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By Peeters K.

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Thus, a diagram with E external lines, I internal lines and v vertices has a power h¯ E+ I −v . 55) On the other hand, if there are v vertices in the diagram, then there must be at least v − 1 internal lines to glue them together, otherwise the diagram would be 3 If you instead consider a 1 → 3 or 3 → 1 process, the only difference in the expression is an overall sign and a different distribution of signs for the momenta. In general, ingoing particles lead to eikx factors with positive exponent, while outgoing ones have a negative exponent (cf.

34), we can flip each of the loops of the “figure eight” around the vertical axis, and we can also flip the eight upside down. This gives 2 · 2 · 2 = 8 for the symmetry factor. 35) we have only the symmetry which mirrors the loop, so a factor of 2. , as stated above. e. with fields depending on space-time points. 44). 39) This expression shows that each line in a Feynman diagram has a momentum variable associated to it, which carries a direction. Moreover, a vertex, which involves an integration over a space-time point (like the z in the example above), leads to a Dirac delta function.

E. 76) where GF ( x − y) is the single-field scalar propagator? e. 1 to the more complicated system under consideration here). (c) Compute, at order λ, the transition amplitude for the 2 → 2 scattering process in which one particle of type i and one of type j scatter to one particle of type k and one of type l. 53). 6. g. 66)), that a scattering cross section always contain integration measures for the outgoing momenta which take the form d3 k . g. 43). 77) is invariant under Lorentz transformations, despite the fact that it only involves a three-dimensional volume element in momentum space.

### A guide to quantum field theory by Peeters K.

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