 Solution

The conjugate complex scalar field has the opposite charge. Thus the derivative on ${\varphi }^{\ast }$ in the kinetic term must mean $(D_{\mu }\varphi {)}^{\ast }=({\partial }_{\mu }-ie{A}_{\mu }){\varphi }^{\ast }$, while the derivative on $\varphi$ is $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. This is the conjugate gauge covariant derivative. Applying the same plus-charge differential operator to both fields would not give the stated gauge-invariant theory. For example, take $A=0$, $\varphi =x^0$ and $\alpha =c{x}^0$; the literal plus-charge kinetic product changes from $1$ to $1+2iec{x}^0$. The opposite-charge interpretation removes this term.

For an arbitrary real function $\alpha (x)$, takeThen $D_{\mu }^{\prime }{\varphi }^{\prime }=e^{-ie\alpha }{D}_{\mu }\varphi$ and the conjugate derivative transforms oppositely. Also $F_{\mu \nu }^{\prime }=F_{\mu \nu }$ because mixed derivatives commute. The kinetic contraction and $|\varphi {|}^2$ are unchanged, proving local U(1) gauge symmetry.

Expanding the covariant kinetic term identifies the interactions:The first term gives the scalar electrodynamics three-point vertex, and the second gives the seagull vertex. For scalar charge flow from an incoming particle of momentum $p$ to an outgoing one of momentum $p^{\prime }$, the factors areThe factor two in the second rule comes from the two identical photon fields. Equivalently, with all momenta incoming, the three-point rule is $ie(r-q{)}^{\mu }$ for a $\varphi$ leg of momentum $q$ and a ${\varphi }^{\ast }$ leg of momentum $r$. An antiparticle line has the opposite charge-flow sign. The vertices are shown with dashed scalar lines and wavy photon lines:

For particle-antiparticle scattering, use incoming $p_1,p_2$ and outgoing $p_3,p_4$, with $p_1,p_3$ particle momenta. At order $e^2$, there are exactly two tree-level Feynman diagrams: $t$-channel photon exchange and $s$-channel annihilation. The seagull vertex has only two scalar legs and cannot alone provide four external scalar legs.

Set $q_t=p_1-p_3=p_4-p_2$ and $q_s=p_1+p_2=p_3+p_4$. The external currents areAll are transverse to the corresponding exchanged momentum: for instance $q_t\cdot J_t=p_1^2-p_3^2=0$ and $q_s\cdot J_s=p_1^2-p_2^2=0$. This is the on-shell scalar quantum electrodynamics Ward identity. With the vertex convention above,The relative sign comes from the opposite particle/antiparticle charge in the exchange diagram and the two equal annihilation-vertex signs.

To see how the Coulomb gauge expression becomes covariant, let $J,K$ be either pair, $q=(q^0,\mathbf{q})$, and initially take $|\mathbf{q}|\ne 0$. The displayed photon propagator givesCurrent conservation gives $\mathbf{J}\cdot \mathbf{q}=q^0{J}^0$ and the same identity for $K$. Since $q^2=(q^0{)}^2-|\mathbf{q}{|}^2$, the temporal coefficient simplifies asThereforeThe common Feynman pole prescription is understood in these formulas. This Coulomb-gauge propagator between conserved currents identity proves equality of the physical amplitudes despite the noncovariant individual propagator components. It should be formed before taking limits such as $\mathbf{q}\to 0$: individual instantaneous and transverse terms can be undefined separately in that limit while their sum has a finite covariant limit away from a physical pole.

Substitution gives the final scattering amplitudeHere $s,t,u$ are the Mandelstam variables for the equal-mass external scalars. This is scalar particle-antiparticle tree scattering; overall external-state phases do not affect its relative channel sign.