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Use $K^{-}=s\bar{u}$ and $p=uud$. In the first reaction a proton $u$ can join the incoming $\bar{u}$ to form the $u\bar{u}$ component of the neutral pion, leaving $uds$. In the second, a $u\bar{u}$ pair can annihilate into a gluon which creates $s\bar{s}$; the remaining $u$ joins $\bar{s}$ in the outgoing kaon, leaving $dss$. Thusbaryon number is one in both cases. Strong strangeness conservation provides a separate check: the first baryon has strangeness $-1$; the second has strangeness $-2$ because the outgoing $K^{+}$ carries $+1$. The Feynman diagrams show representative quark-line flavour rearrangement and pair-annihilation topologies. Bound hadrons require nonperturbative binding, so a single drawn gluon is not an exclusive cross-section calculation.

The original Feynman diagrams below distinguish elementary weak and electromagnetic vertices from an effective hadronic vertex. fermion arrows track fermion number; time runs left to right.

Continuum amplitudes. The leading quantum electrodynamics diagram is electron-positron annihilation into a virtual photon, followed by conversion to a muon pair. Replacing the outgoing muons with a quark-antiquark pair gives the quark-level hadronic diagram. The first two panels below show these processes.

The Feynman rules give, up to signs and factors of $i$,The two vertices supply $e^2$, and the photon propagator supplies $1/s$. Each external current scales as $\sqrt{s}$, so the complete amplitude is of order $e^2$, and its squared magnitude is of order $e^4\propto {\alpha }^2$. The two-body relativistic scattering cross-section supplies the remaining $1/s$ from flux and phase space. Integrating the angular distribution yields ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$. Keeping only the squared propagator while ignoring the external currents would incorrectly predict $1/s^2$.

A quark of charge $Q_{f}e$ contributes $N_c{Q}_f^2$ times the massless muon rate. With $u,d,s,c,b$ open and $N_c=3$, the hadronic R ratio isThis is the requested massless leading-QED continuum approximation. Strong corrections, physical hadron thresholds, resonance contributions and Z exchange are separate effects; the pure photon formula is not a complete high-energy electroweak cross section throughout the entire stated range.

Efficiency normalization. Read the original PDF plot as a continuum of about $0.8\mathrm{nb}$ and a maximum around $1.7\mathrm{nb}$ at $E\simeq 10.58\mathrm{GeV}$. The supplied conversion constant givesHence near the peak,If $C$ multiplies the displayed cross sections, the required scale isHere $ϵ$ is the implied detection efficiency under the common-scale assumption. Using continuum readings $0.75$–$0.85\mathrm{nb}$ gives approximately $C=3.4$–$3.8$, illustrating the graph-reading precision. The nearby open-bottom threshold also limits the precision of the ideal continuum approximation.

Resonance quantum numbers and widths. Direct formation through a single photon requiresThese are the quantum numbers of the electromagnetic current. This concerns single-photon formation; associated production or two-photon reactions need not obey this restriction.

For a Breit-Wigner resonance cross section, subtract the continuum before finding the full width at half maximum. The plotted excess is about $0.9\mathrm{nb}$, so its half-height occurs at a total displayed cross section near $1.25\mathrm{nb}$, not $0.85\mathrm{nb}$. The crossings are roughly $10.57$ and $10.59\mathrm{GeV}$, with a separation around $0.02$–$0.025\mathrm{GeV}$. ThusFor spin-one formation from unpolarized electrons and positrons, $g=3/4$ and $p\simeq M/2$. With ${\Gamma }_{\mathrm{had}}\simeq \Gamma$, the peak resonance cross section and branching fractions relation becomesThe corrected excess is $C(1.7-0.8)\simeq 3.2\mathrm{nb}$. Taking $\Gamma \simeq 24\mathrm{MeV}$ as a representative graph reading givesThus the plot supportsThe width and electronic partial width are estimates from the printed graph under the prescribed Breit-Wigner and efficiency assumptions; these are not precision world-average values. A scale correction changes the peak height and inferred partial width, but not the peak position or width.

Why the lower Upsilon states are narrow. In electromagnetic production and decay of vector quarkonium, every $nS$ state can annihilate as $b\bar{b}\to {\gamma }^{\ast }\to e^{+}{e}^{-}$. The electronic width probes the wavefunction at the origin, ${\Gamma }_{ee}\propto {\alpha }^2{Q}_b^2|{\psi }_{nS}(0){|}^2/M_n^2$. The radial overlaps differ, but the same short-distance annihilation mechanism operates, so electronic widths remain of comparable order.

The hadronic mechanism changes sharply at $2m_B\simeq 10.560\mathrm{GeV}$. The first three Upsilon mesons lie below it and cannot produce an on-shell B-meson pair. Their leading purely strong annihilation channel is $b\bar{b}\to ggg$, followed by hadronization: a color-singlet pair cannot annihilate to one gluon, and two gluons cannot match its negative charge-conjugation parity. Three-gluon annihilation carries three powers of ${\alpha }_s$ in the rate and requires quark-antiquark overlap. Electromagnetic annihilation to light quarks and, for excited states, transitions to lower bottomonium plus pions or photons also contribute.

The fourth state lies above the pair threshold and can instead undergo ${\rm Y}(4S)\to B\bar{B}$ through light-pair creation, keeping the b and anti-b in separate outgoing mesons. The final panel depicts this quark-line topology; it represents a nonperturbative strong decay, not a free on-shell b quark radiating a real gluon. For two pseudoscalar B mesons the pair is in a P wave to give $1^{-}$. Although close to threshold, this open strong channel has a much larger rate than the annihilation channels. Consequently a total width of tens of MeV can coexist with an electronic width below a keV; below threshold the total widths are smaller by hundreds or more, while their electronic widths remain of similar order.

Hyperfine estimate. For two spin-half constituents,The meson hyperfine mass splitting therefore raises the vector mass above the pseudoscalar mass. To infer its coefficient from only one measured mass, adopt the intended simple constituent ansatz that the spin-independent baseline is $m_b+m_q=5.31\mathrm{GeV}$. ThenFor the spin-one partner,The estimate is the same for $\bar{b}u$ and $\bar{b}d$ with the supplied equal light-quark masses. The splitting is below $m_{\pi }$, so the strong decay $B^{\ast }\to B\pi$ is closed. An electromagnetic transition is open and dominates over weak decay:It preserves electric charge, giving the corresponding charged or neutral B meson; the photon energy is approximately $40\mathrm{MeV}$, slightly reduced by recoil. Strictly, the statement that the mass merely includes a spin-spin term does not fix $A$: an independent spin-independent binding energy would leave the baseline unknown. The numerical prediction above explicitly uses the additive constituent-mass approximation.

Count strangeness before drawing a weak interaction diagram. An Omega baryon has $S=-3$, a Lambda baryon has $S=-1$, a negatively charged kaon has $S=-1$, and a pion has $S=0$. Thus the two nonleptonic baryon decays requireThe strong interaction cannot supply either net change of strangeness.

In the upper Feynman diagram, one initial strange quark undergoes the weak charged current transitionThe remaining two strange quarks are spectator quarks. Combine the outgoing $u,d$ and one spectator $s$ into the Lambda baryon, and the remaining $s$ with $\bar{u}$ into the kaon. This uses one W boson propagator and two weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{V}_{ud}|\simeq \sin {\theta }_C\cos {\theta }_C$.

For the lower Feynman diagram, let the upper initial $s$ first emit a $W^{-}$ and become a virtual $u$, then emit a $W^{+}$ and become a $d$. The $W^{+}$ is absorbed by a second initial $s$, turning it into a $u$; the third initial $s$ remains a spectator quark. The emitted $W^{-}$ produces $d\bar{u}$. The complete quark flavour flow is thereforeThe final $d,u,s$ form the Lambda baryon and the other $d\bar{u}$ form the pion. Electric charge conservation holds at every vertex. Internal lines are virtual; this is a contribution to one decay amplitude, not a sequence of freely propagating on-shell constituent-quark decays.

This is a possible lowest-weak-order diagram for a second-order strangeness-changing weak decay: it has two W boson propagators and four weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{|}^2|V_{ud}{|}^2$ in the displayed light-quark topology. Other contributions can exist, but a single charged-current insertion cannot turn two net strange quarks into nonstrange ones. No extra loop Feynman diagram is compulsory: the displayed two-W tree already realizes the required quark flavour change.

At low energies, one-W exchange produces one power of the Fermi constant in an effective decay amplitude; the second process requires an additional weak insertion. Hadronic matrix elements and propagators supply the accompanying dimensions, so a numerical ratio cannot be obtained by simply squaring a dimensionful constant. The robust comparison isThe dominant reason is the extra weak order, with additional Cabibbo suppression in this representative topology. It is not a comparison between a strong decay and a weak decay.

Read the line styles and fermion-number arrows in the original PDF: a wavy line is a photon, and a coiled line is a gluon. The valid fundamental interaction vertices areBoth (a) and (f) have one electron-line arrow entering and one leaving a photon vertex; changing the drawing's orientation does not change the quantum electrodynamics interaction. The Quantum chromodynamics vertices (b) and (g) are the cubic and quartic gluon self-interactions.

The remaining cases fail for different reasons. Diagram (c) couples an isolated electron line to gluons and is not an allowed fermion vertex. Diagram (d) changes an electron to a muon at a photon vertex, violating the flavour-diagonal electromagnetic coupling. Diagram (e) couples a colourless electron to a gluon. Diagram (h) changes an up quark into a down quark at a gluon vertex, whereas Quantum chromodynamics preserves quark flavour; that assignment also fails electric charge conservation.

One measurement of the strong coupling constant uses the fraction of resolved three-jet events in $e^{+}{e}^{-}\to q\bar{q}g$: with a fixed jet-resolution definition, its leading ratio to two-jet production is proportional to ${\alpha }_s$, with calculable higher-order corrections.

The intermediate current is ${\gamma }^{\ast }$ or $Z^{\ast }$, and the extra gluon can be emitted from either the quark or the antiquark. Both Feynman diagrams contribute coherently. The observed jets arise from hadronization; a resolved-jet definition is needed because arbitrarily soft or collinear radiation is not a separate measurable third jet.

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.

At energies large compared with fermion masses, the gauge interactions preserve chirality on each fermion line. A photon couples to charged fermions with equal left and right coefficients $e{Q}_f$. A gluon couples to either quark chirality with coefficient $g_s{T}^a$. A Z boson couples to both, with coefficients $(g/\cos {\theta }_W)(T_3-Q_f{\sin }^2{\theta }_W)$ on left-chiral fields and $(g/\cos {\theta }_W)(-Q_f{\sin }^2{\theta }_W)$ on right-chiral fields. A W boson couples only to left-chiral fermions and the associated right-helicity antifermions: $W^{-}\to {\ell }_L^{-}{\bar{\nu }}_R$ or $d_L{\bar{u}}_R$, and the charge-conjugate $W^{+}$ channels. For quarks the coefficient includes the CKM matrix. There is no Standard Model right-chiral neutrino gauge vertex.

For outgoing pairs at a neutral vertex, the allowed helicities are $f_L^{-}{\bar{f}}_R^{+}$ and $f_R^{-}{\bar{f}}_L^{+}$; the latter labels helicity, not the chirality of the corresponding antifield. Only the $s$-channel photon and Z boson diagrams contribute to $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$ at tree level among these gauge exchanges. A W boson changes charged leptons into neutrinos, and gluons do not couple to leptons.

In a dense beam dump, long-lived pions and kaons usually interact and lose energy before decaying. Their residual stopped decays yield low-energy neutrinos, below the useful high-energy tau-interaction sample. Prompt charmed mesons have much shorter lifetimes and can decay before reinteraction. This suppresses the conventional muon neutrino background relative to the prompt component. The production sequence is Quantum chromodynamics creation of $c\bar{c}$, principally $gg\to c\bar{c}$ with $q\bar{q}\to g^{\ast }\to c\bar{c}$ also allowed, followed by hadronization $c\bar{s}\to D_s^{+}$ and its charge conjugate. The $gg$ process includes both quark-exchange diagrams and the three-gluon diagram. Fragmentation also produces other hadrons; the proton does not turn directly into one isolated Ds meson.The charge-conjugate chain gives a prompt tau antineutrino and a secondary tau neutrino. Thus each leptonic Ds meson chain produces both tau-flavour signs. The corresponding Feynman diagrams are shown in a.

For the CKM matrix, diagonalizing the up/down quark mass matrices by $U_{uL},U_{dL}$ leaves $V=U_{uL}^{†}{U}_{dL}$ in the charged current:The three-generation unitary matrix has three physical mixing angles and one CP violation phase after field rephasings. At fixed kinematics and hadronic matrix element, rates are proportional to $|V_{ij}{|}^2$; phase space, parent lifetimes, form factors and helicity suppression must also be included when comparing branching fractions.

For a charged pseudoscalar meson $P^{+}=(c\bar{q})$, leptonic pseudoscalar decay width givesThe Ds meson channel is Cabibbo-favoured, $|V_{cs}|\simeq 0.974$, whereas the charged D meson channel uses $|V_{cd}|\simeq 0.224$. Taking $m_{D_s}\simeq 1.968$, $m_D\simeq 1.870$ and $m_{\tau }\simeq 1.777$ in GeV givesIncluding the approximate lifetime ratio ${\tau }_{D_s}/{\tau }_D\simeq 0.48$ and the given production ratio $5.2/11$ gives a tau yield ratio about $17(f_{D_s}/f_D{)}^2$, already much greater than one even with equal decay constants. This is an illustrative estimate using approximate additional mass/lifetime inputs, not a determination from the production cross sections alone. A neutral D meson cannot annihilate to the charged $\tau \nu$ pair by electric charge conservation; its semileptonic tau modes have insufficient phase space once the required strange or nonstrange recoil hadron is included. Light mesons are below threshold; bottom production is much smaller at this fixed-target energy. Consequently neither the larger $D^0$ production rate nor ordinary light-hadron production outweighs the Ds meson tau source. ^[ref]

The basic Feynman diagrams have ${\tau }^{-}\to {\nu }_{\tau }{W}^{-\ast }$ followed by $W^{-\ast }\to e^{-}{\bar{\nu }}_e$, ${\mu }^{-}{\bar{\nu }}_{\mu }$, or $d\bar{u},s\bar{u}$, with the latter pairs undergoing hadronization. See the charged-current diagrams in a. Neglect all daughter masses, QCD corrections and inaccessible charm final states, and assume lepton universality. Each leptonic mode then has one unit of decay width; the two accessible quark combinations have three color charge states and total CKM weight $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$. ThereforeThe result is an inclusive parton model estimate. Perturbative QCD corrections increase the hadronic contribution, while muon mass, nonperturbative dynamics and radiative corrections also matter; the physical fraction is roughly $0.65$.

Every ordinary hadronic mode already contains a tau neutrino, so it contains a lepton as well as hadrons if invisible particles are counted. For an additional visible charged lepton, a virtual photon can internally convert in a hadronic mode, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{\gamma }^{\ast }\to {\nu }_{\tau }{\pi }^{-}{e}^{+}{e}^{-}$. The extra electromagnetic vertices suppress such modes; they are not a large fraction of the inclusive width. Alternatively a daughter pion or kaon can decay semileptonically or leptonically in flight, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{K}^0$ followed by a semileptonic neutral-kaon decay. Those are secondary decay chains, not independent primary tau lepton branching fractions. Lepton-flavour-violating neutrinoless modes such as ${\tau }^{-}\to {\mu }^{-}{\pi }^{+}{\pi }^{-}$ do not contribute appreciably in the Standard Model.

A leading valence-level process is $u\bar{d}\to W^{+}\to e^{+}{\nu }_e$, with the up quark supplied by the proton and the down antiquark by the antiproton. The same Feynman diagram works with $e^{+}{\nu }_e$ replaced by ${\mu }^{+}{\nu }_{\mu }$. Both vertices are weak charged current vertices; the production scattering amplitude contains the CKM matrix element $V_{ud}$ or its conjugate according to the chosen fermion-flow convention. The intermediate W boson is produced near its mass shell.

At leading order in QED, the electron exchanges one spacelike virtual photon with a charged parton inside the proton.

The two Feynman vertices contribute $e$ and $e{e}_q$, while the photon propagator contributes $1/q^2=-1/Q^2$. Hence the scattering amplitude scales as $e^2{e}_q/Q^2$ and its squared absolute value as $e^4{e}_q^2/Q^4$. Since the fine-structure constant is $\alpha =e^2/(4\pi )$, the electromagnetic coupling dependence is ${\alpha }^2/Q^4$. The numerical $4\pi$ in the stated differential scattering cross-section additionally uses the flux, Lorentz-invariant phase space, azimuthal integration and the convention defining the deep-inelastic structure function; it cannot be deduced from the propagator alone.

For each measured $x,Q^2$ bin form the reduced quantityPlot this against $Q^2$ at several fixed $x$ values, normally using a logarithmic $Q^2$ axis. Bjorken scaling predicts an approximately horizontal curve at each $x$, or equivalently collapse of different-$Q^2$ data onto one $F_2(x)$ curve. Raw cross sections do not collapse because their explicit kinematic factors vary. QCD predicts logarithmic scaling violations, so exact independence is only the ideal parton model limit. The stated extraction also assumes negligible longitudinal deep-inelastic structure function, as in the leading massless spin-half parton approximation.

The favoured weak charged current transition is $\bar{c}\to \bar{s}{W}^{-}$ followed by $W^{-}\to d\bar{u}$. The spectator quark $u$ combines with $\bar{u}$ to form the $u\bar{u}/\sqrt{2}$ component of the neutral pion, while $d\bar{s}$ forms $K^0$. The product of CKM matrix factors is $V_{cs}{V}_{ud}^{\ast }$. The doubly Cabibbo suppressed transition is $\bar{c}\to \bar{d}{W}^{-}$ followed by $W^{-}\to s\bar{u}$, with factors $V_{cd}{V}_{us}^{\ast }$; the final $s\bar{d}$ is the neutral antikaon. Both are colour-suppressed internal-emission Feynman diagrams, with the same spectator routing.

In the two-generation approximation to the CKM matrix, $V_{ud}=V_{cs}=\cos {\theta }_C$, $V_{us}=\sin {\theta }_C$ and $V_{cd}=-\sin {\theta }_C$. For equal magnitudes of the hadronic matrix elements,The rate ratio is about $2.8\times 10^{-3}$ with the supplied CKM matrix magnitudes. The minus sign of a particular product of CKM matrix elements must be combined with the phases of the hadronic matrix elements and kaon basis states; it is not by itself the observable interference sign.

For a connected cubic Feynman diagram with $V$ internal vertices, $I$ internal lines and three external lines, $3V=2I+3$. Hence $V$ is odd and the loop order is $L=I-V+1=(V-1)/2$. Through order ${\lambda }^3$, there is one tree-level Feynman diagram at $V=1$ and seven labelled one-loop Feynman diagrams at $V=3$:

For an explicit check, use the propagator notation of solution and define, with $\{i,j,k\}=\{1,2,3\}$,There is no ordering of $j,k$ in this notation. The complete connected correlation function through this order isFor completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct Wick contraction counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by $3!(3!{)}^3=1296$.

Connected does not mean one-particle irreducible. After amputation of external propagators, only the triangle is one-particle irreducible at this order; the six other loop graphs belong to the requested connected answer. The tadpole diagrams cannot be discarded without an additional subtraction prescription.

Let $D_{xy}=⟨0|T\varphi (x)\varphi (y)|0{⟩}_0$ be the free Feynman propagator, and abbreviate $v_3=-i\kappa$, $v_4=-i\lambda$. Expand the normalized Dyson series, so vacuum bubbles cancel between numerator and denominator, and retain the connected correlation function. Wick's theorem yieldsThese are respectively the free line, the quartic tadpole diagram, the two-cubic-vertex bubble diagram, and an tadpole contribution to a cubic-theory two-point function, attached by a bridge to the vertex carrying both external legs. The last graph is connected but is a one-particle-reducible Feynman diagram; removing its bridge disconnects it. It must be included when all connected graphs are requested. It disappears if an additional vanishing-one-point-function renormalization condition and its corresponding tadpole subtraction are imposed; that convention was not specified here.

The factors can be counted directly. The quartic term has $4\cdot 3=12$ attachments for the two labelled external legs, divided by $4!=24$. At second order in the cubic interaction the expansion denominator is $2!(3!{)}^2=72$. There are 36 Wick contractions for the bubble, and 36 for the attached-tadpole graph, giving $1/2$ for each. Thus each loop topology shown has symmetry factor $S=2$. Disconnected products of one-point functions and vacuum bubbles are not part of $G_c$.^[ref]

The one-loop cubic-quartic scalar vertex classification follows without truncating arbitrarily in powers of the couplings. A connected graph with $E$ external legs, $V_3$ cubic vertices, $V_4$ quartic vertices and loop number $L$ satisfiesFor $E=2,L=1$, the only possibilities are $(V_3,V_4)=(2,0)$ or $(0,1)$, exactly those just listed.

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.

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.

Label the external four-momenta by $\Psi (p_1)+\chi (p_2)\to \Psi (p_3)+\chi (p_4)$. The only tree-level Feynman diagram exchanges a scalar field particle between the transitions $\Psi (p_1)\to \chi (p_4)$ and $\chi (p_2)\to \Psi (p_3)$. Its internal four-momentum is $q=p_1-p_4=p_3-p_2$. Thus its scattering amplitude, with a consistent external-state ordering, isIt is the crossed, or $u$-channel, exchange with these labels. A direct $t$-channel scalar field exchange would need a flavour-preserving Yukawa coupling, which the Lagrangian density does not contain. There is no annihilation Feynman diagram for two incoming fermions.