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The heavier physical isoscalar member of the light pseudoscalar meson nonet. It mixes the octet and singlet flavor directions, and is distinct from the lighter eta meson; both occupy the central flavor weight.

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.

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.

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 meson with total angular momentum one and negative parity. In the ordinary quark model, a quark-antiquark pair with orbital angular momentum zero and total spin one has these quantum numbers; its neutral self-conjugate states have charge conjugation $C=-1$. The light-flavour tensor product gives an octet and a singlet, forming a vector nonet, just as spin-zero negative-parity pairs form a pseudoscalar meson nonet. Nonzero orbital excitations can also produce vector states.