= Solution
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 $\epsilon_{abc}|abc\rangle$. 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 $3\otimes3\otimes3=10_S\oplus8_M\oplus8_M\oplus1_A$. 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 $56=10\times4+8\times2$. 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 $\Sigma^*$ states, two $\Xi^*$ 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 $\mathbf S_i\cdot\mathbf S_j/(m_im_j)$: each magnetic moment supplies one inverse <mass>. The common colour and spatial-overlap factors are absorbed into $A'$. These are effective constituent <masses>, not short-distance current-quark <masses>. Use dimensionless <spin> operators with $\hbar=1$, 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, $S_{12}=1$. Thus
$$
\langle\mathbf S_1\cdot\mathbf S_2\rangle
=\frac12\left[1(1+1)-\frac34-\frac34\right]=\frac14.
$$
Coupling the strange <spin> gives
$$
\langle(\mathbf S_1+\mathbf S_2)\cdot\mathbf S_3\rangle
=\frac12\left[J(J+1)-2-\frac34\right]
=\begin{cases}-1,&J=1/2,\\1/2,&J=3/2.\end{cases}
$$
The two up–strange denominators are identical. Hence
$$
\boxed{M_{1/2}=2m_u+m_s+A'\left(\frac1{4m_u^2}-\frac1{m_um_s}\right)},\qquad
\boxed{M_{3/2}=2m_u+m_s+A'\left(\frac1{4m_u^2}+\frac1{2m_um_s}\right)}.
$$
With the supplied <masses> expressed in GeV units, the common <mass> is $1.261\,\mathrm{GeV}/c^2$. The predicted values are
$$
\boxed{M(\Sigma^+)=1.177\,\mathrm{GeV}/c^2,\qquad
M(\Sigma^{*+})=1.377\,\mathrm{GeV}/c^2}.
$$
They 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 $s\to uW^-$, $W^-\to d\bar u$. 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>.
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2013/paper-1-sigma-decays.png]
{title=Original weak and strong quark-level contributions to the two charged Sigma-baryon decays; final groupings show the hadrons}
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 $\tau=\hbar/\Gamma$,
$$
\boxed{\tau(\Sigma^+)\simeq8.2\times10^{-11}\,\mathrm s,\qquad
\tau(\Sigma^{*+})\simeq1.8\times10^{-23}\,\mathrm s}.
$$
The widths differ by about $4.5\times10^{12}$, 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.
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