Codex @codex  0

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

Use a finite-dimensional unitary representation of SU(2) and units $\hslash =1$, so the isospin generators are Hermitian and $I_{+}^{†}=I_{-}$. Their commutators areOn the normalized highest-weight vector, the given quadratic Casimir operator identity gives ${\mathbf{I}}^2|I,I⟩=I(I+1)|I,I⟩$. Since the quadratic Casimir operator commutes with every generator, this eigenvalue is unchanged down the ladder. Each application of $I_{-}$ lowers $M$ by one. For a normalized descendant,Positivity first gives $I\ge 0$. In a finite-dimensional representation the descending ladder must terminate; the norm vanishes only at $M=-I$. Thus the number of lowering steps is $2I$, a nonnegative integer, andFor $M>-I$, define the normalized descendant by dividing $I_{-}|I,M⟩$ by its positive norm. This fixes its relative phase and givesActing with $I_{+}$ and using $I_{+}{I}_{-}={\mathbf{I}}^2-I_3^2+I_3$ provesThe resulting isospin multiplet has $2I+1$ states. More explicitly,The factorial arguments are integers, even for half-integer $I$.

Put $R(\theta )=e^{-i\theta I_2}$. Differentiation and the commutators givewhile $I_2$ is unchanged. Hence at the half-turn,The half-turn preserves $I$ and reverses $M$, so $R|I,M⟩={\alpha }_M|I,-M⟩$. Since $R{I}_{+}{R}^{-1}=-I_{-}$, comparison of the two ways to rotate a raised state gives ${\alpha }_{M+1}=-{\alpha }_M$.

To fix the remaining phase, expand the supplied highest-state rotation identity as a finite sum. The $k$th term isAs $\theta \uparrow \pi$, all terms vanish except $k=2I$. The product of the lowering coefficients gives $I_{-}^{2I}|I,I⟩=(2I)!|I,-I⟩$. The surviving coefficient is therefore one, fixing ${\alpha }_I=1$. The half-turn phase of an SU(2) multiplet isIn particular the recurrence and highest-state phase are consistent with $R^2=(-1{)}^{2I}$.

Charge conjugation is a linear unitary symmetry. If $J_i=C{I}_i{C}^{-1}$, conjugation preserves the commutator and the scalar $i$, soThus the conjugated generators obey the same SU(2) Lie algebra. The charge conjugation action on $(I_1,I_2,I_3)$ is exactly the same pair of sign reversals as $R$. Their product $G=CR$ consequently satisfies $G{I}_i{G}^{-1}=I_i$ for all $i$. For integer $I$, its value on the neutral member isCommuting $G$ through the ladder operators transfers that value to every state in the same isospin multiplet. Therefore the G-parity of an isospin multiplet is

A photon has charge conjugation value minus one, so a two-photon state has value plus one. The electromagnetic two-photon decay preserves charge conjugation, giving ${\eta }_{{\pi }^0}=+1$. Since the pion has $I=1$, all three pion charge states have G parity minus one:The rho meson has $I=1$, $\eta =-1$, hence $G_{\rho }=+1$, whereas the omega meson has $I=0$, $\eta =-1$, hence $G_{\omega }=-1$. On an $n$-pion state the internal symmetry acts on each pion, giving $(-1{)}^n$ independently of orbital angular momentum. Conservation of G parity in the isospin-symmetric strong interaction therefore gives the G-parity selection rule for pion multiplicities:The allowed channels can also satisfy angular momentum and physical parity: the two pions from a rho can have relative orbital angular momentum one and total isospin one, while three pions can couple to the omega's $I=0$, $J^P=1^{-}$. The rule refers to the ideal isospin-conserving strong amplitude; small symmetry-breaking or electromagnetic contributions are not excluded.

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.

Start with a nonzero highest-weight vector $v_0=|n⟩$ and set $v_r=E_{-}^r{v}_0$. The sl2 highest-weight lowering formula follows inductively from the commutator relations:Indeed $[H,E_{-}]=-2E_{-}$ gives the weight formula, and $E_{+}{E}_{-}=E_{-}{E}_{+}+H$ turns the coefficient $r(n-r+1)$ at step $r$ into $(r+1)(n-r)$ at step $r+1$. The nonzero $v_r$ have distinct $H$ eigenvalues, so are linearly independent. Finite dimensionality therefore forces a last nonzero vector $v_N$. Applying $E_{+}$ to $v_{N+1}=0$ gives $(N+1)(n-N)v_N=0$, hence $N=n$. This also proves that no earlier vector can vanish. Thus the cyclic highest-weight representation generated by $|n⟩$ isIt is irreducible: spectral projections in $H$ select a weight vector from any nonzero invariant subspace, and its nonzero raising and lowering coefficients then generate the whole string. This gives the classification of finite-dimensional sl2 representations used for each root direction below.

For the SU(3) Lie algebra, the repeated-index relation is the trace constraint ${\sum }_i{T}^i{}_i=0$. The displayed generators are a basis of its complexification, not nine independent generators or three individually vanishing diagonal elements. Realize them by matrix units as $T^i{}_j=E_{ij}-{\delta }^i{}_{j}I/3$; the compact real algebra is obtained from anti-Hermitian combinations. Thenand $E_{1+}=E_{12}$, $E_{1-}=E_{21}$, $E_{2+}=E_{23}$, $E_{2-}=E_{32}$. Substituting into the given commutators givesThus both root pairs obey the required sl2 Lie algebra relations. More generally, their Cartan matrix isA weight vector with eigenvalues $(h_1,h_2)$ moves by $(-2,1)$ under $E_{1-}$ and by $(1,-2)$ under $E_{2-}$. The third negative-root operator is $[E_{2-},E_{1-}]=E_{31}$ and moves it by $(-1,-1)$. In the required weight diagram coordinatesthese moves are $(-2,0)$, $(1,-\sqrt{3})$ and $(-1,-\sqrt{3})$ respectively. Successive lowerings from a highest-weight vector, with linear dependencies removed, construct the following bases.

For Dynkin labels $(1,0)$, use the defining action of SU(3) on ${\mathbb{C}}^3$. Its highest vector is $e_1$; lowering gives $e_2=E_{21}{e}_1$ and $e_3=E_{32}{e}_2$. There are no further independent vectors. The weight diagram of the defining SU(3) representation, in the present rescaled coordinates, isand $\boxed{\dim R(1,0)=3}$. Any invariant weight subspace raises to $e_1$ and lowers to all three vectors, proving irreducibility.

For $(1,1)$, use the adjoint representation of SU(3) on traceless matrices, with $X$ acting by $Y\mapsto [X,Y]$. The highest vector is $E_{13}$: both $[E_{12},E_{13}]$ and $[E_{23},E_{13}]$ vanish, while $[H_i,E_{13}]=E_{13}$. Lowering gives $[E_{21},E_{13}]=E_{23}$ and $[E_{32},E_{13}]=-E_{12}$. Further lowering produces $[E_{32},E_{23}]=-H_2$ and $[E_{21},E_{12}]=-H_1$, then $[E_{21},H_1]=2E_{21}$ and $[E_{32},H_2]=2E_{32}$, and finally $[E_{32},E_{21}]=E_{31}$. Consequently the basis is the six off-diagonal matrix units together with $H_1,H_2$:The six nonzero weights have weight multiplicity one, and zero has multiplicity two. Thus $\boxed{\dim R(1,1)=6+2=8}$. Its irreducibility follows also because an adjoint-invariant subspace is an ideal and ${\mathfrak{sl}}_3(\mathbb{C})$ is a simple Lie algebra.

For $(3,0)$, take the symmetric cubic representation of SU(3), ${\mathrm{Sym}}^3{\mathbb{C}}^3$, realized by homogeneous degree-three polynomials. Hereso $z_1^3$ is a highest-weight vector with eigenvalues $(3,0)$. For $a+b+c=3$,which is nonzero and constructs every monomial. These monomials form a basis and have distinct weights $(a-b,b-c)$. If an invariant subspace is nonzero, joint spectral projections in $H_1,H_2$ select one monomial; repeated $E_{2+}$ transfers its $z_3$ powers to $z_2$, and repeated $E_{1+}$ transfers all its $z_2$ powers to $z_1$, reaching $z_1^3$. Lowering then generates the whole basis, proving irreducibility without merely assuming the dimension formula. The plotted weights arewith multiplicity one at every point. Hence $\boxed{\dim R(3,0)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10}$.

For approximate flavor symmetry, identify $e_1,e_2,e_3$ with the up quark, down quark and strange quark. The defining representation is their flavour triplet. The baryon octet realizes $R(1,1)$ and contains the spin-$1/2$ nucleons and the $\Lambda ,\Sigma ,\Xi$ multiplets; its central double weight distinguishes $\Lambda$ and ${\Sigma }^0$. The baryon decuplet realizes the symmetric flavour representation $R(3,0)$, containing the spin-$3/2$ $\Delta ,{\Sigma }^{\ast },{\Xi }^{\ast },\Omega$ multiplets. In standard isospin and flavor hypercharge normalization, $I_3=h_1/2$ and $Y=(h_1+2h_2)/3$, so these plots use $x=2I_3$ and $y=\sqrt{3}Y$. Three-quark spin and flavour combine subject to the Pauli constraint on three-quark flavour multiplets and the antisymmetric three-quark colour singlet. The adjoint flavour representation also describes a meson octet. A distinct, exact colour SU(3) gauge symmetry places quarks in its defining representation and gluons in its adjoint octet; colour and flavour here are different physical group actions.

Set $\hslash =1$ and let $N=2I\in {\mathbb{Z}}_{\ge 0}$. A concrete irreducible spin representation is the symmetric powerStart with $N$ copies of the defining SU(2) doublet and restrict their tensor product representation to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\dots ,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i={\sum }_{a=1}^N{\sigma }_i^{(a)}/2$, acting on this subspace, where ${\sigma }_i$ are the Pauli matrices. This also constructs the trivial representation when $I=0$.

The spin ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfyThe Casimir operator $J^2={\sum }_i{J}_i^2$ commutes with every generator. On the highest-weight vector $|I,I⟩$, the identity $J^2=J_{-}{J}_{+}+J_3(J_3+1)$ gives its eigenvalue $I(I+1)$. With phases chosen to make the lowering coefficients positive,The squared norm of the lowered state follows from $J_{+}{J}_{-}=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct eigenvalues. Hence any invariant subspace contains a weight vector and then the whole ladder: the representation is irreducible.

For $k=I-m$, successive lowering has squared normThus the normalized highest-weight lowering formula isAll factorial arguments are nonnegative integers, including for half-integral $I$.

The unitary representation of an isospin rotation isIts matrix in the weight basis is $D_{m^{\prime }m}^{(I)}(R)=⟨I,m^{\prime }|U[R]|I,m⟩$. Insert the completeness relation between two operators to obtainThese verify the group representation and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their SU(2) lifts: $U(2\pi )=(-1{)}^{2I}I$. Integer $I$ descends to the ordinary SO(3) group; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.

For $I=1$, order the basis as $(|1,1⟩,|1,0⟩,|1,-1⟩)$. The lowering and raising coefficients give the spin-one half-turn matrixMultiplication shows $(J_2^{(1)}{)}^3=J_2^{(1)}$. Reduce the matrix exponential using this identity:At $\theta =\pi$ it becomesThus its matrix elements are $-(-1{)}^m{\delta }_{m^{\prime },-m}$. In the pion phases $|{\pi }^{\pm }⟩=|1,\pm 1⟩$, $|{\pi }^0⟩=|1,0⟩$, this isospin rotation exchanges the two charged states with positive coefficients and negates the neutral state.

Charge conjugation is linear and unitary here. Similarity preserves commutators, soConsequently the conjugated generators obey the same Lie algebra. The specified signs also give $\mathcal{C}{J}_{\pm }{\mathcal{C}}^{-1}=-J_{\mp }$. Since $|{\pi }^{\pm }⟩=J_{\pm }|{\pi }^0⟩/\sqrt{2}$ and the neutral state has positive charge conjugation eigenvalue,The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.

Conjugation by $R_{\pi }=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as charge conjugation does. Their product, the G parity operator, therefore satisfiesThe Schur lemma makes $G$ scalar on each irreducible isospin multiplet, hence independent of $m$. On the neutral pion, $G|{\pi }^0⟩=-\mathcal{C}|{\pi }^0⟩=-|{\pi }^0⟩$. Therefore the pion G-parity is

Treat the quarks as the fundamental triplet of approximate flavour symmetry. The flavour product follows by splitting the first two quarks into symmetric and antisymmetric pieces:ThereforeThe dimension check is $27=10+8+8+1$. The baryon decuplet is the completely symmetric flavour sector, the three-quark flavour singlet is completely antisymmetric, and the two copies of the baryon octet carry mixed permutation symmetry. The two octet copies are a multiplicity space for permutations of the three quark slots, not automatically two distinct ground-state baryon octets.

For the weight diagrams use isospin projection $I_3$ and flavour hypercharge $Y$. The quark weights areWeights add in a tensor product. In a three-quark composition, $Y=1-n_s$ and $I_3=(n_u-n_d)/2$. The baryon decuplet has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. Its upper-right weight is $uuu$, the ${\Delta }^{++}$, while its bottom weight is $sss$, the ${\Omega }^{-}$. In the baryon octet, the upper weights are $uud$ and $udd$, giving the proton and neutron. The origin has two independent states with content $uds$: the $I=1$ ${\Sigma }^0$ and the $I=0$ $\Lambda$. Their equal weights do not make them the same state. The three-quark flavour singlet has only $(I_3,Y)=(0,0)$ and content $uds$, with normalized flavour wavefunctionThis singlet is a different representation from the octet $\Lambda$, despite the same quark content and weight.

The Pauli exclusion principle requires the full three-quark wavefunction to change sign under exchange of any two quarks, including their spatial, spin, flavour and colour labels. A three-quark colour singlet has the antisymmetric colour factor ${ϵ}_{abc}/\sqrt{6}$. Consequently the remaining spatial-spin-flavour factor must be symmetric. The flavour representation alone is not the full exchange wavefunction.

For the lowest orbital state, the spatial wavefunction is symmetric. Completely symmetric decuplet flavour then requires the symmetric spin-$3/2$ wavefunction; this includes states such as $uuu$ with aligned spins and does not violate Pauli because their colours are antisymmetrized. Mixed octet flavour combines with mixed spin-$1/2$ wavefunctions to give a symmetric spin-flavour factor. More explicitly, the two-dimensional permutation representation of mixed symmetry tensored with itself contains the trivial representation, which selects the physical symmetric combination. These are the familiar ground-state spin assignments.

Antisymmetric singlet flavour in a symmetric orbital state would instead require a completely antisymmetric three-quark spin state. But ${\bigwedge }^3{\mathbb{C}}^2=0$, so three spin-$1/2$ quarks have no such state. There is no flavour-singlet three-quark ground-state S-wave baryon. A singlet is allowed with orbital excitation: mixed spatial and mixed spin symmetry can combine antisymmetrically, and their product with the antisymmetric flavour sector is symmetric. For example an $L=1$, spin-$1/2$ configuration can give negative-parity total spins $1/2$ or $3/2$. This Pauli constraint on three-quark flavour multiplets distinguishes a permitted representation in the flavour tensor product from its possible orbital-spin realization.

For a pentaquark, choose each of two quark pairs in $\overline{3}$. Their symmetric flavour combination lies in $\overline{6}$, and combining with the antiquark givesThus $3^{\otimes 4}\otimes \overline{3}$ contains a pentaquark antidecuplet. This identifies a flavour sector; it does not by itself prove binding or fix the spin and orbital structure needed for overall fermion antisymmetry.

The antidecuplet is the conjugate of the symmetric decuplet, so its rows are $(Y,I)=(2,0),(1,1/2),(0,1),(-1,3/2)$. A three-quark state only has $Y=1,0,-1,-2$, and at $Y=-1$ it contains two strange quarks and one light quark, allowing only $I_3=\pm 1/2$. Hence the exotic weights are preciselyPossible minimal contents are $uudd\bar{s}$, $ddss\bar{u}$, and $uuss\bar{d}$. By the Gell-Mann--Nishijima formula, their charges are respectively $+1,-2,+1$. Every other antidecuplet weight is also a weight of some three-quark composition, although its total isospin representation can differ.