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For $\Psi (p_1)+\varphi (k_1)\to \Psi (p_2)+\varphi (k_2)$, there are two tree-level Feynman diagrams, both with an internal $\chi$ fermion. Their internal four-momenta are $p_1+k_1$ and $p_1-k_2$. The scattering amplitude isThese are the two possible absorption/emission orderings along the fermion flow. There is no third tree-level Feynman diagram using the cubic scalar field interaction vertex, since attaching its remaining scalar field line to the external $\Psi$ line would require a forbidden $\varphi \bar{\Psi }\Psi$ interaction vertex.

For $\Psi (p_1)\to \chi (p_2)+\varphi (k_1)+\varphi (k_2)$, the sole tree-level Feynman diagram has a Yukawa coupling interaction vertex producing an off-shell scalar field particle of four-momentum $q=p_1-p_2=k_1+k_2$, followed by its cubic scalar field interaction vertex. Its invariant decay amplitude isTwo successive Yukawa coupling interaction vertices would change the fermion species twice and leave a $\Psi$, so they cannot produce this final state. Interchanging the two identical outgoing bosons does not create another Feynman diagram; the cubic factorial is already included in the Feynman rule. The integrated decay width instead includes the identical final-state symmetry factor $1/2!$.

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.

For the two-point function, the original Feynman diagrams below show both one-vertex graphs and every two-vertex topology with external momentum flowing between distinct interaction vertices. All lines represent the same real scalar field. The lower row labels the number of parallel connecting lines; a small closed line at a vertex is a tadpole diagram.

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

The coefficient of the mixed graph includes the reflected external attachment as well as its reciprocal Feynman-diagram symmetry factor. If both external lines attach to the same interaction vertex, no external momentum enters the connecting internal lines, and that Feynman diagram cannot contribute to wave-function renormalization.

For incoming external momentum $p$, assign the connecting lines momenta $k_1,\dots ,k_{r-1},p-{\sum }_j{k}_j$. DefineThe momentum space Feynman rules give, in the order listed,Each $C$ is a reciprocal Feynman-diagram symmetry factor, including any equivalent external attachments. There are respectively $2,3,4,4$ independent loop momenta.

If “contribute to wavefunction renormalization” means requiring a divergent kinetic counterterm after subtraction of all subdivergences, only the last topology qualifies. The first three still have nontrivial external momentum dependence and give finite contributions to a normalization condition; their raw divergent factors arise from mass counterterms and interaction vertex counterterms. Keeping all four here makes this distinction explicit. The ultraviolet divergence analysis in d establishes it directly.

With a single quartic interaction vertex, the two unused line ends close into one tadpole diagram. With a single sextic interaction vertex, the four unused ends pair into two tadpole diagrams. These are the only one-vertex one-particle irreducible Feynman diagrams for the two-point function.

Define the Minkowski Feynman integralsThe amputated two-point function contributions arewhere $C_4,C_6$ are the reciprocals of the corresponding Feynman-diagram symmetry factors. Both are independent of external momentum, and hence affect the mass counterterm rather than wave-function renormalization. No numerical Feynman-diagram symmetry factors are needed here.

A dimensionless action in six spacetime dimensions requires the Lagrangian density to have mass dimension six. The kinetic term then fixes $2+2[\varphi ]=6$, and each field interaction term must have the same mass dimension. ThereforeEquivalently, in powers of a length unit $L$, $\varphi$ has units $L^{-2}$, $g_1$ has units $L^0$, and $g_2$ has units $L^2$. Thus literal length exponents would be $(-2,0,2)$; the $d$'s in the supplied superficial degree of divergence formula must instead mean mass dimensions, namely $(2,0,-2)$. This distinction resolves the wording's convention mismatch.

To check the convention independently, let $I,V_3,V_4,\ell$ count internal lines, cubic interaction vertices, quartic interaction vertices, and independent loop momenta. A connected Feynman diagram obeys $2I+n=3V_3+4V_4$ and $\ell =I-V_3-V_4+1$. Hence its superficial degree of divergence is $6\ell -2I=6-2n+2V_4$, agreeing with the supplied formula only with the mass dimensions above.

In momentum space, the Feynman rules for the real scalar field areEach interaction vertex supplies $(2\pi {)}^6{\delta }^{(6)}(\sum p)$, with all its momenta counted inward. Impose momentum conservation, integrate each independent loop momentum with $d^{6}p/(2\pi {)}^6$, and divide by the Feynman-diagram symmetry factor. For a connected amputated scattering amplitude, remove the external scalar propagators and retain a single overall momentum conservation factor. These signs follow by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$; the factorials in the Lagrangian density cancel the corresponding attachments in the Wick theorem. The Feynman i-epsilon prescription specifies the scalar propagator poles.

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.

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.

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.

Use incoming fermion $p_1$ and antiparticle $p_2$, outgoing fermion $p_3$ and antiparticle $p_4$. The two tree-level Feynman diagrams are the $t$ channel, with $q_t=p_1-p_3=p_4-p_2$, and the $s$ channel, with $q_s=p_1+p_2=p_3+p_4$. They occupy the lower row of the figure. In the $t$ channel the fermion flow follows $1\to 3$ and $4\to 2$; in the $s$ channel it follows $1\to 2$ at the incoming interaction vertex and $4\to 3$ at the outgoing interaction vertex.

The two contractions again have a relative fermionic sign. With the same harmless common-phase convention as i, their sum isThe auxiliary-field propagator is the same constant $-i$ in both channels. There is no $u$ channel preserving the external fermion number assignments.

Label the incoming fermion momenta $p_1,p_2$ and outgoing fermion momenta $p_3,p_4$, with $p_1+p_2=p_3+p_4$. There are two tree-level Feynman diagrams: the $t$ channel connects $1$ to $3$ and $2$ to $4$, with exchanged momentum $q_t=p_1-p_3=p_4-p_2$; the $u$ channel connects $1$ to $4$ and $2$ to $3$, with $q_u=p_1-p_4=p_3-p_2$. Both internal lines are auxiliary-field propagators. The upper row in the figure gives these labelled connections; the outgoing labels are exchanged in its second panel to avoid a visually unnecessary crossing.

Each tree-level Feynman diagram contains two factors $ig$ and one factor $-i$, giving $i{g}^2$. The exchange of the two identical outgoing fermions gives the relative fermionic sign. Up to a common external-state phase, the corresponding scattering amplitude isThere is no annihilation channel: both incoming particles carry the same fermion number, whereas a $\sigma \bar{\psi }\psi$ interaction vertex couples one fermion to one antiparticle when both are incoming.

Use $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1)$, and $p^2=(p^0{)}^2-(p^1{)}^2$. The momentum-space Feynman rules for the real scalar field areAt each interaction vertex impose two-dimensional momentum conservation; before removing the overall conservation factor this is $(2\pi {)}^2{\delta }^{(2)}({\sum }_j{p}_j)$. Integrate each independent loop momentum with $d^{2}k/(2\pi {)}^2$, and divide a Feynman diagram by its symmetry factor. External lines carry the same scalar propagator in an unamputated two-point correlation function; they are removed in an amputated scattering amplitude.

The quadratic action is inverted with the Feynman i-epsilon prescription to obtain the scalar propagator. Expanding the Dyson series gives $-i{g}_4/4!$ per insertion; the $4!$ attachments in the Wick theorem cancel the factorial. Fourier integration over the insertion point supplies the momentum conservation factor. Automorphisms that leave the labelled external attachments fixed account for the remaining symmetry factor. In two spacetime dimensions the mass dimensions are $[\varphi ]=0$ and $[g_4]=2$.

Let $I,L,V$ count internal lines, independent loops, and interaction vertices of a connected Feynman diagram. The graph identity $L=I-V+1$ and the two inverse powers from each scalar propagator give its superficial degree of divergence:The same formula applies to every connected subgraph. Graphs with $V\ge 2$ have negative superficial degree of divergence, but this alone does not remove a possible ultraviolet subdivergence. A one-vertex subgraph with external legs in the quartic scalar field theory can diverge only by a tadpole diagram, whose subtraction was fixed in c. Subtract every such tadpole diagram; every remaining loop subgraph then has $V\ge 2$ and negative superficial degree of divergence. Its ultraviolet divergence is absent. Equivalently, normal ordering the quartic interaction removes all self-contractions at a single interaction vertex. Thus no new divergent field-dependent counterterm structure is needed, and in a normal-ordering subtraction scheme in two dimensions the one-loop mass subtraction exhausts these divergent coefficients.

The literal assertion that there are no further counterterm contributions in the on-shell renormalization scheme is stronger and generally false. A concrete counterexample is the two-loop sunset diagram. After analytic continuation below its three-particle threshold, the magnitude of its self-energy at $p^2=m^2$ contains the strictly positive finite integralThe integrable corners do not spoil convergence. Its nonzero value shifts the pole mass, requiring an additional finite mass counterterm if $m$ is to remain the exact pole mass; its derivative can likewise change the pole residue. These are compatible with the power counting result. If vacuum diagrams are retained rather than divided out in normalized correlation functions, a separate constant vacuum energy counterterm is also required: the one-vertex double-loop vacuum diagram is already divergent. The claim about the mass subtraction concerns field-dependent counterterms and normalized correlation functions.

The additional six-leg interaction vertex has Feynman rule $-i{g}_6$. Since both interactions lack derivatives, the graph identity still gives $\omega =2(1-V)$, with $V=V_4+V_6$. The difference is that a one-vertex contraction of the phi-six theory can leave four external legs: choosing the contracted pair gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$. Its quartic interaction correction isConsequently a one-loop quartic interaction counterterm is now necessary:The one-loop mass counterterm from c is unchanged. Contracting four legs of a single six-leg interaction vertex gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)3=45$ double contractions, hence a two-loop raw self-energy $g_6{T}_{\Lambda }^2/8$. Its subtraction must include the quartic interaction counterterm insertion, rather than subtracting this raw expression in isolation.

All divergent field-dependent coefficients are most simply organized by the normal-ordering subtraction scheme in two dimensions. With $T=T_{\Lambda }$, express a potential normal ordered at mass $m$ in ordinary fields:Thus its divergent subtraction coefficients, in this scheme, areIndeed the two-loop self-energy terms are $g_6{T}^2/8+\delta g_{4}T/2=-g_6{T}^2/8$, which the displayed mass counterterm cancels. The constant terms only affect vacuum energy. After removing all one-vertex self-contractions, every remaining subgraph is ultraviolet finite. Additional finite on-shell renormalization scheme adjustments are subject to the same qualification as d.

The two complex scalar fields have scalar propagatorscontractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive conserved charge flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical momentum; an antiparticle arrow opposes it.

There are two conjugate three-leg interaction vertices, both with Feynman rule $-ig$: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms ${\varphi }^2{\Phi }^{†}$ and ${\varphi }^{†2}\Phi$. The two identical $\varphi$ attachments cancel the $1/2$ in the Lagrangian density. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no interaction vertex with three equal-sign charges.

Each interaction vertex conserves four-momentum, supplying $(2\pi {)}^4{\delta }^4(\sum p_j)$ with all momenta entering. Integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. No minus sign is assigned to a closed scalar loop. These rules are for $i\mathcal{M}$ after amputating external lines and removing the overall four-momentum conservation factor.

Take $\varphi (p_1)+\bar{\varphi }(p_2)\to \varphi (p_3)+\bar{\varphi }(p_4)$ with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed $u$-channel $\Phi$ exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge $+2$; the other end absorbs that charge. The zero-charge $s$- and $t$-channel assignments are forbidden by the interaction vertices. ThusIn the center-of-mass frame with $|\mathbf{p}|\ll m$, $s=4m^2+O(|\mathbf{p}{|}^2)$ and $u=O(|\mathbf{p}{|}^2)$. For $M\gg m$, both scattering amplitudes approach $g^2/M^2$. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Usingfor distinguishable final particles, and an extra factor $1/2!$ when integrating the identical-particle final state over the whole sphere, givesThere is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking $M/m\to \infty$, the zero-momentum ratio is $\frac{1}{2}[M^2/(M^2-4m^2){]}^2$, provided the threshold is away from the heavy pole.

Near $s=M^2$, the bare $s$-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of ${\Sigma }_{\Phi }(s)/(s-M^2)$. The $\Phi \to \varphi \varphi$ channel is open for $M>2m$, giving the leading decay widthResumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator $s-M_{\mathrm{pole}}^2+i{M}_{\mathrm{pole}}{\Gamma }_{\Phi }$ near the pole, rather than an infinite scattering cross-section. Also, $s\simeq M^2$ with $M\gg m$ lies outside the assumed nonrelativistic incoming kinematics; the ratio $1/2$ is not a resonance prediction.

Label the incoming four-momenta $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$. Two incoming $\varphi$ particles have conserved charge $+2$, so the only tree-level Feynman diagram is an $s$-channel $\Phi$ exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

Cross the second incoming heavy particle and one outgoing heavy particle in i to obtain $\Phi \bar{\Phi }\to \Phi \bar{\Phi }$. Panel (e) of b's figure shows the resulting box diagram. Its four light-charge arrows again alternate source and sink corners; it is the only connected order-$g^4$ topology. There is no tree exchange because every cubic interaction vertex has only one heavy leg. With $|\mathbf{p}|\ll M\ll m$, crossing does not affect the zero-external-momentum loop integral, soAn independent check is to integrate out the complex scalar field $\varphi$ at constant $\Phi$. Its two real-component squared masses are $m^2\pm g|\Phi |$. The one-loop quantum effective potential is thereforeExpanding the logarithms to fourth order in $\Phi$ givesThus the effective Lagrangian has the positive local coefficient $g^4/(384{\pi }^2{m}^4)$. Its $2!2!$ external attachments give the common scattering amplitude $g^4/(96{\pi }^2{m}^4)$ in both channels. This also checks the box multiplicity and sign.