Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

[]iT 1 = ū(q 1 )(igT a /ɛ ∗ i(k))/q 1 + /k − m + iɛ (−ieQγµ )v(q 2 ) × (223a)−i(p 1 + p 2 ) 2 + iɛ [¯v(p 2)(ieγ µ )u(p 1 )][]iT 2 = ū(q 1 )(−ieQγ µ i)−/q 2 − /k − m + iɛ (igT a/ɛ ∗ (k))v(q 2 ) × (223b)−i(p 1 + p 2 ) 2 + iɛ [¯v(p 2)(ieγ µ )u(p 1 )]Solution 21.∣T ∣ɛ 1 ∗ µ (k)=k µ= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 1sū(q 1)T a (/k + /q 1 − m)/q 1 + /k − m + iɛ γµ v(q 2 )on the other hand= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)T a γ µ v(q 2 ) (224)T 2∣∣ɛ∗ µ (k)=k µ= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)γ µ 1−/q 2 − /k − m + iɛ (/k + /q 2 + m)T a v(q 2 )= −e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)T a γ µ v(q 2 ) = −T 1∣∣ɛ∗ µ (k)=k µ(225)Solution 22.2q 1 q 2 = (q 1 + q 2 ) 2 = (p − k) 2 = s(1 − x 3 )2q 1 k = (q 1 + k) 2 = (p − q 2 ) 2 = s(1 − x 2 )2q 2 k = (q 2 + k) 2 = (p − q 1 ) 2 = s(1 − x 1 )(226a)(226b)(226c)Solution 23.four traces:H µν (q 1 , q 2 , k) =∑ɛ+ ∑ ɛtr[/q 1 T a /ɛ ∗ (/q 1 + /k)γ µ /q 2 γ ν (/q 1 + /k)/ɛT a ]+ ∑ (2q 1 k) 2 ɛtr[/q 1 γ µ (−/q 2 − /k)T a /ɛ ∗ /q 2 γ ν (/q 1 + /k)/ɛT a ]+ ∑ (2q 1 k)(2q 2 k)ɛtr[/q 1 T a /ɛ ∗ (/q 1 + /k)γ µ /q 2 /ɛT a (−/q 2 − /k)γ ν ](2q 1 k)(2q 2 k)tr[/q 1 γ µ (−/q 2 − /k)T a /ɛ ∗ /q 2 /ɛT a (−/q 2 − /k)γ ν ](2q 2 k) 2 (227)45