Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

Solution 6.ū k (p)γ µ u l (p) = ū k ( ⃗ 0)/p + m√p0 + m γ /p + mµ √p0 + m u l( ⃗ 0)=1p 0 + mūk( ⃗ 0)2p µ (/p + m)u l ( ⃗ 0)= 2p µp 0 + mūk( ⃗ 0)(p 0 + m)u l ( ⃗ 0) = 2p µ ū k ( ⃗ 0)u l ( ⃗ 0) = 2p µ δ kl(200a)¯v k (p)γ µ v l (p) = ¯v k ( ⃗ 0)/p − m√p0 + m γ /p − mµ √p0 + m v l( ⃗ 0)=1p 0 + m ¯v k( ⃗ 0)2p µ (/p − m)v l ( ⃗ 0)= −2p µp 0 + m ¯v k( ⃗ 0)(p 0 + m)v l ( ⃗ 0) = −2p µ¯v k ( ⃗ 0)v l ( ⃗ 0) = 2p µ δ kl(200b)ū k (⃗p)γ 0 v l (−⃗p) = ū k ( ⃗ 0)/p + m√p0 + m γ /p † − m0 √p0 + m v l( ⃗ 0)= ū k ( ⃗ 0)/p + m√p0 + m/p − m√p0 + m γ 0v l ( ⃗ 0) = 0(201a)¯v k (−⃗p)γ 0 u l (⃗p) = 0(201b)Solution 7.γ 5 γ 2 = iγ 0 γ 1 γ 2 γ 3 γ 2 = −iγ 0 γ 1 γ 2 γ 2 γ 3 = iγ 0 γ 2 γ 1 γ 2 γ 3 = −iγ 2 γ 0 γ 1 γ 2 γ 3 = −γ 2 γ 5 (202)since every γ µ appears exaclty once and anti-commutes with the other three, we have for every µ[γ 5 , γ µ ] + = 0 (203)Solution 8.product rulei∂ [ µ ¯ψ 1 (x)γ µ ψ 2 (x) ] (= ¯ψ 1 (x) i ←− /∂ + i −↛ )( )∂ ψ 2 (x) = ¯ψ 1 (x) −m + m ψ 2 (x) = 0 . (204)41