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Let ${\varphi }_0=\sqrt{Z_2}\varphi$, $A_{0\mu }=\sqrt{Z_3}{A}_{\mu }$, and define the renormalized coupling by $e_0=e_D{Z}_1/(Z_2\sqrt{Z_3})$. Write $Z_2{m}_0^2=m^2+\delta m^2$ and $Z_2^2{\lambda }_0={\mu }^{ϵ}(\lambda +\delta \lambda )$. The complete renormalized Lagrangian density, with its counterterms already included, isSeparating off the classical renormalized terms leaves the counterterm LagrangianThe seagull vertex factor is thus $2i{e}_D^2{Z}_4{g}^{\mu \nu }$. The bare covariant derivative fixes $Z_4=Z_1^2/Z_2$, rather than supplying an unrelated fourth charge parameter. The scalar QED Ward identity, for a symmetry-preserving regularization, gives $Z_1=Z_2$ and hence $Z_4=Z_2$.

For an invertible quadratic gauge field operator, also include ${\mathcal{L}}_{\mathrm{gf}}=-(\partial \cdot A{)}^2/(2\xi )$. One may choose ${\xi }_0=Z_3\xi$, so this term needs no independent counterterm. The strict transverse limit is $\xi \to 0$. Even if the renormalized quartic coupling is set to zero at one scale, its counterterm need not vanish: photon graphs generate a four-scalar interaction. The renormalized Lagrangian density must retain this allowed operator.

Let the bare fields obey the wave-function renormalization definitions ${\psi }_0=Z_2^{1/2}\psi$, $A_{0\mu }=Z_3^{1/2}{A}_{\mu }$, and put $m_0=Z_{m}m$. Define the vertex renormalization factor by $e_0={\mu }^{ϵ/2}e{Z}_1{Z}_2^{-1}{Z}_3^{-1/2}$. Substitution into the bare QED Lagrangian density givesEquivalently, write $Z_i=1+{\delta }_i$ and ${\delta }_m=Z_2{Z}_m-1$. The counterterm Lagrangian isFor the subsequent Feynman gauge calculation include $-(\partial \cdot A{)}^2/2$ in the gauge-fixed action. The Ward identity for QED renormalization constants implies $Z_1=Z_2$ when the regularization preserves gauge invariance; this is a consequence, rather than an extra independent definition of the renormalization constants.

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.

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.

Continue with the cubic scalar field theory of c, and use precisely its counterterm signs. Take $m^2>0$ to separate ultraviolet divergences from infrared divergences. Make a Wick rotation and impose the rotationally invariant cutoff regularization $|q_E|<\Lambda$. PutFor example, spherical coordinates give $I_1=C{\int }_0^{{\Lambda }^2}{t}^2/(t+m^2)dt$. Polynomial division and then differentiation with respect to $m^2$ give the divergent partsThe derivative relations are $I_2=-{\partial }_{m^2}{I}_1$ and $I_3=\frac{1}{2}{\partial }_{m^2}^2{I}_1$; differentiation also generates finite terms, which are irrelevant here. The first supplied integral identity is consistent with these divergent parts after Wick rotation.

The second supplied expansion is incorrect. Write $A=p^2-m^2$. Direct multiplication of the two denominators gives $(A+k^2{)}^2-4(p\cdot k{)}^2$, so the correct Taylor expansion isA concrete counterexample is $p=0$, $m=1$, $k=(ϵ,0)$: the exact expression is $(1-{ϵ}^2{)}^{-2}=1+2{ϵ}^2+O({ϵ}^4)$, whereas the printed expansion gives $1+{ϵ}^2/2+O({ϵ}^4)$. We therefore compute the self-energy without that identity.

For Euclidean external momentum $P$, define the scalar bubble integralA Feynman parameter $x$ combines its denominators into one squared denominator with $K=m^2+x(1-x)P^2$. Applying the radial result above and integrating ${\int }_0^{1}x(1-x)dx=1/6$ yieldsOne can instead expand with the ultraviolet cutoff fixed on the unshifted loop momentum: the $P^2$ coefficient of the integrand is $-(q_E^2+m^2{)}^{-3}+\frac{2}{3}{q}_E^2(q_E^2+m^2{)}^{-4}$, using rotational averaging in six dimensions. Its leading radial term is $-1/(3q_E^6)$, giving the same logarithm. Shifting a sharp integration boundary changes finite terms in this scalar bubble integral, and does not change the displayed divergent coefficients.

To fix all signs and symmetry factors together, expand the one-loop Euclidean effective action around a background scalar field:These terms are the tadpole diagram, the two-point bubble diagram, and the triangle Feynman diagram. The tadpole diagram has factor $1/2$; the two-point bubble diagram also has factor $1/2$ after two differentiations with respect to the background scalar field. The local three-point interaction vertex is $g^3{I}_3$. Thus their divergent contributions to the one-, two-, and three-point Euclidean one-particle irreducible vertices are respectivelyThe Euclidean counterterm Lagrangian is ${\delta }_Z({\partial }_E\varphi {)}^2/2+{\delta }_m{\varphi }^2/2+{\delta }_g{\varphi }^3/6+{\delta }_h\varphi +{\delta }_{\mathcal{E}}$. Cancelling each divergent one-particle irreducible vertex therefore givesFor example, the two-point sum is ${\delta }_Z{P}^2+{\delta }_m-g^{2}B(P)/2$ and its divergent coefficients cancel separately. Finite counterterms depend on the chosen renormalization conditions. The zero-mass limit can use a positive subtraction scale in the logarithms after infrared regularization.

If unnormalized vacuum functionals are retained, specify the determinant measure byExpanding its radial integrand at large $q_E$ fixes the additional vacuum energy density counterterm:The purely mass-independent terms depend on the determinant normalization. Normalized Green functions eliminate this entire constant.

Use one-particle-irreducible correlation functions to identify independent local counterterms. The exact internal symmetry $\varphi \mapsto -\varphi$ excludes odd $b$, and Dirac fermion number conservation requires equal numbers of $\psi$ and $\bar{\psi }$, so $f$ is even. From b, the complete list of potentially overall divergent vertices with external legs is:

The zero-leg vacuum diagrams, with $\omega =3-n_2$, may also require a vacuum energy density subtraction. Four-fermion vertices have $\omega \le -1$ and do not require a new four-fermion counterterm. These are superficial bounds: a particular allowed diagram can vanish or be convergent.

Locality and Lorentz invariance restrict the divergent parts to a finite basis. The scalar two-point vertex can contain a constant and $p^2$; there is no Lorentz scalar linear in a single momentum. The fermion two-point vertex can contain $I$ and $/p$. The remaining listed vertices have no allowed derivative divergence. A complete counterterm Lagrangian, up to total derivatives, is thereforeThe coefficients supply wavefunction renormalization, mass counterterms and renormalizations of the three interactions already present. Recursive subtraction of ultraviolet subdivergences then leaves only these local overall divergences. This finite counterterm closure is why the complete theory is expected to have perturbative renormalizability. It does not mean every listed coefficient first becomes nonzero at one loop, or that setting an allowed coupling to zero is stable under renormalization.

The mass dimensions are $[\varphi ]=1$, $[\psi ]=3/2$, $[g]=0$. Up to total derivatives, power counting allows local counterterms of mass dimension at most four. A convenient basis for the counterterm Lagrangian compatible with the actual parity symmetry in quantum field theory and Dirac fermion number conservation isThe optional constant $\delta \Lambda$ renormalizes the vacuum energy density. The wavefunction renormalization, mass counterterms and Yukawa coupling renormalization correct the terms already present. The quartic interaction must also be included: the fermion loop with four external bosonic legs is logarithmically divergent, even if the renormalized quartic coupling is set to zero at one chosen renormalization scale. The displayed original interactions alone do not form a closed set of counterterms.

If “most general” is interpreted before imposing the parity symmetry in quantum field theory examined in d, Lorentz invariance, a Hermitian action and Dirac fermion number conservation also allowHere all coefficients can be real in the paper's gamma matrix convention; the last term is Hermitian in the action up to a total derivative. It is an axial kinetic normalization, which could alternatively be removed by a suitable perturbative field redefinition. Every term in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ is odd under the actual parity symmetry in quantum field theory, so its coefficient vanishes in a symmetry-preserving renormalization scheme. Listing these terms makes explicit which assumption excludes a tadpole counterterm, a cubic scalar term, a scalar Yukawa interaction, a pseudoscalar Dirac mass term and an axial kinetic term.

With renormalized Dirac fields, the coefficient of $-\bar{\psi }\psi$ in the counterterm Lagrangian is $\delta m_L=Z_2{m}_0-m$, whereas the bare-parameter shift is $m_0-m=\delta m_L-m\delta Z_2$ at first order. In Feynman gauge and $d=4-ϵ$, QED has $\delta Z_2=-e^2/(8{\pi }^2ϵ)$ and $\delta m_L=-4m{e}^2/(8{\pi }^2ϵ)$. Their difference gives $m_0-m=-3m{e}^2/(8{\pi }^2ϵ)$. Defining the insertion and the mass coefficient before quoting its sign avoids confusing these two counterterms.

For a real scalar field in six spacetime dimensions with interaction $-g{\varphi }^3/6$, the mass dimension of $g$ is zero. With Euclidean sharp cutoff regularization, $C=1/(128{\pi }^3)$ and $L=\log ({\Lambda }^2/m^2)$, choose the Minkowski counterterm Lagrangian ${\delta }_Z(\partial \varphi {)}^2/2-{\delta }_m{\varphi }^2/2-{\delta }_g{\varphi }^3/6-{\delta }_h\varphi$. Its one-loop divergent coefficients areThe proof uses the Euclidean effective action $\frac{1}{2}\mathrm{Tr}\log (-{\partial }_E^2+m^2+g\varphi )$. Its first three background derivatives have divergent parts $g{I}_1/2$, $-g^{2}B(P)/2$, $g^3{I}_3$. Radial integration gives $I_1=C({\Lambda }^4/2-m^2{\Lambda }^2+m^{4}L)$, $I_3=CL$, and $B(P)=C({\Lambda }^2-2m^{2}L-P^{2}L/3)$ modulo finite terms. Subtraction proves the coefficients. The tadpole counterterm is essential despite the absence of a linear term in the original Lagrangian density. A separate constant handles the vacuum energy density.