University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.5

Why a massless vector forces gauge invariance

Gauge invariance usually arrives as a property of a Lagrangian you might have written differently. Here it stops being optional. For a massless particle of helicity ±1 the little group slides every polarisation along k, and the S matrix stays Lorentz invariant only if each amplitude vanishes under the replacement ε → k.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 1.04 showed that a massless particle's states carry one helicity, acted on by the rotation about k as a phase, while the two translation-like generators of the little group ISO(2), the null rotations S(α, β), do nothing to them. Describe the photon with a four-vector field and the null rotations cause trouble: they fix k and every photon state, but they add a multiple of k^μ to each polarisation vector. A field A^μ built from photon operators alone is therefore a four-vector only up to a gauge term ∂^μΩ, and no cleverer polarisation repairs it, because the only vector the null rotations leave alone is k itself, which has helicity 0.

The field strength F_μν never sees the shift and is a genuine tensor. The cost falls on amplitudes. Lorentz covariance makes the amplitude a second observer computes equal to the first one's with ε moved along k, so the S matrix is invariant only if every amplitude with an external photon vanishes when ε is replaced by k, and only if the photon couples to a conserved current.

Weinberg's soft photons turn the same demand into charge conservation. The argument needs m = 0: a vector of mass M has the little group SO(3), a true four-vector field and no gauge symmetry, though its longitudinal polarisation is k/M up to terms of order M/E. It also explains Gupta-Bleuler: insist on a covariant A^μ and unphysical photons come with it.

Simple definition: For a massless particle of helicity ±1, Lorentz invariance of the S matrix forces gauge invariance: the little group shifts each polarisation vector by a multiple of k^μ, so every amplitude ε*_μM^μ must satisfy k_μM^μ = 0.
Example: For k = (5 GeV)(1, 0, 0, 1) the null rotation S(0.6, 0) sends ε₁ = (0, 1, 0, 0) to (0.6, 1, 0, 0.6) = ε₁ + (0.12 GeV⁻¹)k, still of norm −1. Against an M^μ with ε₁⋅M = 2.00 and k⋅M = 1.00 GeV it gives 2.12 instead of 2.00, so Lorentz invariance demands k⋅M = 0.

Little-group action on a polarisationW(θ, α, β) ε^μ(k, ±) = e^(∓iθ) [ε^μ(k, ±) + ((α ± iβ)/(√2 κ)) k^μ]

The rotation supplies the helicity phase; the null rotations add a multiple of k. The norm −1 and k⋅ε = 0 survive, since k is null and orthogonal to ε.

W = S(α, β)R(θ); k = κ(1, 0, 0, 1), κ in GeV; ε(k, ±) = (0, 1, ±i, 0)/√2 dimensionless; θ in rad, α, β pure numbers. Derived.

Only k survives every null rotationS(α, β) u = u for all α, β ⟹ u^μ ∝ k^μ

R(θ)k = k, so u ∝ k carries helicity 0. No four-vector field exists for helicity ±1, while F_μν, in (1, 0) ⊕ (0, 1), carries ±1 freely (quoted).

u^μ: the coefficient a true vector field would attach to a(k, σ); S the null rotations of topic 1.04's ISO(2). Shown here.

The photon field, up to a gauge termU(Λ) A^μ(x) U(Λ)⁻¹ = (Λ⁻¹)^μ_ν A^ν(Λx) + ∂^μ Ω(x, Λ)

Each ε's k^μ shift becomes ∂^μ acting on e^(−ik⋅x). □Ω = 0, so the term even respects Lorenz gauge, yet A^μ is still not a four-vector.

A^μ has mass dimension 1, built from a(k, ±1) and a†; Ω is dimensionless and linear in a and a†. Derived in outline.

F_μν does not see the shiftk^μ ε′^ν − k^ν ε′^μ = k^μ ε^ν − k^ν ε^μ, with ε′ = ε + c k

E and B transform as a true tensor although A^μ does not: the helicity ±1 field is the field strength, and the potential is a convenience with a price.

A plane wave's F^μν ∝ k^μ ε^ν − k^ν ε^μ; c is any number, of dimension mass⁻¹. Shown here.

Lorentz invariance demands the ε → k testM_±(kSp) = M_±(kp) − ((α ∓ iβ)/(√2 κ)) k_μ M^μ(kp)

One process seen from two frames must have one amplitude, so k_μM^μ = 0 for every external photon. For a massless vector, Lorentz invariance has become gauge invariance.

M_± = ε*_μ(k, ±)M^μ, M^μ covariant; S = S(α, β) fixes k; p: the other momenta, on shell. Shown here.

Soft photons demand charge conservationε* → ε* + c k shifts e Σₙ ηₙqₙ (pₙ⋅ε*)/(pₙ⋅k) by c e Σₙ ηₙqₙ

Lorentz invariance needs Σₙ ηₙqₙ = 0 in every process: Weinberg's charge conservation. Borexino finds τ(e⁻ → νγ) above 6.6 × 10²⁸ yr.

ηₙ = +1 outgoing, −1 incoming; qₙ in units of e; soft factor in GeV⁻¹, quoted; the shift derived.

The little group of a null momentum, written as matrices

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Rotate any photon momentum to k = κ(1, 0, 0, 1). Its little group, quoted from topic 1.04, is ISO(2): rotations R(θ) about z and two null rotations S(α, β), each a boost combined with a rotation. With ζ = (α² + β²)/2 the rows of S are (1 + ζ, α, β, −ζ), (α, 1, 0, −α), (β, 0, 1, −β) and (ζ, α, β, 1 − ζ). Multiplying out gives three facts, all derived: SᵀgS = g, so S is a Lorentz transformation; Sk = k, so it belongs to k's little group; and S(α₁, β₁)S(α₂, β₂) = S(α₁ + α₂, β₁ + β₂), so null rotations compose like translations of a plane. Now act on the transverse vectors ε₁ = (0, 1, 0, 0) and ε₂ = (0, 0, 1, 0): Sε₁ = (α, 1, 0, α) = ε₁ + (α/κ)k and Sε₂ = ε₂ + (β/κ)k, derived. For α = 0.4, Sε₁ = (0.4, 1, 0, 0.4), whose norm is 0.16 − 1 − 0.16 = −1: the vector moved, but only along a null direction orthogonal to it, so neither its norm nor k⋅ε notices.

Helicity states shift, and no vector can stand still

The helicity polarisations are ε(k, ±) = (ε₁ ± iε₂)/√2, up to a phase convention. A rotation by θ multiplies them by e^(∓iθ), and the null rotations add ((α ± iβ)/(√2κ))k, so a general element W = S(α, β)R(θ) gives W ε(k, ±) = e^(∓iθ)[ε(k, ±) + ((α ± iβ)/(√2κ))k], derived. Could a better choice of vector avoid the shift? A field with a true vector index that annihilates helicity-σ photons through a coefficient u^μ(k, σ) is covariant only if Wu = (helicity phase) × u for every W, because the photon operators themselves pick up only that phase (topic 1.04). Take θ = 0. The second and third rows of Su = u give α(u⁰ − u³) = β(u⁰ − u³) = 0, so u⁰ = u³; the first then gives αu¹ + βu² = 0 for all α and β, so u¹ = u² = 0. Only u ∝ k survives, and R(θ)k = k gives it helicity 0, shown. So no four-vector field exists for helicity ±1; the lone candidate makes A^μ = ∂^μφ and F = 0. The general rule, quoted from Weinberg: a massless field in the (A, B) representation annihilates only helicity B − A, in his convention, so (½, ½) gives 0 while F_μν, in (1, 0) ⊕ (0, 1), gives ±1.

The photon field is a vector only up to a gauge term

Since no true vector works, build A^μ from what exists: the two transverse polarisations of Advanced QM's Coulomb-gauge field (UAQM-11.03), A^μ(x) = Σ± ∫ d³k/((2π)³ 2|k|) [ε^μ(k, ±) a(k, ±) e^(−ik⋅x) + h.c.]. Write a general Λ as L(Λk) W L(k)⁻¹, with L the standard boosts of topic 1.04; the previous section then gives Λε(k, σ) equal to ε(Λk, σ) times a helicity phase, plus c(Λ, k) times Λk. The operators pick up the compensating phase, U(Λ)a(k, σ)U(Λ)⁻¹ ∝ a(Λk, σ) (quoted), so after k → Λk in the invariant measure everything reassembles into (Λ⁻¹)^μ_ν A^ν(Λx) except the k pieces. As k^μ e^(−ik⋅x) = i∂^μ e^(−ik⋅x), those collect into ∂^μΩ(x, Λ), with Ω linear in a and a†: derived in outline. Dimensions agree: A has mass dimension 1 and ∂ supplies one, so Ω is dimensionless and c(Λ, k) is an inverse mass, as it must be to add k to a dimensionless ε. The term is precisely a gauge transformation: Λ spoils A⁰ = 0 and ∇⋅A = 0, and the shift restores them in the new frame. F_μν = ∂_μA_ν − ∂_νA_μ loses it, since ∂_μ∂_νΩ is symmetric, so the field strength is a true tensor.

Lorentz invariance of S now demands k_μM^μ = 0

Couple the photon to matter through −∫ d⁴x J^μA_μ. Under U(Λ) the extra term contributes −∫ d⁴x J^μ∂_μΩ = +∫ d⁴x (∂_μJ^μ)Ω, after integrating by parts with fields that fall off at infinity, and that vanishes for every Ω only if ∂_μJ^μ = 0: the current must be conserved. In amplitudes, write an outgoing photon's amplitude as M_± = ε*_μ(k, ±)M^μ, where M^μ, built from covariant fields and propagators, obeys M^μ(Λk; Λp) = Λ^μ_ν M^ν(k; p). Choose Λ = S(α, β): it leaves k and the photon state alone and moves only the other momenta, taken spinless, from p to Sp. Then M_±(k; Sp) = (S⁻¹ε)*⋅M(k; p) = M_±(k; p) − ((α ∓ iβ)/(√2κ)) k_μM^μ(k; p), shown. Lorentz invariance of S demands the two be equal for every α and β, so k_μM^μ = 0 for each external photon, the other external lines on shell. The stakes are numerical: if k⋅M/κ were 0.2 times ε₁⋅M, the null rotation with α = −1 acts as the shift ε₁ → ε₁ + k/κ and multiplies the rate by 1.2² = 1.44, while α = +1 acts as ε₁ → ε₁ − k/κ and gives 0.8² = 0.64, for the same physical collision. The figure applies S(α, 0) to ε₁ itself, so there the shift is +αk/κ and the same factors appear at the opposite sign of α. The condition binds the full amplitude, not each diagram, as topics 8.06 and 8.09 show.

Weinberg's soft photons turn the test into charge conservation

The test bites hardest on a photon of small k attached to any process. As k → 0 the dominant diagrams hang the photon on an external line, whose propagator supplies 1/(2pₙ⋅k), and the amplitude becomes M₀ times the soft factor e Σₙ ηₙqₙ (pₙ⋅ε*)/(pₙ⋅k), quoted here and derived with QED's infrared cancellation later in the course: M₀ is the amplitude without the photon, ηₙ = +1 for outgoing and −1 for incoming particles, qₙ each charge in units of e. Replace ε* by ε* + ck. In every term pₙ⋅k cancels, so the factor shifts by c e Σₙ ηₙqₙ whatever the momenta, derived. Lorentz invariance therefore needs Σₙ ηₙqₙ = 0 in every process that happens: the couplings to soft photons, which is what charge means, are conserved. Dimensions check: the soft factor and the shift both go as 1/ω, so the test survives the soft limit ω → 0 unchanged. The measured check is sharp: Borexino saw no 256 keV line from e⁻ → νγ and set τ above 6.6 × 10²⁸ yr. A helicity-2 graviton run through the same argument forces equal gravitational couplings for every particle, the equivalence principle, quoted.

Where the argument stops, and the price of a covariant A^μ

Everything above used m = 0. A vector of mass M has little group SO(3): its three polarisations rotate into one another with no k term, Proca's field (topic 8.08) is a true four-vector, and nothing forces gauge invariance. The two cases meet only through conserved currents. Moving along z, ε_L = (|k|, 0, 0, E)/M = k/M + (M/(E + |k|))(−1, 0, 0, 1), derived, so as M/E → 0 the third polarisation turns into the gauge direction; its coupling to a conserved current falls like M/E, while a non-conserved one sees it grow like E/M. Measurement keeps the photon on the massless side: m_γc² is below 1 × 10⁻¹⁸ eV (quoted), a range ħ/(m_γc) = (197.327 MeV fm)/(10⁻¹⁸ eV) ≈ 2 × 10¹¹ m, beyond the Earth's orbit. Two honest routes remain. Keep two helicities and an A^μ that is not a vector, as Coulomb gauge does, with invariance resting on current conservation (topic 8.04). Or demand a covariant A^μ: then it carries four polarisations, a timelike one of negative norm, and needs the gauge fixing of topic 8.02 and the Gupta-Bleuler condition of topic 8.03, which admits the timelike and longitudinal photons only in zero-norm combinations that no physical quantity sees. Those extra states are the price of covariance.

Change one variable at a time

Make the relationship visible.

Interactive model

Null rotation α0.8

Ward violation δ = k⋅M/(κ ε₁⋅M)0.3

Set δ = 0 and drag α: ε′ gains equal time and z parts, the parallelogram shears without changing its area, and the rate stays put. Raise δ to 0.3 and the same photon's rate depends on the frame, which is why every amplitude must satisfy k⋅M = 0.

TIME PART ε′⁰ = ε′³0.8

NORM ε′⋅ε′-1.00

AMPLITUDE ε′⋅M / ε₁⋅M1.24

RATE RATIO |M′|²/|M|²1.54

Live interpretationTIME PART ε′⁰ = ε′³: 0.8. NORM ε′⋅ε′: −1.00. AMPLITUDE ε′⋅M / ε₁⋅M: 1.24. RATE RATIO |M′|²/|M|²: 1.54

Catch the common trap

Explain before calculating.

Misconception check

The tempting belief is that ε^μ(k, σ) is a four-vector, so ε*_μM^μ is a Lorentz scalar for free and gauge invariance is an optional redundancy of the Lagrangian. It is wrong because a physical polarisation is defined relative to a frame: Λε(k, σ) is the polarisation of Λk, times a phase, plus a multiple of Λk, and the Easy example finds that multiple to be 0.6/κ at β = 0.6. No better choice of ε removes it, since the only vector every null rotation fixes is k itself, of helicity 0. So ε*⋅M is frame independent only if k⋅M = 0: for a massless vector, gauge invariance is what Lorentz invariance means, not a symmetry added on top. A second slip grows from the first. Lorenz gauge ∂⋅A = 0 is a covariant condition, but it does not make A^μ a four-vector: the gauge term has □Ω = 0 and stays inside Lorenz gauge.

A photon moves along z with k = κ(1, 0, 0, 1). Covariant Feynman rules give an amplitude with ε₁⋅M = 2.0 and k_μM^μ/κ = 0.5, in the same units. For an observer related to the first by a null rotation of k, Lorentz covariance makes the amplitude ε′⋅M with ε′ = ε₁ + 0.4k/κ. What does that observer find, and what follows?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA photon of energy κ = 2.0 eV moves along x, linearly polarised along z: k = κ(1, 1, 0, 0), ε = (0, 0, 0, 1). Apply a boost Λ along z with β = 0.6. Find Λk and Λε, compare Λε with the boosted photon's transverse polarisation, and check that F^μν is unaffected.

With γ = 1/√(1 − 0.36) = 1.25 and γβ = 0.75, Λ mixes only t and z: t′ = γ(t + βz), z′ = γ(z + βt). So Λk = κ(1.25, 1, 0, 0.75) = (2.50, 2.00, 0, 1.50) eV, still null: 6.25 − 4.00 − 2.25 = 0.
Λε = (γβ, 0, 0, γ) = (0.75, 0, 0, 1.25). Its norm is 0.5625 − 1.5625 = −1 and Λk⋅Λε = κ(1.25 × 0.75 − 0.75 × 1.25) = 0, but its time component is not zero: it is no longer a Coulomb-gauge polarisation.
The new direction is k̂′ = (1, 0, 0.75)/1.25 = (0.8, 0, 0.6). The unit vector in the x-z plane orthogonal to it is (−0.6, 0, 0.8), so the transverse polarisation is ε_C = (0, −0.6, 0, 0.8), with norm −1 and Λk⋅ε_C = 0.
Subtract: Λε − ε_C = (0.75, 0.6, 0, 0.45) = 0.6 × (1.25, 1, 0, 0.75) = 0.6 Λk/κ. The boost has produced the transverse vector plus a pure gauge piece, with c = 0.6/κ = 0.30 eV⁻¹.
F^μν is built from k′^μ ε^ν − k′^ν ε^μ with k′ = Λk. In units of κ the 03 component is 1.25 × 1.25 − 0.75 × 0.75 = 1.00 with Λε and 1.25 × 0.8 − 0 = 1.00 with ε_C; the 01 component is −0.75 both ways and the 13 component 1.25 both ways.

AnswerΛk = (2.50, 2.00, 0, 1.50) eV and Λε = (0.75, 0, 0, 1.25) = ε_C + 0.6Λk/κ, a gauge shift with c = 0.30 eV⁻¹; F⁰³, F⁰¹, F¹³ = 1.00, −0.75, 1.25 (× κ) either way.

MediumA π⁺ at rest decays to μ⁺ν, the μ⁺ leaving at 60° to z in the x-z plane. A soft photon of energy ω = 1.0 MeV is emitted along z with ε₁ = (0, 1, 0, 0). Using the soft factor Σₙ ηₙqₙ(pₙ⋅ε)/(pₙ⋅k), in units of e, evaluate it with ε₁ and with ε₁ + 0.5k/ω. Repeat for a hypothetical neutral X of the same mass decaying to μ⁺ and a neutral particle with the same momenta.

Two-body kinematics with m_π = 139.57 MeV and m_μ = 105.66 MeV: |p| = (m_π² − m_μ²)/(2m_π) = 29.79 MeV and E_μ = (m_π² + m_μ²)/(2m_π) = 109.78 MeV, so p_μ = (109.78, 25.80, 0, 14.90) MeV.
With k = ω(1, 0, 0, 1): p_μ⋅k = ω(E_μ − p_z) = 94.88 MeV² and p_μ⋅ε₁ = −pₓ = −25.80 MeV. The pion at rest has p_π⋅ε₁ = 0 and p_π⋅k = ωm_π = 139.57 MeV²; the neutrino carries no charge.
Soft factor with ε₁: the μ⁺ (η = +1, q = +1) gives −25.80/94.88 = −0.2719 MeV⁻¹ and the π⁺ term vanishes, so the factor is −0.272 MeV⁻¹.
With ε₁ + 0.5k/ω each charged leg gains ηq × 0.5/ω, since pₙ⋅k cancels: +0.5 MeV⁻¹ from the μ⁺ and −0.5 MeV⁻¹ from the π⁺. The sum is unchanged, −0.272 MeV⁻¹, because Σηq = 1 − 1 = 0.
For X⁰ → μ⁺ + neutral the factor with ε₁ is the same, −0.272 MeV⁻¹, but only the μ⁺ shifts: −0.272 + 0.500 = +0.228 MeV⁻¹. The emission probability ∝ |factor|² drops from 0.0739 to 0.0520 MeV⁻², a factor 0.70, for the same physical photon.

Answerπ⁺ → μ⁺ν: −0.272 MeV⁻¹ both ways. The charge-violating decay: −0.272 → +0.228 MeV⁻¹, rate × 0.70 by frame, so Lorentz invariance demands Σηₙqₙ = 0, i.e. charge conservation.

HardA vector boson of mass M = 1.0 GeV moves along z with E = 20 GeV. Split its longitudinal polarisation into k/M plus a remainder, then evaluate ε_L⋅J for a conserved current along z with J⁰ = 1 (any unit) and for the non-conserved J = (1, 0, 0, 0). Explain what happens as M → 0 and why the massless little group cannot keep this third state.

|k| = √(E² − M²) = √399 GeV = 19.975 GeV, and ε_L = (|k|, 0, 0, E)/M = (19.975, 0, 0, 20.000). Checks: ε_L² = (399 − 400)/1 = −1 and k⋅ε_L = (E|k| − |k|E)/M = 0.
ε_L − k/M = ((|k| − E)/M)(1, 0, 0, −1). As E − |k| = M²/(E + |k|), the remainder is (M/(E + |k|))(−1, 0, 0, 1) = 0.02502 × (−1, 0, 0, 1): ε_L is k/M, of size 20, plus a piece of size 0.025.
Conserved current along z: k⋅J = EJ⁰ − |k|J³ = 0 gives J³ = (E/|k|)J⁰ = 1.00125. Then ε_L⋅J = (|k|J⁰ − EJ³)/M = −J⁰M/|k| = −0.0501: only the remainder contributes, since k⋅J = 0.
Non-conserved J = (1, 0, 0, 0): k⋅J = 20 GeV and ε_L⋅J = |k|J⁰/M = 19.975, dominated by the k/M term. The two differ by |k|²/M² = 399, so the rates differ by 1.59 × 10⁵.
As M → 0 the conserved-current coupling falls as M/|k| and the longitudinal mode decouples smoothly; the non-conserved one grows as |k|/M without limit. As M → 0, Mε_L → k: the would-be third state points along k, the very shift the null rotations generate, null and of helicity 0, so ISO(2) keeps only ±1.

Answerε_L = k/M + 0.0250(−1, 0, 0, 1); ε_L⋅J = −0.0501 for the conserved current and 19.975 for the non-conserved one (rates 1.59 × 10⁵ apart). Only a conserved current gives a smooth M → 0 limit.

The little group of a null momentum, written as matrices

Helicity states shift, and no vector can stand still

The photon field is a vector only up to a gauge term

Lorentz invariance of S now demands kμMμ = 0

Weinberg's soft photons turn the test into charge conservation

Where the argument stops, and the price of a covariant Aμ

Extra challenging questions

Lorentz invariance of S now demands k_μM^μ = 0

Where the argument stops, and the price of a covariant A^μ