University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.4

Wigner's little group: mass, spin and helicity

Every Feynman rule in this course labels an external line by a momentum and a spin or a helicity. This topic earns those labels: the two numbers all observers agree on, the 2j + 1 states a massive particle carries and boosts rotate among, and the single helicity of a massless particle, which no boost can change.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A one-particle state needs labels that every observer can agree on, and topic 1.03's unitary Poincaré action decides which ones exist. Two operators commute with all ten generators: P², which gives the mass, and W², the square of the Pauli–Lubanski vector W^μ that reduces at rest to m times the angular momentum, so W² = −m² j(j + 1), derived. What remains is the label σ that tells apart states of one momentum, and Wigner's construction finds it: fix a standard momentum k, reach every p by a chosen boost L(p), and any Lorentz transformation then acts on σ only through the residual W = L(Λp)⁻¹ΛL(p), which leaves k alone.

For m > 0 that little group is the rotation group, so σ takes 2j + 1 values, and a sideways boost turns the spin away from the momentum: a massive particle's helicity belongs to the observer. For m = 0 there is no rest frame, the little group of a null momentum is ISO(2), the symmetry group of a plane, and once its two translation-like generators are set to zero only a phase is left: one helicity h that no proper orthochronous Lorentz transformation changes, while parity pairs h with −h, giving the photon two states, not three. The classification labels free states and stops there: which fields create them begins in topic 1.05, how they interact in Unit 2.

Simple definition: The little group of a four-momentum k is the set of Lorentz transformations that leave k unchanged; its unitary representations supply the spin label of the states |p, σ⟩: 2j + 1 rotation states when m > 0, one Lorentz-invariant helicity when m = 0.
Example: A proton with |p| = m along x, seen by an observer moving at 0.60 along −y, has its momentum turned through 46.7° but its spin through only 15.7°, so helicity +½ survives with probability cos²(15.5°) = 0.93. A photon's helicity would survive with probability 1.

Pauli–Lubanski vectorW^μ = −½ ε^μνρσ J_νρ P_σ, with ε⁰¹²³ = +1

Derived: W⁰ = J⋅P, W⋅P = 0, and at rest W = (0, mJ). Unlike J^μν it commutes with P^μ and so measures spin with no orbital part.

J^μν the dimensionless Lorentz generators, rotations J and boosts K; P_σ in GeV, so W^μ is in GeV

Spin from a Casimir invariantW² = W^μ W_μ = −m² j(j + 1)

Derived in the rest frame. With P² = m² it commutes with all ten generators, so it labels an irreducible representation for every observer.

m in GeV and W² in GeV², j = 0, ½, 1, …; the minus sign is the metric's, since W^μ is spacelike for m > 0

The Wigner rotation acts on the spin labelU(Λ)|p, σ⟩ = Σ_τ D_τσ(W) |Λp, τ⟩, with W = L(Λp)⁻¹ Λ L(p)

Derived from |p, σ⟩ = U(L(p))|k, σ⟩: every Lorentz transformation reaches the spin label only through the little group of k.

L(p) the chosen transformation with L(p)k = p; W fixes k; D is (2j + 1)-dimensional for m > 0, a phase for m = 0

Spin lag after a sideways boosttan δ = βm/|p|, where δ = α − θ_W

Derived from tan α = γβE/|p| and UAQM-05.01's cos θ_W = (γ₁ + γ)/(1 + γ₁γ). Helicity +j survives with probability cos^(4j)(δ/2).

|p| the momentum before, β the observer's speed across it; α the turn of p̂, θ_W the Wigner rotation; δ in rad

Little group of a null momentum[J_z, A] = iB, [J_z, B] = −iA, [A, B] = 0

Derived: the ISO(2) algebra of a plane's rotations and translations. Non-zero A or B would need a continuous label at fixed k; that case is set aside, so A = B = 0.

k = (κ, 0, 0, κ), κ in GeV; A = Jₓ + K_y and B = J_y − Kₓ in UAQM-05.01's generators, all dimensionless

One invariant helicity when m = 0W^μ = h P^μ on every massless state, with h = J⋅p̂ = 0, ±½, ±1, …

Derived in outline: the little group acts by a phase e^(−ihθ), so no boost changes h. Parity sends h to −h; that h is a multiple of ½ is quoted.

h dimensionless (units of ħ); P^μ is null, so W² = 0, the m → 0 limit of −m² j(j + 1)

Two numbers every observer agrees on

Work in natural units, ħ = c = 1, with the metric (+, −, −, −) and ε⁰¹²³ = +1, kept for the lesson. Topic 1.03 built U(Λ, a) on Fock space from the charges P^μ and J^μν; that is quoted. A label every observer reads alike must commute with all ten. P² does: translations commute and P² is a scalar. J^μν does not, because its orbital part shifts with the origin. The Pauli–Lubanski vector W^μ = −½ ε^μνρσ J_νρ P_σ repairs that. [J_νρ, P^λ] is a sum of metric factors times P; contracted with ε^μνρσ P_σ it leaves two P's in antisymmetric slots, so W^μ commutes with every P^λ; W⋅P = 0 for the same reason. W^μ is a four-vector, so W² is a scalar. Both invariants commute with every generator, derived, and by Schur's lemma, quoted, each is a number on an irreducible set of states. At rest W⁰ = J⋅P = 0 and W = mJ, so W² = −m² j(j + 1), derived. Dimensions: J is a pure number and P a mass, so W² is a mass squared.

One standard momentum, and a boost to every other

P² fixes m but neither the direction of p nor the spin, so Wigner adds a construction. Pick a standard momentum k with k² = m² and k⁰ > 0, and for each p on the mass shell one definite L(p) with L(p)k = p; for m > 0 take k = (m, 0, 0, 0) and L(p) the pure boost along p̂. Define |p, σ⟩ = U(L(p))|k, σ⟩, which inherits topic 1.03's invariant normalisation. Then U(Λ)|p, σ⟩ = U(L(Λp)) U(W)|k, σ⟩ with W = L(Λp)⁻¹ΛL(p), and Wk = L(Λp)⁻¹(Λp) = k: W lies in the little group of k, the Lorentz transformations that fix it. If the states at k carry a unitary representation D of that group, U(Λ)|p, σ⟩ = Σ_τ D_τσ(W)|Λp, τ⟩, derived. Every question about σ has become a question about a smaller group, and which group depends only on whether k² is positive or zero. P² below zero describes no physical particle and P^μ = 0 is the vacuum; both are set aside.

Massive particles: 2j + 1 states and an observer's helicity

For k = (m, 0, 0, 0) the little group is the rotation group, so σ is a J_z label with 2j + 1 values and D = D^(j), quoted from angular momentum in Quantum Mechanics II; half-integer j needs the double cover, in which a 2π turn is −1, quoted from UAQM-05.04. The Wigner rotation has a size you can compute. Let a particle move along x with |p| = um while an observer moves at β along −y, with γ = 1/√(1 − β²). Its momentum turns towards +y through α, tan α = γβ√(1 + u²)/u, while its spin label turns through θ_W, the rotation UAQM-05.01 found in two perpendicular boosts: cos θ_W = (γ₁ + γ)/(1 + γ₁γ), γ₁ = √(1 + u²), quoted. The tangent of α − θ_W then simplifies to tan δ = βm/|p|, derived: only the mass lets the spin lag. A boost along the line is blunter. A μ⁺ from π⁺ decay at rest has |p| = 29.79 MeV, E = 109.78 MeV and helicity −½, fixed by the left-handed neutrino; an observer faster than 0.2714 along its line reverses p but not J, and sees +½.

Massless particles: the little group is ISO(2)

No frame brings a massless particle to rest, so take k = (κ, 0, 0, κ). Acting on four-vectors, J_z annihilates k, and so do A = Jₓ + K_y and B = J_y − Kₓ, derived: a rotation about k̂ and two blends of a rotation with a boost. Their algebra, [J_z, A] = iB, [J_z, B] = −iA and [A, B] = 0, is that of ISO(2), the rotations and translations of a plane, with A and B as translations. A and B commute, so take a common eigenstate with values (a, b). The rotation exp(−iθJ_z) carries (a, b) round a circle, so if either is non-zero a continuous label sits at one momentum: continuous-spin states, never observed and set aside. With A = B = 0 only rotations about k̂ act, and on J_z = h they give a phase, U(W)|k, h⟩ = e^(−ihθ)|k, h⟩. The same components give W^μ = κ(J_z, A, B, J_z), which is h k^μ once A = B = 0, derived, so W² = 0. One helicity h = J⋅p̂, which no boost or rotation changes, derived in outline; that a 2π turn is ±1, making h a multiple of ½, is quoted.

Parity pairs h with −h: the photon's two states

Parity reverses p and leaves J alone, so it sends h to −h, and it lies outside the proper orthochronous group: no boost or rotation links the two. A theory that conserves parity and contains a massless particle of helicity h must contain −h as well, and the pair is two states, not 2j + 1. That is the photon, with helicities +1 and −1. UAQM-11.04 found them in the Coulomb-gauge field from k⋅ε = 0 and quoted Wigner for their invariance; the reason is the group, not the gauge, and topic 8.05 turns the argument round to show that helicity ±1 forces gauge invariance on any vector field made of photons. A graviton would carry ±2, again two states. The weak interaction does not conserve parity, and its neutrinos are left-handed: Goldhaber, Grodzins and Sunyar measured a negative neutrino helicity in 1958, with antineutrinos at +½ by CPT, quoted. The count is measured as well: Stefan's constant π²k_B⁴/(60ħ³c²) counts two states per photon mode, and radiometry agrees with it; a third state would raise it by half.

Where the two little groups meet

With tan δ = βm/|p|, a particle whose momentum dwarfs its mass has its spin dragged round with its momentum: at |p| = 10m and β = 0.60, δ = 3.4°, and the helicities barely mix. In the limit the 2j + 1 states split, in outline, into separate representations of fixed helicity, and a massless theory may keep only some; the photon keeps ±1. The bounds show how close nature runs. The photon's mass is below 1 × 10⁻¹⁸ eV (quoted), so for a 1 eV photon tan δ is at most 10⁻¹⁸, and the chance of meeting helicity 0, sin²δ/2, is below 10⁻³⁶. Neutrinos are not exactly massless, but if one sat at KATRIN's 90% bound, 0.45 eV (quoted), reversing the helicity of a 1.00 MeV neutrino would need an observer within 1.0 × 10⁻¹³ of light speed, a Lorentz factor above 2.2 × 10⁶. The classification ends there: it labels free one-particle states, says nothing about the fields that create them, from topic 1.05, or about interactions, from Unit 2.

Change one variable at a time

Make the relationship visible.

Interactive model

Particle momentum |p|/m1.5

Observer speed β along −y0.80

Spin j0.5

Hold β at 0.80 and slide |p|/m from 0.1 to 10: the dashed spin curve climbs onto the solid momentum curve and δ falls from 82.9° to 4.6°, the massless limit. At |p|/m = 0.5, raise j: the same 58° lag keeps helicity +j with probability 0.765 for spin ½ but 0.342 for spin 2.

MOMENTUM TURN α58.0 °

WIGNER ROTATION θ_W30.0 °

SPIN LAG δ28.1 °

P(HELICITY STAYS +j)0.941

Live interpretationMOMENTUM TURN α: 58.0 °. WIGNER ROTATION θ_W: 30.0 °. SPIN LAG δ: 28.1 °. P(HELICITY STAYS +j): 0.941

Catch the common trap

Explain before calculating.

Misconception check

The tempting count treats a massless spin-j particle as a massive one with states struck out: the photon has spin 1, so 2j + 1 = 3 states, and transversality or a gauge choice removes the longitudinal one, which a cleverer frame might bring back. The little group says otherwise. A null momentum has no rest frame, its little group is ISO(2), not the rotation group, and once the translation-like generators vanish its unitary representations are one-dimensional: a single helicity, fixed under every proper orthochronous Lorentz transformation. The photon has two states because parity pairs +1 with −1, not because one of three was removed. The same count gives a graviton two states, not 5, and a left-handed neutrino one. A massive vector's helicity-0 state is real, but as m → 0, tan δ = βm/|p| → 0 and no boost mixes it with ±1 any more: it becomes a representation of its own.

A spin-½ particle of mass m moves along +x with |p| = 0.75m and helicity +½. An observer moves past at speed 0.60 along −y. With what probability does that observer find helicity +½, and what is the answer for a massless spin-½ particle in the same situation?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe Δ⁺⁺(1232) has m = 1.232 GeV and j = 3/2. Find P² and W², count its states at fixed momentum, and test the count against the measured peak of σ(π⁺p), about 200 mb, with the unitarity limit σₘₐₓ = (4π/k²)(2j + 1)/[(2s_π + 1)(2sₚ + 1)] quoted from Quantum Mechanics II. Take m_π = 139.57 MeV and mₚ = 938.27 MeV.

P² = m² = (1.232 GeV)² = 1.518 GeV². At rest W = (0, mJ), so W² = −m² j(j + 1) = −1.518 × (3/2)(5/2) GeV² = −5.692 GeV². Both are masses squared, and both are the same for every observer.
The little group of k = (m, 0, 0, 0) is the rotation group, so each momentum carries 2j + 1 = 4 states, σ = +3/2, +½, −½, −3/2.
At the peak s = (1.232 GeV)² = 1.5178 GeV². The centre-of-mass momentum is k = √([s − (mₚ + m_π)²][s − (mₚ − m_π)²])/(2√s) = √(0.35608 × 0.87990)/2.464 GeV = 0.2272 GeV.
The pion has spin 0 and the proton spin ½, so σₘₐₓ = (4π/k²)(4/2) = 8π/k² = 487.0 GeV⁻². Restoring ħc with 1 GeV⁻² = 0.3894 mb gives 189.6 mb.
Measured: about 200 mb, within about 5%, so the four states are seen. A Δ of spin ½ would cap the peak at 94.8 mb and one of spin 5/2 would allow 284 mb. Check: 4π/k² has dimension GeV⁻², an area in natural units, as a cross-section must.

AnswerP² = 1.518 GeV² and W² = −5.692 GeV², with 4 states per momentum; the unitarity peak 8π/k² = 190 mb agrees with the measured value of about 200 mb, while j = ½ would allow only 95 mb.

MediumA ρ⁰ meson (m = 775.26 MeV, j = 1) moves along +z with |p| = 1.200 GeV in the helicity state λ = +1. Starting from W = (0, mJ) at rest, find the eigenvalues of W⁰ and W³ on this state, check W⋅P = 0, split W² = −m² j(j + 1) into its longitudinal and transverse parts, and show how W^μ approaches λP^μ as m/|p| → 0.

E = √(1.200² + 0.77526²) GeV = 1.4286 GeV. The state is U(L)|k, λ⟩, with L the boost along z of rapidity η: sinh η = |p|/m = 1.5479 and cosh η = E/m = 1.8428.
At rest W³|k, λ⟩ = mλ|k, λ⟩ and W⁰|k, λ⟩ = 0. W^μ is a four-vector, U(L)⁻¹ W^μ U(L) = L^μ_ν W^ν; on the moving state W⁰ = mλ sinh η = λ|p| = 1.200 GeV and W³ = mλ cosh η = λE = 1.4286 GeV. The first agrees with W⁰ = J⋅P = λ|p| directly.
W⋅P = W⁰E − W³|p| = λ(|p|E − E|p|) = 0, as the antisymmetry of ε demands.
Longitudinal part: (W⁰)² − (W³)² = λ²(|p|² − E²) = −λ²m² = −0.6010 GeV². The invariant is W² = −2m² = −1.2021 GeV², so the transverse components supply ⟨(W^x)² + (W^y)²⟩ = m²[j(j + 1) − λ²] = 0.6010 GeV². A boost along z leaves W^x and W^y alone, and at rest they are mJₓ and mJ_y, which do not vanish on a J_z eigenstate.
Massless form: W^μ − λP^μ has time and z components ∓λ(E − |p|) = ∓λm²/(E + |p|) = ∓0.2286 GeV, 16% of E here, and 3.0 MeV at |p| = 100 GeV. At m = 0 this and the transverse part m²[j(j + 1) − λ²] both vanish, leaving W^μ = λP^μ and W² = 0, the massless result.

AnswerW⁰ = 1.200 GeV and W³ = 1.429 GeV, with W⋅P = 0; W² = −1.202 GeV², half from (W⁰)² − (W³)² = −0.601 GeV² and half from the transverse components; W^μ − λP^μ shrinks as m²/(E + |p|), 0.229 GeV here, and vanishes with m.

HardA Z boson (m = 91.1876 GeV) moves along +x with |p| = 60.0 GeV and helicity +1. An observer moves past at 0.600 along −y. Find the Z's four-momentum in the observer's frame, the turn α of its momentum, the Wigner rotation θ_W of its spin, the probabilities of helicity +1, 0 and −1, and the mean helicity. Repeat δ and P(0) for |p| = 911.9 GeV, and say what the trend implies at m = 0.

Before: E = √(60.0² + 91.1876²) GeV = 109.16 GeV, so γ₁ = E/m = 1.1971. The observer's boost has γ = 1.250 and γβ = 0.750 and adds momentum along +y: E′ = γE = 136.45 GeV and p′ = (60.0, γβE, 0) = (60.0, 81.87, 0) GeV. Check: E′² − |p′|² = 18 617 − 10 302 = 8315 GeV² = m².
Momentum turn: tan α = 81.87/60.0, so α = 53.76°. Wigner rotation, quoted from UAQM-05.01: cos θ_W = (γ₁ + γ)/(1 + γ₁γ) = 2.4471/2.4963 = 0.98027, so θ_W = 11.40° about +z, in the same sense.
The spin, which pointed along the momentum, now leans δ = α − θ_W = 42.36° from it. Check: tan δ = βm/|p| = 0.600 × 91.1876/60.0 = 0.9119 gives δ = 42.36°.
The state is |1, +1⟩ about an axis at δ to p̂′, so the helicity amplitudes are the spin-1 rotation-matrix elements d¹(δ) from +1 to λ′, quoted: P(+1) = [(1 + cos δ)/2]² = 0.756, P(0) = sin²δ/2 = 0.227 and P(−1) = [(1 − cos δ)/2]² = 0.017. They sum to 1.000, and the mean helicity is cos δ = 0.739.
At |p| = 911.9 GeV = 10.0m: tan δ = 0.0600, so δ = 3.43° (α = 37.01°, θ_W = 33.57°), P(0) = 0.0018 and P(−1) = 8 × 10⁻⁷. The spin is now carried round with the momentum.
As m/|p| → 0, δ → 0 for every boost: the helicities stop mixing, so a massless particle may carry +1 alone, or ±1 without 0, as the photon does. Checks: δ is a pure number, it vanishes at β = 0, and as |p|/m → 0 it tends to 90°, a particle at rest seen moving sideways.

AnswerE′ = 136.45 GeV and p′ = (60.0, 81.87, 0) GeV; α = 53.76°, θ_W = 11.40°, δ = 42.36°; P(+1) = 0.756, P(0) = 0.227, P(−1) = 0.017, mean helicity 0.739. At |p| = 10m, δ = 3.43° and P(0) = 0.0018: as m → 0 the helicity freezes.