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There are five vertex types:

The quartic terms arise from the square of the quadratic-field piece of the non-Abelian gauge field strength. Every electroweak quartic gauge boson self-interaction contains a $W^{+}{W}^{-}$ pair. The absence of a fundamental purely neutral quartic vertex does not forbid amplitudes such as light-by-light scattering: those can be generated by loop diagrams, rather than by a single fundamental interaction vertex.

There areFor a Yang-Mills theory, schematically $F=\partial A+gAA$. The gauge kinetic Lagrangian is quadratic in $F$, so its terms are at most quartic in the gauge fields:Thus there is no elementary five-leg pure-gauge interaction vertex in the renormalizable Standard Model. A five-boson scattering amplitude can still be assembled from several lower-valence vertices, or generated by loop diagrams. The number of external legs of a complete Feynman diagram is not the valence of a fundamental vertex.

The parton model treats a fast proton as an ensemble of approximately collinear, nearly free quarks and antiquarks. A hard deep inelastic scattering collision resolves one parton on a time scale short compared with its binding motion. The probabilities are parton distribution functions; the simplest approximation neglects intrinsic transverse momentum, quark masses and the proton mass in the hard collision. Quantum chromodynamics introduces logarithmic violations of Bjorken scaling, so the parton distribution functions depend on a resolution scale as well as a parton momentum fraction.

Write the incident electron and proton four-momenta as $k,P$, the scattered electron four-momentum as $k^{\prime }$, and set $q=k-k^{\prime }$. With the $(+---)$ Minkowski metric, the Bjorken scaling variable and the inelasticity areAn incident parton with four-momentum $\xi P$ emerges on shell, giving $(\xi P+q{)}^2\simeq 2\xi P\cdot q-Q^2=0$. Thus $\xi =x_B$: the observable Bjorken scaling variable measures the struck parton momentum fraction in this approximation.

For clarity, measure the scattering angle $\vartheta$ from the incoming electron direction, and denote the outgoing electron energy by $E_e^{\prime }$. Direct four-momentum products giveThe expression uses the known beam energies and two measured quantities. An electromagnetic calorimeter measures $E_e^{\prime }$; a tracking detector, the collision interaction vertex and the position of the electromagnetic shower determine $\vartheta$. Coverage close to the incoming electron direction matters at small $Q^2$. A hadronic calorimeter supplies an independent consistency check and rejects backgrounds. If the angle is instead measured from the incoming proton, its sine and cosine half-angle factors interchange.

A charged lepton in a semileptonic decay directly identifies the decaying charm quark flavour: ordinarily $D^0\to K^{-}{\mu }^{+}{\nu }_{\mu }$ and ${\bar{D}}^0\to K^{+}{\mu }^{-}{\bar{\nu }}_{\mu }$. With an independently tagged initial quark flavour, the wrong-sign charged lepton requires neutral meson mixing, rather than a competing doubly Cabibbo suppressed amplitude. Its leading time dependence is $e^{-\tau }(x^2+y^2){\tau }^2/4$, which looks theoretically cleaner.

That method measures only the quadratic combination $x^2+y^2$ at leading order, with no direct access to their signs. The missing neutrino also worsens kinematic and decay-time reconstruction. The fully reconstructed $K\pi$ channel has substantial yield, a clear invariant-mass peak and a well-measured decay interaction vertex. Most significantly, quantum interference with the small direct amplitude produces a term linear in $x$ and $y$, enhancing sensitivity to small mixing. Its cost is the unknown strong phase $\delta$, so separating $x$ and $y$ needs further channels or an independent strong-phase measurement. Neither one unqualified rate measurement determines both parameters by itself.

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.

The single-vertex scalar two-point graph at first order in $g_{2k}$ has one $2k$-valent interaction vertex. Attach the two external lines to it and pair its remaining $2k-2$ legs into $k-1$ tadpole diagrams. The graph is connected, has $L=k-1$ independent loop momenta, and is momentum independent after amputation apart from the overall Dirac delta function. In particular, first order in the coupling need not mean one loop.

The number of contractions with fixed external labels isDividing by the interaction factorial $(2k)!$ gives the inverse Feynman-diagram symmetry factor $1/[2^{k-1}(k-1)!]$. Use $m>0$ for the mass in the lowest-order propagator and define the tadpole integralThe linear-order correction to the amputated function is consequentlyThis is the correction to the paper's amputated connected correlator, not a definition of an inverse-propagator self-energy with an unspecified sign. At this order using $m_0$ instead of $m$ gives the same lowest-order integral.

For completeness, Schwinger parameterization and a Gaussian integral giveThis first converges for $0<D<2$ and then defines its dimensional regularization continuation. At $D=4-2ϵ$, the Gamma function expansion yieldsThus its highest ultraviolet pole isFor $k=2$ the remainder means finite terms. For $k>2$ the unrenormalized graph has a higher-order pole from its independent ultraviolet subdivergences, rather than only a simple pole. A massless tadpole integral is scaleless and vanishes in dimensional regularization; that does not justify discarding the massive ultraviolet divergence by setting $m=0$ before separating ultraviolet and infrared effects.

The change of field and parameters separates the Lagrangian density in Euclidean space intoThe two-leg counterterm interaction vertex is $-(\delta Z{p}^2+\delta m^2)$, with the overall momentum conservation delta function understood. With $D_0(p)=1/(p^2+m^2)$, the explicit first-order contribution to the propagator isThe minus sign follows either from expanding $e^{-S_{\mathrm{ct}}}$ or from expanding $(p^2+m^2+\delta Z{p}^2+\delta m^2{)}^{-1}$.

For $k=2$, the loop correction is $-{\tilde{\lambda }}_4{T}_D(m)D_0(p{)}^2/2$. Its ultraviolet pole from c isThere is no external $p^2$ dependence in the amputated tadpole diagram, so no one-loop wave-function renormalization is required. The minimal subtraction scheme therefore givesOnly the pole is subtracted: including $1-{\gamma }_E+\log 4\pi$ would impose a different finite subtraction prescription. The defined $\delta m^2$ is the coefficient in $Z{m}_0^2=m^2+\delta m^2$, not automatically the difference $m_0^2-m^2$ beyond this order.

A phi-fourth theory interaction has mass dimension $4-D$, so its bare coefficient must contain a renormalization scale factor ${\mu }^{4-D}=({\mu }^2{)}^{2-D/2}$ when expressed through a dimensionless renormalized coupling. The tree-level four-leg interaction vertex is linear in that coupling. At the next order, two quartic vertices form the three one-loop bubble diagram graphs shown in c, with external pairings $12|34$, $13|24$ and $14|23$.

For each pairing there are two identical internal lines and Feynman-diagram symmetry factor 2. The loop integral has the large-momentum behaviourIt is logarithmically divergent in four dimensions. Dimensional regularization represents that local divergence by a simple pole proportional to $1/(D-4)$. Its residue is independent of external momenta, so a local quartic counterterm cancels it. The three graphs sum to one constant coefficient; this argument establishes its structure without requiring its value.

The relation consequently has the formThe sign of $C$ incorporates the chosen pole denominator and vertex convention. In particular, a negative $C$ here corresponds to a positive coefficient of $1/(4-D)$ in the bare coupling. Finite pieces depend on the renormalization scheme; a pure-pole relation is natural in minimal subtraction scheme. One-loop two-point tadpole diagrams contribute mass renormalization, but not the momentum-dependent field renormalization that could change this coupling relation at order $g^2$. The four-point object relevant here is the amputated one-particle-irreducible vertex; external-leg self-energy insertions are dealt with separately.

Let $C(u,v)=2D_F(u-v)$ be the action-consistent free propagator established in a. Write $S=S_0+S_I$ and putThe perturbative expansion isDivide numerator and denominator by the same free vacuum functional $Z_0$. For $\mathcal{O}=\Phi (y_1)\cdots \Phi (y_N)$, they becomeFor $\mathcal{O}=\varphi (y_1){\varphi }^{\ast }(y_2)$, abbreviate $C_{12}=C(y_1,y_2)$ and $C_{zz}=C(z,z)$. Since only opposite-charge fields contract, Wick's theorem givesandThe first multiplicity is $2!$ internal pairings. In the connected term, there are two choices for the vertex's ${\varphi }^{\ast }$ contracted to the external $\varphi$ and two for its $\varphi$ contracted to the external ${\varphi }^{\ast }$; the remaining pair forms a tadpole diagram. ThereforeExpanding the inverse denominator cancels the free-line times vacuum bubble term, leavingThe local quartic interaction vertex has factor $-i\lambda$, since $2!2!$ cancels its denominator 4. The oriented tadpole diagram has Feynman-diagram symmetry factor 1; the isolated two-loop vacuum vertex has factor $1/2$ in its coefficient. Coincident kernels and the integrals require a common regularization; none is being assigned a finite numerical value here.

More generally, vacuum bubbles exponentiate and factor from every numerator. Here global U(1) symmetry also makes all one-point functions zero. Thus a surviving graph with the two external insertions in separate connected components would have a component with one unbalanced charge and vanish. This proves that only connected graphs contribute to this particular two-point function to all perturbative orders. Vacuum normalization alone would not remove disconnected components containing external legs in a general four-point function. If one formally chooses a different free kernel, the displayed Wick multiplicities are unchanged; replacing $C$ by $C_{\mathrm{lit}}$ gives a formal oscillator series, not the Gaussian integral of the action as printed.

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.

Label the initial species-1 and species-2 fermions by $(p_1,s_1)$ and $(p_2,s_2)$, and the final species-1 and species-2 fermions by $(p_3,s_3)$ and $(p_4,s_4)$. There is exactly one tree-level Feynman diagram: the $t$-channel scalar propagator joins the continuous species-1 line $1\to 3$ to the continuous species-2 line $2\to 4$. Its exchanged four-momentum $q_t$ determines $t$, in the notation of Mandelstam variables:The top row of the figure displays this Feynman diagram. A crossed attachment $1\to 4$ would change species at an interaction vertex and is forbidden. There is no initial antifermion for an annihilation channel.

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.

For the interacting vacuum state defined perturbatively by the same Feynman i-epsilon prescription, the normalized time-ordered product isEach functional derivative inserts $i\varphi$, accounting for $i^{-m}$. The denominator implements cancellation of vacuum bubbles, while disconnected pieces carrying external insertions still contribute. The lower limit $i=i$ in the printed product is a typographical error; the product here runs from 1 to $m$.

At order $V$ in the cubic scalar field interaction term, the expansion of the generating functional supplies $1/[V!(3!{)}^V]$. The $V!$ counts permutations of indistinguishable integrated interaction vertices, and $3!$ counts permutations of the three field slots at each interaction vertex. Applying the Wick theorem pairs fields into Feynman propagators. Several pairings produce the same Feynman diagram; their number $N_G$ multiplies the expansion coefficient. For fixed labelled external points, the net combinatorial weight isHere $S_G$ is the order of the graph's automorphism group when the external labels are fixed, with permutations of identical parallel lines and reversal of a self-loop included as applicable. This is the Feynman-diagram symmetry factor convention in which a Feynman diagram is divided by $S_G$. Each cubic interaction vertex then contributes $-i\lambda$, each line contributes ${\Delta }_F$, and each internal vertex is integrated over its position. External labelled points are not divided by $m!$.

Hold the external endpoints fixed and distinguish the three Feynman diagrams by their left-to-right order in the source. Write $S$ for the divisor and $w=1/S$ for the multiplying combinatorial weight. Each line below includes external Feynman propagators; amputation leaves the same $w$.

Left: two-point bubble. There are two cubic interaction vertices and two parallel internal lines. Interchanging those lines preserves the labelled external endpoints, giving $S=2$. Direct Wick contraction counting gives two assignments of the external endpoints to the vertices, $3^2$ choices of their field slots, and $2!$ pairings of the remaining slots:With $\Delta ={\Delta }_F$ and external points $x_1,x_2$, its contribution is

Middle: three-point triangle. The three labelled external endpoints fix their three adjacent cubic interaction vertices, so no nonidentity graph automorphism remains. Counting gives $3!$ assignments of endpoints to vertices, $3^3$ choices of external field slots, and $2^3$ assignments of remaining slots to the two specified neighbours:Its contribution isThe apparent rotational symmetry of an unlabelled triangle does not survive fixing three distinct external endpoints.

Right: two-point graph with a chord. There are four cubic interaction vertices: the left and right attachment vertices $A,B$ and the two chord ends $C,D$. Its internal edges are $AC,AD,BC,BD,CD$. Reflection interchanges $C,D$, fixes $A,B$, and preserves the two external labels. This is the only nontrivial graph automorphism, so $S=2$.

For a separate Wick contraction count, choose the ordered external attachment vertices in $4\cdot 3$ ways and their field slots in $3^2$ ways. Assign their two remaining slots to $C,D$ in $2^2$ ways. At each of $C,D$, choose the slot joining $A$ and then the slot joining $B$ in $3\cdot 2$ ways; the last slots join each other. ThusIts contribution isConsequently the three multiplying weights are $\boxed{(1/2,1,1/2)}$, or divisors $\boxed{(2,1,2)}$. The central chord is a single internal line joining two vertices, not a vertex at the centre of the circle.

The momentum space Feynman rules use the scalar propagator and two Dirac propagatorsThe cubic scalar field interaction vertex is $-i\lambda$. Each of the two flavour-changing Yukawa coupling interaction vertices, $\varphi \bar{\Psi }\chi$ and $\varphi \bar{\chi }\Psi$, has factor $-ig{\delta }_{ab}$ in Dirac spinor indices. A fermion flow arrow continues through each interaction vertex, but its species changes. There is no $\varphi \bar{\Psi }\Psi$ or $\varphi \bar{\chi }\chi$ interaction vertex.

Every interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ with inward four-momenta. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, divide by the Feynman-diagram symmetry factor, and include a minus sign and a gamma-matrix trace for each closed fermion loop. The $i0$ is the Feynman i-epsilon prescription.

For an amputated scattering amplitude, incoming fermions supply $u^s(p)$ and outgoing fermions supply ${\bar{u}}^s(p)$; incoming antifermions supply ${\bar{v}}^s(p)$ and outgoing antifermions supply $v^s(p)$. An external real scalar field particle supplies $1$. Use the standard relativistic normalization of a one-particle state; the external factors $1/\sqrt{2E}$ belong to that normalization and the Lorentz-invariant phase-space measure, rather than an additional factor in the invariant scattering amplitude.

The original tree-level Feynman diagrams below use solid arrowed fermion lines and dashed scalar field lines. The labelled internal momenta run in the indicated directions; $g$ and $\lambda$ label the corresponding interaction vertices.

For $\Psi (p_1)\to \chi (p_2)+\varphi (p_3)$, the sole tree-level Feynman diagram is a single flavour-changing Yukawa coupling interaction vertex, with $p_1=p_2+p_3$. Its invariant decay amplitude isThe solid arrow passes from the incoming $\Psi$ to the outgoing $\chi$; the dashed outgoing line carries $p_3$.