Baryon Multiplets, Spin, and the Color Puzzle

A baryon is a bound state of three quarks. The flavor $S U (3)$ decomposition $3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$ from the Eightfold-Way lesson fixes the flavor multiplets, but it leaves open which spin accompanies each. Fixing that requires the full three-quark wavefunction and the Pauli principle, and the answer forced a new quantum number into the theory: color.¹

The four-part wavefunction

Quarks are fermions, so the total wavefunction of a baryon must be antisymmetric under exchange of any two of them. The wavefunction factors into four independent pieces,

Ψ = ψ_{space} \times χ_{spin} \times ϕ_{flavor} \times ξ_{color},

and the exchange symmetry of the whole is the product of the symmetries of the parts. For the ground-state baryons the spatial wavefunction has no orbital angular momentum ( $L = 0$ ) and is therefore symmetric. The requirement is then a statement about the remaining three factors.

The baryon wavefunction as a product of four exchange factors. For the L = 0 ground states the spatial part is symmetric, so the product of spin, flavor, and color must be antisymmetric to satisfy Fermi statistics.

The Δ⁺⁺ puzzle

The sharpest case is the $Δ^{++}$ , a spin- $\frac{3}{2}$ baryon of charge $+ 2$ . Its charge and spin fix it uniquely: three up quarks with all spins aligned. Examine the three factors.

Flavor. All three quarks are $u$ , so $ϕ_{flavor} = uuu$ is symmetric under any exchange.
Spin. The spin- $\frac{3}{2}$ state with $J_{z} = + \frac{3}{2}$ has all three spins up, $χ = ↑↑↑$ , symmetric.
Space. The ground state has $L = 0$ , symmetric.

The product of these three is totally symmetric. For three identical spin- $\frac{1}{2}$ fermions this is forbidden: the wavefunction must be antisymmetric, yet every available factor is symmetric. Either the quark model is wrong, or a fourth, antisymmetric factor exists.

The statistics problem. The Delta-plus-plus is three up quarks with all spins aligned in the L = 0 ground state: symmetric in flavor, spin, and space. For identical fermions this totally symmetric state violates the Pauli principle, unless a further antisymmetric degree of freedom is present.

Color to the rescue

The resolution assigns each quark a further three-valued charge, color, taking values labelled red, green, blue. A baryon is built as the totally antisymmetric combination in the three color labels,

ξ_{color} = \frac{1}{6} ϵ_{ab c} ∣ a ⟩_{1} ∣ b ⟩_{2} ∣ c ⟩_{3},

with $ϵ_{ab c}$ the totally antisymmetric symbol on the three colors. This is antisymmetric by construction, and it is a color singlet — invariant under the color $S U (3)$ . With $ξ_{color}$ antisymmetric and the space–spin–flavor product symmetric, the total wavefunction is antisymmetric, and Fermi statistics is satisfied. The $Δ^{++}$ exists precisely because color exists.

The antisymmetric color factor. The three quarks carry distinct color charges (red, green, blue) combined through the epsilon symbol into a color singlet. This factor is antisymmetric under exchange of any two quarks, so the otherwise-symmetric Delta-plus-plus wavefunction becomes antisymmetric overall.

The same color factor removes the fractional-charge embarrassment of the light baryons and, applied to the meson case, explains why $q \overset{q}{ˉ}$ and $q q q$ are the only simple combinations seen: they are the smallest color singlets. The independent, quantitative confirmation of exactly three colors comes from the $R$ -ratio and the $π^{0} \to γ γ$ rate, developed in the final lesson.

Octet and decuplet spin content

With color fixed as antisymmetric, the space–spin–flavor product is symmetric, and the allowed spins follow from how flavor and spin combine.

Spin content of the two ground-state baryon multiplets. The flavor- symmetric decuplet (ten states) must carry symmetric spin, giving spin three-halves; the mixed-symmetry octet (eight states) carries spin one-half. The apex of the decuplet is the triple-strange Omega-minus.

Magnetic moments

The quark model makes a quantitative, parameter-light prediction for baryon magnetic moments. Each quark is a point Dirac particle with moment $μ_{q} = q_{q} ℏ/ (2 m_{q})$ (in units where $g = 2$ ), pointing along its spin. A baryon's moment is the expectation value of the sum $\sum_{i} μ_{i} σ_{i z}$ in its spin-flavor wavefunction.

The proton's spin-up state is the totally symmetric combination of two up quarks and one down. Coupling the two like quarks to spin $1$ (they sit in the symmetric pair) and adding the down quark to reach total $J_{z} = + \frac{1}{2}$ gives, after normalization,

∣ p ↑ ⟩ = \frac{1}{18} (2 u ↑ u ↑ d ↓ - u ↑ u ↓ d ↑ - u ↓ u ↑ d ↑ + permutations) .

Taking $⟨ \sum_{i} μ_{i} σ_{i z} ⟩$ in this state, the up quarks contribute with net weight $+ \frac{4}{3}$ and the down with $- \frac{1}{3}$ , so the proton (uud) moment is

μ_{p} = \frac{1}{3} (4 μ_{u} - μ_{d}), μ_{n} = \frac{1}{3} (4 μ_{d} - μ_{u}),

and, since $μ_{u} / μ_{d} = q_{u} / q_{d} = - 2$ when $m_{u} \approx m_{d}$ , the ratio is predicted with no free parameters:

\frac{μ _{n}}{μ _{p}} = \frac{4 μ _{d} - μ _{u}}{4 μ _{u} - μ _{d}} = \frac{4 μ _{d} + 2 μ _{d}}{- 8 μ _{d} - μ _{d}} = \frac{6 μ _{d}}{- 9 μ _{d}} = - \frac{2}{3} .

The measured ratio is $- 1.913/2.793 = - 0.685$ , within a few percent of $- \frac{2}{3}$ . Fixing the two light-quark moments from $μ_{p}$ and the strange moment from $μ_{Λ} = μ_{s}$ , the model predicts the entire octet.²

Predicted versus measured magnetic moments for the spin-one-half baryon octet, in nuclear magnetons. Bars above the axis are positive moments, below are negative (the sign is shown by direction, magnitudes on the vertical scale). The proton and lambda are used to fix the quark moments; the rest are predictions, agreeing with data to about ten percent.

The agreement is the model's most direct quantitative success at the level of individual particles. It also carries a physical message: the moments come out right only if the quarks behave as nearly free spin- $\frac{1}{2}$ Dirac particles inside the baryon, with effective (constituent) masses of a few hundred MeV. The constituent mass is larger than the current-quark mass that enters the QCD Lagrangian because it includes the energy of the surrounding gluon and quark–antiquark field, a distinction developed in the QCD module. Where the model misses — the $Σ^{+}$ prediction is high by about ten percent — the discrepancy marks the limits of treating the baryon as three static constituent quarks.³

Griffiths, Introduction to Elementary Particles, §5.7–5.8, gives the three-quark wavefunction, the $Δ^{++}$ statistics argument, and the color resolution. The spin assignments of the octet and decuplet follow the exchange-symmetry analysis there. ↩
Griffiths, §5.8, derives $μ_{p} = (4 μ_{u} - μ_{d}) /3$ and the octet magnetic moments; predicted and measured values are tabulated there, using constituent masses fit to $μ_{p}$ . Measured moments from the Particle Data Group, pdg.lbl.gov. ↩
Tong, The Standard Model (Cambridge Part III), §3.3.3, stresses that the baryon mass is set by the QCD scale $Λ_{QCD}$ rather than the bare quark masses, and that the constituent quark carries the energy of the surrounding strongly interacting fields. ↩