Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

Solution 9.product rule and [γ 5 , γ µ ] + = 0i∂ [ µ ¯ψ 1 (x)γ µ γ 5 ψ 2 (x) ] = ¯ψ 1 (x)(i ←− /∂ γ 5 + i −↛ )(∂ γ 5 ψ 2 (x) = ¯ψ 1 (x) i ←− /∂ γ 5 − γ 5 i −↛ )∂ ψ 2 (x)()= ¯ψ 1 (x) −mγ 5 − γ 5 m ψ 2 (x) = −2m ¯ψ 1 (x)γ 5 ψ 2 (x) . (205)Solution 10.( ) ( )0 σk 0 σl− 2iσ kl =−σ k 0 −σ l − (k ←→ l)0( )( )−[σ=k , σ l ] − 0σ0 −[σ k , σ l = −2iɛ klm m 0] − 0 σ m(206)therefore( ) σσ kl = ɛ klm m 00 σ m(207)Solution 11.(i/∂ − m) S(x, m) = ( −∂ 2 − m 2) D(x, m) = δ 4 (x) (208)Solution 12.∫ d 4 p1S(x, m) = (i/∂ + m)(2π) 4 e−ipx p 2 − m 2 +iɛ∫ d 4 ∫p /p + m d 4=(2π) 4 e−ipxp 2 − m 2 +iɛ = p 1(2π) 4 e−ipx /p − m+iɛ (209)Solution 13.tr (/a/b/c/d) = + tr (/b/c/d/a) = − tr (/b/c/a/d) + 2 · ad · tr (/b/c)= + tr (/b/a/c/d) − 2 · ac · tr (/b/d) + 2 · ad · 4 · bc= − tr (/a/b/c/d) + 2 · ab · tr (/c/d) − 2 · ac · 4 · bd + 2 · ad · 4 · bcthereforetr (/a/b/c/d) = 4 · (ab · cd − ac · bd + ad · bc) (133 ′ )42