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The antiquark $\bar{d}$ in the specified ${\overline{B}}_d^0=b\bar{d}$ is a spectator. For the charmed final state the leading external-W boson tree iswith $c\bar{d}=D^{+}$ and $d\bar{u}={\pi }^{-}$. For the charmless final state replace the active transition by $b\to u+W^{-}$, giving $u\bar{d}={\pi }^{+}$ and the same $d\bar{u}={\pi }^{-}$. Their weak amplitudes have factors $V_{cb}{V}_{ud}^{\ast }$ and $V_{ub}{V}_{ud}^{\ast }$, respectively, up to an irrelevant overall conjugation convention.

The diagrams determine the leading weak flavour factors. Strong-interaction matrix elements and, particularly for the charmless channel, additional loop diagrams can also contribute to physical branching fractions.

Use the parton model, neglect antiquarks, binding and target mass corrections, quark masses, muon mass, QCD scaling violation and the W boson propagator correction. The struck quark is effectively free during the short interaction and carries parton momentum fraction $x$, giving $\hat{s}\simeq xs$. The charged current reaction is ${\bar{\nu }}_{\mu }{u}_i\to {\mu }^{+}{d}_j$. Its helicity amplitude has the same $(1-y)$ factor as in a. Define the effective parton distribution function by $q(x)={\sum }_{i,j}|V_{ij}{|}^2{u}_i(x)$, summing only kinematically accessible final quarks; for valence-only protons and accessible $d,s$ this is approximately $u(x)$. It is not the unweighted sum $u+d$.Here $y=p\cdot (k-k^{\prime })/(p\cdot k)$ is the inclusive deep inelastic scattering variable, and $x=Q^2/(2p\cdot (k-k^{\prime }))$ in the scaling limit. Integrating $0<y<1$ gives $1/3$, henceThe moment measures the momentum fraction in the active quarks. Approximate linearity is evidence for scattering from pointlike spin-one-half constituents with approximately scale-independent parton distribution functions, rather than coherent scattering from an extended proton. It is an intermediate-energy prediction: thresholds, experimental deep inelastic scattering cuts, QCD scaling violation and eventually the finite W boson propagator modify it.

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.

At the parton level, the dominant valence reactions are $e^{-}u\to {\nu }_{e}d$ through W boson exchange and $e^{+}d\to {\bar{\nu }}_{e}u$ through W boson exchange. The first also acts on $c$, while the second acts on $s$ with the appropriate Cabibbo-Kobayashi-Maskawa matrix factors; crossing supplies the antiquark reactions. The lines in the figure show fermion number flow; the Positron and antineutrino arrows point opposite to their momenta.

For unpolarized beams, the leading charged current cross sections, suppressing Cabibbo-Kobayashi-Maskawa matrix factors, have the formHere each parton distribution function is evaluated at $(x,Q^2)$. The proton has two valence up quarks and one valence down quark; moreover, its valence contribution in the Positron reaction carries the chiral angular factor $(1-y{)}^2$. Both effects favour $e^{-}p$. Increasing $Q^2$ at fixed collider energy enforces $x\ge Q^2/s$, progressively suppressing the sea relative to the valence quarks; large $y$ further suppresses the Positron valence term. There is no universal factor of two valid at every $(x,y)$.

The escaping neutrino prevents direct reconstruction from a scattered charged lepton. Instead use the Jacquet-Blondel reconstruction. With the $+z$ axis along the proton beam, measure the complete hadronic final state and formThese relations follow from four-momentum conservation, with the unobserved neutrino transverse momentum opposite to the measured hadronic sum. A nearly hermetic hadronic calorimeter, supplemented by tracking detectors, must measure jets, their energies and their directions. Forward losses, initial-state radiation and small $y_h$ cause significant reconstruction uncertainties.

Represent a quark-antiquark state by a matrix $F$ with $q_i{\bar{q}}_j\leftrightarrow E_{ij}$. Under flavor symmetry, $F\mapsto UF{U}^{†}$, so an infinitesimal generator acts by $[T,F]$. The minus sign in this commutator is the antiquark transformation law; acting on the quark alone would miss states.

Start from ${\pi }^{+}=E_{ud}$. Three useful first steps areFurther actions give $[I_{-},E_{us}]=E_{ds}=K^0$, $[I_{-},E_{sd}]=-E_{su}=-K^{-}$ and $[I_{-},{\pi }^0]=\sqrt{2}{E}_{du}=\sqrt{2}{\pi }^{-}$. Overall phases are conventional. FinallyThus the second neutral octet state is obtained by removing the known ${\pi }^0$ component. Equivalently it is the normalized diagonal state orthogonal to both ${\pi }^0$ and the singlet. The full meson octet isTheir flavour matrices are traceless; the independent singlet is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The source uses ${\eta }^{\prime }$ for this ideal singlet. That notation is a basis convention here: the physical eta and eta prime mesons is not an exactly pure singlet because physical neutral pseudoscalar mesons mix.^[ref]

Treat a nucleon as incoherent collinear spin-half quarks and antiquarks carrying fractions $x$ of its momentum. Neglect their masses and transverse motion, use a charged-current four-fermion interaction, and take $Q^2\ll m_W^2$. The subprocess energy is $\hat{s}\simeq xs$. Antineutrino scattering on a quark has the same selected-current helicity factor $(1-y{)}^2$ as in part a; crossing to an antiquark changes the allowed spin pairing and gives a flat distribution. Multiplying the elementary strength by the respective parton densities givesHere $q,\bar{q}$ are the effective weak flavour combinations with the isoscalar proton-neutron average and CKM weights absorbed into their definitions. With only light flavours and charge symmetry, the relevant up and down combinations coincide on an isoscalar target. Rescaling those density definitions requires rescaling the prefactor consistently.

The equation is a leading-order scaling result: it neglects logarithmic QCD evolution, heavy-quark thresholds, higher-twist and target-mass corrections, nuclear modifications, and weak-propagator suppression. The current produces ${\mu }^{+}$ for an incident muon antineutrino. At higher momentum transfer the weak propagator supplies a factor approximately $[m_W^2/(m_W^2+Q^2){]}^2$, and at low energy resonance and quasielastic scattering invalidate a pure DIS description.

For a nucleon at rest and beam energy well above its mass scales, $s\simeq 2m_{N}E$. If the leading parton distributions are approximately independent of beam energy, integration over $x,y$ therefore makes both total cross sections approximately proportional to $E$. The plotted plateaus are evidence for scattering on approximately pointlike constituents in this range, with the departures expected from QCD evolution and threshold effects.

Define the momentum-weighted contents $Q={\int }_0^{1}xq(x)dx$ and $\bar{Q}={\int }_0^{1}x\bar{q}(x)dx$. Since ${\int }_0^1(1-y{)}^{2}dy=1/3$, the two cross sections obeyRead the original high-energy plateaus as approximately $0.67$ and $0.33$ in the plotted $10^{-38}\mathrm{c}{\mathrm{m}}^2/GeV$ units. With $R={\sigma }_{\bar{\nu }}/{\sigma }_{\nu }\simeq 1/2$ and $r=\bar{Q}/Q$,Using the literal readings $0.33/0.67$ gives $r=0.1905$, so twenty percent is an appropriate graphical estimate, not a precision fit. A purely valence target would have $R=1/3$, while a fully symmetric sea would give $R=1$.

The neutrino-antineutrino DIS sea momentum ratio is a ratio of momentum fractions in the effective flavour combinations, not an unweighted count of antiquarks. The cross-section integral contains $xq(x)$: it cannot determine $\int \bar{q}/\int q$ without additional assumptions about their shapes. A distribution growing at small $x$ can have a large number integral while its momentum integral remains finite.

Use the orthonormal basis $|u\bar{u}⟩,|d\bar{d}⟩,|s\bar{s}⟩$. The normalized flavour singlet isThe SU(3) invariant contraction pairs each fundamental flavour with its conjugate. Applying any supplied raising operator to the singlet gives zero because its quark and antiquark contributions cancel.

To fix the neutral isospin triplet phase consistently with the source's charged states, apply $I_{+}$ to $-|d\bar{u}⟩$:Thus one missing member of the meson octet isIndeed $I_{+}|T_0⟩=\sqrt{2}|u\bar{d}⟩$, the correct ladder operator strength for the $I=1$ triplet.

The other missing state is orthogonal to both $|1⟩$ and $|T_0⟩$. For coefficients $(a,b,c)$, orthogonality requires $a=b$ and $a+b+c=0$. Choose the convenient overall phase $c>0$ and normalize:It can also be obtained by adding the supplied ladder operations $V_{+}[-|s\bar{u}⟩]=|s\bar{s}⟩-|u\bar{u}⟩$ and $U_{+}[-|s\bar{d}⟩]=|s\bar{s}⟩-|d\bar{d}⟩$. The three displayed states are mutually orthonormal. The meson octet here is the traceless adjoint component of $\bar{3}\otimes 3$; the source's word “antisymmetric” does not impose antisymmetry on all its quark-antiquark flavour wavefunctions.

The observed dominance of kaons supports a predominantly $s\bar{s}$ phi meson. In a connected strong decay, the original strange quark and antiquark survive into different kaons, while a gluon produces a light $q\bar{q}$ pair:This gives $K^{-}{K}^{+}$ or ${\bar{K}}^0{K}^0$ and is allowed by the OZI rule. A pionic final state contains no strange valence constituents, so the initial $s\bar{s}$ pair must annihilate. The strong annihilation of a colour-singlet $1^{--}$ state requires at least three gluons: one gluon cannot carry the singlet colour, and two have the wrong charge-conjugation quantum number. This disconnected flavour topology is suppressed by the OZI rule, although it need not vanish.

The leptonic mode proceeds through $s\bar{s}\to {\gamma }^{\ast }\to {\ell }^{+}{\ell }^{-}$ and is suppressed by electromagnetic couplings; its width is proportional to ${\alpha }^2$ times the vector-meson overlap. Unlike a strong hadronic decay, it cannot create its final particles through quark rearrangement. Hence a branching fraction of order $10^{-4}$ is qualitatively consistent with electromagnetic annihilation competing with strong modes.

Kaon dominance is especially suggestive because phase space is limited: using the supplied neutral-kaon and phi masses,The two spin-zero kaons have orbital angular momentum $L=1$ to match the vector parent, so their partial width has the additional near-threshold factor $p_K^3$. Pionic modes have more phase space yet remain subdominant because of their annihilation topology. The observations therefore support, but do not establish exact, ideal vector-meson mixing: a small nonstrange admixture, electromagnetic amplitudes and nonperturbative strong dynamics can all contribute to the pionic rate.

Use the leading-order parton model: scatter incoherently from an approximately on-shell collinear quark carrying nucleon momentum fraction $x$, neglect quark and lepton masses, use an unpolarized target, and neglect nuclear binding and higher-twist corrections. The condition $Q^2\gg m_p^2$ suppresses target-mass effects, while $Q^2\ll M_W^2$ permits the local Fermi interaction instead of the full W propagator. CKM-weighted accessible final flavours are approximated by their leading light-quark sums. The struck-quark centre-of-mass energy is $\hat{s}\simeq xs$, with $s=(k+p{)}^2\simeq 2m_N{E}_{\nu }$.

A neutrino scatters through $\nu d\to {\mu }^{-}u$ or $\nu s\to {\mu }^{-}c$, and through the corresponding crossed antiquark channels such as $\nu \bar{u}\to {\mu }^{-}\bar{d}$. The chiral charged current makes the massless quark contribution independent of $y$ and the antiquark contribution proportional to $(1-y{)}^2$. The scattering probability is weighted by the parton distribution functions; the explicit factor $x$ comes from $\hat{s}=xs$. In the usual convention $q(x)dx$ is a number probability density, while $xq(x)$ is a momentum density. Approximate Bjorken scaling leaves mainly $x$ dependence; QCD evolution adds a logarithmic $Q^2$ dependence.

The printed proton-density formula is not an isoscalar formula in the standard PDF convention. For a proton, neglecting incoming charm and taking accessible CKM sums as unity, the conventional leading-order result isFor an isoscalar target per nucleon, average proton and neutron and use charge symmetry of nucleon parton distributions: $d_n=u_p$, ${\bar{u}}_n={\bar{d}}_p$, $s_n=s_p$. This givesThe standard normalization follows equivalently by inserting $F_2^{\nu p}=2x(d+s+\bar{u})$ and $x{F}_3^{\nu p}=2x(d+s-\bar{u})$ into the charged-current structure-function formula.^[ref]

A concrete discrepancy remains even with a simple valence model: take $u_p=2h(x)$, $d_p=h(x)$ and all sea densities zero. The correctly averaged isoscalar bracket is $3h(x)$, while the printed bracket is only $h(x)$ with the same prefactor. Thus the intended helicity and $x$ dependence can be explained, but the literal printed isoscalar claim cannot be derived. If the symbols were instead defined as momentum densities, the extra explicit $x$ would also need removal.

For the stationary target, $p=(m_N,0)$. The invariant definition immediately givesNo massless approximation is needed for this laboratory identity when binding and target recoil conventions are neglected.

For the angular relation, specify the neutrino-parton centre-of-momentum frame and neglect masses there. The incoming and outgoing leptons then each have energy $\sqrt{\hat{s}}/2$. Their momentum transfer satisfiesSince $Q^2=xys$ and $\hat{s}=xs$, cancellation givesThis identifies the antiquark weight $(1-y{)}^2$ with the corresponding squared angular factor. The relevant massless condition is sufficient energy in the lepton-parton subprocess; $E_{\nu }\gg m_{\mu }$ alone does not control every endpoint configuration.

The frame qualifier matters: the complete neutrino-nucleon centre-of-mass final state contains a variable-mass hadronic system. In that frame, still neglecting masses but retaining $x$, direct two-body kinematics for $\mu +X$ gives insteadFor example $x=1/2$ and ${\theta }_{\nu N}=\pi /2$ give $y=2/3$, not $1/2$. Thus the simple angle-only relation implicitly uses the partonic frame.

Useful examples with no net quark flavor are the eta c meson ${\eta }_c=c\bar{c}$ and the eta b meson ${\eta }_b=b\bar{b}$. Examples with nonzero quark flavor are the kaon $K^{+}=u\bar{s}$ and the D meson $D^{+}=c\bar{d}$. These are pseudoscalar mesons: in the lowest quark model configuration the constituent spins combine to $S=0$, the relative orbital angular momentum is $L=0$, and the total angular momentum and parity are $J^P=0^{-}$. For the self-conjugate examples, meson parity and charge conjugation also gives $C=(-1{)}^{L+S}=+1$. Electrical neutrality alone would not imply zero net quark flavor; for example $K^0=d\bar{s}$ has strangeness.

Restoring reduced Planck constant, decay width and mean lifetime satisfy $\Gamma \tau =\hslash$. Thus the representative numbers correspond toThe mean lifetimes differ by at least $3.0\times 10^{10}$. Hidden-heavy-flavor quarkonium can undergo quark-antiquark annihilation into gluons, followed by hadronization. A $0^{-+}$ color-singlet state can annihilate into two gluons. A lightest open-flavor meson cannot simply annihilate through the strong interaction, which conserves net quark flavor. With no lighter accessible final state carrying that flavor, it must undergo a weak interaction changing a constituent's flavor. The small weak interaction amplitude, CKM matrix factors and available Lorentz-invariant phase space produce a much smaller decay width.

Qualification. A decay width of order tens of MeV is not a universal property of flavor-neutral pseudoscalar mesons. The neutral pion has $\Gamma \simeq 7.8\mathrm{eV}$ and decays predominantly by the electromagnetic interaction, a concrete counterexample. Its mass lies below the lightest allowed hadronic final state. The examples above illustrate the intended annihilation-versus-open-flavor comparison; the numerical scale depends on thresholds and selection rules.^[ref]

Assume the charmonium state annihilates through a virtual photon, with lepton universality and negligible higher-order corrections. For a vector meson of mass $M$, the partial decay width into a charged lepton pair contains both the relativistic two-body decay factor ${\beta }_{\ell }$ and a vector-current spin factor:^[ref] For example, contracting the vector polarization sum with the Dirac trace for the two final fermions gives a factor proportional to $M^2+2m_{\ell }^2$, explaining why retaining only ${\beta }_{\ell }$ is insufficient. Using $m_{\tau }=1.777\mathrm{GeV}$, $m_{\mu }=0.1057\mathrm{GeV}$ and $M=3.686\mathrm{GeV}$ givesThe supplied total decay width also gives ${\Gamma }_{\mu \mu }=2.352\mathrm{keV}$ and ${\Gamma }_{\tau \tau }\simeq 0.91\mathrm{keV}$. It cancels from the branching fraction ratio.

The decisive change at $3.770\mathrm{GeV}$ is crossing the open-charm D meson-pair threshold, approximately $3.73\mathrm{GeV}$. The strong interaction can now create a light quark-antiquark pair and rearrange $c\bar{c}$ into $D\bar{D}$, without annihilating the heavy pair. This increases the total decay width from hundreds of keV to tens of MeV. The leptonic decay partial decay width is additionally smaller because $\psi (3770)$ is predominantly a $1{}^3{D}_1$ state, whereas $\psi (3686)$ is predominantly $2{}^3{S}_1$; the short-distance annihilation overlap is suppressed for nonzero orbital angular momentum. A roughly hundredfold increase in total decay width together with an order-ten reduction in leptonic partial decay width explains the order-thousand reduction in branching fraction. Attributing all three orders of magnitude to the threshold alone would miss the second effect.^[ref]

This labelled part supplies a definition rather than a separate task. Its notation is interpreted as an incoherent charge-squared sum: the photon couples to a quark with electric charge $e{e}_q$, and the squared scattering amplitude weights its contribution by $e_q^2$. To make the antiquark convention explicit, letThen, restricting the displayed example to three active light quark flavors, the deep-inelastic structure function isThus the source's sum must either include antiquarks as separate partons or define each $q^p$ to include its antiquark distribution. It must not silently omit the sea quarks, especially in the small-$x$ limit used below.

Subtracting the deep-inelastic structure functions in e givesThus at moderate and large $x$, the charge-weighted up quark content of the proton exceeds its down quark content. In the valence quark dominated regime this is consistent with the $uud$ valence structure, compared with $udd$ for the neutron. The larger $2/3$ up quark electric charge makes the proton scatter more strongly. This observation alone does not establish $u^p(x)=2d^p(x)$ pointwise: the valence-number sum rules constrain integrals, not every $x$ bin.

If the small-$x$ observation means equality to good relative accuracy, it supports dominance by approximately charge-symmetric sea quarks, with $U\simeq D$ and negligible relative valence excess. Gluon splitting creates quark-antiquark pairs, and the leading strong interaction treats light quark flavors similarly. It does not follow that every sea quark flavor has exactly the same distribution at every $x$.

Logical qualification. Equality of the limiting values alone is weaker than equality of their ratio. For example $U=2$, $D=1$, $S=0$ near $x=0$ gives $F_2^p=x$ and $F_2^n=2x/3$: both tend to zero, although $U\ne D$ and $F_2^n/F_2^p=2/3$. The usual inference about a symmetric sea needs $F_2^n/F_2^p\to 1$ or equivalent information about relative differences, rather than merely coincident zero limits.

Away from narrow resonances and quark thresholds, the QED annihilation cross section into each accessible quark flavour is its squared fractional electric charge times the muon-pair cross section, multiplied by the number $N_c$ of distinguishable color charges. Thus the inclusive hadronic ratio isWith $N_c=3$, the continuum predictions are $R_{uds}=2$, $R_{udsc}=10/3$ and $R_{udscb}=11/3$ as successive charm quark and bottom quark channels become accessible. The measured normalization and steps support three color charges. At leading order these hadronic events form two back-to-back jets from the quark and antiquark; quantitative comparisons include QCD corrections and avoid the nonperturbative threshold region.

The extra resolved jet is explained by $e^{+}{e}^{-}\to {\gamma }^{\ast }\to q\bar{q}g$, with gluon radiation from either the quark or antiquark. One possible color charge routing is initial $q_r{\bar{q}}_{\bar{r}}$ followed by $q_r\to q_b+g_{r\bar{b}}$; the final $q_b$, ${\bar{q}}_{\bar{r}}$ and $g_{r\bar{b}}$ combine to a net colour singlet.

The three-jet cross section contains one additional strong interaction vertex in the amplitude, so its ratio to the two-jet rate is of order ${\alpha }_s=g_s^2/(4\pi )$ times a resolution-dependent kinematic coefficient. A resolved fraction of order $15\%$ is consistent with perturbative QCD and a strong coupling constant much larger than the fine-structure constant. It is not an equality ${\alpha }_s=0.15$: the jet definition and cuts are needed. Photon radiation has a much smaller electromagnetic rate. The large hadronic third-jet rate supports gluon radiation; identifying the radiated object specifically as a colour-octet gluon also uses the QCD vertex and colour-flow description, not merely the existence of three visible jets.

Separate the center-of-mass motion from the relative motion. Two equal quark masses give reduced mass $\mu =m_q/2$. Replacing the hydrogen Coulomb strength by $a>0$ gives the Coulomb binding energies and length scaleThe factor of two in the reduced mass matters: using $m_q$ directly would double the binding energy. These are relative binding energies measured from the free quark–antiquark threshold; the rest energy $2m_q{c}^2$ is excluded.

The neutron has valence content $udd$ and total spin $J=1/2$; its ground-state parity is positive. The two down quarks contribute charge $-2/3$ and the up quark contributes $+2/3$, giving net charge zero. “Three valence quarks” specifies the net flavor quantum numbers, rather than excluding gluons or additional sea quark–antiquark components from the full Quantum chromodynamics state.

In the approximate light-flavor SU(3) classification, the neutron belongs to the ground-state baryon octet, containing eight hadrons:These are eight particle species, without counting their antiparticles again as extra members of the same octet. The corresponding antibaryons form the conjugate particle set. Within the smaller SU(2) isospin classification, the neutron instead belongs to the two-member nucleon doublet $(p,n)$. Stating the symmetry distinguishes the usual eight-member flavor multiplet answer from the also valid two-member isospin multiplet. The baryon decuplet has different ground-state spin and is not the neutron's octet.

The light pseudoscalar meson nonet has $J^P=0^{-}$ and flavor structure $3\otimes \bar{3}=8\oplus 1$. Its members are the three pions, four kaons, and the two neutral isospin-zero states eta meson and eta prime meson. For mesons the baryon number is zero, so the Gell-Mann--Nishijima formula reduces toHere $S$ is strangeness, with an $s$ quark carrying $-1$ and an antiquark $\bar{s}$ carrying $+1$. Thus the upper row is $K^0=d\bar{s}$ at $(I_3,S)=(-1/2,1)$ and $K^{+}=u\bar{s}$ at $(1/2,1)$. The lower row is $K^{-}=s\bar{u}$ at $(-1/2,-1)$ and ${\bar{K}}^0=s\bar{d}$ at $(1/2,-1)$. The charged pions lie at $(\pm 1,0)$, while ${\pi }^0,\eta ,{\eta }^{\prime }$ share $(0,0)$ as flavor-weight coordinates.

The neutral pion direction is $(u\bar{u}-d\bar{d})/\sqrt{2}$. The octet and singlet isoscalar directions mix to make the physical eta and eta-prime states; counting three central states is essential to obtaining nine rather than seven particles. Horizontal lines in the diagram have constant $S$, while lines $I_3+S/2=Q$ mark constant charge. Meson exchange describes part of the residual nuclear force; the underlying quark-level strong interaction is mediated by gluons, not by treating pions as elementary gauge bosons.

A baryon with constituent content four quarks and one antiquark, whose net baryon number is one. Such content admits more flavour representations than three quarks alone. A representation in the tensor product is not evidence by itself of a bound state.