= Solution
\b[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,
$$
|1_c\rangle=\frac1{\sqrt6}\epsilon_{abc}|abc\rangle,
$$
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> is
$$
\boxed{S_{uu}=1,}
$$
a <triplet state>. This holds whether the complete baryon has total <spin> $1/2$ or $3/2$.
\b[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 $1/(m_im_j)$. The constant $A'$ absorbs the <strong coupling constant>, colour factors and the spatial overlap. A common $A'$ is a constituent-model approximation.
\b[Spin algebra and the two masses]. Set $\hbar=c=1$, 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 $\mathbf S_{12}=\mathbf S_1+\mathbf S_2$ gives
$$
\langle\mathbf S_1\cdot\mathbf S_2\rangle
=\frac12\left[S_{12}(S_{12}+1)-\frac34-\frac34\right]
=\frac14.
$$
Next square the total <spin> $\mathbf J=\mathbf S_{12}+\mathbf S_3$:
$$
\langle\mathbf S_1\cdot\mathbf S_3+\mathbf S_2\cdot\mathbf S_3\rangle
=\frac12\left[J(J+1)-2-\frac34\right]
=\begin{cases}-1,&J=1/2,\\1/2,&J=3/2.\end{cases}
$$
Substituting these pair sums into the <baryon hyperfine mass formula> yields
$$
\boxed{M_{uus,1/2}=2m_u+m_s+\frac{A'}4
\left(\frac1{m_u^2}-\frac4{m_um_s}\right),}
$$
$$
\boxed{M_{uus,3/2}=2m_u+m_s+\frac{A'}4
\left(\frac1{m_u^2}+\frac2{m_um_s}\right).}
$$
Their difference is $3A'/(2m_um_s)$. Repeating the argument for the identical $ss$ pair in $uss$ gives
$$
M_{uss,1/2}=m_u+2m_s+\frac{A'}4\left(\frac1{m_s^2}-\frac4{m_um_s}\right),\qquad
M_{uss,3/2}=m_u+2m_s+\frac{A'}4\left(\frac1{m_s^2}+\frac2{m_um_s}\right).
$$
Thus the model predicts <equal Sigma and Xi hyperfine splittings>:
$$
\boxed{\frac{M(\Sigma^*)-M(\Sigma)}{M(\Xi^*)-M(\Xi)}=1.}
$$
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 $A'$ values, can account for departures from the common-coefficient approximation.
\b[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 $1/2^+\to1/2^+$.
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2018-baryon-decay-diagrams.png]
{title=Quark diagrams for Sigma-to-Lambda photon emission and an example strong Delta decay}
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 $E_\gamma$, so
$$
M_\Sigma=E_\gamma+\sqrt{M_\Lambda^2+E_\gamma^2},
\qquad
\boxed{E_\gamma=\frac{M_\Sigma^2-M_\Lambda^2}{2M_\Sigma}
=74.5151\,\mathrm{MeV}\simeq74.5\,\mathrm{MeV}.}
$$
The 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.
\b[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 are
$$
\boxed{\Delta^-,\Delta^0,\Delta^+,\Delta^{++};\qquad
\Sigma^{*-},\Sigma^{*0},\Sigma^{*+};\qquad\Xi^{*-},\Xi^{*0}.}
$$
The <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 $\Sigma^*\to\Lambda\pi$ or $\Sigma\pi$, and the excited <Xi baryons> as $\Xi^*\to\Xi\pi$, 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.
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