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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.

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.

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.

Let the charge parameter depend on position temporarily. Using the gauge-covariant derivative of a charged scalar field, the variation of the matter action is $\delta S=\int d^{4}x{j}^{\mu }{\partial }_{\mu }\alpha$. The resulting Noether current isHere ${\nabla }_{\mu }\bar{\varphi }={\partial }_{\mu }\bar{\varphi }+i{A}_{\mu }\bar{\varphi }$. The potential does not contribute, and $j^{\mu }=-\delta S_{\mathrm{matter}}/\delta A_{\mu }$. Because $A$ has not been rescaled to a canonical kinetic term, there is no factor $e$ in this Noether current or in the matter interaction vertices.

A local change of variables in the path integral gives $\delta ⟨\varphi (y)\bar{\varphi }(z)⟩=⟨\delta (\varphi (y)\bar{\varphi }(z))-(\delta S)\varphi (y)\bar{\varphi }(z)⟩=0$. Integration by parts in the term containing $\partial \alpha$ produces the two opposite contact terms of the Ward identity.

For clarity about phases, transform the current correlator with $e^{i(p-p^{\prime })\cdot x-ip\cdot y+i{p}^{\prime }\cdot z}$ and suppress its overall momentum conservation Dirac delta function. Denote the result by $C^{\mu }(p,p^{\prime })$. The coordinate Ward identity becomesAmputation of the two external quantum field theory propagators then fixes the longitudinal part of the current interaction vertex. Since the action couples $A_{\mu }$ to $-j^{\mu }$, the gauge interaction vertex has the opposite sign. The phase convention for the paper's gauge interaction vertices is ${\Gamma }^{\mu }=i{G}^{\mu }$, where $G^{\mu }$ is the corresponding derivative of the Euclidean quantum effective action.

One can derive the exact one-particle-irreducible correlation function directly, without assuming the entire connected current correlator is one-particle irreducible. Local gauge invariance of the quantum effective action impliesA linear covariant gauge fixing adds a breaking term independent of the scalars, which disappears after the scalar differentiations below. The identity requires a regularization in quantum field theory and counterterms preserving the Abelian gauge theory Ward identity; scalar quantum electrodynamics has no gauge anomaly.

Differentiate with respect to $\varphi (y)$ and $\bar{\varphi }(z)$, then set all background fields to zero. The two-point derivative is the inverse exact quantum field theory propagator. Use the Fourier transform convention $\varphi (x)={\int }_p{e}^{ip\cdot x}\varphi (p)$, $\bar{\varphi }(x)={\int }_{p^{\prime }}{e}^{-i{p}^{\prime }\cdot x}\bar{\varphi }(p^{\prime })$, with the photon momentum $p^{\prime }-p$ incoming. This givesAt tree level $G^{\mu }=-(p+p^{\prime }{)}^{\mu }$, so ${\Gamma }^{\mu }=-i(p+p^{\prime }{)}^{\mu }$, and the identity reduces to $-i(p^2-p^{\prime 2})=i(p^{\prime 2}-p^2)$.

To obtain the two-photon scalar Ward identity, differentiate the same functional Ward identity also with respect to $A_{\nu }(w)$. In terms of Euclidean action derivatives it readsThis construction automatically includes the seagull vertex: the Noether current depends on $A$, with $\delta j^{\mu }(x)/\delta A_{\nu }(w)=-2{\delta }^{\mu \nu }{\delta }^4(x-w)|\varphi (x){|}^2$. It cannot be discarded when differentiating a current insertion.

Take the first photon momentum to be incoming $k$, and the second incoming $p^{\prime }-p-k$. On the first scalar leg the contact term shifts $p$ to $p+k$; on the other it shifts $p^{\prime }$ to $p^{\prime }-k$. ConsequentlyWith ${\Gamma }^{\mu \nu }=i^2{G}^{\mu \nu }=-G^{\mu \nu }$, the result isThere is a momentum-routing sign error in the printed second identity. With the scalar momentum convention of the first identity, its first shifted argument must be $p^{\prime }-k$, not $k-p^{\prime }$. This is already forced at tree level: the seagull vertex has $G^{\mu \nu }=2{\delta }^{\mu \nu }$ and hence ${\Gamma }^{\mu \nu }=-2{\delta }^{\mu \nu }$. The corrected right-hand side is $-2k^{\nu }$, matching the left-hand side, whereas the printed right-hand side is $-2p^{\prime \nu }$. Changing to an all-incoming scalar convention would also change the first identity, so it does not fix both printed formulas simultaneously.

Promoting $\alpha$ to a spacetime-dependent parameter makes the scalar kinetic term vary by both terms linear and quadratic in the gauge field. Merely adding $e{J}^{\mu }{A}_{\mu }$ accounts for the linear term but is not invariant because the scalar current itself changes under a local transformation.

Withminimal coupling givesThe missing term is the seagull vertex $4e^2{A}^2|\Phi {|}^2$. This agrees exactly with replacing partial derivatives by gauge covariant derivatives.

The Dirac kinetic term is first order in derivatives:Its expansion is only linear in $A_{\mu }$, so the current coupling already completes the gauge-invariant fermion kinetic term and no $A^2\overline{\Psi }\Psi$ term arises.

With all momenta directed into each vertex, the momentum-space Feynman rules are:

Each vertex also carries its momentum-conserving delta function; integrate every independent loop momentum and attach the appropriate external wavefunctions and photon polarization vectors.

Three connected tree-level Feynman diagrams contribute at order $e^2$: a scalar-exchange diagram in which photon $k_1$ is emitted before $k_2$ along the charged scalar line; the crossed scalar-exchange diagram with $k_1$ and $k_2$ interchanged; and the local seagull vertex joining both incoming scalars and both outgoing photons. The exchange denominators are $(p_1-k_1{)}^2-m^2$ and $(p_1-k_2{)}^2-m^2$.

Expanding the covariant kinetic term of scalar electrodynamics gives a one-photon two-scalar vertex and a two-photon two-scalar seagull vertex. Their tree amplitudes and Ward identities depend on a consistent choice of charge-flow and gauge covariant derivative convention.

Expanding the covariant scalar kinetic term gives $ie{A}_{\mu }(\varphi {\partial }^{\mu }{\varphi }^{\ast }-{\varphi }^{\ast }{\partial }^{\mu }\varphi )$. With $D=\partial +ieA$, the vertex for an incoming charged particle of momentum $p$ and an outgoing one of momentum $p^{\prime }$ is $-ie(p+p^{\prime }{)}^{\mu }$. The antiparticle has the opposite charge-flow sign. The quadratic term $e^2{A}^2{\varphi }^{\ast }\varphi$ gives the separate seagull vertex $2i{e}^2{\eta }^{\mu \nu }$. On equal-mass external shells, contracting the three-point momentum with the photon momentum gives zero, the tree scalar quantum electrodynamics Ward identity.

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.