× ( − ieγ $^{λ}$) iS$_{0}$ ( l ) e − ip ∧ $^{q}$, = 2 e sin( p ∧ q )Λ µνλ a 2 ( p, q, k ) (105) = − ∫ d $^{4}$l (2 π ) $_{4}$tr iS$_{0}$ ( − l )( − ieγ $^{λ}$) iS$_{0}$ ( − p − q − l ) × ( − ieγ $^{µ}$) iS$_{0}$ ( − q − l )( − ieγ $^{ν}$) e ip ∧ $^{q}$. 2 e sin( p ∧ q )Λ µνλ a 2 ( p, q, k ) = e 3 ∫ d $^{4}$l (2 π ) $_{4}$tr S T 0 ( l ) γ T $^{λ}$S T 0 ( p + q + l ) γ T $^{µ}$S$_{0}$ ( q + l ) γ T $^{ν}$eip ∧ q = e 3 d $^{4}$l (2 π ) $_{4}$tr[ γ $^{ν}$S$_{0}$ ( q + l ) γ $^{µ}$S$_{0}$ ( p + q + l ) γ $^{λ}$S$_{0}$ ( l )] $^{T}$e ip ∧ q = e 3 ∫ d $^{4}$l (2 π ) $_{4}$tr γ $^{ν}$S$_{0}$ ( q + l ) γ $^{µ}$S$_{0}$ ( p + q + l ) γ $^{λ}$S$_{0}$ ( l ) e ip ∧ $^{q}$. (106) 2 e sin( p ∧ q )Λ µνλ a ( p, q, k ) = 2 e sin( p ∧ q ) [ Λ µνλ a 1 ( p, q, k ) + Λ µνλ a 2 ( p, q, k ) ] (107) = 2 ie 3 ∫ d $^{4}$l (2 π ) $_{4}$tr γ $^{ν}$S$_{0}$ ( q + l ) γ $^{µ}$S$_{0}$ ( p + q + l ) γ $^{λ}$S$_{0}$ ( l ) sin( p ∧ q ) , Λ µνλ a ( p, q, k ) = ie 2 ∫ d $^{4}$l (2 π ) $_{4}$tr γ $^{ν}$S$_{0}$ ( q + l ) γ $^{µ}$S$_{0}$ ( p + q + l ) γ $^{λ}$S$_{0}$ ( l ) . (108) p$_{µ}$ Λ µνλ a ( p, q, k ) = Π νλ a ( q ) − Π λν a ( k ) , (109) i Π νλ a ( q ) = − ∫ d $^{4}$l (2 π ) $_{4}$tr( − ieγ $^{ν}$) iS$_{0}$ ( q + l )( − ieγ $^{λ}$) iS$_{0}$ ( l ) (110) These diagrams are different only in the circulation of the momentum integration. Since Cγ $^{µ}$C − 1 = − γ T µ and CS$_{0}$ ( l ) C − 1 = S T 0 ( − l ), we can rewrite the above expression as Thus, by summing the two diagrams, we obtain i.e., Therefore, using also ( / p + / q + / l − m ) − ( / q + / l − m ), we get where is the photon self-energy, with a fermion loop. From now on, differently for the previous calculations, the contributions to the ST identity turns out to be very involved and a complete verification is unfeasible. In this situation we restrict ourselves in to verify the matching of the divergent parts of the two sides of Eq. (103). 17