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Use dimensionless spin angular momentum operators with $\hslash =1$ and the common-coefficient baryon hyperfine mass formula. The identical pair has total spin one, so its pair expectation is $1/4$. For total baryon spin $1/2$, the two mixed-flavour pair expectations sum to $-1$; for total spin $3/2$, they sum to $1/2$. Their difference is $3/2$, giving the title formula for either repeated-flavour choice. The equality assumes the same effective coefficient $A^{\prime }$ in the two flavour configurations; different spatial overlaps can change it.

For the lowest spatial state, orbital angular momentum is zero and the spatial wavefunction is symmetric. A three-quark colour singlet is antisymmetric, proportional to ${ϵ}_{abc}|abc⟩$. The Pauli exclusion principle for quarks therefore requires a symmetric spin–flavour factor. Colour and flavour are different degrees of freedom even though both use three-dimensional group representations.

The tensor product flavour decomposition is $3\otimes 3\otimes 3=10_S\oplus 8_M\oplus 8_M\oplus 1_A$. Three spin-one-half particles have a symmetric spin-$3/2$ quartet and mixed-symmetry spin-$1/2$ sectors. The symmetric spin quartet combines with symmetric flavour factors to give a baryon decuplet. Mixed spin and mixed flavour factors instead combine to a symmetric spin–flavour state, giving one physical ground-state baryon octet, not two independent ground-state octets. Equivalently the symmetric spin-flavour SU6 representation has dimension $56=10\times 4+8\times 2$. The antisymmetric flavour singlet would require a fully antisymmetric three-spin state, which does not exist for only two single-quark spin components.

The spin-one-half baryon octet contains $(p,n)$, $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$, $\Lambda$ and $({\Xi }^0,{\Xi }^{-})$. The spin-three-halves baryon decuplet contains the four $\Delta$ charge states, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and ${\Omega }^{-}$. Both have positive parity in this $L=0$ approximation. Strange-quark mass breaks the approximate flavour symmetry and splits multiplets.

In the baryon hyperfine mass formula, the constituent-mass sum accounts phenomenologically for common confinement energy. A colour-magnetic spin–spin interaction contributes a contact-type term proportional to ${\mathbf{S}}_i\cdot {\mathbf{S}}_j/(m_i{m}_j)$: each magnetic moment supplies one inverse mass. The common colour and spatial-overlap factors are absorbed into $A^{\prime }$. These are effective constituent masses, not short-distance current-quark masses. Use dimensionless spin operators with $\hslash =1$, as implicit in the supplied formula and coefficient.

Label the identical up quarks 1 and 2. Their flavour and $L=0$ spatial factors are symmetric, and their colour pair is antisymmetric, so their spin is a triplet, $S_{12}=1$. ThusCoupling the strange spin givesThe two up–strange denominators are identical. HenceWith the supplied masses expressed in GeV units, the common mass is $1.261\mathrm{GeV}/c^2$. The predicted values areThey are about $13$ and $3\mathrm{MeV}/c^2$ below the stated measurements, respectively. The predicted splitting is $201\mathrm{MeV}/c^2$, compared with $190\mathrm{MeV}/c^2$. The agreement is useful for such a simple model, but a common overlap constant neglects flavour-dependent spatial structure and other residual interactions.

For the weak decay, one contribution is $s\to u{W}^{-}$, $W^{-}\to d\bar{u}$. The two spectator up quarks combine with $d$ into the proton, and the new $u\bar{u}$ supplies the neutral pion's corresponding component. This changes strangeness. In the strong decay, a gluon creates $d\bar{d}$; rearrangement gives $uds$ and $u\bar{d}$ without changing strangeness.

The strong channel conserves parity: the positive-parity daughter baryon and negative-parity pion require odd relative orbital angular momentum, with a $P$ wave sufficient. The weak channel need not conserve parity. Using the mean lifetime relation $\tau =\hslash /\Gamma$,The widths differ by about $4.5\times 10^{12}$, principally because one is weak and flavour changing and the other is an open strong decay. They are not comparable transitions differing only by a small spin factor.

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.

The baryon hyperfine mass formula is a phenomenological constituent-quark model. The three constituent quark masses include effects of strong-interaction dressing and an approximate spin-independent confinement energy; they are not the small running quark masses in a short-distance quantum chromodynamics calculation.

The residual correction has the form of a colour-magnetic spin-spin interaction. A nonrelativistic reduction of one-gluon exchange produces a contact contribution proportional toAfter taking its expectation in the common spatial wavefunction and the colour singlet, the colour factor, coupling, and contact density are absorbed into $A^{\prime }$. Summing the three quark pairs gives the supplied formula. Positive $A^{\prime }$ raises aligned-spin configurations relative to lower-spin configurations. In natural units, the spin angular momentum operators are dimensionless, so $A^{\prime }$ has units of energy cubed, as in the given $\mathrm{Ge}{\mathrm{V}}^3$.

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: