Casimir labels and the particle taxonomy

Start from what every inertial observer must agree on. The Poincaré group — four translations generated by P^μ, six rotations and boosts generated by the antisymmetric M^μν — acts on the state space through unitary operators, because a change of frame cannot change a probability. Wigner's definition is then forced: a free particle is a subspace closed under all ten generators with no smaller closed subspace inside it, an irreducible unitary representation. Reducible would mean two particles hiding under one name; non-unitary would mean a boost changing a norm. The natural basis is the joint eigenbasis of the four commuting P^μ, written |p, σ⟩, with p confined to one orbit of the Lorentz group and σ a discrete index for whatever the momentum leaves undecided. The whole topic is two questions: what fixes the orbit, and how many values σ takes.

An operator commuting with all ten generators is, by Schur's lemma, a multiple of the identity on an irreducible space, so its eigenvalue is a label the whole particle shares. The Poincaré algebra has two independent ones. P² = P_μP^μ is obvious. The second is built from the Pauli–Lubański vector W_μ = ½ ε_μνρσ M^νρ P^σ, which commutes with P^μ and satisfies W⋅P = 0. Evaluate both where they are simplest, in the rest frame p = (m, 0, 0, 0): P² = m², W⁰ = 0, and the spatial part is m times the rotation generator J, so W² = −m²J² = −m²s(s+1). The residual symmetry that fixes the rest momentum — the little group — is SU(2), and its irreducible representations are the familiar spin-s multiplets with σ = mₛ running over 2s + 1 values. For the proton, m = 938.3 MeV and s = ½ give W² = −¾m² = −0.660 GeV² and two states per momentum; for the Z, m = 91.19 GeV and s = 1 give W² = −1.66 × 10⁴ GeV² and three.

With m = 0 there is no rest frame, so the standard momentum becomes k = (κ, 0, 0, κ) and its little group is no longer SU(2) but the Euclidean group of the plane, ISO(2). P² = 0, and every massless particle seen in nature also has W² = 0; the two vectors are then no longer independent: a null vector orthogonal to another null vector is parallel to it, so W_μ = h P_μ. The number h = J⋅p̂ is the helicity, and it is Lorentz-invariant only here — a massive particle of speed β can be overtaken by a collinear boost faster than β, which reverses p while leaving J_z alone and so flips h; for m > 0 helicity is a frame choice and s is the label. Requiring the two translation-like generators of ISO(2) to act trivially — the alternative, continuous spin, is not seen in nature — leaves one state per representation, and parity pairs h with −h. So the photon has h = ±1 and two states, the gluon ±1 per colour, the graviton ±2, while the W with m = 80.4 GeV keeps 2s + 1 = 3, its longitudinal state supplied by a Goldstone mode of the Higgs doublet.

The abstract argument becomes a matrix calculation in the four-vector representation, where a spin-1 particle's states are polarisation vectors ε^α. Write the Lorentz generators as 4 × 4 matrices (M^μν)^α_β = i(η^μα δ^ν_β − η^να δ^μ_β), pick a numerical momentum p^σ, and assemble W_μ = ½ ε_μνρσ M^νρ p^σ with numpy.einsum over the Levi-Civita symbol. Then form W² = Σ_μ η^μμ W_μ W_μ, a single 4 × 4 matrix. For p = (91.19, 0, 0, 0) GeV it sends every ε with ε⋅p = 0 to −2m²ε = −1.663 × 10⁴ GeV² ε, and it annihilates ε ∝ p. Boost to p = (104.0, 0, 0, 50.0) GeV and the same two eigenvalues appear, on transformed vectors. That is the decomposition 4 = 3 ⊕ 1 of the vector representation into a spin-1 irrep, on which W² reads −m²⋅1⋅2, and the spin-0 component along p, on which it reads −m²⋅0⋅1 = 0 — and it is why a massive vector field carries three physical polarisations, not four. Set p null and W² reduces to ε ↦ 2(ε⋅p)p: every eigenvalue is zero, yet it is not the zero matrix — it annihilates each ε with ε⋅p = 0 and sends the rest along p, the pure-gauge direction. That is the massless case.

The taxonomy used in every table sits on top of the labels. Fermion or boson is the one question W² settles: the spin–statistics theorem makes half-integer s antisymmetric under exchange and integer s symmetric, so leptons, quarks and baryons obey Pauli exclusion while mesons, gauge bosons and the Higgs do not. Elementary or composite is a statement about the Lagrangian, not the representation: the Standard Model's elementary content is six leptons and six quarks at s = ½, the photon, eight gluons, W± and Z at s = 1, and the Higgs at s = 0, while every hadron is a bound state of the quark fields. Wigner's labels are blind to the difference — a proton and a hypothetical elementary fermion of 938.3 MeV would be the same irrep. Baryon or meson is a count: three quarks give B = 1 and half-integer spin, a quark and an antiquark give B = 0 and integer spin, and the tetraquark and pentaquark states established at LHCb obey the same rule with B = 0 and B = 1. Ask the questions in order: read m and s from the Casimirs, statistics from s, then what the state is made of.

Wigner's representations are for asymptotic states, eigenstates of P² with a real mass that persist forever. An unstable particle is not one. Its signature is a pole of the scattering amplitude, A(s) ∝ 1/(s − M² + iMΓ), at √s ≈ M − iΓ/2 on the unphysical sheet; on the real axis it shows as a Breit–Wigner bump of full width Γ, and in time as an amplitude decaying like e^{(−Γt/2ℏ)}, so τ = ℏ/Γ. The ρ(770) has Γ = 149 MeV, τ = 4.4 × 10⁻²⁴ s, and completes M/(2πΓ) = 0.83 rest-frame phase cycles before it is gone: it exists only as a feature of the ππ P-wave amplitude, which is what fixes its spin at 1. The Δ(1232) with Γ = 117 MeV is the same kind of object. The Z, with Γ = 2.495 GeV but Γ/M = 0.027, and the J/ψ, with Γ = 92.6 keV, are narrow enough to treat as particles on the timescale of their production. The mass and spin of a pole are still meaningful — its position and the partial wave it sits in — but the state count, the exchange symmetry and the whole taxonomy apply to it only by extension from the stable limit.