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The zero-isospin, zero-flavor hypercharge state ${\eta }_8=(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$ in the pseudoscalar meson octet. It is a pure octet basis state; the physical eta may mix it with the flavor-singlet state.

For the light quark flavors, electric charge is $Q=I_3+Y/2$, with flavor hypercharge $Y=B+S$. On $(u,d,s)$ it gives $(2/3,-1/3,-1/3)$, and additivity extends it to the baryon octet and meson octet.

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.

Compared with the complete three-flavour meson tensor product, the additional flavour states are exactlyThus there are seven new flavour states. For the pseudoscalar mesons, the six open-charm states are ${\bar{D}}^0,D^{-},D_s^{-}$ and $D^0,D^{+},D_s^{+}$, respectively, using the D meson and Ds meson conventions. A pure $c\bar{c}$ pseudoscalar meson is the idealized flavour state associated with the eta c meson. The quark model flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other meson species.

If comparing the traceless SU(4) $15$ with the traceless SU(3) meson octet, the extra seven states are instead the same two triplets and the normalized SU(3) singletIts coefficient matrix is traceless and has squared norm one. In particular, $c\bar{c}$ by itself is not a vector in the traceless $15$; its coefficient matrix has trace one.

To relate these two descriptions, let $S_4=(u\bar{u}+d\bar{d}+s\bar{s}+c\bar{c})/2$ be the normalized SU(4) singlet and $S_3=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ the old SU(3) singlet. ThenThis orthonormal change of basis shows explicitly why the complete tensor product contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.

For an A2 root system weight $w=(w_1,w_2{)}^T$, the specified electric charge and flavour hypercharge are the linear functionalsApply these to the meson octet weights in solution. The results, with electric charges measured in units of the positive elementary electric charge, areThe table illustrates the meson octet of pseudoscalar mesons; the same weight calculation applies to another meson species arranged in the same representation. Here flavour hypercharge is $Y=B+S$, with $B$ the baryon number and $S$ the strangeness, rather than weak hypercharge. If the full quark model tensor product $3\otimes \overline{3}$ is retained, its additional trivial representation is a ninth meson with $(q,y)=(0,0)$. This singlet is separate from the octet and need not share its mass.

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.

When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by SU(3) leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate flavour symmetry, distinct from the exact colour gauge symmetry. The quark transforms as $3$ and the antiquark as $\overline{3}$.

For mesons, a quark-antiquark tensor is a $3\times 3$ matrix. Its trace and traceless parts are invariant subspaces, givingThe trace is the eta singlet state; the traceless part is the meson octet. For baryons, first separate a pair of quarks into symmetric and antisymmetric tensors:The antisymmetric pair transforms as $\overline{3}$ by contraction with the invariant epsilon tensor. In $6\otimes 3$, complete symmetrization gives ${\mathrm{Sym}}^{3}3$ of dimension $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{3}\otimes 3$, trace and traceless parts give a singlet and another octet. Thus the tensor cube of the defining SU(3) representation isThe two octets carry the two mixed permutation-symmetry components. Their weight diagrams are identical; the extra copy is a multiplicity, not a new arrangement of weights.

The printed charge relation fixes the half-hypercharge convention, rather than the usual Flavor hypercharge normalization:The quark weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the baryon weight diagrams in the half-hypercharge convention. The baryon octet has $(p,n)$ at $Y=1/2$, the isotriplet $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$ and the isosinglet ${\Lambda }^0$ at $Y=0$, and $({\Xi }^0,{\Xi }^{-})$ at $Y=-1/2$. The baryon decuplet has rows $\Delta$ at $Y=1/2$, ${\Sigma }^{\ast }$ at zero, ${\Xi }^{\ast }$ at $-1/2$, and ${\Omega }^{-}$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.

The fully symmetric flavour representation is precisely the baryon decuplet, $10={\mathrm{Sym}}^{3}3$. For a spatial ground state, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ quarks, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the Pauli exclusion principle for the three fermions.

An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state isIt is completely antisymmetric and an SU(3) colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third exterior power: ${\bigwedge }^3{\mathbb{C}}^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the quark model. This argument motivates three colours rather than deriving their exact number without further physical input.

The Pauli constraint on three-quark flavour multiplets also explains the absence of a flavour-singlet baryon in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. Butso three spin-$1/2$ quarks have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state baryon octet with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full Pauli constraint on three-quark flavour multiplets is applied. Excited orbital wavefunctions can permit three-quark flavour singlet; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every baryon at every energy.

A flavour-singlet meson is allowed: a quark and an antiquark are not three identical fermions, and the invariant flavour combination $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no meson exclusion.

Finally, ${\pi }^{+}p$ has baryon number one, strangeness zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ Delta baryon ${\Delta }^{++}$, the low-energy resonance with spin $3/2$. For $K^{+}p$, the kaon $u\bar{s}$ contributes strangeness plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ baryon can have positive strangeness, since every $s$ quark contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic quark content, for example $uuud\bar{s}$, and is not mandated by this spectrum. This is strangeness selection in pion-proton and kaon-proton resonances.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.

For the quark model, the relevant approximate flavour symmetry rotates the three light quarks $u,d,s$ in the fundamental $3$. The up quark, down quark, and strange quark basis states have the weight diagram of the defining SU(3) representation. In physical coordinates define isospin $I_3=H_1$ and flavor hypercharge $Y=2H_2/\sqrt{3}$. ThenThe last row follows from the Gell-Mann--Nishijima formula $Q=I_3+Y/2$, in units of the positive elementary electric charge. The antiquarks transform in $\overline{3}$ and have the negatives of these additive quantum numbers. Each quark has baryon number $B=1/3$; the strange quark has strangeness $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing flavor hypercharge from electroweak hypercharge. The Weyl group geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.

A conventional meson has valence content $q\bar{q}$, so its flavour symmetry gives a meson octet and a singlet. For the light pseudoscalar mesons, the octet contains the pions ${\pi }^{+},{\pi }^0,{\pi }^{-}$ at $Y=0$ with isospin $I=1$, the kaons $K^{+}=u\bar{s},K^0=d\bar{s}$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ Eta octet state. The central weight multiplicity two is realized byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The eta and eta prime mesons are mixtures of ${\eta }_8,{\eta }_1$, rather than exact octet and singlet mass eigenstates; this gives the pseudoscalar meson nonet. A different spin coupling of the same flavour product gives the vector mesons, again an octet and singlet, including the $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ mixtures. The lowest orbital states have $J^P=0^{-}$ and $1^{-}$ respectively.

A conventional baryon has three valence quarks, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest baryon octet has spin $1/2$ and positive parity. Its isospin rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the nucleons $p,n$, the three Sigma baryons, the Lambda baryon, and the two Xi baryons. The central states ${\Sigma }^0$ and $\Lambda$ have the same weight $(I_3,Y)=(0,0)$ but different total isospin; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity baryon decuplet is the triangle with rowsThe Delta baryons form its top row, and the Omega baryon is the bottom $sss$ state. All ten weights have multiplicity one. Their electric charges follow directly from the Gell-Mann--Nishijima formula; for example ${\Delta }^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and ${\Omega }^{-}$ has $I_3=0,Y=-2$, so $Q=-1$.

The two algebraic octets do not mean that two independent ground-state octets must occur. A three-quark colour singlet has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the Pauli exclusion principle therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the symmetric spin-flavour SU6 representation hasThe three-quark flavour singlet is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but ${\bigwedge }^3{\mathbb{C}}^2=0$. The Pauli constraint on three-quark flavour multiplets thus excludes it from this ground-state sector; orbital excitation permits other multiplets.

Finally, this flavour symmetry is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The weight diagrams classify flavour states, rather than asserting exact equality of their masses. Colour SU(3) is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the eightfold way. The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.

The commuting Hermitian operators $H_1,H_2$ admit a weight-space decomposition. The six root vectors shift their simultaneous eigenvalues byor their negatives. Put $h_i=2{\alpha }_i\cdot H$, $e_i=\sqrt{2}{E}_{+}^i$ and $f_i=\sqrt{2}{E}_{-}^i$. The supplied relations giveThese are three sl2 triples, coming from the complexifications of the three SU(2) Lie algebra root-pair subalgebras. They act through homomorphisms into $\mathrm{End}(V)$. For a nontrivial representation, simplicity of the SU(3) Lie algebra makes its kernel zero, so these images are actual ${\mathfrak{sl}}_2(\mathbb{C})$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $i{h}_i$, $i(E_{+}^i+E_{-}^i)$ and $E_{+}^i-E_{-}^i$.

Decomposing under any one of these subalgebras gives finite weight strings in its root direction. The two simple-root highest weights must be nonnegative integers. In the SU(3) highest-weight coordinates used here, this condition isHere $(a,b)$ are the Dynkin labels; $(p,q)$ are the eigenvalues of $H_1,H_2$. An irreducible representation is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the orthogonal complement of an invariant subspace remains invariant.

The weight diagram has Weyl group symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary weight multiplicities are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min (a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the dominant weight multiplicity formula for sl3: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and setIts multiplicity is zero unless $r,s$ are integers, and otherwise is $\max \{0,1+\min (a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.

For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_{+}+E_{-})/\sqrt{2}$ and $(E_{+}-E_{-})/(i\sqrt{2})$. In the defining representation these eight generators satisfy $\mathrm{tr}(t_a{t}_b)={\delta }_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their structure constants of a Lie algebra antisymmetric, sothe antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a quadratic Casimir operator and is scalar on an irreducible representation by the Schur lemma.

Let $v$ be a highest-weight vector, annihilated by all $E_{+}^i$. On $v$, each $E_{-}^i{E}_{+}^i$ vanishes and each $E_{+}^i{E}_{-}^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt{3}{H}_2$. Henceand therefore the SU(3) quadratic Casimir eigenvalue isFor a reducible representation the same calculation applies separately to its irreducible blocks.

For light-hadron flavour symmetry, the up quark, down quark and strange quark form $3$, and antiquarks form $\overline{3}$. Conventional mesons have $3\otimes \overline{3}=1\oplus 8$. Conventional three-quark baryons haveobtained from $3\otimes 3=6\oplus \overline{3}$, $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$. Their Dynkin labels are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional meson octet, baryon octet, baryon decuplet and singlet multiplets give $\boxed{X=0,3,6}$, with the same values for conjugate antibaryon multiplets. More generally, the zero flavour triality of a light-quark hadron imposes $a+2b\equiv 0(\mathrm{mod}3)$ on multiplets constructed entirely from light quarks, antiquarks and gluons. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $27$, of labels $(2,2)$, has $X=8$.

A weight of a representation of $SU(3)$ is a joint eigenvalue of two commuting generators spanning a Cartan subalgebra. Equivalently it describes the character by which the diagonal maximal torus acts on a weight vector. Use the Hermitian generatorsTheir multiples by $i$ are in the SU(3) Lie algebra. A vector with weight $(a,b)$ acquires the phase $e^{i(a\theta +b\phi )}$ under $\exp (i\theta I_3+i\phi Y)$. The conventional generator $H_8$ is related by $Y=2H_8/\sqrt{3}$; using $H_8$ instead just rescales the vertical weight coordinate.

In the defining group representation, the coordinate vectors $e_u,e_d,e_s$ are joint eigenvectors. The weight diagram of the defining SU(3) representation therefore hasComplex conjugation reverses all torus phases, so the weights of $\overline{3}$ are $-{\mu }_u,-{\mu }_d,-{\mu }_s$, each with multiplicity one. The conjugation bar is essential: the tensor product here is $3\otimes \overline{3}$.

Weights add in a tensor product of group representations. Thus the nine weights of this product are ${\mu }_i-{\mu }_j$. There are three zero weights from $i=j$, and the remaining six areTo turn this weight calculation into a direct-sum decomposition, identify the tensor product with $\mathrm{End}({\mathbb{C}}^3)$ by $v\otimes \overline{w}\mapsto v{w}^{†}$. The group acts by $M\mapsto UM{U}^{-1}$. This gives two invariant spaces,The scalar line has weight zero and is the trivial group representation $1$. The traceless space has the six nonzero weights just computed, represented by the off-diagonal matrix units $E_{ij}$, and two zero weights, represented by traceless diagonal matrices. Its dimension is eight and it is the complex Adjoint representation $8$. ThereforeThis proves the octet and singlet in a fundamental SU3 tensor product with the correct central weight multiplicities.

The singlet is irreducible because it is one-dimensional. For the octet, an invariant complex subspace of ${\mathfrak{sl}}_3$ is stable under commutators with the complexified Lie algebra. Commuting Cartan generators project it into weight spaces. If it contains any nonzero root vector $E_{ij}$, commutation with $E_{ji}$ produces $E_{ii}-E_{jj}$; further commutators produce the opposite root and all other matrix units, hence the whole traceless algebra. If it contains only a nonzero diagonal traceless matrix $D$, two diagonal entries differ, so $[E_{ij},D]=(D_{jj}-D_{ii})E_{ij}$ supplies a root vector and reduces to the preceding case. There is no nonzero proper invariant subspace. The octet is irreducible, not a sum of six one-dimensional weight spaces and two singlets: the nondiagonal generators connect those spaces.

In the quark model, take $u,d,s$ as the defining flavour triplet and their antiquarks as the conjugate triplet. A colour-singlet quark-antiquark state with relative orbital angular momentum $L=0$ and total spin $S=0$ has $J=0$ and parity $(-1{)}^{L+1}=-1$. Thus this flavour decomposition classifies a pseudoscalar meson nonet: a meson octet and a singlet. The use of approximate flavor symmetry is important; the strange quark is heavier, so these states need not have equal masses.

In the $(I_3,Y)$ diagram, $Y$ is flavor hypercharge, not electroweak hypercharge. Mesons have baryon number zero, so $Y$ equals their strangeness, and electric charge is $Q=I_3+Y/2$. The outer six weight states areThe pions form an isospin triplet, completed at the centre by ${\pi }^0=(u\overline{u}-d\overline{d})/\sqrt{2}$. The four kaons occupy the two hypercharge-one and two hypercharge-minus-one weights. In particular electrically neutral kaons are not at the centre of the weight diagram.

An orthonormal basis of the three central flavour combinations isThe Eta octet state is the second zero weight of the octet; the eta singlet state is the separate invariant scalar. Both have isospin zero, whereas the neutral pion has isospin one. All have $I_3=Y=Q=0$, so the location of a weight alone does not determine either isospin or irreducible multiplet. Their neutral flavour-diagonal $L=S=0$ states have charge conjugation $C=(-1{)}^{L+S}=+1$ and hence $J^{PC}=0^{-+}$.

The physical eta and eta prime mesons are mixtures of the octet and singlet combinations, conventionally described at this level byFlavour breaking and the singlet axial anomaly affect their masses and mixing. The neutral pion is much lighter, about $135\mathrm{MeV}$, and decays predominantly to two photons. The eta has mass about $548\mathrm{MeV}$ and important two-photon and three-pion decay modes; the eta prime has mass about $958\mathrm{MeV}$ and important decays to $\eta \pi \pi$. The singlet axial anomaly explains why the eta prime is not an additional light Goldstone boson merely because the flavour tensor product contains a singlet. The centre contains two octet directions and one singlet direction; physical eta mixing combines the latter two, leaving the isospin-one neutral pion separate to a good approximation.

Use flavor hypercharge $Y=B+S$, where $B$ is baryon number and $S$ is strangeness. The isospin coordinate is $I_3$. The baryon octet has coordinatesHere the nucleons have $S=0$, the Sigma baryons and Lambda baryon have $S=-1$, and the Xi baryons have $S=-2$. The central Sigma baryon belongs to an isospin triplet while the central Lambda baryon is an isospin singlet; equal coordinates do not identify the states.

The pseudoscalar meson octet isThe pions have $S=0$, the upper kaons have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral pion and the Eta octet state. This is the octet basis of flavor symmetry; the physical eta can also mix with the flavor-singlet state.

For the flavor SU(3) Cartan generators, choose the Hermitian physics convention and the inner product $⟨A,B⟩=\mathrm{tr}(AB)$. Then an orthonormal diagonal basis isStrictly, $i{h}_1,i{h}_2$ are elements of the anti-Hermitian SU(3) Lie algebra; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down quarks form an isospin doublet and the strange quark is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each quark has baryon number $1/3$, and their strangeness values are $(0,0,-1)$. It follows thatThese flavor hypercharge conventions differ from the electroweak hypercharge convention. If instead the inner product is $2\mathrm{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt{2}$, $t_8=h_2/\sqrt{2}$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt{3}$.

In the ordinary quark model, the proton has valence content $uud$ and charge $+1$, while the neutron has content $udd$ and charge zero. Additivity of electric charge gives $2q_u+q_d=1$ and $q_u+2q_d=0$, hence $q_u=2/3$, $q_d=-1/3$. The Sigma baryon ${\Sigma }^{-}$ has content $dds$ and charge $-1$, giving $2q_d+q_s=-1$. Thus the quark triplet's electric charges, in units of the positive elementary charge, areThe Gell-Mann--Nishijima formula is consequentlyIt also reproduces every baryon octet charge from the first diagram. On antiquarks the additive quantum numbers reverse sign, and combining a quark with an antiquark reproduces the meson octet charges.

Because the down and strange quarks have identical electric charge, $Q$ commutes with the U-spin generatorsThey satisfy $[U_a,U_b]=i{ϵ}_{abc}{U}_c$; equivalently the anti-Hermitian matrices $i{U}_a$ span an $\mathfrak{su}(2)$ subalgebra. The entries on the $d,s$ block of $Q$ are equal, so $[Q,U_a]=0$ for all three U-spin generators. The electric charge is therefore constant within each irreducible U-spin multiplet. For example, U-spin relates ${\pi }^{+}$ and $K^{+}$, and relates $p$ and ${\Sigma }^{+}$, without changing their charge. It does not imply exact mass degeneracy: unequal down- and strange-quark masses break U-spin.

For pion-nucleon octet channels, assume the collision is governed by the strong interaction. The initial baryon number is one and strangeness is zero, so an outgoing meson-baryon pair must preserve $B=1$, $S=0$, and electric charge. Thus its total flavor hypercharge is $Y=1$. The allowed types areA kaon of $S=+1$ can accompany a Lambda baryon or Sigma baryon of $S=-1$. An antikaon cannot balance the nonpositive strangeness of an octet baryon. A Xi baryon would require a meson of $S=+2$, which the meson octet does not contain.

Resolving these types by electric charge gives all possible pairs:In the isospin-symmetric approximation, total isospin is conserved as well: the incoming $1\otimes \frac{1}{2}$ contains $I=\frac{1}{2},\frac{3}{2}$. The $\pi N$ and $K\Sigma$ channels contain both values, while ${\eta }_{8}N$ and $K\Lambda$ contain only $I=\frac{1}{2}$. The extreme-charge initial states are pure $I=\frac{3}{2}$, consistently excluding ${\eta }_{8}N$ and $K\Lambda$. Clebsch-Gordan coefficients relate amplitudes in different charge channels; the table establishes permission, not equal probabilities. Electromagnetism and unequal up- and down-quark masses introduce small violations of isospin symmetry.

Finally, energy and momentum conservation require $\sqrt{s}\ge m_M+m_B$ for a particular pair, where $s$ is the squared total four-momentum. Only channels above their own threshold can occur. Total angular momentum and parity symmetry in quantum field theory constrain the partial waves of the meson-baryon scattering: a pseudoscalar meson and a positive-parity spin-one-half baryon have pair parity $-(-1{)}^L$ and total angular momentum $J=L\pm \frac{1}{2}$ (only $J=\frac{1}{2}$ for $L=0$). Initial and final partial waves must have matching $J$ and parity. Sufficient energy can open a channel, but cannot remove the electric charge conservation, baryon number, or strangeness constraints.

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]

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.