University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.8

Proca field: propagator and three polarisations

Proca's field is the control experiment for the photon. Give the vector a mass and every device of this unit, from gauge fixing to negative-norm states and the Gupta–Bleuler condition, becomes unnecessary; the cost is a third polarisation. Watching which results survive M_V → 0 shows what masslessness actually buys QED.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Add ½M_V² A_μA^μ to Maxwell's Lagrangian and the gauge symmetry is gone, and with it the trouble it caused. The field equation ∂_μF^μν + M_V²A^ν = j^ν has an antisymmetric first term, so its divergence leaves M_V² ∂⋅A = ∂⋅j: for a conserved current the Lorenz condition is an equation of motion rather than a choice, and every component obeys the Klein–Gordon equation with mass M_V. A⁰ still has no conjugate momentum, but its own equation now solves for it in terms of E and the charge density, so three canonical pairs remain: three polarisations, each spacelike with ε⋅ε* = −1 and of positive norm, a positive Hamiltonian, and no condition on the states.

The kinetic operator is invertible with no gauge fixing, which gives the propagator −i(g_μν − k_μk_ν/M_V²)/(k² − M_V² + iε) and, in the static limit, Yukawa's potential of range ħ/(M_V c). The price sits in two places: the longitudinal polarisation, whose components grow as |k|/M_V, and the k_μk_ν/M_V² term, which does not fall at large k. Against a conserved current both are harmless, because k_μk_ν drops out and the longitudinal coupling falls as M_V/E; the limit M_V → 0 is then smooth, and a small photon mass becomes a legitimate infrared regulator.

Coupled to a current that is not conserved, the same pieces grow with energy and the model stops being predictive, which is a problem for a later course.

Simple definition: The Proca field is a real vector field with Lagrangian −¼F_μνF^μν + ½M_V²A_μA^μ. It describes a spin-one particle of mass M_V with three polarisation states, all of positive norm, and it has no gauge freedom to fix.
Example: A Z boson (M_V = 91.19 GeV) moving along z with |k| = 200 GeV has E = 219.8 GeV and longitudinal polarisation ε_L = (2.193, 0, 0, 2.410). Then ε_L⋅ε_L ≈ −1 and k⋅ε_L = 0, yet both nonzero components exceed 2.

Divergence of Proca's equationM_V² ∂_μA^μ = ∂_μ j^μ

A conserved current forces ∂⋅A = 0 and leaves (□ + M_V²)A^ν = j^ν. The Lorenz condition is an equation of motion here, so no gauge is ever fixed.

From ∂_μF^μν + M_V²A^ν = j^ν, since ∂_ν∂_μF^μν = 0. A^μ, j^μ: mass dimension 1 and 3; M_V in GeV (M_V c² in J)

Longitudinal polarisation, k along zε_L^μ = (|k|, 0, 0, E)/M_V

At rest it is the unit vector along z. For E ≫ M_V it approaches k^μ/M_V, with a remainder M_V/(E + |k|) times (−1, 0, 0, 1).

E = √(|k|² + M_V²), all in GeV; ε_L is dimensionless, with ε_L⋅ε_L = −1 and k⋅ε_L = 0

Sum over the three polarisationsΣ_λ ε_μ(k, λ) ε_ν*(k, λ) = −g_μν + k_μk_ν/M_V²

Exact for every k, with no Ward identity needed. It replaces the photon's −g_μν of topic 8.07, and its k_μk_ν piece drops only against a conserved current.

λ = +, −, L with k² = M_V²; dimensionless. Its trace with g^μν is −3, one unit per state

Proca propagatorD_μν(k) = −i(g_μν − k_μk_ν/M_V²)/(k² − M_V² + iε)

The inverse of the kinetic operator with no gauge fixing. Its residue at k² = M_V² is i times the polarisation sum, and its k_μk_ν term does not fall at large k.

k: exchanged four-momentum in GeV; D has mass dimension −2, in GeV⁻²

Static potential between two chargesV(r) = q₁q₂ exp(−M_V r)/(4πr)

Like charges repel. M_V → 0 returns Coulomb's q₁q₂/(4πr), and at r = 1/M_V the exponential has already cut V to 37% of it.

q: Heaviside–Lorentz charges, dimensionless; r in fm and V in fm⁻¹ (ħcV in MeV), with range ħ/(M_V c) = (197.327 MeV fm)/(M_V c²)

Longitudinal coupling to a conserved currentε_L⋅J = −(M_V/E) J³

The large components of ε_L cancel against each other. The longitudinal state decouples as M_V/E, which is why the massless limit is smooth.

k along z and k⋅J = 0, so J⁰ = (|k|/E)J³; M_V and E in GeV, and J carries the units of the vertex that produces it

A mass makes the Lorenz condition an equation of motion

Work in natural units with the metric (+, −, −, −) and start from L = −¼F_μνF^μν + ½M_V²A_μA^μ − j_μA^μ. Varying A_ν gives ∂_μF^μν + M_V²A^ν = j^ν (derived). Now take ∂_ν of both sides. F^μν is antisymmetric and ∂_ν∂_μ is symmetric, so the first term vanishes identically and M_V² ∂_νA^ν = ∂_νj^ν remains. In Maxwell's theory the same step gave only 0 = ∂⋅j, a condition on the source that says nothing about A, which is why topic 8.04 had to choose a gauge. Here a conserved current forces ∂⋅A = 0, and since ∂_μF^μν = □A^ν − ∂^ν(∂⋅A), the field equation collapses to (□ + M_V²)A^ν = j^ν: four Klein–Gordon equations tied by one constraint. Dimensions check: A has mass dimension 1, so M_V²A_μA^μ has dimension 4, like F_μνF^μν. Nothing was imposed by hand, and no transformation A → A + ∂Λ survives, because the mass term is not invariant under it.

A⁰ is solved for, so three canonical pairs remain

At first nothing changes canonically: L contains no ∂₀A⁰, so π⁰ = 0 just as in topic 8.01, and the momenta of the spatial components are the electric field up to sign. The difference lies in the ν = 0 field equation. For Maxwell it was Gauss's law, ∇⋅E = j⁰, a constraint on the canonical data that generated gauge transformations and left A⁰ free. With a mass it reads ∇⋅E + M_V²A⁰ = j⁰, so A⁰ = (j⁰ − ∇⋅E)/M_V² is fixed at each instant by the other fields (derived). The independent pairs are (A^i, Eⁱ), three of them, and eliminating A⁰ gives the free Hamiltonian H = ½∫d³x [E² + B² + M_V²(A⁰)² + M_V²|A|²], a sum of squares (derived in outline). No state has negative norm or negative energy, so there is no Gupta–Bleuler condition to impose. With a source the same elimination adds a contact term (j⁰)²/(2M_V²) to the interaction Hamiltonian, and it cancels the non-covariant piece of the time-ordered product: topic 2.04's T* remark at work.

Three polarisations, every one of positive norm

A plane wave ε^μ e^(−ik⋅x) solves the free equations when k² = M_V² and k⋅ε = 0: four components and one condition leave three solutions. In the rest frame, k = (M_V, 0, 0, 0), they are the three spatial unit vectors, the three states of spin one. Boost along z to momentum |k| and energy E. The transverse pair ε_± = ∓(0, 1, ±i, 0)/√2 is unchanged, and the longitudinal vector becomes ε_L = (|k|, 0, 0, E)/M_V (derived). All three are spacelike with ε⋅ε* = −1, so expanding A^μ in a(k, λ) and a†(k, λ) and imposing the canonical commutators gives (in outline) [a(k, λ), a†(k′, λ′)] = (2π)³ 2E δ_λλ′ δ³(k − k′) with a plus sign for every λ. Compare topic 8.02, where four a_μ obeyed a commutator proportional to −g_μν and the timelike one had negative norm. The cost appears at high energy. A Z boson with |k| = 200 GeV has ε_L = (2.193, 0, 0, 2.410), components above 2 although ε_L⋅ε_L = −1, because ε_L = k/M_V + (M_V/(E + |k|))(−1, 0, 0, 1) and the first term grows without limit.

Summing the three gives −g_μν + k_μk_ν/M_V²

Rates need Σ_λ ε_μ ε_ν*. In the rest frame the three spatial unit vectors sum to diag(0, 1, 1, 1) in lower-index components. The only tensor built from g_μν and k_μ that reduces to this is −g_μν + k_μk_ν/M_V², because at rest k_μk_ν/M_V² = diag(1, 0, 0, 0) and −g_μν = diag(−1, 1, 1, 1). Both sides are tensors, so the identity holds in every frame (derived). Three checks follow. Contracting with k^μ gives −k_ν + k_ν = 0, as k⋅ε = 0 demands. The trace, g^μν(−g_μν + k_μk_ν/M_V²) = −4 + 1 = −3, counts three states with ε⋅ε* = −1. And the tensor is dimensionless. Now set it beside topic 8.07. There two transverse photons summed to −g_μν only after the Ward identity removed terms in k_μ and a reference vector; here the sum is exact, needs no identity, and keeps its own k_μk_ν piece, which leaves a rate only when the vector couples to a conserved current. A decay divides this sum by the three spin states it counts, not by two, and the Hard example shows what the wrong divisor would do to Z → νν̄.

Inverting the operator: propagator and Yukawa potential

Integrate by parts to write the free action as ½A_μK^μνA_ν with, in momentum space, K^μν = (M_V² − k²)g^μν + k^μk^ν. In topic 8.04's projectors, P_T^μν = g^μν − k^μk^ν/k² and P_L^μν = k^μk^ν/k², this is K = (M_V² − k²)P_T + M_V²P_L. For Maxwell the longitudinal coefficient was zero and K had no inverse until a gauge-fixing term supplied one; here M_V² does that job. Inverting each projector and multiplying by i gives D_μν = −i(g_μν − k_μk_ν/M_V²)/(k² − M_V² + iε) (derived). At the pole the numerator is i times the polarisation sum, as unitarity requires, while at large k the k_μk_ν/M_V² piece tends to a constant instead of falling as 1/k². For static charges the ν = 0 equation is (−∇² + M_V²)A⁰ = ρ, solved by A⁰ = q exp(−M_V r)/(4πr), so V(r) = q₁q₂ exp(−M_V r)/(4πr) (derived), repulsive between like charges. The range ħ/(M_V c) is 2.46 × 10⁻³ fm for the W, and the measured bound on the photon mass, below 1 × 10⁻¹⁸ eV (quoted), puts its range beyond 2.0 × 10¹¹ m, about 1.3 AU.

Against a conserved current the massless limit is smooth

Exchange between two currents reads J₁^μD_μνJ₂^ν. If both are conserved, k⋅J = 0 removes the k_μk_ν term and leaves −iJ₁⋅J₂/(k² − M_V² + iε), which becomes Feynman-gauge photon exchange as M_V → 0 (derived). Emission works the same way. With k along z, k⋅J = 0 sets J⁰ = (|k|/E)J³, and then ε_L⋅J = −(M_V/E)J³ (derived). For a Z with |k| = 300 GeV the two terms of ε_L⋅J are 3.1476J³ and −3.4385J³, and they cancel to −0.291J³. So the longitudinal state decouples as M_V/E, the three-state sum acts on conserved currents exactly like the photon's −g_μν, and −ig_μν/(k² − μ_γ² + iε) is a legitimate infrared regulator: Units 12 to 15 use μ_γ to tame soft photons and show that it drops out of inclusive rates. The smoothness is conditional. If k⋅J ≠ 0, the k_μk_ν/M_V² term grows like E²/M_V² and ε_L⋅J like E/M_V, amplitudes with longitudinal vectors outgrow unitarity, and the model loses predictive power at high energy.

Change one variable at a time

Make the relationship visible.

Interactive model

Vector mass M_V20 GeV

Non-conservation δ0.020

Momentum |k|200 GeV

Drag M_V down towards 1 GeV: the solid curve slides left, so a conserved current decouples the longitudinal state at every fixed |k| and the massless limit is smooth. Then raise δ to 0.05 and watch the dashed curve climb as M_V falls.

ENERGY E201.0 GeV

COMPONENT ε_L⁰ = |k|/M_V10.00

CONSERVED M_V/E0.0995

WITH δ: |ε_L⋅J|/|J³|0.299

Live interpretationENERGY E: 201.0 GeV. COMPONENT ε_L⁰ = |k|/M_V: 10.00. CONSERVED M_V/E: 0.0995. WITH δ: |ε_L⋅J|/|J³|: 0.299

Catch the common trap

Explain before calculating.

Misconception check

The tempting conclusion is that a massive vector is dominated at high energy by its longitudinal state, because ε_L grows as |k|/M_V, and so the massless limit must be singular: a photon with any mass at all would carry a third polarisation coupling at full strength, raising Planck's radiation law by a factor 3/2 and changing every QED rate. The growth is real in the components and absent from the amplitude. Write ε_L = k/M_V + O(M_V/E). The large piece is proportional to k^μ, and k_μJ^μ = 0 for a conserved current, so what survives is ε_L⋅J = −(M_V/E)J³. Longitudinal photons of mass μ_γ are emitted and absorbed at rates suppressed by (μ_γ/E)², of order 10⁻³⁷ for a visible photon at the present bound, so they are effectively never produced and Planck's law with two polarisations stands. The growth matters only where the current is not conserved.

A vector boson of mass M_V = 10 GeV moves along z with |k| = 100 GeV and is emitted by a conserved current, k_μJ^μ = 0. Its longitudinal polarisation ε_L = (|k|, 0, 0, E)/M_V has components close to 10. What is the longitudinal emission amplitude ε_L⋅J, in terms of the current component J³?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA vector meson of mass 782.66 MeV is exchanged between two nucleons, each carrying the same vector charge with g²/(4π) = 10. Find the range of the force and the potential at separations of 0.50 fm and 1.00 fm, and say whether it attracts or repels.

Restore units for the range: λ = ħ/(M_V c) = ħc/(M_V c²) = 197.327 MeV fm ÷ 782.66 MeV = 0.2521 fm. Dimensions check: MeV fm over MeV leaves fm.
With ħc restored the static potential is V(r) = (g²/4π) ħc exp(−r/λ)/r, positive because vector exchange between like charges repels.
At r = 0.50 fm: r/λ = 1.983 and exp(−1.983) = 0.1376, so V = 10 × 197.327 MeV fm ÷ 0.50 fm × 0.1376 = 543 MeV.
At r = 1.00 fm: r/λ = 3.966 and exp(−3.966) = 0.01894, so V = 10 × 197.327 MeV fm ÷ 1.00 fm × 0.01894 = 37.4 MeV. Doubling r cut V by a factor 14.5, where 1/r alone would give 2.
Limit check: as M_V → 0 the exponential goes to 1 and V returns to the Coulomb form (g²/4π)ħc/r.

AnswerRange 0.252 fm; V = +543 MeV at 0.50 fm and +37.4 MeV at 1.00 fm. The force is repulsive and short-ranged.

MediumA Z boson (M_V = 91.19 GeV) travels along z with |k| = 300 GeV. Write its longitudinal polarisation, check it, and find ε_L⋅J for a conserved current and for the non-conserved current J = (0, 0, 0, J³), in units of J³.

E = √(300² + 91.19²) GeV = 313.55 GeV.
ε_L = (|k|, 0, 0, E)/M_V = (3.2898, 0, 0, 3.4385). Checks: ε_L⋅ε_L = 3.2898² − 3.4385² = −1.00, and k⋅ε_L = (E|k| − |k|E)/M_V = 0.
A conserved current has k⋅J = E J⁰ − |k|J³ = 0, so J⁰ = (|k|/E)J³ = 0.95678J³.
ε_L⋅J = ε_L⁰J⁰ − ε_L³J³ = (3.2898 × 0.95678 − 3.4385)J³ = (3.1476 − 3.4385)J³ = −0.291J³, which is −(M_V/E)J³ = −(91.19/313.55)J³.
For J = (0, 0, 0, J³), k⋅J = −|k|J³ ≠ 0 and nothing cancels: ε_L⋅J = −3.4385J³, larger by E²/M_V² = 11.8.

Answerε_L = (3.290, 0, 0, 3.438); ε_L⋅J = −0.291J³ for the conserved current and −3.44J³ for the non-conserved one.

HardTake the Zνν̄ vertex as −i(g_Z/4)γ^μ(1 − γ⁵) with g_Z² = 4√2 G_F M_Z² (quoted), G_F = 1.16638 × 10⁻⁵ GeV⁻² and M_Z = 91.1876 GeV, with massless neutrinos. Find Γ(Z → νν̄) for one flavour at tree level, and the number of flavours implied by the measured invisible width, 499.0 ± 1.5 MeV.

Spin sum by Advanced QM's trace identities: Σ|M|² = (g_Z²/8)Tr[p̸₁γ^μp̸₂γ^ν(1 − γ⁵)]ε_με_ν* = (g_Z²/2)[p₁^μp₂^ν + p₁^νp₂^μ − g^μν p₁⋅p₂]ε_με_ν*, plus an antisymmetric γ⁵ piece that vanishes against the symmetric polarisation sum.
Sum the three polarisations with −g_μν + k_μk_ν/M_Z². With k = p₁ + p₂ and p₁² = p₂² = 0, k_μ[p₁^μp₂^ν + p₁^νp₂^μ − g^μν p₁⋅p₂] = (p₁⋅p₂)(p₁ + p₂ − k)^ν = 0: the current is conserved and the k_μk_ν term drops.
The −g_μν contraction gives 2p₁⋅p₂ = M_Z², so the sum is g_Z²M_Z²/2 = 0.5486 × 8315.2/2 GeV² = 2281 GeV², and averaging over three spin states gives 760.3 GeV².
Two-body rate with |p| = M_Z/2: Γ = |p|⟨|M|²⟩/(8πM_Z²) = g_Z²M_Z/(96π) = G_F M_Z³/(12√2 π) = 1.16638 × 10⁻⁵ × 7.582 × 10⁵ ÷ 53.31 GeV = 0.1659 GeV. Dimensions: GeV⁻² × GeV³ = GeV.
N_ν = 499.0 MeV ÷ 165.9 MeV = 3.01 ± 0.01, from the experimental error alone; radiative corrections omitted here shift Γ by about 1%. Dividing the three-state sum by two, the photon's count, would give 248.8 MeV and N_ν = 2.0: the average must run over the same three states the sum includes, so the width measures N_ν, not the number of polarisations.

AnswerΓ(Z → νν̄) ≈ 166 MeV per flavour at tree level, so N_ν = 3.01 ± 0.01: three light neutrinos.