# Connected and strongly connected Feynman diagrams

+ 3 like - 0 dislike
58 views
1. Recently I read, that only connected Feynman diagrams give contribution of nonzero values into the scattering amplitude. Why it is so and what is the physical sense of connected diagrams (due to their definition in Wikipedia)?

2. Also, I don't understand why strongly connected (=one-particle irreducible) Feynman diagrams are so important in scattering theory. By the other words, I don't understand why do we cut off one of the internal lines in the Diagram and does it relate to some physical process.

This post imported from StackExchange Physics at 2014-03-24 04:02 (UCT), posted by SE-user Andrew McAddams
asked Dec 4, 2013

## 1 Answer

+ 4 like - 0 dislike

The fact that only connected Feynman diagrams contribute to the scattering amplitude can be interpreted in terms of the vacuum of the theory. Omitting disconnected diagrams amounts to a shift of the vacuum: the vacuum of the interacting theory differs from that of the free theory.

Regarding your second question: strongly connected (also called one-particle irreducible) diagrams are needed in order to calculate loop corrections to the propagator. The exact propagator is given by a geometric series consisting of one-particle irreducible diagrams. Furthermore, they play a role in the calculation of the exact vertex function.

I can recommend two excellent and free sources for more information on the subject: David Tong's lectures on QFT and Mark Srednicki's book.

This post imported from StackExchange Physics at 2014-03-24 04:02 (UCT), posted by SE-user Frederic Brünner
answered Dec 4, 2013 by (920 points)
To be clear, reducible diagrams contribute to the self-energy? but are included as "products" of 1PI diagrams? and we needn't compute them?

This post imported from StackExchange Physics at 2014-03-24 04:02 (UCT), posted by SE-user innisfree
@innisfree The self-energy is given only in terms of 1PI diagrams, see chapter 14 of Srednicki.

This post imported from StackExchange Physics at 2014-03-24 04:02 (UCT), posted by SE-user Frederic Brünner
Thanks, Fig 14.2 shows the point nicely.

This post imported from StackExchange Physics at 2014-03-24 04:02 (UCT), posted by SE-user innisfree

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.