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TODO (source gap): Figure 1(a), including its internal connections and four-momentum labels, is absent from the recoverable source. The local PDF renderer reports missing image objects, and the matching official PDF's figure could not be retrieved for visual inspection. Therefore the individual Feynman diagram called (a) cannot be assigned an integrand and its separate independence of $\xi$ cannot be verified from the available diagram. A guessed replacement drawing would not resolve this gap.

An independent determination of the complete ultraviolet pole, given in part (iv), remains possible from the scalar QED Ward identity. Together with the textually supplied integral in part (iv), it fixes the sum of the contributions outside that integral to $3i{e}^3{C}_{ϵ}{k}^{\mu }$ in the stated scalar electrodynamics three-point vertex convention. That sum alone does not identify the contents of panel (a).

The superficial degree of divergence of this scalar electrodynamics three-point vertex is one. Its ultraviolet pole is consequently a local expression at most linear in the external four-momenta. The allowed counterterm Lagrangian, with the charge-conjugation and Hermiticity properties of scalar electrodynamics, contains a single one-derivative scalar electrodynamics three-point vertex:Thus the coefficient ${\delta }_1$ is independent of the chosen external four-momenta. Setting $k^{\prime }=0$ leaves $-ie{\delta }_1{k}^{\mu }$, so a nonzero generic $k$ still determines that coefficient. It does not determine the full finite scattering amplitude at other four-momenta.

Keep $m\ne 0$ and generic off shell $k$ while extracting the ultraviolet pole. These choices avoid spurious infrared divergences; setting all scales to zero would instead produce scaleless integrals that can conceal the ultraviolet pole.

The integral printed in the text can be evaluated without reconstructing Figure 1. Put $A=l^2-m^2+i0$ and $B=(l+k{)}^2-m^2+i0$, and let $I^{\mu }$ denote that integral without its factor $\xi$. The numerator contraction responsible for its covariant gauge dependence isWith the scalar electrodynamics three-point vertex $-ie(p+p^{\prime }{)}^{\mu }$, three such vertices, two scalar propagators and one photon propagator have overall factor $(-ie{)}^3{i}^2(-i)=-e^3$. Thus the supplied integrand gives $V_{\mathrm{supplied}}^{\mu }=-e^3\xi I^{\mu }$.

First isolate the numerator term $l^2$. Its contribution isUnder $l\mapsto -l-k$ the two denominators exchange and the numerator changes sign. A shift-invariant dimensional regularization therefore gives $\boxed{I_0^{\mu }=0}$, including the finite part.

For the remaining numerator $2(l\cdot k)(2l+k{)}^{\mu }$, only its term $4l^{\mu }(l\cdot k)$ contributes to the ultraviolet pole. Corrections from expanding the denominators, and the numerator $2k^{\mu }(l\cdot k)$, are ultraviolet finite by power counting. Use auxiliary-mass extraction of ultraviolet poles with any $M>0$ rather than literally integrating a massless scaleless integral. Lorentz covariance then givesTODO (source gap): The algebra of the supplied expression is complete, but its graphical identification with Figure 1(b), including the routing at each internal vertex, cannot be visually verified from the unreadable panel.

The complete renormalization constant can nevertheless be obtained independently. Here is a short scalar QED wave-function pole calculation, to avoid assuming either missing panel's value. For an external four-momentum $p$, the exchange contribution to the scalar self-energy has numeratorIts denominator is $l^2[(p-l{)}^2-m^2+i0]$. The necessary ultraviolet poles, with the measure $d^{d}l/(2\pi {)}^d$, are respectivelyThe first two follow by a Feynman parameter and a shift of $l$; the third is a tadpole integral; the last follows by the same large-$l$ extraction used above. The seagull vertex contributes only a photon tadpole integral, which is a scaleless integral, and in any case cannot affect the coefficient of $p^2$. Multiplication by the exchange graph's factor $-e^2$ gives the scalar self-energy insertionThe kinetic counterterm $i{\delta }_2{p}^2$ therefore requires ${\delta }_2=e^2(3-\xi )C_{ϵ}$. The scalar QED Ward identity sets ${\delta }_1={\delta }_2$, soIndeed the full unrenormalized scalar electrodynamics three-point vertex has ultraviolet pole $i{e}^3(3-\xi )C_{ϵ}{k}^{\mu }$, which is cancelled by $-ie{\delta }_1{k}^{\mu }$. Subtracting the supplied contribution gives the gauge-independent sum stated in part (ii), without assigning either unreadable panel an integrand.

Expand the gauge covariant derivative in the scalar kinetic term:The cubic interaction is thereforeFor an incoming scalar momentum $p$ and outgoing momentum $q$, derivatives give $-ip$ on the incoming field and $+iq$ on its outgoing conjugate. Multiplication by the factor $i$ from the interaction action gives the scalar electrodynamics three-point vertex $-ie(p+q{)}_{\mu }$.

The other vertex comes fromDifferentiating twice with respect to the identical photon fields gives a factor of two, so the seagull vertex is $+2i{e}^2{\eta }_{\mu \nu }$. There is no additional factor of two from the scalar fields, since the complex scalar field and its conjugate are distinct legs.

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.

At order $e^2$, charged-scalar particle-antiparticle scattering has $s$ annihilation and $t$ photon exchange. For the convention $D=\partial +ieA$ and incoming momenta $p_1,p_2$, outgoing $p_3,p_4$, its amplitude isEqual external masses turn the numerators into $u-t$ and $s-u$. The scalar electrodynamics three-point vertex gives conserved currents, allowing the Coulomb-gauge propagator between conserved currents identity to replace each propagator by the covariant one. The seagull vertex has only two scalar legs and does not give a four-scalar tree at this order.