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For an overall logarithmically divergent one-loop Feynman integral, its leading large-momentum term fixes its ultraviolet pole. Giving its denominators a positive auxiliary mass $M$ regulates the artificial small-momentum singularity produced by setting external scales to zero. The difference from the original leading expansion must be ultraviolet convergent. This justifies extracting its ultraviolet pole without mistaking a scaleless integral for an ultraviolet-finite one; multiloop subdivergences require additional treatment.

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.

There is one four-gluon tadpole diagram. Temporarily replace its internal gluon propagator by $-i{g}_{\rho \sigma }{\delta }^{cd}/(l^2-m_g^2+i0)$. Contract the supplied four-gluon Feynman vertex with $g_{\rho \sigma }{\delta }^{cd}$. Its first colour factor vanishes because $f^{cce}=0$. Each of the other two terms contributes $C_A{\delta }^{ab}(D-1)g^{\mu \nu }$, soInclude the symmetry factor $1/2$ and the factor $-i$ from the internal gluon propagator:The exact regulated tadpole integral isIn particular,The entire expression is proportional to $(m_g^2{)}^{1-ϵ/2}$ and vanishes as the auxiliary mass tends to zero near four dimensions. Equivalently, its expanded terms are proportional to $m_g^2/ϵ$, $m_g^2\log m_g^2$ and $m_g^2$, all of which disappear when the regulator is removed. HenceThis is the massless scaleless integral result obtained directly in dimensional regularization. The artificial mass is only an intermediate device for this graph; retaining it in the full theory would modify the gauge symmetry and cannot be used when applying the massless transversality identity in part (iv).

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

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.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

The positive denominators in this part are Euclidean. Starting with the Minkowski spacetime Feynman rules, first perform a Wick rotation; a bare Minkowski metric integral without a pole prescription would not equal the displayed positive expression. Work initially with $m^2>0$ and $\Re n>D/2$, where the Euclidean massive loop integral converges.

Use Schwinger parameterization followed by the Gaussian integral:The remaining Gamma function integral givesThe factorial form is for positive integral $n$. Dimensional regularization defines the result elsewhere by analytic continuation.

For the scalar bubble integral, combine the two denominators with a Feynman parameter $x$ and shift the integration variable to $\ell =k-(1-x)p_E$. ThenFor $D=4-ϵ$ this isThe Gamma function recurrence gives $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral is $1+O(ϵ)$. ConsequentlyThis ultraviolet divergence is independent of the external momentum. The assumption $m^2>0$ avoids an infrared divergence. In particular the completely massless zero-momentum integral is a scaleless integral and vanishes in dimensional regularization; its ultraviolet and infrared poles must be separated before assigning it the displayed ultraviolet residue.

Every interaction contains an even number of $b$ fields, so the exact discrete symmetry $b\mapsto -b$ forbids $\Phi \to bbb$ to every perturbative order. There is no such prohibition for $\Phi \to bb$.

The latter decay channel is absent at tree level in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b{)}^2$ vertex produces a tadpole diagram with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi (\partial b{)}^2$ multiplied by an integral proportional to $\int d^4\ell /({\ell }^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed effective field theory operators. With a ultraviolet cutoff, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or counterterms.

There is a relevant qualification to a claim that the one-loop decay width must be nonzero. Since $b$ is exactly massless here, the one-vertex tadpole integrals have no external scale and are scaleless integrals. They vanish in dimensional regularization. Thus the correct conclusion is allowed, and generically generated by quantum corrections or matching, not a regulator-independent nonzero one-loop prediction from ${\lambda }_b$ alone. No exact symmetry protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.

The two identical photon fields at a scalar electrodynamics seagull give a vertex multiplicity two. Joining them into one closed photon line gives Feynman-diagram symmetry factor $1/2$. Directly expanding $-e^2{A}^2{\varphi }^{†}\varphi$ gives only one photon Wick contraction, independently checking the compensation. In a transverse gauge the remaining tensor trace is $d-1$; a massless photon tadpole is a scaleless integral in dimensional regularization.