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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.

The Grassmann field $c$ is odd and the adjoint scalar field $\varphi$ is even. The two printed signs are consistent with a right-acting BRST differential, whose graded Leibniz rule isThe bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary Lie bracket $[c,\varphi ]=c\varphi -\varphi c$ givesThe second bracket here is graded because both its entries are odd. The graded Jacobi identity gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$. ThereforeThe side of the odd derivation is essential. With the usual left graded Leibniz rule and the same two printed signs, the result would be $-[[c,c],\varphi ]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent BRST symmetry formulas here use the right-acting convention.

Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}{T}^a$, an invariant positive invariant bilinear form on a Lie algebra, and the adjoint covariant derivative $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Couplings are absorbed into this convention; restoring $D_{\mu }={\partial }_{\mu }+g[A_{\mu },\cdot ]$ multiplies each ghost interaction vertex below by $g$. The associated BRST charge acts as $Q{A}_{\mu }=-D_{\mu }c$, $Q\bar{c}=ih$, and $Qh=0$.

Write $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$. For the gauge-fixing fermion $\Psi =\int {\bar{c}}^a(F^a-i{h}^a/2)$, the right graded Leibniz rule givesThe Gaussian functional integral over the Nakanishi-Lautrup field produces the positive gauge fixing term $F^a{F}^a/2$. The ghost operator is $-\partial \cdot D-(n\cdot \partial ){\mathrm{ad}}_{\varphi }/\sqrt{2}$.

Now use the canonical free kinetic terms $F_{\mu \nu }^a{F}_{\mu \nu }^a/4+(\partial {\varphi }^a{)}^2/2$. At nonzero momentum in Euclidean space $p$, put $\eta =n\cdot p$. The quadratic kernel for $(A_{\mu }^a,{\varphi }^a)$, per color, isThe transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2{\delta }_{\mu \nu }$. The scalar quantum field theory propagator is the inverse Schur complement, not merely the inverse of the scalar diagonal entry:Hence the free adjoint-scalar propagator in scalar-dependent gauge fixing isFor completeness the mixed quantum field theory propagator is $⟨A_{\mu }^a{\varphi }^b{⟩}_0=-{\delta }^{ab}\eta p_{\mu }/(\sqrt{2}{p}^4)$; ignoring this mixing would give an incorrect scalar answer.

The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$. For a unit $n$, $p^2=p_{\perp }^2+p_{\parallel }^2$, and the free scalar quantum field theory propagator still depends on $p_{\parallel }$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless zero mode in field theory at $p=0$ needs a separate infrared prescription.

Finally, integration by parts gives the ghost action in an unambiguous convention:Use $e^{ip\cdot x}$ for the Fourier transform of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the ghost vertices in scalar-dependent gauge fixingThe free Faddeev-Popov ghost field propagator is ${\delta }^{ab}/p^2$ and every closed ghost loop contributes a minus sign. There are no further ghost interaction vertices in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the Fourier transform and ghost-ordering conventions; the displayed ghost action fixes both here.