Baryons, colour and exchange antisymmetry

Take the Δ⁺⁺ at S_z = +3/2, mass 1232 MeV/c² and width 117 MeV. It is the lowest uuu state, so its spatial wavefunction is nodeless with L = 0: symmetric under exchange of any two quarks. Its flavour content is uuu: symmetric. Its spin state is |↑↑↑⟩, the top member of the S = 3/2 quadruplet: symmetric. So P̂₁₂Ψ = +Ψ, and three identical spin-½ particles are not permitted to do that. The corners of the decuplet make this unavoidable rather than exotic — Δ⁻(ddd) and Ω⁻(sss) are built the same way — and no rearrangement of space, spin or flavour rescues them, because each factor is already as symmetric as a factor can be. Two exits were open in 1964–65: abandon Fermi statistics for quarks, which is Greenberg's parastatistics, or give each quark one more degree of freedom whose state can carry the minus sign, which is Han and Nambu's. The second is the one nature took.

Give each quark a label i = 1, 2, 3 and call the values R, G, B. In three dimensions there is exactly one totally antisymmetric rank-three invariant tensor, εᵢⱼₖ, so there is exactly one totally antisymmetric three-quark colour state: |1⟩_c = (1/√6) εᵢⱼₖ|qᵢ qⱼ qₖ⟩ = (1/√6)(RGB − RBG + GBR − GRB + BRG − BGR). Three is also the smallest number of colours admitting such a state for three quarks, which is the first argument for N_c = 3. Two properties matter downstream. It is an SU(3)_c singlet: every one of the eight generators annihilates it, T̂ᵃ|1⟩_c = 0, so the total colour Casimir vanishes and there is no colour charge left for a gluon to couple to. And it is entangled rather than factorised — trace out any two quarks and the third is left in (1/3)𝟙, maximally mixed in colour. That is the precise sense in which no quark inside a proton is red.

If only colour singlets reach a detector, then 'which hadrons exist' is a question about tensor products. 3 ⊗ 3̄ = 8 ⊕ 1 contains a singlet, so qq̄ mesons exist; 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 contains one, so qqq baryons exist; 3 ⊗ 3 = 6 ⊕ 3̄ contains none, which is why nothing is built from two quarks. One-gluon exchange already favours the surviving channels: with T̂₁⋅T̂₂ = ½[C₂(total) − C₂(1st) − C₂(2nd)] and C₂(3) = 4/3, C₂(8) = 3, C₂(1) = 0, the qq̄ singlet gives −4/3, attractive, and the octet +1/6, repulsive. Two cautions. The rule does not forbid exotics: qqqq̄q pentaquarks and qqq̄q̄ tetraquarks are colour singlets too, and LHCb has reported both. And confinement itself is not derived here — it is a lattice result and an experimental fact, not a theorem.

Write Ψ = ψ_space ⊗ χₛₚᵢₙ ⊗ φ_flavour ⊗ c_colour. Exchange acts on all four at once, and colour is now fixed antisymmetric, so space ⊗ spin ⊗ flavour must be symmetric. In the ground band L = 0 makes space symmetric, so spin ⊗ flavour alone must be totally symmetric: the 56 in 6 ⊗ 6 ⊗ 6 = 56 ⊕ 70 ⊕ 70 ⊕ 20. Its content is 56 = (10, 4) ⊕ (8, 2), which is the observed spectrum — decuplet with S = 3/2, octet with S = 1/2, never the reverse. Two negative predictions ride along, and both hold. There is no S = 1/2 uuu state, because symmetric flavour demands symmetric spin. And there is no L = 0 flavour singlet: a flavour-singlet uds is antisymmetric, which would need a totally antisymmetric spin state of three spin-½ objects, and in a two-dimensional spin space no such state exists. The lightest flavour singlet, Λ(1405), is conventionally an L = 1 member of the 70, some 290 MeV above the octet Λ.

The symmetric spin-flavour state is explicit: |p↑⟩ = (1/√18)(2u↑u↑d↓ − u↑u↓d↑ − u↓u↑d↑ + the cyclic permutations of each). Sandwiching Σᵢ μᵢ σ̂_z, i in it returns spin fractions Δu = 4/3 and Δd = −1/3, hence μₚ = (4μᵤ − μ_d)/3 and, by u ↔ d, μₙ = (4μ_d − μᵤ)/3. Put μ_q = q_q eħ/2m_q with a single constituent mass m: in nuclear magnetons μₚ = mₚ/m and μₙ = −(2/3)(mₚ/m), so the ratio is −2/3 with m cancelled. Experiment gives −1.913/2.793 = −0.6849, a 2.7% miss from a wavefunction with nothing adjustable in it. Magnitudes are another matter: μₚ = 2.793 μ_N fixes m = 938.27/2.793 = 335.9 MeV/c², which is a fit, not a prediction. And Δu + Δd = 1 claims the quark spins carry all of the proton's spin, where polarised deep inelastic scattering measures about 0.30.

Colour never appears in a final state, so N_c is counted indirectly. In e⁺e⁻ annihilation R = σ(hadrons)/σ(μ⁺μ⁻) = N_c Σ_q q_q², which is 3(4/9 + 1/9 + 1/9) = 2 below the charm threshold and 10/3 between charm and bottom; measured values, with the QCD factor 1 + αₛ/π folded in, sit near 2.1 and 3.6. The π⁰ → γγ width scales as N_c², giving 7.76 eV against a measured 7.80 ± 0.12 eV, where N_c = 1 would put it under 1 eV. The bill is just as sharp. The Lagrangian masses are mᵤ ≈ 2.2 and m_d ≈ 4.7 MeV/c², so the three quarks are about 1% of the proton's 938 MeV/c² and the rest is gluon field energy the constituent model never writes down. Splittings come from the fitted term A Σ Ŝᵢ⋅Ŝⱼ/(mᵢ mⱼ), where ⟨Σ_(i<j) Ŝᵢ⋅Ŝⱼ⟩ = +3/4 ħ² for S = 3/2 and −3/4 ħ² for S = 1/2. The model fits well; it derives nothing.