PI. 1PR diagrams can be constructed out of 1PI graphs as the building blocks. Therefore, while considering the divergences of a theory it suffices to regularize and renormalize the 1PI diagrams. But what about scattering?

1. Can one forget about the contribution of 1PR diagrams to a scattering? My understanding is that we cannot neglect 1PR diagrams while computing scattering amplitudes and they have to be taken into account in addition to the 1PI diagrams. But I'm not quite sure.

2. What is the utility of the classification of connected diagrams into 1PR and 1PI classes?

Qmechanic
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SRS
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2 Answers2

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You should use all connected diagrams with external legs amputated. This means that you throw away Feynman propagators of ingoing/outgoing particles and disregard all diagrams with any decorations on external legs. Therefore some $1$PR diagrams are thrown away, but not all of them. This is explained in detail in Peskin, Schroeder.

As for your second question, the first big utility of this classification is mentioned already in your post - classification of divergences. Note also that once you know how to do integrations in $1$PI diagrams you can get values of all diagrams relatively easily. Once values of all $1$PI diagrams in a theory are known, values of all diagrams can be obtained by computing tree-level diagrams with values of the original $1$PI diagrams used as vertices. This is all nicely organized by the fact that there exists a generating functional for $1$PI diagrams - the so called quantum effective action. To read more about it, c.f. e.g. Zinn-Justin, Weinberg or Srednicki.

Blazej
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You do not neglect the reducible diagrams. But a reducible diagram is a chain of irreducible diagrams. So when you sum up the contributions of all 1PI diagrams, this includes all reducible diagrams at "higher 1PI orders". See also this question and my answer there.

ACuriousMind
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