Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

• the Dirac equation is obviously linear and its solutions satisfy the proper relativisticdispersion relation.• can we construct “objects” γ µ , that satisfy (31)?• a sufficient condition is[γ µ , γ ν ] +:= γ µ γ ν +γ ν γ µ = 2g µν · 1 (34)because partial derivatives commute ∂ µ ∂ ν = ∂ ν ∂ µ .• using a useful and ubiquitous notation, the Feynman slash/a = γ µ a µ = γ µ a µ (35)this reads equivalently [/a, /b] +:= /a/b+/b/a = 2 · ab = 2 · a µ b µ2.5 Gamma Matrices• recall the Pauli matrices with the defining property]3∑[σ k , σ l = σ k σ l − σ l σ k = 2i ɛ klm σ m(σk ) †= σkm=1(36a)(36b)using the totally antisymmetric tensor ɛɛ 123 = ɛ 231 = ɛ 312 = 1, ɛ 213 = ɛ 321 = ɛ 132 = −1 (37)• concrete realisationσ 1 =( ) 0 1, σ 2 =1 0( ) ( )0 −i 1 0, σ 3 =i 0 0 −1(38)• with• in particular3∑σ k σ l = δ kl 1 + i ɛ klm σ m (39)[σ k , σ l ]+m=1= 2δ kl 1 (40)• Dirac realisation of the gamma (a. k. a. Dirac) matrices:( ) ( )1 00 σγ 0 = , γ i i=0 −1 −σ i 0(41)• there are (infinitely) many more realisations, but no smaller one8