“…It is now trivial to check that, in the limit eB → 0, the above equation exactly boils down to the 1 2 times pure vacuum contribution given in Eq. (10). Thus extracting the pure vacuum contribution from the above equation we get, In this appendix we will calculate the quantities N µν π,nl (q , k ) = d 2 k ⊥ (2π) 2 N µν π,nl (q , q ⊥ = 0, k) (E1) N µν p,nl (q , k ) = d 2 k ⊥ (2π) 2 N µν p,nl (q , q ⊥ = 0, k) .…”