Quantum Field Theory

where, as usual, all fields are in the interaction picture. Just as in the bosonic calculation,the contribution to nucleon scattering comes from the contraction: ¯ψ(x 1 )ψ(x 1 ) ¯ψ(x 2 )ψ(x 2 ) :{ }} {φ(x 1 )φ(x 2 ) (5.41)We just have to be careful about how the spinor indices are contracted. Let’s startby looking at how the fermionic operators act on |i〉. We expand out the ψ fields,leaving the ¯ψ fields alone for now. We may ignore the c † pieces in ψ since they give nocontribution at order λ 2 . We have∫: ¯ψ(x 1 )ψ(x 1 ) ¯ψ(x d2 )ψ(x 2 ) : b s †⃗p br †3 k 1 d 3 k 2⃗q|0〉 = − [(2π) ¯ψ(x 6 1 ) · u m ( ⃗ k 1 )] [ ¯ψ(x 2 ) · u n ( ⃗ k 2 )]e −ik 1·x 1 −ik 2·x 2√4E⃗k1 E ⃗k2b m ⃗ k1b n ⃗ k2b s †⃗p br †⃗q|0〉 (5.42)where we’ve used square brackets [·] to show how the spinor indices are contracted. Theminus sign that sits out front came from moving ψ(x 1 ) past ¯ψ(x 2 ). Now anti-commutingthe b’s past the b † ’s, we get=−12 √ E ⃗p E ⃗q([ ¯ψ(x1 ) · u r (⃗q)] [ ¯ψ(x 2 ) · u s (⃗p)]e −ip·x 2−iq·x 1− [ ¯ψ(x 1 ) · u s (⃗p)] [ ¯ψ(x 2 ) · u r (⃗q)]e −ip·x 1−iq·x 2)|0〉 (5.43)Note, in particular, the relative minus sign that appears between these two terms. Nowlet’s see what happens when we hit this with 〈f|. We look at〈0|b r′⃗q ′ bs′ ⃗p [ ¯ψ(x ′ 1 ) · u r (⃗q)] [ ¯ψ(x 2 ) · u s e+ip′·x 1 +iq ′·x 2(⃗p)] |0〉 =2 √ E ⃗p ′E ⃗q ′[ū s′ (⃗p ′ ) · u r (⃗q)] [ū r′ (⃗q ′ ) · u s (⃗p)]e+ip′·x 2 +iq ′·x 1−2 √ [ū r′ (⃗q ′ ) · u r (⃗q)] [ū s′ (⃗p ′ ) · u s (⃗p)]E ⃗p ′E ⃗q ′The [ ¯ψ(x 1 ) · u s (⃗p)] [ ¯ψ(x 2 ) · u r (⃗q)] term in (5.43) doubles up with this, cancelling thefactor of 1/2 in front of (5.40). Meanwhile, the 1/ √ E terms cancel the relativisticstate normalization. Putting everything together, we have the following expression for〈f|S − 1 |i〉(−iλ) 2 ∫ d 4 x 1 d 4 x 2 d 4 k(2π) 4 ie ik·(x 1−x 2 )k 2 − µ 2 + iǫ([ū s′ (⃗p ′ ) · u s (⃗p)] [ū r′ (⃗q ′ ) · u r (⃗q)]e +ix 1·(q ′ −q)+ix 2·(p ′ −p))− [ū s′ (⃗p ′ ) · u r (⃗q)] [ū r′ (⃗q ′ ) · u s (⃗p)]e ix 1·(p ′ −q)+ix 2·(q ′ −p)– 116 –