Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 50 / 3 / d / Solution

At order $e^2$, scalar particle–antiparticle scattering has a $t$-channel photon exchange and an $s$-channel annihilation and recreation diagram. They are shown below, with incoming momenta $p,q$ and outgoing momenta $p^{\prime },q^{\prime }$. The seagull vertex has only two scalar legs and does not yield a four-scalar tree diagram at this order.

For the exchange graph, let $k=p-p^{\prime }=q^{\prime }-q$. Apart from overall charge signs, the two scalar currents are $J=p+p^{\prime }$ and $K=q+q^{\prime }$. The external mass shell conditions giveFor the annihilation graph, $k=p+q=p^{\prime }+q^{\prime }$, with currents $J=p-q$ and $K=p^{\prime }-q^{\prime }$, again up to vertex signs. Their contractions are $p^2-q^2=0$ and $p^{\prime 2}-q^{\prime 2}=0$. Thus each graph couples the photon to conserved currents, the relevant scalar quantum electrodynamics Ward identity.

Write $k=(\omega ,\mathbf{k})$ and ${\kappa }^2=|\mathbf{k}{|}^2$. conserved current means $\mathbf{k}\cdot \mathbf{J}=\omega J^0$ and $\mathbf{k}\cdot \mathbf{K}=\omega K^0$. For $\kappa \ne 0$, contraction with the Coulomb-gauge photon propagator gives, displaying the rational identity before the pole boundary limit,We used $k^2={\omega }^2-{\kappa }^2$. Restoring the common Feynman i-epsilon prescription yields the Coulomb-gauge propagator between conserved currents resultHence each on-shell tree amplitude can use the Lorentz-invariant propagator. This is equality after contraction, not equality of the gauge-dependent tensors. At $\mathbf{k}=0$, the separate Coulomb-gauge terms are ill-defined; combine them first and take the conserved-current limit. The combined expression has only the physical photon pole.