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The two-point function Feynman integral with two interaction vertices joined by five parallel scalar propagators has four independent loop momenta:This formula uses Euclidean momenta. In three dimensions the convolution theorem and $G(x)=e^{-m|x|}/(4\pi |x|)$ givewith cutoff regularization required at $\rho =0$. With a lower radial cutoff $a$, the coefficient of $p^2$ has ultraviolet divergence $-\log (1/(ma))/(1536{\pi }^4)$. It therefore generates a divergent wave-function renormalization in three-dimensional sextic scalar field theory. By contrast the three-line sunset diagram has only a momentum-independent logarithmic ultraviolet divergence in three dimensions. This illustrates why a superficial degree of divergence of two allows both a mass counterterm and a kinetic counterterm, while degree zero allows only the former in a two-point function.

A mass counterterm cancels the divergent part of a self-energy that has the tensor structure of a mass term. Its finite part depends on the renormalization scheme, while an on-shell scheme chooses it so that the renormalized mass equals the pole mass.

With additive counterterms $\frac{1}{2}\delta Z(\partial \varphi {)}^2+\frac{1}{2}\delta m^2{\varphi }^2$, their Euclidean propagator insertion is $-(\delta Z{k}^2+\delta m^2)$. These minimal subtraction scheme coefficients cancel the cubic bubble insertion. A bare mass shift also subtracts $m^2\delta Z$ after wavefunction renormalization; it is not the same coefficient as the additive mass counterterm.

The self-energy bubble diagram at loop order one in six-dimensional cubic scalar field theory has Feynman-diagram symmetry factor two. With positive Euclidean scalar propagators, its amputated Green's function is the displayed integral. Its superficial degree of divergence is two. Expanding at large internal momentum and taking the angular average shows that its divergent part is a constant plus a multiple of $p^2$, precisely the mass counterterm and wave-function renormalization structures.

For a spherical ultraviolet cutoff $|k|\le \Lambda$ and $m^2>0$, its value at $p=0$ isThe divergent $p^2$ coefficient is $-g^2\log ({\Lambda }^2/m^2)/(768{\pi }^3)$. Finite terms depend on the regularization convention. Tadpole diagram insertions are excluded from the one-particle-irreducible Feynman diagram, and are removed from the connected two-point function by imposing a vanishing one-point function.

In a common finite-volume regularization in quantum field theory, the oscillator contribution is $\Delta M_{\mathrm{osc}}=\frac{1}{2}{\sum }_n({\Omega }_n-{\Omega }_n^{(0)})$, including every discrete and continuum-derived mode and treating translation as a collective coordinate. Physical frequencies are ${\Omega }_n=m{\omega }_n$ when coordinates are measured in units of $m^{-1}$. A mass counterterm is added only afterwards. The vacuum subtraction removes the extensive vacuum energy but can leave an ultraviolet divergence.

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.

To keep the paper's coupling convention explicit, write $g=\sqrt{\beta }$ and use physical coordinates $X=x/m$, $T=t/m$. Assume $\beta >0$ and weak coupling $g\ll 1$. The canonical real scalar field is $\phi =\varphi /g$, and the physical potential isThe symbol $g$ is the coupling usually appearing in canonical Sine-Gordon theory; the paper's $\beta$ is its square. The vacua have $g\phi =2\pi n$. A static Sine-Gordon kink has $g{\phi }_K=4\arctan e^{m(X-X_0)}$ and satisfies $\frac{1}{2}{\phi }_K^{\prime 2}=U({\phi }_K)$. Since $1-\cos (g{\phi }_K)=2{\mathrm{sech}}^2[m(X-X_0)]$, its classical mass isThe topological sector is important: this energy is measured relative to a vacuum, and the kink joins distinct vacua at the two spatial ends. Expanding around a spatially constant vacuum cannot construct this state by any finite-order perturbation in $g$.

For the general one-loop soliton mass correction, start with a canonical scalar field potential $U$ and a stable static soliton ${\phi }_K$. Write $\phi ={\phi }_K+\eta$. The term linear in $\eta$ vanishes by the classical Euler-Lagrange field equation. The quadratic fluctuation Hamiltonian isChoose a common large box and a common finite-mode regularization in quantum field theory. Expand the nonzero normal modes as $\eta ={\sum }_n{q}_n{f}_n$, with ${\mathcal{H}}_K{f}_n={\omega }_n^2{f}_n$ and normalized eigenfunctions. Each pair $(q_n,p_n)$ is a quantum harmonic oscillator contributing ground-state energy $\hslash {\omega }_n/2$. In the vacuum, replace ${\mathcal{H}}_K$ by ${\mathcal{H}}_0=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{vac}})$. Subtract the two ground-state energies and add the local counterterms evaluated on the soliton relative to the vacuum. This derives the general formulaThe prime excludes exact zero modes in field theory from oscillator quantization; their zero frequencies contribute no zero-point energy, but their role in mode counting must not be forgotten. The two sums mean a paired finite regulator, not separate subtractions of divergent answers. Equivalently the nonzero-mode term is the regulated difference of the square-root traces of the two fluctuation operators. Discrete bound states and continuum modes both contribute. This is a vacuum-subtracted soliton mass and the first term in the semiclassical soliton mass expansion.

Translation gives a zero mode in field theory because differentiating the static equation yields ${\mathcal{H}}_K{\phi }_K^{\prime }=0$. A Gaussian oscillator or an unprimed functional determinant is inappropriate along this flat direction. Replace its amplitude by the position collective coordinate $X_0(T)$ and require the residual fluctuation to obey $\int \eta {\phi }_K^{\prime }dX=0$, preventing double counting. The associated change-of-variables Jacobian supplies the zero-mode normalization. At low speed the collective-coordinate effective Lagrangian for a soliton is $-M_{\mathrm{cl}}+\frac{1}{2}{M}_{\mathrm{cl}}{\dot{X}}_0^2+\cdots$; quantizing the position gives the soliton momentum and its translational states, not an extra oscillator rest energy. More generally, every physical continuous modulus needs a collective coordinate; gauge directions require gauge fixing rather than additional physical states.

The Sine-Gordon kink fluctuation operator is particularly simple:while $D{D}^{†}=-{\partial }_X^2+m^2={\mathcal{H}}_0$. This supersymmetric factorization of the one-soliton potential shows stability and generates all nonzero eigenfunctions from vacuum plane waves. The sole normalizable bound state is the translational zero mode of a sine-Gordon kink, proportional to $\mathrm{sech}m(X-X_0)$; there is no positive-frequency internal bound oscillator. The continuum has $\omega (k)=\sqrt{k^2+m^2}$ and no reflection. Applying $D^{†}$ to $e^{ikX}$ gives a transmission amplitudeThis scattering phase shift changes the density of continuum modes. The high-frequency vacuum subtraction cancels the extensive vacuum contribution but still leaves a logarithmic ultraviolet divergence.

The finite part also requires consistent mode-number regularization of soliton masses. The periodic-box phase-shift quantization condition is $k_{n}L+\delta (k_n)=2\pi n$. Match $2N+1$ modes: the kink has its translation mode plus the two continuum modes at each $n=1,\dots ,N$, while the vacuum has the $k=0$ oscillator of frequency $m$ and the corresponding continuum pairs. With $\hslash =1$ and $\Lambda =2\pi N/L$, expanding $k_n-k_n^{(0)}=-\delta (k_n^{(0)})/L+o(L^{-1})$ givesIntegration by parts uses $\delta (0)=\pi$ and yieldsThe cutoff surface term for a Sine-Gordon kink tends to $-m/\pi$. It cannot be dropped merely because $\delta (\Lambda )\to 0$: $\omega (\Lambda )$ grows at the same time.

A renormalization condition must specify which mass and coupling are held fixed. For vacuum normal ordering, or cancellation of the vacuum tadpole diagram with the elementary mass fixed at $m$, the quartic interaction gives the mass countertermEvaluating this Sine-Gordon vacuum tadpole counterterm on the kink givesThe logarithmic ultraviolet divergence cancels, leaving the renormalized one-loop Sine-Gordon kink mass in the stated vacuum scheme:This example illustrates why a zero mode in field theory must be treated as a collective coordinate, why vacuum subtraction alone need not remove ultraviolet divergences, and why the finite relation between the two regulators matters. Different finite counterterms amount to different definitions of the renormalized parameters; an unregulated frequency difference without a renormalization condition is not a physical mass prediction.

In the convention implied by the printed expansion, the truncated two-point quantity is the inverse connected propagator, or one-particle-irreducible two-point vertex:It is the second functional derivative of the quantum effective action (or statistical Legendre effective action) about a translationally invariant zero-field equilibrium. The relation follows from the inverse Hessian relation for a connected two-point function. The subscript $c$ in $G$ already removes disconnected one-point products; the word “truncated” here is not a request to subtract that product a second time. Nor is a general amputated connected correlation function interchangeable with a one-particle-irreducible correlation function.

Use the zero-momentum renormalized mass $R=m^2(0,T)$ in the free propagator ${\tilde{G}}_0(p)=(p^2+R{)}^{-1}$, and split the quadratic coupling into $R$ plus a mass counterterm $\delta m^2=m^2(\Lambda ,T)-R$. The self-energy $\Sigma (p)$ is the sum of loop 1PI insertions, excluding the separately displayed counterterm. Summing repeated insertions by Dyson resummation givesThe sign convention is that a positive tadpole shifts the inverse propagator upwards. At first order, the connected correlation function correction is $-G_0^2\Sigma$, consistent with this inverse-propagator convention. A different split between the reference mass and counterterm produces the same renormalized result.

For the positive quartic interaction $g{\varphi }^4/4!$, the one-loop tadpole diagram has no external-momentum dependence. Its symmetry factor is $1/2$: assigning the two external legs to four vertex fields gives $4\cdot 3$ contractions, and division by $4!$ gives $1/2$. With the dimensionless statistical-action convention,The physical zero-momentum condition $\tilde{\Gamma }(0)=R$ sets $\delta m^2+{\Sigma }_1(0)=0$, hence the one-loop mass relation isWriting the internal line with $R$ is a renormalized or self-consistent one-loop convention. Away from critical infrared singularities it differs from a bare-mass insertion only at higher perturbative order. This equation does not by itself provide exact critical exponents once loop corrections become large.

For $D>2$, subtract the critical-temperature condition $0=m^2(\Lambda ,T_C)+(g/2)I_D(0)$. Take $g$ and the regular coefficients at their critical values, absorbing smooth changes into a coefficient $A>0$. Sincethe one-loop critical-mass subtraction becomesLet $K_D$ be the area of the unit $(D-1)$-sphere divided by $(2\pi {)}^D$. Radial integration gives $J_D(R)=K_D{\int }_0^{\Lambda }{k}^{D-3}(k^2+R{)}^{-1}dk$.

For $D>4$, $J_D(0)=K_D{\Lambda }^{D-4}/(D-4)$ is infrared finite, so it merely renormalises the coefficient and $R\propto T-T_C$ is consistent. For $2<D<4$, setting $k=\sqrt{R}x$ yieldswith a finite positive dimensionless integral. The correction is singular relative to the term $R$, invalidating the finite-coefficient linear-mass assumption. At $D=4$,so the boundary is logarithmically marginal. For $D\le 2$, even the subtraction using $I_D(0)$ needs an infrared regulator; it cannot be used to restore a finite linear critical expansion. Thus the ordinary upper critical dimension isA fixed-coupling self-consistent one-loop formula is not the full marginal renormalization group analysis, but its logarithm already shows why an uncorrected linear power law is not generic at $D=4$.

At a tricritical point, both the quadratic and quartic scaling directions must be tuned; the leading stabilising interaction is sextic. With canonical scalar-field engineering dimension $(D-2)/2$, the sextic coupling has eigenvalue $D-6(D-2)/2=6-2D$. It becomes marginal at $D=3$, givingLower even couplings generated by coarse-graining must remain tuned. This is why using an untuned quartic tadpole to diagnose a tricritical point would give the wrong boundary. The tricritical sextic beta function supplies marginal logarithmic corrections at three dimensions.

For a general multicritical even Landau potential, assume the lower stabilising even terms have been tuned away and the first remaining one is $A_{2n}{M}^{2n}$, with $n\ge 2$ and $A_{2n}>0$. Minimising the potential on its ordered branch givesIts curvature at the minimum is $4(n-1)a_2|t|$. For a finite positive gradient stiffness $c$, the longitudinal correlation length therefore scales as $\xi \propto |t{|}^{-1/2}$. The Ginzburg criterion compares the order-parameter fluctuation averaged over a correlation volume with this squared mean-field value. Keeping momenta of order ${\xi }^{-1}$ or less,Consequently the multicritical Ginzburg ratio behaves asOnly for a positive exponent do these relative fluctuations vanish on approaching the critical point. Thus the general upper critical dimension isEquivalently the interaction eigenvalue $y_{2n}=2n-(n-1)D$ vanishes there. The cases $n=2$ and $n=3$ reproduce 4 and 3 respectively.

For $D<D_C$, the ratio diverges and the mean-field assumptions lose self-consistency arbitrarily close to the transition. At $D=D_C$ the marginal Ginzburg criterion is scale-independent at this leading estimate, rather than tending to zero; the criterion alone does not prove a divergence or force new power indices. Marginal interactions require a renormalization group calculation and generally give logarithmic corrections, as for the quartic and sextic cases above. The boundary case is marginal, not a strict power-law divergence. This qualifies the printed wording at equality while recovering the requested upper critical dimensions.

Use a mass counterterm and a wavefunction renormalization counterterm, defining their additive coefficients byThe sign follows from how the Euclidean path integral expands. A quadratic counterterm inserts $-\delta K$ into the propagator, whereas the loop defined in the question inserts $+I$. To this order,Thus cancellation requires $\delta K=I_{\mathrm{div}}$, rather than its negative. In the minimal subtraction scheme, with no finite parts added,These are the minimal-subtraction two-point counterterms in cubic scalar theory. They absorb respectively the $k^2$ and constant terms in the two-point pole.

The additive mass coefficient is distinct from the shift of a bare mass when the bare field also includes wavefunction renormalization. If ${\varphi }_B=\sqrt{1+\delta Z}\varphi$ and $(1+\delta Z)m_B^2=m^2+\delta m^2$, then to one-loop orderThis last relation specifies the convention; the boxed coefficients are those multiplying the local counterterms displayed above.

For every interacting $n\ge 3$, the one-loop scalar effective action already contains counterexamples. On a constant background, with a positive auxiliary mass $\mu$ to control infrared divergences, its field-dependent part isThe term of order ${\lambda }^V$ has $E=(n-2)V$ external scalar fields and is a one-loop polygon one-particle-irreducible Feynman diagram. Choose $V>n/(n-2)$ and $2V>d$. Then $E>n$, its loop momentum integral is ultraviolet convergent, and its coefficient is nonzero. Removing the ultraviolet cutoff does not remove this finite higher-point scalar vertex. For $d>2$, generic nonexceptional external momenta give the same conclusion without retaining the auxiliary mass. In $d\le 2$, the massless theory still needs an infrared prescription; removing the ultraviolet cutoff does not remove that need or the induced vertices. Hence there are no interacting pairs $(n,d)$ under the literal wording. For $n=2$ the functional determinant is field independent; $n=1$ merely shifts a Gaussian integral when an infrared prescription exists.

There is a different conventional interpretation: absence of new independent divergent counterterms. The superficial degree of divergence of a connected diagram isThe standard perturbative counterterm closure criterion is $[\lambda ]\ge 0$, with the same pairs as in part ii, provided one includes all symmetry-allowed lower-degree potential terms, the kinetic term, and the vacuum energy. A pure monomial family need not itself have counterterm closure: a sextic interaction in three dimensions generates a quartic counterterm, and a quartic interaction generates a mass counterterm. This interpretation concerns the local divergent part or the continuum defining action, not the complete quantum effective action with its finite interaction vertices and derivative expansion.

For the massive four-dimensional quartic scalar field theory, write ${\lambda }_B$ for the bare quartic coupling, ${\lambda }_R$ for its renormalized value, and $m>0$ for the mass held fixed by a renormalization condition. Define the scalar bubble integralThe quartic one-particle-irreducible correlation function, defined as a derivative of the Euclidean quantum effective action, isThere are three bubble diagrams, one for each pairing of external momenta, and each has Feynman-diagram symmetry factor $2$. The minus sign follows equivalently from the quadratic term in the functional determinant $\frac{1}{2}\mathrm{Tr}\log [1+({\lambda }_B/2)(-{\partial }^2+m^2{)}^{-1}{\varphi }^2]$.

Radial integration at zero external momentum givesThe bare quartic vertex is not finite at fixed bare coupling. Its logarithmic ultraviolet divergence must be subtracted. Impose the momentum-subtraction scheme condition ${\Gamma }_R^{(4)}(0,0,0,0)={\lambda }_R$. To one loop this requiresFor the finite difference, a Feynman parameter and a shift of loop momentum giveThe boundary error due to shifting a sharp cutoff vanishes in this limit. Substitution yields the finite renormalized quartic scalar vertexThus finiteness requires holding the renormalized coupling fixed and allowing the bare coupling to depend on ${\Lambda }_0$. The unsubtracted assertion would be false.

The same one-loop scalar effective action explains the other effective interactions. For a constant background its expansion isThe $r=1$ tadpole diagram gives a mass correction $\delta m^2={\lambda }_R{I}_1/2$, withIt requires a mass counterterm. The field-independent vacuum energy also requires subtraction. The $r=2$ term is the quartic logarithm already treated. There is no one-loop wave-function renormalization from the quartic tadpole diagram.

For $r\ge 3$ the polygon diagrams generate finite higher-point scalar vertices, withFor example, the induced sextic term in the effective potential is $+{\lambda }_R^3{\varphi }^6/(768{\pi }^2{m}^2)$, or a six-point interaction vertex $15{\lambda }_R^3/(16{\pi }^2{m}^2)$ when normalized by $6!$. These finite terms survive the removal of the ultraviolet cutoff at fixed $m,{\lambda }_R$.

The external-momentum dependence of the bubble diagrams also generates a derivative expansion of quartic interactions. At small $P$, the subtracted scalar bubble integral is $-P^2/(96{\pi }^2{m}^2)+O(P^4/m^4)$, giving finite derivative couplings. At order ${\lambda }_R^2$ these are the new field-dependent interactions beyond the mass and quartic terms; the sextic and higher interaction vertices require higher powers of ${\lambda }_R$, though they still occur at one loop. Such coefficients are suppressed by powers of the physical mass or external momentum, not by powers of ${\Lambda }_0$. The calculation establishes an order-by-order perturbative limit; it does not establish a nonperturbative interacting continuum limit of a quantum field theory in four dimensions.

Restore the dimensional-regularization factor ${\mu }^{ϵ}$ and defineUsing $\Gamma (ϵ/2)=2/ϵ-\gamma +O(ϵ)$ givesThe first term is the ultraviolet divergence. In the modified minimal subtraction scheme, subtraction of $2/ϵ-\gamma +\log 4\pi$ leavesOn the tree-level mass shell, $p^2=-m^2$ and $i/p=-m$, so $\Delta =x^2{m}^2$. The mass counterterm isThe pole condition $m_{\mathrm{phys}}=m-{\Sigma }_{\overline{\mathrm{MS}}}{|}_{i/p=-m}+O(e^4)$ therefore givesEvaluating the elementary parameter integral yields the equivalent expression

For a connected amputated Feynman diagram, write $L$ for the number of independent loop momenta. Each loop momentum integration scales as ${\lambda }^4$, each internal scalar propagator as ${\lambda }^{-2}$, and each internal Dirac propagator as ${\lambda }^{-1}$ when both four-momenta and mass are scaled. The Yukawa coupling has mass dimension zero. ThusCounting the two fermion ends and one scalar end at each interaction vertex, and using connectedness, givesConsequently $n_f=n_v$, $L=n_s+1$, andThe requested exponent requires amputation and removal of the overall momentum delta function. An ordinary unamputated connected Green function has an additional $n$ external scalar propagators, giving mass dimension $4-3n$ after stripping that delta function, or $-3n$ if it is retained. For a concrete check, the free two-point scalar propagator scales as ${\lambda }^{-2}$, not ${\lambda }^2$. The exponent $4-n$ is the engineering result for the amputated object. A fixed renormalization scale introduces logarithms; the displayed homogeneity is a naive power counting statement rather than an exact quantum scaling law at fixed subtraction scale.

More generally, with $E_s$ external scalar ends and $E_f$ external fermion ends, the same counting gives the superficial degree of divergenceThis is an upper bound on the overall ultraviolet divergence. Numerator cancellations and Lorentz invariance may lower it; an apparently convergent whole Feynman diagram can still have ultraviolet subdivergences. Local subtraction of those ultraviolet subdivergences is required first.

For pure scalar ends, $E_s=0,1,2,3,4$ can require counterterms. Lorentz invariance reduces the two-point terms to a mass counterterm and wave-function renormalization; derivatives of a single scalar field are total derivatives, and a one-derivative cubic scalar term cannot be a Lorentz scalar. With two external fermions, $E_s=0$ requires their kinetic term and mass counterterm, while $E_s=1$ requires Yukawa coupling renormalization. Four external fermions give $D=-2$, so no independent four-fermion counterterm is required.

A complete counterterm Lagrangian consistent with the species symmetry can therefore be writtenThe vacuum energy density term matters for an unnormalized vacuum functional and cancels from normalized Green functions. The tadpole counterterm can impose $⟨\varphi ⟩=0$. There is no $\varphi \mapsto -\varphi$ symmetry of the massive Yukawa interaction, so linear, cubic and quartic scalar terms cannot be excluded. For instance, in a constant background the divergent fermion loop contribution contains $(m+g\varphi {)}^4$, exhibiting all these powers. This is counterterm closure of a massive Yukawa theory.

The two species have identical couplings, so their common unitary flavour rotations preserve the Lagrangian density and their counterterms can be chosen species independent. Equality of the scalar and fermion masses, however, is not protected: $\delta m_s^2$ and $\delta m_f$ are independent. Including the displayed finite set of operators yields a renormalizable quantum field theory. Restricting the bare Lagrangian density to the terms printed in the question is not closed under renormalization.

Let $L$ be the number of independent loop momenta in a connected Feynman diagram. Every scalar propagator falls as $k^{-2}$, and every loop momentum integration contributes three powers, so $\omega =3L-2e$. Counting line ends and using connectedness givesEliminating $L$ and $e$ yieldsThus increasing the sextic interaction vertex count does not worsen the superficial degree of divergence; adding a quartic interaction vertex improves it. Only finitely many external-line structures can have an overall ultraviolet divergence, even though arbitrarily high loop order is possible.

After subtracting ultraviolet subdivergences, locality and Lorentz invariance restrict the divergent counterterms to a constant vacuum energy density, a mass counterterm, a wave-function renormalization, and quartic and sextic field interaction terms. Odd powers are forbidden by the $\varphi \mapsto -\varphi$ Z2 symmetry. In particular, $n=2$ permits a constant and $p^2$, $n=4$ permits only a constant, and $n=6$ permits only a constant. This is perturbative renormalizability, not a guarantee that every superficially allowed divergence actually occurs or that every unrenormalized graph with $\omega <0$ lacks divergent subgraphs.

To include all momentum-dependent contributions to the two-point function, put one external line at each interaction vertex. Let $r$ be the number of lines connecting the vertices, and $t_A,t_B$ their numbers of tadpole diagrams. For vertex valences $v_A,v_B\in \{4,6\}$,The integer $r$ is odd, and $r=1$ gives a one-particle-reducible Feynman diagram and is excluded. Thus the complete list is