Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

7 SolutionsSolution 1.∂ µ e −ipx = −ip µ e −ipx(a∂)(b∂)e −ipx = −(ap)(bp)e −ipx∂ 2 e −ipx = −i(p∂)e −ipx = −p 2 e −ipx(190a)(190b)(190c)Solution 2.∂ µ x µ = gµ ν ∂x µ /∂x ν = gµ ν δ µ ν = g µ µ = 4 (191a)( )∂ 2 e −x2 /2 = −∂ µ x µ e −x2 /2= −x µ ∂ µ e −x2 /2 − (∂ µ x µ ) e −x2 /2= x µ x µ e −x2 /2 − g µµ e −x2 /2 = (x 2 − 4)e −x2 /2(191b)Solution 3.( ) ( )[γ k , γ l] 0 σ = k 0 σl+ −σ k 0 −σ l + (k ←→ l) =0=( )−σ k σ l 00 −σ k σ l( ) (−[σ k , σ l ] + 00 −[σ k , σ l = −2δ] kl 1 0+ 0 1+ (k ←→ l))(192a)Solution 4.ū k (p) = u † k (p)γ 0 = u † k (⃗ 0)γ 0 γ 0¯v k (p) = ¯v k ( ⃗ 0)/p − m√p0 + m/p † + m√p0 + m γ 0 = ū k ( ⃗ /p + m0) √p0 + m(193a)(193b)Using the definition and multiplying with γ 0 from the right2∑k=1u k ( ⃗ 0)ū k ( ⃗ 0) = γ 0 + 1,22∑k=1v k ( ⃗ 0)¯v k ( ⃗ 0) = γ 0 − 12(194)39