Quantum Field Theory

The D 00 piece of the propagator doesn’t look a whole lot different from the transversephoton propagator. But wouldn’t it be nice if they were both part of something moresymmetric! In fact, they are. We have the following:Claim: We can replace the propagator D µν (p) with the simpler, Lorentz invariantpropagatorD µν (p) = −i η µνp 2 (6.85)Proof: There is a general proof using current conservation. Here we’ll be more pedestrianand show that we can do this for certain Feynman diagrams. In particular, wefocus on a particular tree-level diagram that contributes to e − e − → e − e − scattering,pp /∼ e 2 [ū(p ′ )γ µ u(p)] D µν (k) [ū(q ′ )γ ν u(q)] (6.86)q/qwhere k = p − p ′ = q ′ − q. Recall that u(⃗p) satisfies the equation( /p − m)u(⃗p) = 0 (6.87)Let’s define the spinor contractions α µ = ū(⃗p ′ )γ µ u(⃗p) and β ν = ū(⃗q ′ )γ ν u(⃗q). Thensince k = p − p ′ = q ′ − q, we havek µ α µ = ū(⃗p ′ )( /p − /p ′ )u(⃗p) = ū(⃗p ′ )(m − m)u(⃗p) = 0 (6.88)and, similarly, k ν β ν = 0. Using this fact, the diagram can be written asα µ D µν β ν = i= i= −(⃗α · ⃗βk 2(⃗α · ⃗β− (⃗α ·⃗k)( β ⃗ · ⃗k) )k 2 | ⃗ + α0 β 0k| 2 | ⃗ k| 2)− k2 0 α 0β 0k 2 k 2 | ⃗ k| + α 0β 02 | ⃗ k| 2(⃗α · ⃗βk 2 − 1k 2 | ⃗ k| 2 (k2 0 − k 2 ) α 0 β 0)= − i (k 2α · β = αµ − iη )µνβ ν (6.89)k 2– 142 –