Quantum Field Theory

We have analogous results for the negative frequency solutions, which we may write as( √p )· σ ηs with vr †v s (⃗p) =− √ (⃗p) · v s (⃗p) = 2p 0 δ rs(4.124)p · ¯σ η s and ¯v r (⃗p) · v s (⃗p) = −2mδ rsWe can also compute the inner product between u and v. We haveū r (⃗p) · v s (⃗p) = ( ξ r † √ p · σ , ξ r †√ p · ¯σ ) γ 0 ( √p · σ ηs− √ p · ¯σ η s )= ξ r†√ (p · ¯σ)(p · σ)η s − ξ r †√ (p · ¯σ)(p · σ)η s = 0 (4.125)and similarly, ¯v r (⃗p) · u s (⃗p) = 0. However, when we come to u † · v, it is a slightlydifferent combination that has nice properties (and this same combination appearswhen we quantize the theory). We look at u r † (⃗p) · v s (−⃗p), with the 3-momentum inthe spinor v taking the opposite sign. Defining the 4-momentum (p ′ ) µ = (p 0 , −⃗p), wehaveu r † (⃗p) · v s (−⃗p) = ( ξ r † √ p · σ , ξ r †√ p · ¯σ ) ( √ )p′ · σ η s− √ p ′ · ¯σ η s= ξ r†√ (p · ¯σ)(p ′ · σ)η s − ξ r †√ (p · ¯σ)(p ′ · σ)η s (4.126)Now the terms under the square-root are given by (p·¯σ)(p ′·σ) = (p 0 +p i σ i )(p 0 −p i σ i ) =p 2 o − ⃗p 2 = m 2 . The same expression holds for (p ′ · ¯σ)(p · σ), and the two terms cancel.We learnOuter Productsu r † (⃗p) · v s (−⃗p) = v r † (⃗p) · u s (−⃗p) = 0 (4.127)There’s one last spinor identity that we need before we turn to the quantum theory. It is:Claim:2∑s=1u s (⃗p) ū s (⃗p) = /p + m (4.128)where the two spinors are not now contracted, but instead placed back to back to givea 4 × 4 matrix. Also,2∑s=1v s (⃗p) ¯v s (⃗p) = /p − m (4.129)– 104 –