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Use the momentum-space Feynman rules with all momenta conserved at each vertex. The free scalar propagator, free Dirac propagator and pseudoscalar Yukawa interaction vertex are respectivelyThe Dirac propagator follows by multiplying the inverse kinetic matrix $/p-m$ by $/p+m$, using the Clifford algebra to obtain $(/p-m)(/p+m)=(p^2-m^2)I_4$. The vertex follows by expanding $e^{i{S}_{\mathrm{int}}}$ once: there is no factorial because it has one scalar and two distinct spinor field factors. An internal scalar has no spinor indices; an internal fermion line carries the displayed matrix with its momentum directed along fermion flow.

These rules use the coefficient exactly as printed. With the standard Hermitian ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$, the Hermiticity of a pseudoscalar Dirac bilinear gives $(\bar{\psi }{\gamma }^5\psi {)}^{†}=-\bar{\psi }{\gamma }^5\psi$. A Hermitian physical interaction therefore requires ${\lambda }^{\ast }=-\lambda$; writing $\lambda =ig$ with real $g$ changes the vertex to $g{\gamma }^5$. This convention issue does not alter the formal rules above, and no unstated real-coupling assumption is needed.

Label the incoming momenta and spin states $(p_1,r_1),(p_2,r_2)$ and the outgoing ones $(p_3,r_3),(p_4,r_4)$, with $p_1+p_2=p_3+p_4$. Write $u_i=u^{r_i}({\mathbf{p}}_i)$ and $v_i=v^{r_i}({\mathbf{p}}_i)$. The Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. At lowest order, each connected Feynman diagram has two pseudoscalar Yukawa interaction vertices and one internal scalar propagator.

Define the scattering amplitude by $S_{fi}^{\mathrm{connected}}=i(2\pi {)}^4{\delta }^4(p_1+p_2-p_3-p_4)\mathcal{M}$. Fix the external Fock state ordering as $b_1^{†}{b}_2^{†}|0⟩$ to $b_3^{†}{b}_4^{†}|0⟩$ for the two-fermion process. The $t$ exchange connects $1$ to $3$ and $2$ to $4$; the $u$ exchange connects $1$ to $4$ and $2$ to $3$. The two Feynman vertices contribute $(-i\lambda {)}^2$, and the internal scalar propagator contributes $i$ divided by its denominator. Exchanging the two identical final fermions supplies a relative fermionic sign. ThusThis changes sign when the two outgoing labels are exchanged, as identical-fermion antisymmetry requires. There is no $s$ annihilation diagram for two incoming fermions: the interaction preserves Dirac fermion number conservation.

For the fermion-antifermion process take initial state $b_1^{†}{d}_2^{†}|0⟩$ and final state $b_3^{†}{d}_4^{†}|0⟩$. The $t$ exchange uses bilinears $({\bar{u}}_3{\gamma }^5{u}_1)({\bar{v}}_2{\gamma }^5{v}_4)$, and the $s$ annihilation uses $({\bar{v}}_2{\gamma }^5{u}_1)({\bar{u}}_3{\gamma }^5{v}_4)$. To fix the sign without a guess, expand the current $J=:\bar{\psi }{\gamma }^5\psi :$ in creation and annihilation operators. Its relevant terms have the signsThe antifermion sign of a normal-ordered bilinear makes the exchange contraction negative, whereas the product of annihilation and creation contractions is positive for the specified external ordering. Multiplying by $(-i\lambda {)}^{2}i=-i{\lambda }^2$ therefore givesA different common phase convention for an external state can reverse the overall amplitude, but never the relative sign of its two contributions. Both results are tree scattering with pseudoscalar exchange and are of order ${\lambda }^2$.

The superficial degree of divergence is the power of a common large-momentum scale obtained by scaling all independent internal momenta together, before cancellations or subtraction of divergent subdiagrams. In a scalar field theory with a standard quadratic kinetic term and non-derivative interactions, each loop momentum integral supplies $d$ powers of momentum and each internal propagator supplies minus two. ThusFor a connected Feynman diagram, counting vertex half-edges and applying the stated identity for the loop order givesIt follows that $D=(d-2)L-2V+2$. Meanwhile, ${\sum }_n(n-4)V_n=2I+E-4V=2L-2V+E-2$. Substitution provesThis power counting in quantum field theory concerns the overall scaling: even a Feynman diagram with $D<0$ can contain divergent subdiagrams, and cancellations may improve convergence when $D\ge 0$.

The canonical mass dimensions are $[\varphi ]=(d-2)/2$ and $[g_n]=d-n(d-2)/2$. A finite polynomial scalar field theory has power-counting renormalizability when its interactions have nonnegative mass dimensions and the allowed local counterterms form a finite set closed under subtraction of divergent subdiagrams. For $d>2$, non-derivative interactions therefore satisfy $n\le 2d/(d-2)$. Equivalently,so adding vertices of such interactions cannot force an indefinitely increasing number of external legs or derivatives in divergent counterterms. In dimensions $d\le 2$ a fixed finite polynomial still has this favorable power counting in quantum field theory; the displayed upper bound on $n$ is intended only for $d>2$.

For six-dimensional cubic scalar field theory, $3V=2I+E$ and $L=I-V+1$ yieldThe coupling $g$ has zero mass dimension. The possible overall divergences have $E\le 3$: vacuum energy, a linear tadpole diagram term, a mass term, a kinetic term, and a cubic interaction. These are finitely many local counterterms. Divergent subdiagrams of higher-point Feynman diagrams are removed by the same counterterms, establishing perturbative renormalizability of cubic scalar theory in six dimensions. A real cubic potential is unbounded below, so this claim is one of perturbative quantum field theory, not a claim that the polynomial defines a stable nonperturbative vacuum.

After Wick rotation, use the Euclidean action with positive kinetic and mass terms and interaction $g{\varphi }^3/3!$. The Euclidean momentum-space Feynman rules are: an internal scalar propagator $1/(k^2+m^2)$, a vertex $-g$, momentum conservation at each vertex, integration $d^{6}k/(2\pi {)}^6$ for each independent loop, and division by the Feynman-diagram symmetry factor. Strip the overall momentum-conserving delta function and external propagators when computing an amputated Green's function.

The one-particle-irreducible two-point vertex at one loop is the bubble:

Interchanging its two internal lines is an automorphism, giving Feynman-diagram symmetry factor two. In the normalization of the printed Minkowski amputated Green's function, the two vertices contribute $(-ig{)}^2$, the two internal propagators contribute $(-i{)}^2$, and Wick rotation supplies the overall $i$ already factored out there. Thus the Euclidean bubble contribution isFor this self-energy calculation we impose a vanishing one-point function, so tadpole diagram insertions are canceled by the linear counterterm. Without that convention a connected but one-particle-reducible Feynman diagram with a tadpole insertion is also possible: both external legs attach to one vertex, a zero-momentum line joins it to a second vertex, and that second vertex has a self-loop. Its amputated contribution is $g^2/(2m^2)\int d^{6}k/[(2\pi {)}^6(k^2+m^2)]$, independent of $p$. It is not the bubble drawn above; keeping it without a tadpole subtraction would add this constant to the full amputated answer.

To find the bubble's local divergence, put $K=k^2+m^2$ and expand at large $|k|$:Odd terms vanish under the angular integral. Rotational symmetry gives $⟨(k\cdot p{)}^2⟩=k^2{p}^2/6$. The terms through order $p^2$ are thereforeand the $p^2$ coefficient approaches $-1/(3k^6)$. Terms of order $p^4$ and higher are integrable at infinity in six dimensions; subtracting the two displayed terms leaves an ultraviolet-convergent integral. Hence the divergent part is a local polynomial $A{p}^2+B$, with momentum-independent coefficients. For example, when $m^2>0$ its logarithmically divergent coefficient isChanging the reference scale in this logarithm only changes a finite local term.

At zero external momentum the spherical ultraviolet cutoff and $\mathrm{Vol}(S^5)={\pi }^3$ giveWith $u=k^2$, integrate $u^2/(u+m^2{)}^2=1-2m^2/(u+m^2)+m^4/(u+m^2{)}^2$. This gives the momentum-cutoff two-point function in six-dimensional cubic scalar theoryIn particular, the divergent part of this constant iswhile the last term in $B_{\Lambda }$ tends to a finite constant. For $m=0$ the zero-momentum integral gives $B_{\Lambda }=g^2{\Lambda }^2/(256{\pi }^3)$ directly. The finite parts of a local subtraction can depend on the regularization prescription; the conclusion about mass and kinetic counterterms does not.

Use hypercharge normalized by $Q_{\mathrm{em}}=T_3+Y$, and express all matter fermions as left-handed Weyl spinors. Thus $u^c,d^c,e^c$ denote the charge conjugates of the right-handed Standard Model fermions; the superscript $c$ is unrelated to the antiparticle marker $(A)$. Each chiral superfield contains a complex scalar field and a Weyl spinor, while each vector superfield contains a gauge boson and a gaugino. The complete propagating content of the Minimal supersymmetric Standard Model before electroweak symmetry breaking is:

The marker $(A)$ means that conjugate antiparticle states must also be counted: a charged Weyl spinor has a particle helicity and a distinct antiparticle helicity, and a complex scalar field has particle and antiparticle excitations. The real adjoint gauge bosons and adjoint Majorana fermions already contain their conjugate states, so no extra marker is attached to them. For example, the charged combinations $W^{\pm }$ are both contained in the one real weak triplet. There are no right-handed-neutrino chiral superfields, no gravitino, and no propagating auxiliary fields in this field table. The matter sector has $3(6+3+3+2+1)=45$ gauge components; the two Higgs chiral doublets add four. Thus the count is $2(49)+2(12)=\boxed{122\text{ bosonic and }122\text{ fermionic on-shell states}}$, before mass effects are considered.

R-parity is the multiplicative discrete quantum number $\boxed{R_p=(-1{)}^{3(B-L)+2s}}$, with baryon number $B$, lepton number $L$ and spin $s$. Ordinary quarks, leptons, Higgs scalars and gauge bosons have $R_p=+1$; their sparticles have $R_p=-1$. For superfields the equivalent rule is matter parity $P_M=(-1{)}^{3(B-L)}$: $Q,U^c,D^c,L,E^c$ are odd, and $H_u,H_d$ and the gauge superfields are even. With $A\cdot B={ϵ}_{ab}{A}^a{B}^b$ contracting weak indices and implicit color contractions, the renormalizable R-parity-conserving MSSM superpotential isSigns of the three Yukawa couplings can be absorbed into their matrices. Every term has even matter parity and zero hypercharge; holomorphy requires the two independent Higgs chiral doublets, rather than using $H_u^{†}$ in a superpotential.

Conserved R-parity forces an even number of odd particles in a process whose incoming particles are all ordinary particles, so colliders produce sparticles in pairs or another even multiplicity. Their decay chains end in the stable lightest supersymmetric particle. A neutral, weakly interacting lightest supersymmetric particle, commonly a neutralino after electroweak symmetry breaking, escapes the detector and gives missing transverse momentum; it is also a possible dark matter constituent. Stability alone does not establish its relic abundance, and a charged or colored stable lightest supersymmetric particle is not the usual viable dark matter candidate. Conserved R-parity forbids the renormalizable interactions responsible for the decay below. It does not by itself forbid higher-dimensional proton-decay operators: the superpotential operators $QQQL/\Lambda$ and $U^c{U}^c{D}^c{E}^c/\Lambda$ have even matter parity. Thus the renormalizable R-parity-conserving MSSM protects the proton, while ultraviolet physics can still induce its decay.

The gauge-invariant renormalizable R-parity-violating superpotential isHere ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$ because of the weak and color antisymmetric contractions. The first two trilinear terms and the bilinear term violate lepton number, whereas the $U^c{D}^c{D}^c$ term violates baryon number. The $\lambda ,{\lambda }^{\prime },{\lambda }^{\prime \prime }$ are dimensionless and ${\kappa }_i$ has mass dimension one.

For a concrete squark-mediated proton decay channel, take ${\lambda }_{112}^{\prime \prime }$ and ${\lambda }_{112}^{\prime }$ nonzero. Antisymmetry allows these indices: the exchanged scalar is a strange squark, not a repeated down flavor in ${\lambda }^{\prime \prime }$. Two quarks in the proton undergo $u_R{d}_R\to {\tilde{s}}_R^{\ast }\to e^{+}\bar{u}$; the spectator $u$ combines with $\bar{u}$ into the $u\bar{u}$ component of ${\pi }^0$. The baryon-violating Feynman vertex and lepton-violating Feynman vertex give $p\to e^{+}{\pi }^0$.

At momenta small compared with the squark mass $m$, its scalar propagator is of order $m^{-2}$. Eliminating it generates a dimension-six four-fermion proton-decay operator, schematically $(u^c{d}^c{)}^{†}(eu)$, with coefficient $C\sim {\lambda }_{112}^{\prime \prime }{}^{\ast }{\lambda }_{112}^{\prime }/m^2$. Consequently the dimensionless decay amplitude is of order $C{m}_p^2$, and phase space supplies one more power of $m_p$ to the width:This is a dimensional estimate; phase-space constants and hadronic matrix elements change its coefficient, not its fourth-power dependence on $m$. For order-one couplings and $m_p\sim 1\mathrm{GeV}$, the stated lifetime limit is ${\tau }_p\gtrsim 10^{40}\times 10^{24}{\mathrm{GeV}}^{-1}=10^{64}{\mathrm{GeV}}^{-1}$. Therefore $m^4\gtrsim 10^{64}{\mathrm{GeV}}^4$, giving $\boxed{m\gtrsim 10^{16}\mathrm{GeV}}$.

Such a mass is incompatible with the intended weak-scale MSSM explanation of the hierarchy problem: squarks that cancel the Higgs-sector sensitivity to high scales must not introduce a new enormous splitting from their light quark partners. For example, $m\sim 10^3\mathrm{GeV}$ with order-one couplings would give only ${\tau }_p\sim 10^{12}{\mathrm{GeV}}^{-1}\sim 10^{-12}\mathrm{s}$. The field content alone does not mathematically prohibit extremely heavy superpartners, but that choice sacrifices the weak-scale naturalness motivation. The usual resolution is conserved R-parity, or sufficiently suppressed simultaneous baryon number and lepton number violation, rather than making the exchanged squark this heavy.

Choose $\varphi (x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ip\cdot x}\tilde{\varphi }(p)$ in Euclidean signature. The inverse quadratic kernel is the smooth-cutoff scalar propagatorThe subscript records the explicit cutoff dependence implied by the condition $f_{\Lambda }(z)={\Lambda }^{2}z$ at small $z$. For $p^2+m^2\le {\Lambda }^2$, this reduces to $1/(p^2+m^2)$, the usual Euclidean scalar propagator. For $p^2+m^2\gg {\Lambda }^2$, the kernel $f_{\Lambda }$ grows rapidly and its inverse tends to zero.

High-momentum modes are strongly suppressed. For a finite smooth regulator they are not literally identically zero; that statement would require a sharp cutoff. The field variance carried by those Fourier transform modes is correspondingly negligible.

The cubic scalar field theory has a one-particle-irreducible Feynman diagram with two cubic vertices, one external leg at each vertex, and two internal lines joining them:

The Euclidean Feynman rule for each cubic interaction vertex is $-g{\mu }^{(6-d)/2}$: the minus sign comes from expanding $e^{-S_{\mathrm{int}}}$, and $1/3!$ in the action cancels the permutations of its three fields. The mass dimension of the cubic coupling is $(6-d)/2$, so ${\mu }^{(6-d)/2}$ permits $g$ to remain dimensionless. Two vertices supply $(-g{)}^2{\mu }^{6-d}$.

Each internal scalar propagator contributes $1/(p^2+m^2)$ with its own momentum. Conservation leaves one independent loop momentum; choose the two propagator momenta to be $p$ and $p+k$. The remaining Fourier integration measure is $d^{d}p/(2\pi {)}^d$. Finally, the factor $1/2$ is the Feynman-diagram symmetry factor for exchanging the two identical internal lines. Direct Wick contraction counting gives the same factor: two choices for which vertex receives a labeled external leg, $3^2$ choices of the incident fields and $2!$ pairings of the remaining fields, divided by $2!(3!{)}^2$, yield $1/2$.

The displayed loop integral is the amputated two-point insertion. For the correction to the full propagator, multiply it by the external scalar propagators, giving $D_0(k{)}^{2}I(k)$ with $D_0(k)=1/(k^2+m^2)$.

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.

For incoming four-momenta, the momentum-space Feynman rules in the chosen Minkowski metric convention are:

There is one connected one-loop one-particle irreducible two-point graph: a single quartic vertex with two external lines and one tadpole diagram. Its symmetry factor is $S=2$ from interchange of the two loop half-edges.

There are three connected one-loop one-particle irreducible four-point graphs: two quartic vertices joined by two parallel internal lines, with external partitions $(12|34)$, $(13|24)$ and $(14|23)$. These are the $s$, $t$ and $u$ scattering channels; each bubble diagram has symmetry factor $S=2$. External-branch insertions are excluded by the one-particle irreducible requirement.

Expand the normalized Euclidean path integral with weight $e^{-S_E}$ and apply Wick's theorem. Use labelled external momenta and a fixed Fourier transform convention throughout. A line carrying momentum $q$ contributes $(q^2+m_0^2{)}^{-1}$. Each $2k$-valent interaction vertex contributes $-{\lambda }_{2k}=-{\mu }^{d_k}{g}_{2k}$: its $(2k)!$ possible contractions cancel the factorial in the Lagrangian density. A vertex also carries $(2\pi {)}^D{\delta }^{(D)}(\sum q)$ with all incident momenta incoming. Integrate each independent loop momentum with $d^{D}q/(2\pi {)}^D$ and divide by the Feynman-diagram symmetry factor for fixed external labels. There are no Minkowski factors of $i$.

For the full $N$-point Green's function, include every graph in which all vertices belong to components attached to external insertions. In particular, disconnected components carrying external insertions must be included, and the zeroth-order result is the sum of free pairings. Pure vacuum bubbles cancel against the denominator. The connected Green's function restricts to connected graphs, while a one-particle-irreducible vertex imposes the stronger restriction that removing one internal line cannot disconnect the graph. These are different objects. Odd $N$ vanishes in the symmetric vacuum because every interaction has even valence. At fixed perturbative order, multiply by the appropriate power of the coupling and sum all allowed graph topologies and external attachments.

If the overall momentum-conservation delta function is removed by convention, report the remaining function as its coefficient. The ordinary unamputated Green's function includes external scalar propagators; dividing them out defines an amputated function. The definition in c performs this amputation but does not, by itself, impose the usual one-particle-irreducibility condition.

Use the Fourier transform convention $\varphi (x)=\int d^{4}p/(2\pi {)}^4{e}^{-ip\cdot x}\tilde{\varphi }(p)$, and similarly for $\Phi$. All four-momenta at an interaction vertex are incoming. The quadratic terms are diagonal, so the scalar propagators areThe field interaction terms are $\lambda \varphi {\Phi }^2/2$ and $g\Phi {\partial }_{\mu }\varphi {\partial }^{\mu }\varphi /2$. The two identical field slots cancel the factor $1/2$ at either cubic interaction vertex. Expanding $\exp (i{S}_{\mathrm{int}})$ therefore givesIn the second rule $p,q$ are specifically the two $\varphi$ four-momenta; there is no derivative acting on the $\Phi$ leg. Indeed the two derivatives give $(-i{p}_{\mu })(-i{q}^{\mu })=-p\cdot q$, fixing the minus sign. This is a mixed scalar derivative cubic vertex. For arbitrary off-shell legs one may equivalently write $-ig(k^2-p^2-q^2)/2$, but may not replace the squared four-momenta by squared masses unless those legs are on their mass shells.

Each interaction vertex also carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. An amputated external scalar leg has factor 1. The Feynman i-epsilon prescription fixes the propagator poles. As a dimensional check in four spacetime dimensions, $[\varphi ]=[\Phi ]=1$, $[\lambda ]=1$ and $[g]=-1$ in mass dimensions; the two powers of momentum in the derivative rule restore mass dimension one for its cubic interaction vertex.

Let $p_1,p_2$ denote the incoming $\varphi ,\Phi$ four-momenta and $p_3,p_4$ the outgoing $\varphi ,\Phi$ four-momenta. Thus $p_1+p_2=p_3+p_4$, $p_1^2=p_3^2=m^2$ and $p_2^2=p_4^2=M^2$. Define the Mandelstam variables and exchanged four-momenta byThe leading connected four-point scattering amplitude uses two cubic interaction vertices. There are four tree-level Feynman diagrams: $s$ and $u$ exchange of $\Phi$ with two $\lambda$ vertices, and $s$ and $u$ exchange of $\varphi$ with two $g$ vertices.

There is no term proportional to $\lambda g$: one vertex of each kind has three field slots of each species altogether, whereas the external state uses two of each. This leaves one $\varphi$ and one $\Phi$ to contract internally, and their mixed scalar propagator is zero. There is also no $t$ channel: grouping both external $\varphi$ legs at one vertex would require a $g$ vertex with an internal $\Phi$, while grouping both external $\Phi$ legs at the other would require a $\lambda$ vertex with an internal $\varphi$.

Define $i\mathcal{M}$ after removing the overall momentum conservation delta function. The nonderivative exchange givesFor the derivative $s$ channel, the incoming $\varphi$ four-momenta at the first vertex are $p_1,-q_s$, and at the second they are $q_s,-p_3$. The two interaction vertices are therefore $ig{p}_1\cdot q_s$ and $ig{p}_3\cdot q_s$. The same reasoning with $q_u$ applies to the $u$ channel, givingThe mass shell conditions and momentum conservation implyCombining the four Feynman diagrams therefore yieldsEach displayed tree-level Feynman diagram has Feynman-diagram symmetry factor 1 for fixed external species and labels. The result is of order ${\lambda }^2,g^2$; if both couplings are scaled by a common parameter $\epsilon$, the next connected corrections begin at order ${\epsilon }^4$. A four-point cubic Feynman diagram has $3V=2I+4$, so its vertex count $V$ is even. The internal lines remain off shell throughout: imposing an on-shell relation on an exchanged four-momentum would incorrectly change the derivative vertices.

For four-momentum $k$ flowing along the fermion flow, the momentum space Feynman rules areThe scalar propagator and Dirac propagator invert their respective quadratic kernels. The interaction vertex follows by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$, fixing its sign. Species is conserved along each open fermion line; there is no vertex changing species 1 into species 2.

An interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum k)$ with inward four-momenta. Impose momentum conservation and integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$. Include the closed fermion loop sign $-1$ for every fermion loop, the relevant fermionic signs when rearranging external states, and the Feynman-diagram symmetry factor. External incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antifermions carry ${\bar{v}}_s(p)$ and $v_s(p)$. An external real scalar field leg has factor 1 in an amputated scattering amplitude. The Feynman i-epsilon prescription specifies the poles.

Let $p_1,p_3$ label the incoming and outgoing fermion, and $p_2,p_4$ the incoming and outgoing antifermion. The bottom row of the figure gives the two tree-level Feynman diagrams: