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Allowed. There are two tree-level Feynman diagrams: $t$-channel Z boson exchange through the weak neutral current, and $s$-channel $W^{+}$ formation through the weak charged current.

Allowed. There are again two tree-level Feynman diagrams: $t$-channel Z boson exchange and the crossed charged-current diagram with a W boson exchanged between the neutrino and electron lines.

The complete tree-level list for (ii), (iv), and (v) is:

The interaction $-g\varphi {\Psi }_i^{\ast }{\Psi }_i$ gives one cubic vertex $-ig$ joining a real-scalar line to a particle-antiparticle pair of either complex scalar species. The second-order term of the Dyson series isWick theorem contracts the incoming ${\Psi }_1{\overline{\Psi }}_1$ pair at one vertex, the outgoing ${\Psi }_2{\overline{\Psi }}_2$ pair at the other, and the two $\varphi$ fields with each other. The two assignments of $x,y$ cancel the factor $2!$. Thus there is one connected tree-level Feynman diagram, the $s$-channel exchangeWith standard relativistic external-state normalization,up to the physically irrelevant common sign convention for $\mathcal{M}$.

In the center-of-momentum frame, $q=(E_1,-\mathbf{p})$ and $p=(E_1,\mathbf{p})$, soPair creation is kinematically possible exactly when $s\ge 4M_2^2$. The threshold incident momentum is therefore

A connected Feynman diagram that is a cubic tree with five external legs has three vertices and two internal lines. Its unique unlabeled topology is a chain: two external legs meet at each end vertex and one external leg attaches to the middle vertex. The five-point tree amplitude in phi cubed theory therefore has fifteen labeled tree-level Feynman diagrams: choose the middle external leg in five ways, then partition the remaining four legs into two unordered pairs in three ways.

At order ${\lambda }^2$, four connected tree-level Feynman diagrams contribute: one four-point contact diagram and three diagrams with two cubic vertices joined by a scalar propagator in the $s$, $t$, and $u$ channels.

Three connected tree-level Feynman diagrams contribute at order $e^2$: a scalar-exchange diagram in which photon $k_1$ is emitted before $k_2$ along the charged scalar line; the crossed scalar-exchange diagram with $k_1$ and $k_2$ interchanged; and the local seagull vertex joining both incoming scalars and both outgoing photons. The exchange denominators are $(p_1-k_1{)}^2-m^2$ and $(p_1-k_2{)}^2-m^2$.

All three interactions respect color charge; quark flavor is conserved by the strong interaction and the electromagnetic interaction, but can change in the weak interaction. These statements concern charges and net numbers, rather than an unchanging inventory of particles.

A quark transforms in the fundamental representation of the color charge gauge group $SU(3{)}_C$. The strong interaction can change an individual quark's colour by exchanging a gluon, which itself carries color charge. For example, the colour flow $q_r\to q_b+g_{r\bar{b}}$ is allowed: demanding that the outgoing quark remain red would be incorrect. At each interaction vertex the full colour flow is conserved. The colourless photon, W boson, and Z boson instead couple to a quark with an identity factor ${\delta }_{ab}$ in colour space. A weak charged current can therefore change $d_r$ to $u_r$, but cannot turn $d_r$ into $u_b$ without another coloured participant. External isolated hadrons are colour singlets; confinement prevents direct observation of a free coloured state.

For each quark flavor $f$, define the net quark flavor number $F_f=N_f-N_{\bar{f}}$. The strong interaction and electromagnetic interaction preserve every $F_f$: pair creation $g\to s\bar{s}$, for example, changes neither net strangeness nor any other $F_f$. Their quark couplings are diagonal in quark flavor. The weak charged current contains ${\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)V_{ij}{d}_j$, where $V$ is the Cabibbo-Kobayashi-Maskawa matrix, and changes these net numbers. Weak neutral currents are diagonal in tree-level Feynman diagrams; flavor-changing neutral currents can occur through higher-order Feynman diagrams.

The observed hierarchy of mean lifetimes and decay channels supplies evidence: associated strange-particle production through the strong interaction preserves net strangeness, whereas a kaon's weak interaction decay to pions changes it. The electromagnetic interaction process ${\pi }^0\to \gamma \gamma$ has zero net quark flavor before and after annihilation. Neutrino oscillations show that individual lepton family numbers are not exact conserved quantities. Within the usual effectively massless-neutrino approximation, charged current decays such as ${\mu }^{-}\to e^{-}{\bar{\nu }}_e{\nu }_{\mu }$ preserve family numbers only when the emitted neutrinos are counted.

Evidence for three colours includes the hadronic R ratio $R=3{\sum }_f{Q}_f^2$, the colour multiplicities in W boson decay widths, and three-jet events revealing gluon radiation. These test quantum chromodynamics and its colour structure together; they are not direct measurements of a named red or blue charge on a free quark.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.

A bare gauge boson mass would spoil the electroweak gauge structure. In the Standard Model a complex Higgs doublet instead has a symmetry-breaking potential. Choose the conventionThe Higgs mechanism rearranges three scalar components into the longitudinal polarizations of the massive W bosons and Z bosons; the unbroken electromagnetic gauge symmetry leaves the photon massless. One physical scalar excitation, the Higgs boson $h$, remains.^[ref] Using tree-level Feynman diagrams,These equations use natural units. Yukawa interactions generate charged-fermion masses and Higgs couplings proportional to those masses; the minimal model does not account for observed neutrino masses. The gauge kinetic term supplies Higgs couplings to W and Z pairs. The physical Higgs exchange also cancels the otherwise excessive high-energy growth of longitudinal vector-boson scattering. The mechanism does not fix the Yukawa couplings or the Higgs self-coupling, and most proton mass arises from strong-interaction dynamics rather than directly from the light-quark Higgs masses.

The mass-generation mechanism is a statement about the field and its vacuum expectation value; the observable boson is the excitation about that vacuum.

There is one tree-level Feynman diagram: the distinguishable electron and muon exchange a photon in the $t$ channel. There is no identical-particle exchange diagram.

Multiplying the external Dirac spinors, the two Feynman vertices and the photon propagator givesThe overall phase of this scattering amplitude is convention dependent. Four external spin labels with two choices each give $2^4=16$ possible configurations before taking a limit. For $E\gg m,M$, the vector QED interaction preserves fermion chirality, which becomes helicity for each external particle. Thus the surviving configurations, written as $(h_e,h_{\mu }{)}_{\mathrm{in}}\to (h_e,h_{\mu }{)}_{\mathrm{out}}$, areHere $L,R$ specify helicity along each particle's own momentum. The other configurations require at least one mass-suppressed flip and vanish in the strictly massless limit.

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 $p_1,p_2$ 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.

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:

Thus the Mandelstam variables in the two denominators areThe two scattering amplitude terms have a relative fermionic sign of minus when written with spinor products $({\bar{u}}_3{u}_1)({\bar{v}}_2{v}_4)$ and $({\bar{v}}_2{u}_1)({\bar{u}}_3{v}_4)$. The overall phase depends on the chosen ordering of the external states. No $u$ channel joins an incoming fermion to an outgoing antifermion through a species-preserving Yukawa interaction.

With all four external fermions of species 1, both final attachments are possible. The middle row of the figure gives the two tree-level Feynman diagrams:

For a fixed ordering of external fermion states, the fermionic sign from exchanging the final identical fermions givesHere $u_j=u_{s_j}(p_j)$ are Dirac spinors. In particular, the scattering amplitude changes sign when the complete final labels $3$ and $4$ are exchanged, as required for identical fermions.

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.

Define $\mathcal{M}$ by removing the factor $i(2\pi {)}^4{\delta }^{(4)}(p_1+p_2-p_3-p_4)$ from the connected transition matrix element. The single tree-level Feynman diagram in i givesThere is no identical-particle exchange term because the species differ. Each incoming fermion has two equally populated spin angular momentum states in an unpolarized beam. Their probabilities multiply to $1/4$, giving a spin average over the four initial choices. The unobserved final spin angular momentum states are distinct outcomes, so their probabilities are added by a spin sum, with no final averaging:Taking the complex conjugate of ${\bar{u}}_3{u}_1$ gives ${\bar{u}}_1{u}_3$. The fermion spin sum therefore converts each squared bilinear into a gamma-matrix trace:Using $\mathrm{tr}1=4$, $\mathrm{tr}{\gamma }^{\mu }=0$, $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$, and the mass shell conditions $p_j^2=m^2$, each gamma-matrix trace isThe second equality follows from $q^2=2m^2-2p_1\cdot p_3$; momentum conservation gives the identical result for $p_2,p_4$. HenceThe printed prefactor $g^2$ is incorrect for the stated coupling. Each of the two Yukawa interaction vertices contributes one $g$ to the scattering amplitude, so its square is proportional to $g^4$. As a concrete counterexample, take $g=2$, $m>0$, and forward elastic kinematics $p_3=p_1$, $p_4=p_2$. Then $q^2=0$ and the spin average is $16g^4=256$, whereas the printed expression gives $16g^2=64$.

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 $p_3$.

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.

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.