Lorentz-covariant tensor integration replaces $q_{\mu }{q}_{\nu }$ by $g_{\mu \nu }{q}^2/d$. The numerator consequently contributes $(d-1)k^2{q}^2/d$. The supplied master integral with powers $a=1,b=3$ givesInserting the corrected parameter integral givesThe pole uses $\Gamma (ϵ/2)=2/ϵ+O(1)$, $D^{-ϵ/2}=1+O(ϵ)$ and ${\int }_0^1(1-x)dx=1/2$. ThusThere is no momentum-independent pole at $\lambda =0$ in this transverse gauge after removing the photon tadpole. The two-point counterterm insertion is $-i(\delta Z_{\varphi }{k}^2+\delta M^2)$, so minimal subtraction requiresThese coefficients cancel the actual one-loop insertion derived from the action. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\epsilon$ convention the same answer is $\delta Z_{\varphi }=3e^2/(16{\pi }^2\epsilon )$. Finally, $\delta M^2=0$ does not mean that the bare mass parameter is unchanged: $Z_{\varphi }{m}_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_{\varphi }+O(e^4)$. Photon, charge and quartic counterterms require their respective Green functions and cannot be determined from this scalar two-point function alone.

First remove the projector denominator explicitly:Let $A=(l+k{)}^2+m^2$ and $B=l^2$. The Feynman parameter for a simple and a squared propagator follows by differentiating the elementary two-denominator Feynman formula with respect to $B$ givesShift $q=l+xk$. Then $xA+(1-x)B=q^2+D$, with $D=x(1-x)k^2+x{m}^2$. The numerator is unchanged by this parallel shift:Therefore the correctly combined integral isThe printed weight $x/2$ is not compatible with this denominator and prefactor. Exchanging $x$ with $1-x$ changes the mass term in $D$ as well and cannot supply the missing normalization. A concrete counterexample is $m=0$ with nonzero spacelike external momentum and $2<d<4$: after Wick rotation, the remaining integral is nonzero and symmetric under $x\leftrightarrow 1-x$. The source's $4e^2\int (x/2)dx$ expression is then exactly one quarter of the correct expression. Thus the claimed equality cannot be imposed when computing the pole.

Apply the dimensionally regulated tadpole integral before removing the infrared regulator:At fixed regulator $0<\mathrm{Re}ϵ<2$ away from gamma-function poles, the power of $m_{\gamma }^2$ has positive real part. ConsequentlyBoth the correctly weighted tadpole and the source's doubled version vanish. This is the near-four-dimensional limit appropriate to the question; the order of limits is important. Equivalently the exactly massless tadpole is a scaleless integral and is zero in dimensional regularization. The statement is not valid for every arbitrary fixed dimension: for example, $ϵ=5/2$ gives a nonzero prefactor and growth as $m_{\gamma }^{-1/2}$. One must keep the regulator in its stated near-four-dimensional domain.

Route the photon momentum as $l$ and the internal scalar momentum as $l+k$. The transverse projector obeys $P_{\alpha \beta }{l}^{\beta }=0$, so the two cubic vertices giveMultiplying the two vertices and propagators therefore givesFor the seagull, $g^{\alpha \beta }{P}_{\alpha \beta }=d-1$. Including the photon-loop symmetry factor gives insteadThe printed seagull coefficient is twice this result. The vertex's factor two differentiates the two photon fields; closing them into one line requires the compensating symmetry factor. Direct expansion of the action gives the same check: the interaction density $-e^2{A}^2{\varphi }^{†}\varphi$ has one photon contraction, yielding coefficient $-e^2$, not $-2e^2$. The identical-photon origin of the vertex factor is also explicit in the standard scalar-electrodynamics rules.^[ref] This discrepancy does not affect the massless-photon limit in part d or the ultraviolet pole in part f.

The one-particle-irreducible scalar two-point function has three one-loop topologies: a scalar-photon exchange loop built from two cubic vertices; a photon tadpole at one seagull vertex; and a scalar tadpole at one quartic vertex. The last is absent when $\lambda =0$. Reducible chains of self-energy insertions belong to the Dyson expansion and are not additional one-loop self-energy graphs. The scalar-QED photon-tadpole symmetry factor is $1/2$ because its two identical photon legs are joined; the oriented complex-scalar tadpole has no such identical-real-line factor.

Gauge invariance and power counting permit the same local operators as in the original renormalizable action. With renormalized fields, a convenient counterterm parametrization isHere $D_{\mu }={\partial }_{\mu }-i{\mu }^{ϵ/2}e{A}_{\mu }$. Expanding the first term supplies kinetic, scalar-scalar-photon, and seagull counterterms with the relative coefficients required by the Ward identity. Equivalently introduce separate vertex constants first and then impose that identity. For ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, $A_0=Z_A^{1/2}A$, and $e_0={\mu }^{ϵ/2}{Z}_{e}e$, it gives $Z_e{Z}_A^{1/2}=1$. Thus charge renormalization can be represented through the photon field factor, rather than adding an independent gauge-noninvariant interaction.

The mass coefficient defined here is $\delta M^2=Z_{\varphi }{m}_0^2-m^2$; it must be distinguished from $m_0^2-m^2$. Gauge fixing may require the corresponding gauge-parameter redefinition, and vacuum diagrams can require an additive vacuum-energy counterterm. No photon mass counterterm is allowed by the unbroken gauge symmetry. The quartic scalar counterterm remains necessary for the full theory even at a scale where the renormalized $\lambda$ is zero, because gauge interactions generate scalar four-point divergences.

For $M>0$, integrate the vector action by parts, dropping the boundary term, and write it as a quadratic functional:The explicit mass makes this Proca kernel invertible: the displayed action alone has no gauge redundancy left to fix. It is interpreted as unitary gauge when obtained from a Higgs or Stückelberg completion with its would-be Goldstone field removed.

Introduce a source $J_{\mu }$ and complete the square, $A=A^{\prime }-K^{-1}J$. The normalized Gaussian path integral isThe source-independent functional determinant cancels in $Z[J]/Z[0]$. For Fourier momentum $k$, the mixed-index kernel is ${\tilde{K}}^{\mu }{}_{\nu }=-(k^2+M^2){\delta }^{\mu }{}_{\nu }+k^{\mu }{k}_{\nu }$. For $k^2\ne 0$, use the projectorsTheir orthogonality givesCombining the projectors removes their apparent massless denominator. Adding the Feynman prescription gives the unitary-gauge propagatorMultiplication by $\tilde{K}$ verifies that it inverts the quadratic operator. On the physical pole $k^2=-M^2$, the numerator is transverse to $k$, as required for the three physical massive-vector polarizations.

The drawback is its off-shell ultraviolet behaviour: at large spacelike momentum the transverse term falls as $1/k^2$, but the longitudinal tensor term remains of order $1/M^2$. Naive loop power counting therefore looks worse and renormalizability is not manifest. In a finite $R_{\xi }$ gauge the corresponding expression iswhich falls as $1/k^2$ at fixed finite $\xi$. Its $\xi \to \infty$ limit gives the unitary-gauge longitudinal term, so taking that limit before the ultraviolet limit changes power counting.

This is not a claim that a renormalizable Higgs theory ceases to be physically renormalizable in this gauge. Nor does it mean that the free Proca Gaussian integral is inconsistent. For conserved external sources $k\cdot J=0$, the longitudinal numerator drops out of $J^{\mu }{D}_{\mu \nu }{J}^{\nu }$ and the tree exchange again falls as $1/k^2$. The difficulty concerns off-shell Green functions and manifest control of loop divergences, not the number of physical polarizations.