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This effective baryon mass model adds a colour-magnetic spin-spin interaction to a sum of constituent quark masses. The common coefficient $A^{\prime }$ absorbs a spatial contact density, colour factors, and the strong coupling. $A^{\prime }$ has dimensions of energy cubed in natural units. A shared coefficient is an approximation because different flavour configurations can have different spatial wavefunctions.

Write $|\uparrow ⟩,|\downarrow ⟩$ for the two $S_3$ eigenstates, of eigenvalues $\pm \hslash /2$. The spin-spin interaction can be evaluated usingFor $|\beta |\gg |\alpha |$, take $H_0=\beta (S_3^{(1)}-S_3^{(2)})/\hslash$ and $V=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$. The unperturbed energy eigenvalues are $\beta ,-\beta ,0,0$ on $|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$, respectively. For an isolated normalized energy eigenstate $|n⟩$, first-order nondegenerate perturbation theory gives $\Delta E_n^{(1)}=⟨n|V|n⟩$. The two antiparallel spins have expectation $-\alpha /4$, soFor the zero-energy degenerate eigenspace, degenerate perturbation theory requires diagonalizing the restriction of $V$. On the ordered basis $(|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩)$ it is $(\alpha /4)I$: the ladder operators cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so the remaining two levels are exactly $\alpha /4$, retaining their twofold quantum degeneracy.

For $|\alpha |\gg |\beta |$, take $H_0=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hslash$. The spin-one-half triplet state basis and spin-one-half singlet state areThe identity ${\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}=({\mathbf{J}}^2-3{\hslash }^2/2)/2$ gives unperturbed energy eigenvalues $E_t^{(0)}=\alpha /4$ and $E_s^{(0)}=-3\alpha /4$. Direct action givesConsequently the restriction of $V$ to the entire triplet degenerate eigenspace is zero, so no first-order splitting is missed. The singlet expectation is also zero. In second-order nondegenerate perturbation theory, $\Delta E_n^{(2)}={\sum }_{r\ne n}|⟨r|V|n⟩{|}^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective matrix is ${\beta }^2|t_0⟩⟨t_0|/\alpha$, already diagonal in this basis. Thus, through second order,The last two expressions have higher corrections beginning at $O({\beta }^4/|\alpha {|}^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the energy denominators above.

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 $\mathrm{Ge}{\mathrm{V}}^3$.

Use GeV consistently with the supplied $A^{\prime }$:The spin-independent sum is $m_u+2m_s=1.436\mathrm{GeV}$. The spin-spin interaction corrections areThusThe displayed decimal places document the calculation, rather than asserting that this simple constituent-quark model has sub-MeV predictive precision.

The predicted-minus-measured differences areRelative to the measured masses, these are $+0.755\%$ and $-0.416\%$. The predicted spin-spin interaction splitting is $200.62\mathrm{MeV}$, compared with the measured $216.93\mathrm{MeV}$, a discrepancy of about $7.5\%$ in the splitting.

This is an effective constituent-quark model, with a common spatial contact factor and a nonrelativistic spin-dependent interaction applied to light quarks. Relativistic motion, different spatial wavefunctions, and omitted strong-interaction corrections prevent an expectation of accuracy at the experimental $0.2$--$0.3\mathrm{MeV}$ scale. Agreement at the level of several percent in absolute baryon masses would already be useful for such a simple formula; the sub-percent absolute-mass agreement here is better than that rough expectation.

No theoretical uncertainty or fitting prescription is supplied, so there is no unique numerical “expected error” or justified statistical significance to attach to the deviations. Small relative errors in the large spin-independent mass sum also need not imply equally small relative errors in the much smaller spin-spin interaction splitting.

For two valence electrons, the central-field shell structure should be the largest scale, residual electrostatic interactions should establish a reasonably definite configuration and atomic term, and spin-orbit coupling should be weaker than the separations between the relevant terms. Then $L$ and $S$ remain approximately good quantum numbers. Within a well-isolated term the leading fine structure interaction can be represented as $A\mathbf{L}\cdot \mathbf{S}$. Its coefficient must be effectively constant for all $J$ levels in that term. Mixing of neighboring terms, spin-spin interactions, higher-order corrections or strong applied fields can spoil the simple interval rule; hyperfine structure must be negligible or separately resolved.

A suitable approximate example is the magnesium $3s3p{}^3{P}_J$ term, with $L=1$, $S=1$ and $J=0,1,2$. The interval-rule prediction isAn arbitrary common term energy may be added. For $A>0$ the levels ascend in the order ${}^3{P}_0,{}^3{P}_1,{}^3{P}_2$, with gap ratio one to two. Each level still contains $2J+1$ degenerate magnetic sublevels in zero external field. The superscript three labels the spin multiplicity $2S+1$, while the different subscripts identify fine structure components of that same term. This is an approximate example of LS coupling, not a claim that every real two-electron atom has exact one-to-two fine splitting.

A spin-spin interaction couples two spin angular momenta. In an isotropic model it is proportional to their dot product, whose expectation in a total-pair-spin state follows from $2{\mathbf{S}}_i\cdot {\mathbf{S}}_j=S_{ij}^2-S_i^2-S_j^2$. Magnetic interactions generate examples in atoms and effective colour-magnetic interactions in hadron models.