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The spin-one-half octet of flavor symmetry contains the nucleons, Sigma baryons, Lambda baryon, and Xi baryons. Its weight diagram has six peripheral states and two states at $(I_3,Y)=(0,0)$: ${\Sigma }^0$ and ${\Lambda }^0$, with different total isospin.

A strong interaction decay must conserve quark flavour and have final rest energy below the initial mass. An excited light baryon can often decay to a lighter baryon plus a pion. A ground-state strange baryon may lack any such open flavour-conserving channel even when its spin permits one. In particular, the Omega baryon cannot decay strongly to a Xi baryon plus a pion because that changes strangeness; its flavour-conserving Xi-kaon channel is closed by the mass threshold. A spin-three-half assignment alone does not imply a strong decay.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

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.

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.

Each quark has baryon number $1/3$, and each antiquark has baryon number $-1/3$. A strange quark has strangeness $-1$, a strange antiquark has strangeness $+1$, and the light $u,d$ have zero strangeness. The isospin projections of $u,d,s$ are $+1/2,-1/2,0$, with the opposite values for their antiquarks. ThusFor three valence quarks, $B=1$ and $Y=B+S=1-n_s$. The baryon octet is reproduced by the following quark model assignments:

Consequently the top-left and top-right states in the source diagram are $\boxed{X=n,Y=p}$. Here the source's label $Y$ means the unknown particle at the top right; the conventional symbol $Y=B+S$ elsewhere in this solution denotes flavor hypercharge.

Colour and identical-quark spin. A spatial ground-state baryon has a symmetric $\ell =0$ spatial wavefunction. For $uuu$, $ddd$ or $sss$, the quark flavour state is also symmetric, and the total-spin $3/2$ state of three spin-one-half quarks is symmetric. Without another degree of freedom, the whole state would be symmetric under exchange of identical fermions, contrary to the Pauli exclusion principle.

A three-quark colour singlet supplies a completely antisymmetric factor,using the three color charge states. The remaining spatial-spin-flavour factor can then be symmetric. These baryons therefore provide evidence for an additional internal degree of freedom with at least three states, realized as colour in quantum chromodynamics.

For the $uu$ pair in $uus$, the spatial and quark flavour factors are symmetric under exchange of the two up quarks, while the colour factor is antisymmetric. The pair's spin state must consequently be symmetric. The identical-quark pair spin in a ground-state baryon isa triplet state. This holds whether the complete baryon has total spin $1/2$ or $3/2$.

Origin of the pair interaction. The terms in the baryon hyperfine mass formula are effective colour-magnetic spin-spin interactions, arising from gluon exchange between the quarks' colour-magnetic moments. Each magnetic moment brings an inverse constituent quark mass, explaining the factor $1/(m_i{m}_j)$. The constant $A^{\prime }$ absorbs the strong coupling constant, colour factors and the spatial overlap. A common $A^{\prime }$ is a constituent-model approximation.

Spin algebra and the two masses. Set $\hslash =c=1$, so each dimensionless spin angular momentum operator satisfies ${\mathbf{S}}_i^2=3/4$. Label the two up quarks 1 and 2 and the strange quark 3. Squaring ${\mathbf{S}}_{12}={\mathbf{S}}_1+{\mathbf{S}}_2$ givesNext square the total spin $\mathbf{J}={\mathbf{S}}_{12}+{\mathbf{S}}_3$:Substituting these pair sums into the baryon hyperfine mass formula yieldsTheir difference is $3A^{\prime }/(2m_u{m}_s)$. Repeating the argument for the identical $ss$ pair in $uss$ givesThus the model predicts equal Sigma and Xi hyperfine splittings:For a like-flavour comparison using exactly the supplied masses, the $uus$ splitting is $1383-1189=194\mathrm{MeV}$ and the $uss$ splitting is $1532-1315=217\mathrm{MeV}$, giving $194/217=0.894$. Averaging within the supplied isospin multiplets instead gives $192/215=0.893$. The predicted equality captures the common scale but misses the observed ratio by about $11\%$ in this table. Different spatial overlaps, and hence different effective $A^{\prime }$ values, can account for departures from the common-coefficient approximation.

Radiative decay and recoil. In a radiative Sigma-to-Lambda transition, both baryons retain valence content $uds$. A photon couples to an electrically charged constituent while the final quarks are projected onto the lower baryon's spin-flavour wavefunction. The leading up- and down-quark diagrams are shown below; their contributions must be combined coherently. In the simplest model, the strange-quark spin operator alone cannot change the $ud$ pair from its Sigma triplet to the Lambda singlet. The transition is an $M1$ multipole radiation transition, consistent with $1/2^{+}\to 1/2^{+}$.

For the photon energy, let $M_{\Sigma }=1193\mathrm{MeV}$ and $M_{\Lambda }=1116\mathrm{MeV}$ in natural units. Four-momentum conservation gives equal opposite final momenta of magnitude $E_{\gamma }$, soThe recoil kinetic energy of the Lambda baryon is $77-74.5151=2.4849\mathrm{MeV}$; assigning the entire mass difference to the photon would miss it.

Which baryons decay strongly? The open strong-decay thresholds for baryons require both sufficient energy and conserved quark flavour, including strangeness. The dominant strong decayers in the supplied table areThe Delta baryon channels $\Delta \to N\pi$ have threshold near $940+140=1080\mathrm{MeV}$, below $1232\mathrm{MeV}$. The excited Sigma baryons can decay as ${\Sigma }^{\ast }\to \Lambda \pi$ or $\Sigma \pi$, and the excited Xi baryons as ${\Xi }^{\ast }\to \Xi \pi$, all with open thresholds and conserved strangeness. Their positive-parity baryon states can emit the negative-parity pion in an odd relative orbital angular momentum, with $L=1$ already allowing total $J=3/2$.

The last diagram illustrates ${\Delta }^{++}(uuu)\to p(uud)+{\pi }^{+}(u\bar{d})$: a gluon creates a $d\bar{d}$ pair and the quarks regroup into the two hadrons. No quark changes flavour at the gluon vertices. This is a representative strong pair-creation topology; quantitative hadronic decay rates require bound-state dynamics.

The Omega baryon is an important exception among the spin-$3/2$ states. Its $S=-3$ cannot become the $S=-2$ of $\Xi \pi$ through the strong interaction. Its flavour-conserving $\Xi K$ threshold is at least $1315+500=1815\mathrm{MeV}$, above its $1673\mathrm{MeV}$ mass, so it decays weakly. The ground-state strange octet also has no open strong channel: for example $\Sigma \to \Lambda \pi$ needs at least $1256\mathrm{MeV}$, and $\Lambda \to NK$ or $\Xi \to \Lambda K$ needs still more than the relevant initial mass. The neutral Sigma decays electromagnetically as above; the neutron and the other unstable octet members decay weakly, while the proton is stable within these processes.

A ground-state baryon contains three quarks in a spatially symmetric $L=0$ wavefunction, giving positive parity. Each quark is a spin one-half fermion, so the complete wavefunction must be antisymmetric under interchange of its complete quark labels. A physical baryon is a colour singlet, with the antisymmetric colour factorConsequently the combined spin-and-flavour factor must be symmetric. The product notation in the question organizes the degrees of freedom; it need not mean that spin and flavour are separately unentangled. In particular the octet requires a correlated sum of products.

For the three light flavours, the flavour symmetry decomposition isThe subscripts describe permutation symmetry, with $M$ denoting mixed symmetry. The addition of angular momentum of three spin one-half particles givesThe symmetric spin quartet has total $J=3/2$; the two spin doublets have $J=1/2$.

For $J=3/2$, the spin factor is symmetric, so the flavour factor must also be symmetric: this gives the spin-$3/2$ baryon decuplet. Its flavour content is four $\Delta$ states, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states, and one $\Omega$ state.

For $J=1/2$, the mixed-symmetry spin functions combine with the mixed-symmetry flavour functions to make one fully symmetric spin-flavour factor for each octet weight: this gives the spin-$1/2$ baryon octet, containing two nucleons, three Sigma baryons, one Lambda baryon, and two Xi baryons. The two mixed-symmetry octets in the raw flavour product are not two distinct ground-state octets; their permutation partners are correlated with the spin factors. An antisymmetric flavour singlet would require a fully antisymmetric spin factor to make a symmetric product, but three spin states drawn from a two-dimensional single-quark spin space have no such factor.

Equivalently, the symmetric three-particle representation of spin-flavour symmetry has dimension $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ and decomposes under flavour and spin intoThese are approximate multiplets: the larger strange quark mass breaks exact flavour symmetry and splits their energies.

Label the identical strange quarks as 1 and 2 and the up quark as 3. In the ground spatial state, the $ss$ pair has symmetric space and flavour factors but an antisymmetric colour factor. The Pauli exclusion principle therefore requires a symmetric spin factor: the pair has $S_{ss}=1$.

The angular momentum identitygivesNext couple ${\mathbf{S}}_{ss}$ to ${\mathbf{S}}_u$, where $S_u=1/2$. For total $J$,For the spin-$1/2$ Xi baryon, this is $(3/4-3/4-2)/2=-1$. By symmetry the two $us$ pair correlations are each $-1/2$. Inserting the pair correlations into the baryon hyperfine mass formula givesFor the spin-$3/2$ state, the same coupling gives $(15/4-3/4-2)/2=1/2$. Each $us$ pair correlation is therefore $1/4$, as expected for a fully symmetric spin state. HenceThe predicted splitting is a useful independent check:

The table gives the Xi baryon mean lifetime ${\tau }_{{\Xi }^0}=2.90\times 10^{-10}\mathrm{s}$. At $t_1=10^{-12}\mathrm{s}$, an initially resting particle has survival probabilityThus the most probable contents are one undecayed ${\Xi }^0$. The initial momentum is unspecified. For a moving parent, time dilation replaces the denominator by ${\gamma }_{\Xi }{\tau }_{{\Xi }^0}$, making the survival probability at least as large. The times below are characteristic proper mean lifetimes; laboratory times depend on the particle velocities.

The principal first step isIt typically occurs on the $10^{-10}\mathrm{s}$ scale. The neutral pion almost immediately converts to two photons, with ${\tau }_{{\pi }^0}=8.43\times 10^{-17}\mathrm{s}$ and $B_{\gamma \gamma }=0.98823$. The Lambda baryon has ${\tau }_{\Lambda }=2.632\times 10^{-10}\mathrm{s}$ and two dominant routes.

In the charged-pion route,The muon lifetime is a standard additional input, not present in the supplied five table pages.^[ref] The successive contents are approximately $\Lambda +2\gamma$, then $p+{\pi }^{-}+2\gamma$, then $p+{\mu }^{-}+{\bar{\nu }}_{\mu }+2\gamma$. After a few muon lifetimes the dominant stable endpoint isThe leading-path probability including the tabulated two-photon pion branch isHere the last factor is the tabulated dominant muonic charged-pion fraction; rare radiative corrections are being neglected in describing a fixed photon count. Before selecting pion decay channels, the probability of reaching the $p{\pi }^{-}{\pi }^0$ history is $0.99524\times 0.639=0.63596$.

In the neutral-pion route,This produces $n+4\gamma$ by roughly the nanosecond scale. It is a long-lived intermediate state, not the final endpoint: the free neutron has ${\tau }_n=879.4\mathrm{s}$ in the original table and eventually undergoes beta-minus decay,Thus after several thousand seconds the dominant stable endpoint isBefore selecting the two ${\pi }^0\to 2\gamma$ decays, the $n{\pi }^0{\pi }^0$ history has probability $0.99524\times 0.358=0.35630$. The familiar rough description is therefore a $64\%$ charged-pion route and a $36\%$ neutron route; including the specified leading pion branches gives approximately $62.8\%$ and $34.8\%$ for the two displayed endpoints. The remainder includes Dalitz decays and other rare channels, not a failure of probability conservation.

These are particle decay cascades, treating the separated final proton, electron, photons and neutrinos as stable asymptotic particles and neglecting subsequent collisions and rare extra radiation. An infinitely large box prevents escaping particles from being removed from the inventory. Neutrino labels specify their production flavours. Each lifetime describes a stochastic waiting time, so there is no finite sharp instant at which every unstable particle has certainly decayed.