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Write $C_{ikjl}={\sum }_a{T}_{ik}^a{T}_{jl}^a$. Hermiticity converts its complex-conjugate factors into $T_{ki}^b{T}_{lj}^b$. Summing all external colors therefore givessince SU(3) has eight generators. This is a sum, before averaging the nine equally populated incoming color states.

For the spinor current, use $\bar{u}=u^{†}{\gamma }^0$, $({\gamma }^0{)}^2=1$ and the stated gamma-matrix adjoint identity:The spin sum completeness relation then closes each product of currents into a trace. Averaging two spins for each incoming quark, in addition to nine incoming colors, givesThe factor is $1/18$, rather than $2/9$: the latter omits the initial spin average.

To derive the four-gamma matrix trace, first anticommutation and cyclicity give $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$, with $\mathrm{Tr}1=4$. If $T=\mathrm{Tr}({\gamma }^{\lambda }{\gamma }^{\mu }{\gamma }^{\sigma }{\gamma }^{\nu })$, move ${\gamma }^{\lambda }$ successively through the next three matrices. The last term is $-T$ by cyclicity, soSubstitution yieldsConsequently the first momentum trace is $4(p_3^{\mu }{p}_1^{\nu }+p_3^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_3\cdot p_1)$. Contracting both traces givesHere $s=(p_1+p_2{)}^2$, $u=(p_1-p_4{)}^2$ and all external masses vanish. Thus the distinct-flavor quark scattering color factor givesWe used $t=-s(1-\cos {\theta }^{\ast })/2$ and $u=-s(1+\cos {\theta }^{\ast })/2$. The forward divergence reflects massless $t$-channel exchange in this parton-level tree approximation; it is not an unrestricted prediction for an experimentally resolved jet cross section.

Finally, completeness of the SU(3) generators gives the fundamental-generator Fierz identityFor each distinct pair $i\ne j$, the allowed final colors are either $k=i,l=j$, with coefficient $-1/6$, or $k=j,l=i$, with coefficient $1/2$. If $i=j$, only $k=l=i$ is nonzero, with coefficient $1/3$. These are 15 ordered configurations: six unchanged-color cases, six exchanged-color cases and three all-equal cases. Their squared coefficients sum to $6/36+6/4+3/9=2$, independently checking the first result. Flavor remains tied to each external line; exchanging colors does not exchange the up and down quark species.

The SU(5) grand unified theory embeds the strong and electroweak groups in a single simple gauge group. Its 24 gauge generators contain the eight gluons, three weak generators and one hypercharge generator, plus twelve additional generators connecting color and weak directions. A common gauge coupling applies at the unification scale with the properly normalized hypercharge generator; low-energy couplings differ through their renormalization group evolution.

Using left-chiral fields throughout, one Standard Model generation fits intoThe dimensions are $3+2=5$ and $6+3+1=10$, so no right-handed neutrino is required in these multiplets. A singlet neutrino can be added. A suitable adjoint scalar vacuum breaks SU(5) to the Standard Model subgroup; a fundamental scalar contains an electroweak doublet and a colored triplet. The extra heavy vector bosons can mediate transitions between quarks and leptons, hence proton decay in the minimal model. This is a proposed unification, not an experimentally established identification of new particles.

The hierarchy problem concerns keeping an elementary weak-scale scalar light in the presence of much heavier physical scales. A heavy state coupled to it generically induces a threshold correction of order $\delta m_H^2\sim c{M}^2/(16{\pi }^2)$, with $c$ a product of couplings. Without protection, a weak-scale result then needs cancellation against a much larger parameter. In SU(5), this is sharpened by doublet-triplet splitting: the electroweak doublet must stay light while its colored partner must be sufficiently heavy to avoid rapid proton decay. Choosing a small tree-level doublet mass is not by itself a mechanism protecting it from subsequent heavy-particle corrections. Supersymmetry and other proposals change this protection problem, but the minimal SU(5) embedding alone does not solve it.

For the usual colour SU(3) fixing the fourth coordinate of the SU(4) defining representation,Taking the exterior square givesThe identification ${\Lambda }^{2}3\cong \overline{3}$ follows by contracting an antisymmetric tensor with the invariant three-index Levi-Civita symbol. Combining this with c gives all colour multiplets:A colour triplet has the color charge transformation law of a quark, an antitriplet that of an antiquark, and a colour singlet is colour-neutral, as a lepton is. Thus both parent representations permit quark-like and antiquark-like colour sectors; the vector has one neutral sector and the spinor two. These are colour assignments only. They do not determine particle spin, statistics, electric charge, weak quantum numbers or a full generation of the Standard Model. In a real parent field the conjugate colour sectors may describe one complex colour field and its conjugate, rather than independent particle species.

Global group qualification. The printed chain is correct at the Lie algebra level, $\mathfrak{su}(3)\subset \mathfrak{su}(4)\cong \mathfrak{so}(6)\subset \mathfrak{so}(7)$. Globally its simply connected form is $SU(3)\subset SU(4)\cong \mathrm{Spin}(6)\subset \mathrm{Spin}(7)$. The standard map of SU(4) into $SO(7)$ through $6\oplus 1$ has a nontrivial kernel: the concrete central element $-I_4$ acts as $+I_6$ on ${\Lambda }^{2}4$ and trivially on the singlet representation. Its image is $SO(6)\cong SU(4)/\{\pm I_4\}$, rather than an injective copy of SU(4) under this map.

The global form obstruction for the Spin7 spinor is explicit: the central element $-1$ of the Spin group $\mathrm{Spin}(7)$ maps to the identity in $SO(7)$ but acts as $-I_8$ on the eight-dimensional spin representation. Equivalently $-I_4$ acts as $-I$ on both $4$ and $\overline{4}$ in its restriction. Therefore the spinor does not descend to a genuine SO(7) group representation. A theory containing that spinor needs the Spin group global form, or must understand the source's notation as specifying only its Lie algebra. The displayed colour branchings remain valid with this correction.

The commuting direction in solution provides the correct candidate Lie algebra for a gauge group, $\mathfrak{su}(3{)}_c\oplus \mathfrak{su}(2{)}_L\oplus \mathfrak{u}(1{)}_Y$. This is a necessary structural condition for the Standard Model, but it does not by itself establish a viable grand unified theory.

The representations in solution do not give a Standard Model fermion generation with the correct weak hypercharges. Use $Q_{\mathrm{electric}}=T_3+Y$. Identifying $(3,2{)}_1$ with the left-handed quark doublet, whose weak hypercharge is $1/6$, fixes $Y=X/6$. But then the doublet in $(1,3)$ has $Y=1/6$, whereas the left-handed lepton doublet requires $Y=-1/2$. Its singlet has $Y=-1/3$, whereas a right-handed electron requires $Y=-1$. The $(\overline{3},1{)}_0$ is a further concrete mismatch: right-handed quarks written as left-handed antiquarks require $Y=-2/3$ or $1/3$, and a rescaling cannot turn zero into either value. These comparisons use the electroweak representation and hypercharge table.

The charge-$-3$ doublet in $(1,8)$ does have the required lepton weak hypercharge under this normalization, but the same Adjoint representation also brings a doublet of the opposite charge and an extra triplet. Therefore merely adding that representation does not supply the missing chiral generation without additional states and a mechanism for their masses.

Furthermore $SU(3)\times SU(3)$ is a product of two Simple Lie groups, permitting two independent gauge couplings; it does not enforce a common strong and electroweak gauge coupling. Thus the embedding exists, but the displayed matter content does not furnish a viable minimal unification of the Standard Model. A more elaborate model would require additional representations, consistent chiral matter, anomaly cancellation, and a specified spontaneous symmetry breaking mechanism. The calculations here do not prove that every possible extension using this product group is impossible.

Take the quarks $q=(u,d,s,c{)}^T$ in the defining $4$ of flavour SU(4) and the antiquarks in $\overline{4}$. The flavour part of a meson is a quark-antiquark pair $M^i{}_j=q^i{\bar{q}}_j$. It transforms as $M\mapsto UM{U}^{-1}$, so the complete four-flavour meson multiplet isIts traceless part $M-\frac{1}{4}(\mathrm{tr}M)I_4$ is the Adjoint representation $15$, the direct four-flavour analogue of the three-flavour meson octet. The trace supplies an additional trivial representation. Thus “the mesons” can refer to the traceless multiplet or to the complete quark model tensor product; keeping both explicit avoids dropping a state.

Using solution, the SU(3) representation contents areDirectly, the upper-left block $q^i{\bar{q}}_j$ with $i,j\le 3$ is the old $3\otimes \overline{3}=8\oplus 1$; the three entries $q^i\bar{c}$ form $3$, the three entries $c{\bar{q}}_j$ form $\overline{3}$, and $c\bar{c}$ is another singlet. The dimensions are $9+3+3+1=16$. This is a classification by flavour symmetry, rather than colour or spin. The much larger charm quark mass breaks the flavour SU(4) approximation strongly; it does not change this representation counting.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger representations containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large representation.

A dual vector must be pulled back by $P^T$, since $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus a G2 root system weight $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet meson retains its values from solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor instance the first three come from the G2 root system weights $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The weight diagram displays the preserved octet and the new states.

The fractional electric charges make these additional states qualitatively different from ordinary quark-antiquark mesons, whose electric charges are integers. This is a consequence of the proposed symmetry extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

A meson has quark model flavour content $q_i{\bar{q}}^j$, so four quark flavours give sixteen flavour states inThis counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states $3\otimes \bar{3}=8\oplus 1$.

Taking the fourth quark to be the charm quark, the first nine combinations $q_{\alpha }{\bar{q}}^{\beta }$, with $\alpha ,\beta \in \{u,d,s\}$, are the light sector. The extra seven are $q_{\alpha }\bar{c}$ in $3$, $c{\bar{q}}^{\alpha }$ in $\bar{3}$, and $c\bar{c}$ in $1$ under light-flavour $SU(3)$. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from $SU(4)$ invariant. Exact flavour $SU(4)$ would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour $SU(4)$ is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour $SU(3)$.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu =P\cdot q>0$ and $x=Q^2/(2\nu )>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the physical momentum support of a deep-inelastic hadronic tensor. The original PDF instead prints ${\delta }^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would requirein the Bjorken scaling limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero deep-inelastic structure functions. Below, use the physically consistent absorption delta function ${\delta }^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.

Work in the leading parton model: the target is unpolarized; scattering is incoherent from massless collinear quarks and antiquarks; intrinsic transverse momentum and target-mass effects are neglected. The parton distribution functions count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.

Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace isIntegrating its one-particle final Lorentz-invariant phase space, with the $1/(4\pi )$ tensor normalization, supplies ${\delta }_{+}(r^2)/2$. Thus the spin-averaged electromagnetic parton tensor isOn physical support, $r^2=0$ gives $\xi =x$ and $p\cdot q=Q^2/2$. Define the transverse tensorsThe bracket in $w_f$ becomesThis follows by writing $p=\xi \tilde{P}-q/2$ on shell and expanding; it also verifies transversality to $q$.

The correct parton convolution normalization for a hadronic tensor isThe $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.

Use the parton on-shell delta identity,Matching the two tensor coefficients now givesIn particular $F_2=2x{F}_1$, the Callan-Gross relation for spin-one-half partons. Scaling here means the naive leading approximation; full Quantum chromodynamics produces scale-dependent distributions and corrections.

Each quark contributes $+1$ and each antiquark contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the quark number sum rule:Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.

Under the separately imposed valence-only approximation, ${\bar{q}}_f=0$ and only up/down distributions remain. The proton has $(N_u,N_d)=(2,1)$ and the neutron has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the valence-only proton and neutron structure-function moments areNo extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.

Use the electroweak hypercharge normalization $Q=T_3+Y$, rather than $Q=T_3+\hat{Y}/2$. In the latter convention all tabulated hypercharges would be doubled. The electroweak representation and hypercharge table for one minimal Standard Model generation isThe quarks are colour triplets, the leptons are colour singlets, and all listed right-handed fields are weak singlets. A sterile right-handed neutrino, if added, has representation $(1,1{)}_0$ and changes none of the anomaly traces.

On each left weak doublet, $T_3=\mathrm{diag}(1/2,-1/2)$. The electric charges are therefore $1/6\pm 1/2=(2/3,-1/3)$ for $Q_L$ and $-1/2\pm 1/2=(0,-1)$ for $L_L$. On each right singlet $T_3=0$, so its electric charge equals its hypercharge. These are the observed quark and lepton charge assignments.

For Standard Model anomaly cancellation, traces must include both components of each doublet and all three colours. Every right weak singlet has $T_3^2=0$, givingOn the left doublets, $T_3^2=I/4$, while ${\mathrm{tr}}_L{T}_3=0$ separately in each doublet. ConsequentlyThe linear trace isEquivalently, ${\mathrm{tr}}_{L}Q=3(2/3-1/3)+(0-1)=0$. This proves cancellation of the mixed weak-hypercharge gauge anomaly within one generation.

The one-generation hypercharge anomaly traces for the cubic condition areThus both required anomaly differences vanish generation by generation. The right-handed subtraction is important: counting all physical fields with the same chirality sign would not give the cubic cancellation.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

Use units in which the proton electric charge is one. The up quark and down quark both have spin angular momentum $1/2$ and form a flavour isospin doublet:Their antiquarks have the opposite charges and third components of isospin, with the same total isospin and spin. This is strong-interaction flavour isospin, not the chiral weak isospin assignment.

The Clebsch-Gordan decomposition for SU2 says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and isospin. ThusBefore exchange symmetry is imposed, three light quarks have total spin $S=1/2$ or $3/2$, and total isospin $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero orbital angular momentum, $J=S$.

Physical baryons must also be colour singlets. The three-quark colour singlet is completely antisymmetric, so the Pauli exclusion principle requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the Pauli constraint on three-quark flavour multiplets; it leavesEquivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20=2\cdot 2+4\cdot 4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state baryons.

For three light quarks, charge addition gives $Q=I_3+1/2$. The observed nucleon doublet and Delta baryon quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:The proton and neutron have $J=1/2$; all four Delta baryons have $J=3/2$. These states have positive parity in the zero-orbital ground-state model.

For a light quark-antiquark meson, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states areEach row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero orbital angular momentum their parity is negative, following the meson parity and charge conjugation rule $P=(-1{)}^{L+1}$. The spin-zero isotriplet is the pion multiplet, and the spin-one isotriplet the rho meson multiplet. The nonstrange spin-one isosinglet is the idealized omega meson flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the eta and eta prime mesons; it is not automatically a physical mass eigenstate once other flavours and mixing are included.

Now introduce an isospin-zero quark $h$ of spin $1/2$ and charge $q_h$. For baryons with one isospin-singlet quark, the two light quarks have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt{2}$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence light-diquark spin-isospin correlation gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin producesThe three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half Lambda baryon, the spin-one-half Sigma baryon triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half Lambda c baryon of charge $1$, and the Sigma c baryon triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate antibaryons have opposite charges and unchanged $I,J$.