SU(3) flavour and multiplet decomposition

Write the Pauli matrices into the u–d block of a 3×3 matrix and you have λ₁, λ₂, λ₃; the same pattern in the u–s block gives λ₄, λ₅, in the d–s block λ₆, λ₇, and the traceless diagonal λ₈ = diag(1, 1, −2)/√3 completes the set, normalised so that Tr(λₐ λ_b) = 2δ_ab. With Tₐ = λₐ/2 the products close on the set: λ₁λ₂ = diag(i, −i, 0) and λ₂λ₁ is its negative, so [T₁, T₂] = iT₃ and f₁₂₃ = 1; the u–s block gives [λ₄, λ₅] = i(λ₃ + √3 λ₈), so f₄₅₃ = ½ and f₄₅₈ = √3/2. No three generators commute with one another; the largest commuting sets have two members, and the conventional choice is the diagonal pair (T₃, T₈): the rank is 2, against 1 for SU(2). A state can therefore carry two simultaneous eigenvalues and no more, I₃ = T₃ and Y = 2T₈/√3 = diag(⅓, ⅓, −⅔), which is the B + S of the previous topic. So u sits at (½, ⅓), d at (−½, ⅓) and s at (0, −⅔). Antiquarks transform with −Tₐᵀ, so every weight changes sign and the 3̄ is the 3 upside down.

The six off-diagonal generators pair into ladder operators: I± = T₁ ± iT₂ inside the u–d doublet, V± = T₄ ± iT₅ between u and s, U± = T₆ ± iT₇ between d and s. Commuting each with the Cartan pair gives its step: [T₃, V₊] = ½V₊ and [Y, V₊] = V₊, so V₊ moves a state by (+½, +1); U₊ moves it by (−½, +1) and I₊ by (+1, 0). These three vectors and their negatives are the roots, and every multiplet is a set of points joined by them. Start from the highest weight, the one state all three raising operators annihilate, and lower. For the irrep (p, q) the top row is an isospin multiplet with I = p/2 at Y = (p + 2q)/3, the bottom row has I = q/2 at Y = −(2p + q)/3, and the boundary is a hexagon whose edges take p, q, p, q, p, q root steps in turn. Multiplicity is 1 on the boundary and rises by one per layer inward until a layer becomes a triangle. The octet (1, 1) is a hexagon of six with two states at the centre; the decuplet (3, 0) is a triangle of rows 4, 3, 2, 1; the formula (p + 1)(q + 1)(p + q + 2)/2 returns 8 and 10.

The weight of a product state is the sum of its factors' weights, so list the nine quark–antiquark pairs: u s̄ at (½, 1), d s̄ at (−½, 1), u d̄ at (1, 0), d ū at (−1, 0), s ū at (−½, −1), s d̄ at (½, −1), and uū, dd̄, ss̄ all at the origin — a hexagon with a triple centre. The highest weight (½, 1) heads a doublet row, so p = 1, and Yₘₐₓ = (p + 2q)/3 = 1 forces q = 1: the octet, which takes the six corners and two of the three central states. One state is left over, the singlet, and 3 ⊗ 3̄ = 8 ⊕ 1. The centre is where the work is. Lowering u d̄ with I₋, remembering that the antiquark doublet is (d̄, −ū), gives dd̄ − uū, so π⁰ = (uū − dd̄)/√2 up to sign. The singlet is the trace Σᵢ qᵢ q̄ᵢ, invariant because Σᵢ Uᵢⱼ U*ᵢₖ = δⱼₖ: η₁ = (uū + dd̄ + ss̄)/√3. Orthogonal to both is η₈ = (uū + dd̄ − 2ss̄)/√6, whose coefficients are proportional to the diagonal of λ₈ — no accident, since the octet is the adjoint representation and its states map onto the generators.

Do it in two stages. 3 ⊗ 3 splits by exchange symmetry: the six symmetric pairs uu, dd, ss, (ud + du)/√2, (us + su)/√2, (ds + sd)/√2 form the 6 = (2, 0), and the three antisymmetric pairs form a 3̄ = (0, 1), because εᵢⱼₖ qⱼ qₖ transforms like an antiquark. Then 6 ⊗ 3 = 10 ⊕ 8 and 3̄ ⊗ 3 = 8 ⊕ 1, so 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 and 27 = 10 + 8 + 8 + 1. Weight counting checks it. A three-quark weight depends only on the counts (nᵤ, n_d, nₛ), and the number of orderings is 3!/(nᵤ! n_d! nₛ!): 1 at the corners uuu, ddd, sss, 3 at each of the six hexagon sites such as uud, 6 at uds. The totally symmetric decuplet takes one state from every site: the Δ quartet at Y = 1, the Σ* triplet, the Ξ* doublet, Ω⁻ = sss at Y = −2. Remove it and the corners are empty, the hexagon sites hold 2 each and the centre 5; two octets take 6 + 2 apiece, leaving one state at uds, the totally antisymmetric singlet. Which octet is which is a question of exchange symmetry, and the next topic needs that answer.

Build the eight λₐ as 3 × 3 complex arrays and check the algebra numerically: f_abc = −2i Tr([Tₐ, T_b] T_c), exact because Tr(Tₐ T_b) = δ_ab/2. A generator on the nine-dimensional quark–antiquark space is Tₐ ⊗ 1 − 1 ⊗ Tₐᵀ, assembled with np.kron; the minus sign and the transpose are the conjugate representation. The quadratic Casimir C₂ = Σₐ Tₐ² commutes with every generator, so it is constant on each irrep, (p² + q² + pq)/3 + p + q: 4/3 on the 3, 3 on the 8, 6 on the 10, 0 on the singlet. np.linalg.eigvalsh on the 9 × 9 Casimir returns 3 eight times and 0 once — that is 8 ⊕ 1 without a diagram drawn. For three quarks the generator is Tₐ ⊗ 1 ⊗ 1 + 1 ⊗ Tₐ ⊗ 1 + 1 ⊗ 1 ⊗ Tₐ on 27 dimensions, and the eigenvalues come back 6 ten times, 3 sixteen times, 0 once. The machine has counted, but the sixteen-fold eigenvalue 3 is two octets it cannot separate, because C₂ commutes with the permutations of the three quarks. Diagonalising the exchange operator P₁₂ inside that subspace splits them, and diagonalising T₃ and Y inside each irrep recovers the weights.

With mᵤ = m_d = m the quark mass matrix diag(m, m, mₛ) equals ((2m + mₛ)/3)⋅1 − ((mₛ − m)/√3)⋅λ₈: an SU(3) singlet plus a piece proportional to λ₈, the I = 0, Y = 0 member of an octet. First-order perturbation theory in an octet operator has, by the SU(3) Wigner–Eckart theorem, at most two reduced matrix elements inside a multiplet, so every mass takes the form M = a + bY + c[I(I + 1) − Y²/4]. In the decuplet I = 1 + Y/2 turns the bracket into 2 + 3Y/2, so M is linear in Y and the rows are equally spaced: 1232, 1385, 1533 MeV, steps of 153 and 148, which is how the Ω⁻ was predicted near 1680 MeV two years before it was seen at 1672. In the octet three parameters fit four masses and leave one relation, 2(m_N + m_Ξ) = 3m_Λ + m_Σ: 4514 against 4540 MeV, 0.6% — first order in mₛ − m ≈ 90 MeV against hadron masses near 1 GeV is about that good. Try the same with charm at 1.27 GeV and the SU(4) multiplets split by more than their lightest member weighs: bookkeeping, no longer a symmetry.