“…We now proceed by presenting two examples of sequences of intermediate states leading to contributions ∝ J 6 . The first example is given by P ab (λγ b,L,↑ c L,k↑ e iφ2/2 )(λγ a,R,↑ c R,q↑ e iφ1/2 )(λc † R,q↑ γ b,R,↑ e −iφ2/2 )(λc † L,k↑ γ a,L,↑ e −iφ1/2 )P ab = P ab λ 4 (γ b,L,↑ c L,k↑ γ a,R,↑ c R,q↑ c † R,q↑ γ b,R,↑ c † L,k↑ γ a,L,↑ )P ab = P ab λ 4 (γ a,R,↑ γ a,L,↑ γ b,R,↑ γ b,L,↑ )(c L,k↑ c R,q↑ c † R,q↑ c † L,k↑ )P ab = −P ab λ 4 ẑa ẑb (c L,k↑ c R,q↑ c † R,q↑ c † L,k↑ )P ab = −P ab λ 4 ẑa ẑb (u q u k ) 2 (γ L,q↑ γ R,q↑ γ † R,q↑ γ † L,k↑ )P ab = −P ab λ 4 ẑa ẑb (u q u k ) 2 P ab (39) leading to a term ∝ ẑa ẑb and the second example is given P ab (λγ b,L,↑ c L,k↑ e iφ2/2 )(λγ a,R,↓ c R,q↓ e iφ1/2 )(λc † R,q↓ γ b,R,↓ e −iφ2/2 )(λc † L,k↑ γ a,L,↑ e −iφ1/2 )P ab = P ab λ 4 (γ b,L,↑ c L,k↑ γ a,R,↓ c R,q↓ c † R,q↓ γ b,R,↓ c † L,k↑ γ a,L,↑ )P ab = P ab λ 4 (γ a,L,↑ γ a,R,↓ γ b,L,↑ γ b,R,↓ )(c L,k↑ c R,q↓ c † R,q↓ c † L,k↑ )P ab = −P ab λ 4 xa xb (c L,k↑ c R,q↓ c † R,q↓ c † L,k↑ )P ab = −P ab λ 4 xa xb (u k u q ) 2 (γ L,k↑ γ L,q↓ γ † L,q↓ γ † L,k↑ )P ab = −P ab λ 4 xa xb (u k u q ) 2 P ab (40) producing a term ∝ xa xb . The energy denominator for both examples is given by −1/[(E k + U ) 2 (E k + E q + 2U )].…”