Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

color part of the quark traces tr(T a T a ) = C F tr(1) = C F N C and polarization sumH µν tr[/q 1 (/q 1 + /k)γ µ /q 2 γ ν (/q 1 + /k)] tr[/q 1 /q 2 γ µ (/q 1 + /k)(−/q 2 − /k)γ ν ](q 1 , q 2 , k) = 2C F N c +2C(2q 1 k) 2 F N c(2q 1 k)(2q 2 k)+ 2C F N ctr[/q 1 γ µ (−/q 2 − /k)(/q 1 + /k)γ ν /q 2 ](2q 1 k)(2q 2 k)+ 2C F N ctr[/q 1 γ µ (−/q 2 − /k)/q 2 (−/q 2 − /k)γ ν ](2q 2 k) 2 (228)contractionH µ tr[/q 1 (/q 1 + /k)/q 2 (/q 1 + /k)]tr[/q 1 /q 2 ]µ(q 1 , q 2 , k) = −4C F N c +8C(2q 1 k) 2 F N c (q 1 +k)(−q 2 −k)(2q 1 k)(2q 2 k)tr[/q 1 /q 2 ]+ 8C F N c (−q 2 − k)(q 1 + k)(2q 1 k)(2q 2 k) − 4C tr[/q 1 (−/q 2 − /k)/q 2 (−/q 2 − /k)]FN c (229)(2q 2 k) 2final tracesH µ 2(q 1 k)(q 2 k) (q 1 q 2 + q 1 k + q 2 k)(q 1 q 2 ) 2(q 1 k)(q 2 k)µ(q 1 , q 2 , k) = −16C F N c −64C(2q 1 k) 2 F N c −16C F N c(2q 1 k)(2q 2 k)(2q 2 k) 21 − x 1(1 − x 3 )−8C F N c −16C F N c1 − x 2 (1 − x 1 )(1 − x 2 ) −8C 1 − x 2x 2 1FN c = −8C F N + x2 2c1 − x 1 (1 − x 1 )(1 − x 2 )(230)resultd 2 σ 4πα 2 Q 2 α s= N cdx 1 dx 2 3s 2π C x 2 1 + x2 2F(1 − x 1 )(1 − x 2 )(231)Solution 24.with p = p + + p − :giT = −i [¯v(p + )γ µ (g e V − g e2 cos θAγ 5 )u(p − )] −igµν + ip µ p ν /M 2 Z(i gM Zgw p 2 − M 2 νρ )ɛ ∗,ρ (q)Zcos θ w(232)and from current conservation [¯v(p + )γ µ (g e V − ge A γ 5)u(p − )] p µ = −2ig e A m e¯v(p + )γ 5 u(p − ) =O(m e )T = − g2 M Z 1[¯v(p2 cos 2 θ w s − M 2 + )/ɛ ∗ (q)(g e V − g e Aγ 5 )u(p − )] (233)ZSolution 25.• momentum conservation is obvious and energy conservation follows fromE H + E Z = √ s = 2E (234)46