Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

phase space (118)dσdΩ (θ e − H) = 1 √1 λ(s, M2H, M 2 Z ) 12s 16π 2 2s 4∑|T| 2 (241)spins, pol.i. e.dσdΩ (θ e − H) =√λ(s, M2H, M 2 Z )sα 2 (g e V 2 + g e A2 )128s sin 4 θ w cos 4 θ w8sM 2 Z + λ(s, M2 H , M2 Z ) sin2 θ(s − M 2 Z )2 (242)Solution 27.with∫dΩ (a + b sin 2 θ) = 4πa + 8π 3 b (243)followsorσ =√λ(s, M2H, M 2 Z )sπα 2 (g e 2 V + g e A2 ) 12sM 2 Z + λ(s, M2 H , M2 Z ) (244)48s sin 4 θ w cos 4 θ w (s − M 2 Z )2σ =√λ(s, M2H, M 2 Z )sπα 2 (1 + (1 − 4 sin 2 θ w ) 2 )192s sin 4 θ w cos 4 θ w12sM 2 Z + λ(s, M2 H , M2 Z )(s − M 2 Z )2 (245)σ/fb25020015010050M H = 100 GeVM H = 120 GeVM H = 140 GeV0150 200 250 300 350 400 450√ s/GeV48