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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.

Set $g=g_1$ and $g_2=0$. The superficial degree of divergence is then $\omega =6-2n$, independently of the number of cubic interaction vertices. After subtracting ultraviolet subdivergences, the divergent part of a one-particle irreducible Feynman diagram is a local polynomial in external momenta of degree at most $\omega$. Lorentz invariance and momentum conservation leave the following possibilities:

Thus a finite set of counterterms closes under perturbation theory. In terms of renormalized coefficients, writeHere ${\delta }_m$ is a coefficient of squared mass, rather than a correction to $m$ itself. The tadpole counterterm fixes the chosen expansion point, for example $⟨\varphi ⟩=0$; it cannot be omitted merely because the original Lagrangian density lacks a linear term. The vacuum energy density term cancels out of normalized Green functions, but is required if the unnormalized vacuum functional is also to be renormalized. This establishes perturbative renormalizability; the unbounded scalar potential of a real cubic scalar field theory does not by itself define a stable nonperturbative vacuum state.

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.

Define the free Feynman propagators byThe interaction preserves the global phase symmetry $\phi \mapsto e^{i\alpha }\phi$, with the real $\varphi$ neutral. Therefore the charge-two correlator vanishes in a symmetry-invariant perturbative vacuum:In Wick contractions, a complex field contracts only with its conjugate; one vertex has one of each and cannot absorb both external $\phi$ fields. The symmetry shows this vanishing holds to all perturbative orders, not only first order.

The neutral connected correlator has one allowed contraction to each external point at order $g$:The propagators are even functions of their separation; their orientations encode the conjugate-field contractions. Equivalently, with every external momentum Fourier transformed as incoming,No factor of two accompanies the vertex: $\phi$ and ${\phi }^{†}$ are distinct fields.

For the full correlator as literally written, a disconnected scalar three-point tadpole is also possible. Contract the two external complex fields with each other and contract the complex pair at the vertex into a loop. Unless normal ordering is imposed on the interaction or a tadpole counterterm fixes $⟨\varphi ⟩=0$, addThe coincident propagator is regulator dependent; in dimensional regularization the massless scaleless tadpole diagram vanishes. More generally this is $⟨T{\phi }^{†}(x)\phi (y){⟩}_0⟨\varphi (z)⟩$ to this order and is removed by taking the connected correlator or imposing the stated vacuum-renormalization condition. Normalizing by vacuum bubbles alone does not remove an external-leg disconnected tadpole diagram. The charged correlator remains zero in every one of these conventions.^[ref]

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.