Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

Solution 18.e − (p 1 )e − (q 1 )iT t =γ[, Z 0 ](218a)e + (p 2 )e − (p 1 )e + (q 2 )e − (q 1 )iT s =γ[, Z 0 ](218b)e + (p 2 )e + (q 2 )Solution 19.∵∴• finallydσdΩ = α22s(1 + cos 2 θ2t = − s ( ) θ2 (1 − cos θ) = −s sin2 (219a)2u = − s ( ) θ2 (1 + cos θ) = −s cos2 (219b)2s 2 + u 2= 1 + ( )cos4 θ2t 2 sin ( ) (220a)4 θ2u ( )2θst = −cos4 2sin ( ) (220b)2 θ+ 1 + ( ) ( ))cos4 θ2sin ( ) − 2 cos4 θ24 θsin ( )2 θ222= α24s( ) 3 + cos 2 2θ(221)1 − cos θSolution 20.¯v(p 2 )ieγ µiT 1 =u(p 1 )¯v(p 2 )iT 2 =ieγ µu(p 1 )−ig µν(p 1 + p 2 ) 2 + iɛi/q 1 + /k − m igT + iɛ a γ ρ−ig µν(p 1 + p 2 ) 2 + iɛi−/q 2 − /k − m + iɛ44igT a γ ρ−ieQγ ν−ieQγ νv(q 2 )ū(q 1 )v(q 2 )ū(q 1 )ɛ ∗ ρ(k)ɛ ∗ ρ(k)(222a)(222b)