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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).

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.

Normalize the generating functional by its value with all sources zero, and take the connected part of the time-ordered correlation function. Source differentiation of the interaction exponential is equivalent to the Dyson series followed by Wick's theorem. Write $D(x-y)=⟨T\varphi (x)\varphi (y){⟩}_0$ and $F_{ij}(x-y)=⟨T{\psi }_i(x){\bar{\psi }}_j(y){⟩}_0$ temporarily, so that every contraction has an unambiguous meaning.

The second-order Dyson series contributesThere are two connected external scalar field attachments:They become equal after exchanging the integration variables $x,y$, cancelling $2!$. The connected fermion contractions have the closed fermion loop sign:The minus sign can also be checked directly: the pairing of the first with the fourth field and the second with the third has positive pairing parity, but $⟨T{\bar{\psi }}_i(x){\psi }_j(y)⟩=-F_{ji}(y-x)$. Multiplying it by $(ig{)}^2=-g^2$ leaves a positive result:A contraction of the two external fields with each other times a vacuum bubble cancels against the normalization. Separate fermion tadpole integrals at $x$ and $y$ belong to a product of one-point functions, removed by taking the connected part.

To match the source's generating functional convention, distinguish its displayed kernels from contractions. The positive $iJ\Delta J/2$ exponent gives $D=-i\Delta$. With the source insertion order in equation (5), the positive $i\bar{\eta }{S}_F\eta$ exponent similarly gives $F=-i{S}_F$. The two scalar propagators and two Dirac propagators supply a total phase $(-i{)}^4=1$. Thus, writing $S=S_F$ as in the requested formula,The matrix trace closes the spinor indices; no extra factor of two or additional symmetry factor remains for a Dirac field with this Yukawa interaction.

It is convenient to evaluate the one-particle irreducible vertices using the Euclidean action and then translate the counterterms back to the Minkowski spacetime Lagrangian density. For the tadpole integral, analytic continuation of solution givesIndeed $\Gamma (-1+ϵ/2)=\Gamma (ϵ/2)/(ϵ/2-1)=-2/ϵ+O(1)$. The negative pole is a property of analytic continuation, not the value of a convergent positive integral.

The Euclidean space inverse two-point vertex has the one-loop correctionThe factor $1/2$ is the symmetry factor weight of the tadpole diagram. A mass counterterm $\delta m^2={\lambda }_0{m}^2/(16{\pi }^2ϵ)$ cancels this pole. There is no momentum-dependent pole at this order, and hence no one-loop wave-function renormalization.

For the four-point vertex, the three bubble diagrams giveOne can check the relative minus sign independently from the one-loop effective action: its $\frac{1}{2}\mathrm{Tr}\log (1+G{\lambda }_0{\varphi }^2/2)$ term has a positive term linear in ${\lambda }_0{\varphi }^2$ and a negative term quadratic in it. Thus $\delta \lambda =3{\lambda }_0^2/(16{\pi }^2ϵ)$ cancels the quartic pole.

In the Euclidean action, these two counterterms are $+\delta m^2{\varphi }^2/2+\delta \lambda {\varphi }^4/4!$. Returning to the Minkowski spacetime Lagrangian density givesFor a direct sign check with Feynman rules, the two amputated loop pole insertions are $+i{\lambda }_0{m}^2/(16{\pi }^2ϵ)$ and $+i3{\lambda }_0^2/(16{\pi }^2ϵ)$, while the counterterm insertions are $-i\delta m^2$ and $-i\delta \lambda$.

As usual, the displayed pole residues suppress renormalization scale powers: $[{\lambda }_0]=ϵ$ away from four dimensions. The next part restores these powers explicitly with a dimensionless renormalized coupling. Replacing the tree coupling by its renormalized counterpart inside a one-loop residue changes only higher-loop terms.

The single-vertex scalar two-point graph at first order in $g_{2k}$ has one $2k$-valent interaction vertex. Attach the two external lines to it and pair its remaining $2k-2$ legs into $k-1$ tadpole diagrams. The graph is connected, has $L=k-1$ independent loop momenta, and is momentum independent after amputation apart from the overall Dirac delta function. In particular, first order in the coupling need not mean one loop.

The number of contractions with fixed external labels isDividing by the interaction factorial $(2k)!$ gives the inverse Feynman-diagram symmetry factor $1/[2^{k-1}(k-1)!]$. Use $m>0$ for the mass in the lowest-order propagator and define the tadpole integralThe linear-order correction to the amputated function is consequentlyThis is the correction to the paper's amputated connected correlator, not a definition of an inverse-propagator self-energy with an unspecified sign. At this order using $m_0$ instead of $m$ gives the same lowest-order integral.

For completeness, Schwinger parameterization and a Gaussian integral giveThis first converges for $0<D<2$ and then defines its dimensional regularization continuation. At $D=4-2ϵ$, the Gamma function expansion yieldsThus its highest ultraviolet pole isFor $k=2$ the remainder means finite terms. For $k>2$ the unrenormalized graph has a higher-order pole from its independent ultraviolet subdivergences, rather than only a simple pole. A massless tadpole integral is scaleless and vanishes in dimensional regularization; that does not justify discarding the massive ultraviolet divergence by setting $m=0$ before separating ultraviolet and infrared effects.

There are three independent nonfactorized momentum integrals, namely a tadpole integral, the three-line sunset diagram integral, and the five-line banana integral. The six displayed Feynman diagrams involve six complete integral expressions:The distinction between independent building blocks and their products removes an ambiguity in counting “different momentum integrals”. For the strict divergent-wave-function renormalization interpretation of ii, the graphs in c reduce to $T,T^2,B_5$, comprising three full integral expressions and two independent nonfactorized building blocks.

Make a Wick rotation and use $m>0$ to separate ultraviolet divergences from infrared divergences. Write $D_E(k)=1/(k^2+m^2)$, and let $T,B_r(P)$ be the Euclidean versions of the integrals above, with their harmless overall factors of $i$ removed. A spherical ultraviolet cutoff givesThus the one-vertex quartic graph is linearly divergent, and the one-vertex sextic graph is quadratically divergent as $T_{\Lambda }^2$.

For the other two building blocks, the spatial Fourier transform is especially transparent. The three-dimensional Euclidean scalar propagator isThe convolution theorem converts the $r$ parallel lines into $G(x{)}^r$. Angular integration gives, with a short-distance cutoff $a={\Lambda }^{-1}$,This is a different convenient cutoff regularization for the connected block; power-divergent coefficients depend on the cutoff convention, whereas the conclusions about powers and the presence of a kinetic logarithm do not. The sinc function is understood to be $1$ at $P=0$.

For $r=3$, the integrand starts as ${\rho }^{-1}+O(1)$, soIts divergent part is independent of external momentum. In particular,The finite difference and its finite derivative show why the pure three-line sunset diagram does not need a divergent wave-function renormalization in three dimensions.

For $r=5$, the small-distance Taylor expansion isConsequentlyIn particular, the nonzero $P^2\log \Lambda$ term requires a genuine divergent wave-function renormalization, in addition to the mass counterterm. The coefficients here refer to the displayed position-space cutoff and have not included Feynman-diagram symmetry factors or bare couplings.

The comparison with the superficial degree of divergence isThus the actual leading power agrees with $\omega$ in every case. The additional logarithms in products come from ultraviolet subdivergences, which uniform power counting alone does not resolve.

For $T{B}_3$ and $T^2{B}_3$, differentiation with respect to $P^2$ removes the logarithmic divergence of $B_3$, but leaves divergent factors $T$ or $T^2$ multiplying a finite, nonpolynomial function of $P^2$. These are subdivergent contributions, not new nonlocal counterterms: subtracting each tadpole diagram subdivergence and the momentum-independent $B_3$ subdivergence removes them. For example, in a factorized subtraction scheme $T{B}_3\mapsto (T-T_{\mathrm{div}})(B_3-B_{3,\mathrm{div}})$, and likewise for $T^2{B}_3$. The remaining $B_5$ kinetic logarithm explains the strict interpretation in ii. A dimensional regularization scheme can suppress power divergences; it does not invalidate this cutoff-based comparison with $\omega$.

Set the renormalized $g_3$ to zero. For a six-scalar vertex, b gives $\omega =-n_2$, so a new overall logarithmic divergence can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.

At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional toThis vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading $/k$ by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega =-3$. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. ThereforeThe six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in $2+1$ dimensions. In this representation,The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega =0$, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put $\varphi =v+\eta$ with constant $v$. Define the background fermion mass $\mu =m+g_1{v}^2/2$, the background scalar squared mass ${\mathcal{B}}^2=M^2+g_2{v}^2/2$, and the induced three-leg Yukawa coupling $y=g_{1}v$. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, isHere the dots and gamma-matrix traces are Euclidean. With $A=k^2+{\mu }^2$, $B=l^2+{\mu }^2$, $C=(k-l{)}^2+{\mathcal{B}}^2$, its numerator is exactlyThe first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term isFor positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to ${\int }^{\Lambda }dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying $H$ containsThe two-loop sextic counterterm in quartic Yukawa theory follows because its $v^6$ coefficient is $g_1^4\ne 0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. ThusA sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to $g_3=0$ at every scale lacks counterterm closure when $g_1\ne 0$. The decoupled exception $g_1=0$ is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.

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.

In scalar electrodynamics with photon propagator $-i[g_{\mu \nu }-(1-\xi )l_{\mu }{l}_{\nu }/l^2]/(l^2+i0)$ and dimensional regularization $d=4-2ϵ$, the minimal subtraction scheme gives $Z_{\varphi }=1+e^2(3-\xi )/(16{\pi }^2ϵ)+O(e^4)$. For an external four-momentum $p$, expand the exchange numerator as $4p^2-4\xi p\cdot l+\xi l^2-4(1-\xi )(p\cdot l{)}^2/l^2$. The logarithmic ultraviolet poles of the scalar, vector and tensor integrals give the $p^2$ coefficient $4-2\xi -(1-\xi )=3-\xi$. A massless photon tadpole integral does not change it. The scalar QED Ward identity gives the same vertex renormalization constant.

For interaction ${\lambda }_{2k}{\varphi }^{2k}/(2k)!$, one vertex with two labelled external legs has $k-1$ independently paired tadpole diagrams. Counting contractions gives symmetry factor $S=2^{k-1}(k-1)!$. In a normalized Euclidean path integral, its amputated contribution is $-{\lambda }_{2k}{T}_D(m{)}^{k-1}/S$, where $T_D$ is the tadpole integral. It is linear in the coupling but has $k-1$ loops and a highest four-dimensional ultraviolet pole of order $k-1$ for nonzero mass.