Feynman Diagrams For Pedestrians - Herbstschule Maria Laach
Feynman Diagrams For Pedestrians - Herbstschule Maria Laach
Feynman Diagrams For Pedestrians - Herbstschule Maria Laach
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4.2 Trace Techniques• consider a matrix elementū(p)Γu(q) =• using the tensor productwe can writeū(p)Γu(q) =4∑ū k (p)Γ kl u l (q) =k,l=14∑Γ kl u l (q)ū k (p) (126)k,l=1⎛⎞u 1 (q)ū 1 (p) · · · u 1 (q)ū 4 (p)⎜u(q) ⊗ ū(p) = ⎝.. ..⎟ . ⎠ (127)u 4 (q)ū 1 (p) · · · u 4 (q)ū 4 (p)4∑Γ kl [u(q) ⊗ ū(p)] lk =k,l=1• using the trace of a matrixtr(A) =4∑(Γ[u(q) ⊗ ū(p)]) kk (128)k=14∑A kk (129)k=1we can express a matrix element equivalently as a traceū(p)Γu(p) = tr(Γ[u(p) ⊗ ū(p)]) (130)• independent of the concrete realization of the Dirac matrices, we can computetheir traces using their anti commutation relations (34) alonetr(1) = 4(131a)tr (/a/b) = 1 () (34)tr (/a/b) + tr (/b/a) = tr(1) · ab = 4 · ab (131b)2tr (/a 1 ) = tr (/a 1 /a 2 /a 3 ) = tr (/a 1 /a 2 · · · /a 2n+1 ) (γ 5γ 5 = 1)= 0 (131c)tr (/a 1 /a 2 · · · /a n ) = tr (/a n · · · /a 2 /a 1 )tr(γ 5 ) = tr(γ 5 /a) = tr(γ 5 /a/b) = tr(γ 5 /a/b/c) = 0tr(γ 5 /a/b/c/d) = 4i · ɛ(a, b, c, d)only (131d) depends on the existence of a charge conjugation matrix(131d)(131e)(131f)• also from the anti commutation relations alone, we can prove contraction formulae:γ µ /aγ µ = −2 · /aγ µ /a/b/cγ µ = −2 · /c/b/aγ µ γ µ = 4γ µ /a/bγ µ = 4 · ab(132a)(132b)(132c)(132d)25