Colour charge, SU(3)_c and the QCD Lagrangian

Colour was not observed; it was demanded. The Δ⁺⁺ is three u quarks with J = 3/2 in the ground state, so its spatial wavefunction is symmetric, its spin state ↑↑↑ is symmetric and its flavour content uuu is symmetric — a totally symmetric state of three identical fermions, which Pauli forbids. The repair is a hidden index i on each quark and a totally antisymmetric combination of it, εᵢⱼₖ qᵢ qⱼ qₖ/√6. That combination exists only if the index takes at least three values, and with exactly three it is unique. So quarks live in a three-dimensional internal space, and the group that rotates a complex triplet while preserving that antisymmetric singlet is SU(3): quarks in the fundamental 3, antiquarks in the conjugate 3̄, whose generators are −(λ^a)*/2. Everything else in the topic is what that one choice forces once the symmetry is gauged.

Gauge the triplet and the covariant derivative is fixed: (D_μ)ᵢⱼ = δᵢⱼ ∂_μ − i gₛ (λ^a/2)ᵢⱼ G^a_μ, one gluon field per generator, so L = Σ_f q̄_f (iγ^μD_μ − m_f) q_f − ¼ G^a_μν G^aμν with a single coupling for all six flavours. Why eight gluons and not nine? A gluon carries a colour and an anticolour, and 3 ⊗ 3̄ = 8 ⊕ 1. The six off-diagonal combinations r b̄, r ḡ, g r̄, g b̄, b r̄, b ḡ change a quark's colour; two diagonal ones, (r r̄ − g ḡ)/√2 and (r r̄ + g ḡ − 2b b̄)/√6, are λ³ and λ⁸. The ninth, (r r̄ + g ḡ + b b̄)/√3, is the identity — the extra generator of U(3). It commutes with the other eight, has no self-coupling, and would be a massless colour-singlet vector coupling to every quark alike: an unconfined, infinite-range strong force between nucleons, which nobody sees. So the group is SU(3), the gluons are eight, and the inputs are gₛ and six masses, plus a θ-term that the neutron's electric dipole moment bounds below 10⁻¹⁰.

Nobody sums eight 3×3 matrices inside a calculation; they use the Casimirs. Normalise tr(T^aTb) = T_F δ^ab with T_F = ½. Since Σₐ T^aTa commutes with every generator, Schur's lemma makes it a multiple of the identity, C_F 𝟙; trace both sides and (N² − 1) T_F = N C_F, so C_F = (N² − 1)/2N = 8/6 = 4/3. Check it on a red quark directly: the four off-diagonal generators that touch row 1 — T¹, T², T⁴, T⁵ — each contribute |½|² = ¼, a total of 1; the diagonal ones add (½)² + (1/2√3)² = ¼ + 1/12 = 1/3; the sum is 4/3. In the adjoint the same construction reads f^acd f^bcd = C_A δ^ab, and for a = b = 1 the six non-zero f¹cd (±1, ±½, ±½) square to 2(1 + ¼ + ¼) = 3, so C_A = N = 3. These are the weights at every vertex once colour is summed: C_F when a quark radiates a gluon, C_A when a gluon radiates a gluon, T_F when a gluon splits into q q̄.

One gluon exchanged between two colour charges gives a Coulomb-like potential V(r) = ⟨T₁⋅T₂⟩ αₛ ħc/r, and the colour factor comes from completing the square: (T₁ + T₂)² = T₁² + T₂² + 2T₁⋅T₂, so ⟨T₁⋅T₂⟩ = ½[C(R) − C(R₁) − C(R₂)] for a pair coupled to representation R. For q q̄ the singlet gives ½[0 − 8/3] = −4/3 and the octet ½[3 − 8/3] = +1/6; for q q the antisymmetric 3̄ gives ½[4/3 − 8/3] = −2/3 and the symmetric 6, with C = 10/3, gives +1/3. Put numbers on it: at r = 0.50 fm with αₛ = 0.30, αₛ ħc/r = 0.30 × 197.3/0.50 = 118 MeV, so the q q̄ singlet sits at −158 MeV, the octet at +20 MeV, the q q antitriplet at −79 MeV and the sextet at +39 MeV. The pattern is the point. The only attractive channels are the ones that build colour singlets: a q q̄ pair in a 1 is a meson, and two quarks in a 3̄ look like an antiquark to the third quark, which is how three quarks close into a singlet baryon. Octets and sextets repel and never bind — confinement's shape, before confinement is even invoked.

In QED the photon is neutral and the field strength is linear in A. In QCD the field strength G^a_μν = ∂_μG^a_ν − ∂_νG^a_μ + gₛ f^abc G^b_μ G^c_ν carries a term with no derivative at all, and −¼G² then holds vertices with three gluons (∝ gₛ f^abc) and with four (∝ gₛ² f^abe f^cde). Physically, a gluon labelled r b̄ is itself a colour charge, so it radiates gluons exactly as a quark does — but with weight C_A = 3 instead of C_F = 4/3. That ratio, 9/4, is measurable: a gluon jet should end in more hadrons than a quark jet of the same energy, the multiplicity ratio tending to C_A/C_F = 2.25 at asymptotic energies and sitting near 1.5 at LEP's 91 GeV once finite-energy corrections are included. The same self-coupling is what makes the one-loop coefficient β₀ = (11/3)C_A − (2/3)n_f positive: the +11 exists only because f^abc ≠ 0. An abelian theory of colour would have no gluon jets to compare, no asymptotic freedom, and no reason for quarks to be confined.

No detector registers colour, because only singlets reach it. What it registers are rates, and colour multiplies them. In e⁺e⁻ → hadrons the photon makes a q q̄ pair in any of N_c colours, so R = σ(hadrons)/σ(μ⁺μ⁻) = N_c Σ Q_f²: with u, d, s that is N_c × 2/3, above the charm threshold N_c × 10/9, above bottom N_c × 11/9. For N_c = 3 the plateaus are 2, 3.33 and 3.67 before the (1 + αₛ/π) correction, and the data sit near 2.2, 3.6 and 3.9; for N_c = 1 they would be 0.67, 1.11 and 1.22. The π⁰ → γγ decay runs through a quark triangle whose amplitude is proportional to N_c(Qᵤ² − Q_d²) = N_c/3, so the rate goes as (N_c/3)²: 7.8 eV predicted at N_c = 3 against 7.8 ± 0.1 eV measured, and a factor of nine too small at N_c = 1. And the τ decays to hadrons N_c times as often as to each lepton, giving R_τ = N_c (1 + αₛ/π + …) ≈ 3.6 against a measured 3.64. Three integers, one answer, none of them read off a colour meter.