University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.1

The Maxwell Lagrangian and the missing momentum of A⁰

Advanced QM quantised the photon only after Coulomb gauge had discarded two of A^μ's four components. Here you find out why something had to go: write down Maxwell's Lagrangian, try to quantise all four components the way Advanced QM quantised a scalar, and watch the step fail at A⁰, whose momentum is identically zero.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The Maxwell Lagrangian, −¼F_μνF^(μν) − j_μA^μ, is the simplest local Lagrangian for a four-vector field whose action is unchanged by A_μ → A_μ + ∂_μΛ, and its Euler-Lagrange equations are Maxwell's. Almost everything follows from the antisymmetry of F_μν. Applied to the field equation, it makes ∂_νj^ν = 0 a condition for any solution to exist.

Applied to the Lagrangian, it removes the velocity ∂₀A₀: the momentum conjugate to Aᵢ is the electric field, but the momentum conjugate to A₀ is zero. A⁰ is then a Lagrange multiplier, and its equation, Gauss's law, constrains the fields at one instant instead of evolving them. Gauge invariance makes the evolution ambiguous, because a gauge function switched on later changes A without touching the initial data, and in momentum space the kinetic operator has a zero mode and no inverse.

So the recipe that quantised the scalar field, one commutator per component, fails twice for A_μ: there is no Π⁰ to pair with A⁰, and no unique evolution to quantise. Coulomb gauge, quoted from Advanced QM, solves the constraint and leaves two transverse oscillators per wavevector, the two polarisations the blackbody spectrum counts, at the cost of manifest Lorentz covariance. The analysis is classical and canonical: it stops before quantisation, naming without using Dirac's theory of constraints and the change of Lagrangian that topic 8.02 makes.

Simple definition: The momentum conjugate to A₀ in the Maxwell Lagrangian is identically zero, because ∂₀A₀ appears only in F₀₀ = 0; A⁰ is therefore not a dynamical coordinate but a Lagrange multiplier enforcing Gauss's law, a constraint on the fields at one instant.
Example: In Coulomb gauge, −∇²A⁰ = ρ fixes a proton's potential at every instant from its charge alone, with no ∂ₜA⁰ ever supplied as initial data: at the Bohr radius the electron's energy in it is −α/a₀ = −27.2 eV, twice hydrogen's 13.6 eV binding.

Maxwell's equations from the Lagrangian (derived)ℒ = −¼F_μνF^(μν) − j_μA^μ ⟹ ∂_μF^(μν) = j^ν

From ∂ℒ/∂(∂_μA_ν) = −F^(μν) and ∂ℒ/∂A_ν = −j^ν. ν = 0 is ∇⋅E = ρ; ν = i is ∇ × B − ∂ₜE = j. ∇⋅B = 0 and Faraday's law hold for any F built from A.

F_μν = ∂_μA_ν − ∂_νA_μ. Mass dimensions [A] = 1, [F] = 2, [j] = 3, [ℒ] = 4 (GeV powers); Heaviside-Lorentz, α = e²/(4π).

Conservation as a consistency condition (derived)∂_νj^ν = ∂_ν∂_μF^(μν) = 0

A precondition, not a consequence: a source with ∂⋅j ≠ 0 cannot be coupled to A_μ at all. For a conserved source it also keeps Gauss's law true once it holds at t = 0.

j^μ = (ρ, j) in GeV³; ∂_ν∂_μ is symmetric in μν and F^(μν) antisymmetric; ∂_νj^ν = ∂ₜρ + ∇⋅j.

Canonical momenta of the four potentials (derived)Π^μ = ∂ℒ/∂(∂₀A_μ) = −F^(0μ), so Πⁱ = Eⁱ and Π⁰ = 0

Π⁰ vanishes identically, before any gauge is chosen, so [A₀(x, t), Π⁰(y, t)] = iδ³(x − y) cannot be imposed: A⁰ is not a canonical coordinate.

Π in GeV² (dimension 2, as E). For the Cartesian Aⁱ = −Aᵢ the momentum is −E^i, UAQM-11.03's −ε₀E⊥.

Gauss's law in potentials is a constraint (derived)−∇²A⁰ − ∂ₜ(∇⋅A) = ρ

It relates the fields at one instant and evolves nothing. In Coulomb gauge (UAQM-11.01, quoted) it fixes A⁰ as the instantaneous Coulomb potential of ρ.

A⁰ and A in GeV, ρ in GeV³. No second time derivative of A⁰, or of any A^i, appears.

The Hamiltonian, with A⁰ as a multiplier (derived)H = ∫d³x [½(E² + B²) − j⋅A − A⁰(∇⋅E − ρ)]

A⁰ enters linearly, so no equation fixes ∂ₜA⁰. Keeping Π⁰ = 0 needs ∂ₜΠ⁰ = −δH/δA⁰ = ∇⋅E − ρ = 0, Gauss's law; on that surface H is blind to A⁰.

H in GeV; E = Π and B = ∇ × A; one integration by parts, the surface term dropped.

The pure-gauge direction is a zero mode of K (derived)K^(μν)(k) k_ν = 0, with K^(μν) = −k²g^(μν) + k^μ k^ν

Because A_μ → A_μ + ∂_μΛ leaves F unchanged. K has no inverse, so A has no propagator yet, and k_μK^(μν) = 0 forces k_μj̃^μ = 0.

Maxwell's equation for A_ν ∝ e^(−ik⋅x) is K^(μν)Ã_ν = j̃^μ; k in GeV, K in GeV².

Maxwell's equations from −¼F² − j⋅A

Natural units ħ = c = 1, the metric (+, −, −, −) and topic 1.01's Heaviside-Lorentz charge, α = e²/(4π), hold throughout. With A^μ = (A⁰, A), the tensor F_μν = ∂_μA_ν − ∂_νA_μ has F⁽ⁱ⁰⁾ = Eⁱ and F^(ij) = −ε^(ijk)B^k, where E = −∇A⁰ − ∂ₜA and B = ∇ × A, so ℒ = ½(E² − B²) − ρA⁰ + j⋅A. Varying A_ν, ∂ℒ/∂(∂_μA_ν) = −F^(μν), the ¼ cancelled by the four places ∂_μA_ν sits in F_μνF^(μν), and ∂ℒ/∂A_ν = −j^ν, so the Euler-Lagrange equation is ∂_μF^(μν) = j^ν, derived. Its ν = 0 component is Gauss's law, ∇⋅E = ρ, and its ν = i components are ∇ × B − ∂ₜE = j. The other two Maxwell equations need no Lagrangian, since they hold for any F built from A. Dimensions: [A] = 1 and [F] = 2 make [ℒ] = 4 and [j] = 3, a charge per volume. Static limit: Gauss's law on a sphere centred on a point charge e, where spherical symmetry makes E radial and equal all over the sphere, gives E = e/(4πr²), and the electron-proton energy −e²/(4πr) = −α/r is hydrogen's.

Antisymmetry makes conservation a condition

Take ∂_ν of the field equation. On the left, ∂_ν∂_μF^(μν) vanishes identically, because the derivatives commute while F^(μν) changes sign when μ and ν swap, so every solution requires ∂_νj^ν = ∂ₜρ + ∇⋅j = 0, derived. This is not Noether's theorem delivering a current from a symmetry of the matter; it is a precondition, and a source that is not conserved cannot be coupled to A_μ at all. Try one: a charge e switched on at the origin at t = 0 with no current flowing in. Gauss's law needs a flux e through every sphere about the origin at once, which spherical symmetry makes a radial E = e/(4πr²), 1.44 × 10⁻⁹ V m⁻¹ already at 1 m in SI, while Ampère's law with j = 0 gives ∂ₜ(∇⋅E) = ∇⋅(∇ × B) = 0; no field does both. The same identity means the ν = 0 equation never has to be solved twice: ∂ₜ(∇⋅E − ρ) = ∇⋅(∇ × B − j) − ∂ₜρ = −(∂ₜρ + ∇⋅j). For a conserved current, Gauss's law true at t = 0 is kept true by the other three equations. It is a condition on the initial data.

The electric field is a momentum; A⁰ has none

Canonical quantisation needs a momentum for every coordinate. From the same derivative, Π^μ = ∂ℒ/∂(∂₀A_μ) = −F^(0μ), derived. For the spatial components Πⁱ = −F⁽⁰ⁱ⁾ = Eⁱ: the electric field is the momentum conjugate to Aᵢ, or −Eⁱ for the Cartesian Aⁱ = −Aᵢ, UAQM-11.03's −ε₀E⊥ in SI. For the time component Π⁰ = −F⁽⁰⁰⁾ = 0, identically. The velocity ∂₀A₀ enters ℒ only through F₀₀ = ∂₀A₀ − ∂₀A₀, so the matrix of second derivatives of ℒ in the four velocities ∂₀A_μ is diag(0, 1, 1, 1), singular, and no Legendre transform can recover ∂₀A₀ from the momenta. The canonical rule [A₀(x, t), Π⁰(y, t)] = iδ³(x − y) would set zero equal to iδ³: no commutator can be imposed on A⁰. Each component of a scalar multiplet, by contrast, had Π = ∂₀φ in Advanced QM's Unit 9. Where the rule does work the dimensions agree, [Aᵢ] + [Πⁱ] = 1 + 2 = 3 = [δ³]. And no gauge has been chosen yet: Π⁰ = 0 is a property of the Lagrangian.

Gauss's law constrains, and A⁰ only enforces it

In potentials the ν = 0 equation reads −∇²A⁰ − ∂ₜ(∇⋅A) = ρ, derived. It holds no second time derivative of anything, so it relates the fields at one instant and moves nothing forward: a constraint, not an evolution. The Hamiltonian shows what A⁰ does. With ∂₀Aᵢ = Eⁱ + ∂ᵢA⁰, ℋ = Πⁱ∂₀Aᵢ − ℒ = ½(E² + B²) + E⋅∇A⁰ + ρA⁰ − j⋅A, and one integration by parts gives H = ∫d³x [½(E² + B²) − j⋅A − A⁰(∇⋅E − ρ)], derived. A⁰ enters linearly, multiplying Gauss's law: it is a Lagrange multiplier. No Hamilton equation fixes ∂ₜA⁰, since Π⁰ is absent from H, while keeping Π⁰ = 0 in time needs ∂ₜΠ⁰ = −δH/δA⁰ = ∇⋅E − ρ = 0, which is Gauss's law again. On that surface H is blind to A⁰, and with no charges it is ½∫(E² + B²) d³x, UAQM-11.03's radiation energy. Dirac's theory of constrained systems calls Π⁰ ≈ 0 the primary constraint and Gauss's law the secondary one, both first class; it is named here only.

A gauge function switched on later leaves A undetermined

Let A_μ solve the field equations, and let Λ vanish for t ≤ t₀ but be any smooth function afterwards. A′_μ = A_μ + ∂_μΛ has the same F_μν, so it is also a solution, with the same A_μ and ∂₀A_μ at t₀, derived: identical initial data, different futures, as the figure plots. The field equations are not an evolution for A_μ. In momentum space the equation for A_ν ∝ e^(−ik⋅x) is K^(μν)Ã_ν = j̃^μ with K^(μν) = −k²g^(μν) + k^μ k^ν, and K^(μν)k_ν = −k² k^μ + k^μ k² = 0. The pure-gauge direction is a zero mode of K, K has no inverse and A no propagator, derived; contracting with k_μ returns 0 = k_μj̃^μ. Mass is the contrast. Adding ½M²A_μA^μ leaves Π⁰ = 0 but turns Gauss's law into ∇⋅E + M²A⁰ = ρ, which fixes A⁰; only as M → 0 is A⁰ set free. So gauge invariance, not Π⁰ = 0 alone, frees A⁰, and the measured photon mass, below 10⁻¹⁸ eV (PDG, quoted), says nature takes M = 0.

Coulomb gauge pays with covariance; two polarisations remain

Count first. Of A^μ's four components, A⁰ has no momentum and is fixed at each instant by Gauss's law once a gauge is chosen, and one more combination is pure gauge, so two degrees of freedom per wavevector survive. In phase space, eight variables per point less two constraints and two gauge conditions leave four: two coordinates and two momenta. Measurement agrees. Blackbody radiation with two polarisations has, quoted, u = (π²/15)T⁴ with k_B = 1, the Stefan-Boltzmann constant 5.670 × 10⁻⁸ W m⁻² K⁻⁴, and 4.17 × 10⁻¹⁴ J m⁻³ in the 2.7255 K microwave background; four independent oscillators would double both. Coulomb gauge (UAQM-11.01, 11.03), quoted, makes the count explicit: ∇⋅A = 0 turns the constraint into −∇²A⁰ = ρ, so A⁰ is the instantaneous Coulomb potential, which alone gives hydrogen's 13.598 eV with the reduced mass, and the two transverse components are quantised with the transverse delta function. The price is manifest covariance, since ∇⋅A = 0 singles out a frame. The covariant cure has to change ℒ so that ∂₀A⁰ appears in it, topic 8.02's job.

Change one variable at a time

Make the relationship visible.

Interactive model

Final gauge amplitude ℓ0.6 ρ₀/k³

Switch-on time kt₀2.0

Ramp duration kΔt2.0

Set ℓ to zero for Coulomb gauge, then raise it: A⁰ bulges, A^z is left shifted for good, and the E_z strip never moves. Halve the ramp time and the bulge doubles. Nothing objects, because A⁰ has no momentum; temporal gauge would be the endless ramp ℓ = −kt.

PEAK δA⁰ = ∂ₜΛ0.47 ρ₀/k²

FINAL δA^z = −∂_zΛ0.60 ρ₀/k²

Π³ = E_z AT THE PEAK1.00 ρ₀/k

Live interpretationPEAK δA⁰ = ∂ₜΛ: 0.47 ρ₀/k². FINAL δA^z = −∂_zΛ: 0.60 ρ₀/k². Π³ = E_z AT THE PEAK: 1.00 ρ₀/k

Catch the common trap

Explain before calculating.

Misconception check

The tempting diagnosis is that Π⁰ = 0 is an artefact of Coulomb gauge, and that in Lorenz gauge, where Maxwell's equations become □A^ν = j^ν, four wave equations, each component can be quantised like a Klein-Gordon field. The momentum comes from the Lagrangian, not from the equations of motion. −¼F_μνF^(μν) contains ∂₀A₀ only inside F₀₀, which is zero by antisymmetry before any gauge is chosen, so Π⁰ = −F⁽⁰⁰⁾ = 0 in every gauge. Imposing ∂_μA^μ = 0 restricts the solutions without touching ℒ, so it cannot supply a momentum. The cure must change the Lagrangian itself, which is topic 8.02's job, and even then the Lorenz condition cannot hold as an operator identity: the momentum the new ℒ gives A⁰ is −∂_μA^μ itself, and setting it to zero would undo the cure.

With ℒ = −¼F_μνF^(μν) and metric (+, −, −, −), at one point and instant ∂ₜA⁰ = 0.50 GeV², ∂ₜA¹ = 0.20 GeV² and ∂₁A⁰ = 0.10 GeV², every other first derivative being zero. What are the momenta Π⁰ and Π¹ conjugate to A₀ and A₁?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWith ℒ = −¼F_μνF^(μν) and no charges, at one point and instant ∂ₜA⁰ = 0.80 keV², ∂ₜA = (0.30, 0, −0.10) keV², ∇A⁰ = (−0.10, 0.20, 0.40) keV² and ∇ × A = (0, 0.25, 0) keV². Find E, Π⁰ and Πⁱ (conjugate to Aᵢ), ℒ and the energy density in J m⁻³. Then a gauge function with ∇(∂ₜΛ) = (0.05, 0, 0) keV² and ∂ₜ²Λ = −2.00 keV² there is applied. What changes?

E = −∇A⁰ − ∂ₜA = (0.10 − 0.30, −0.20 − 0, −0.40 + 0.10) = (−0.20, −0.20, −0.30) keV², and B = (0, 0.25, 0) keV². Restored, |E| = 0.41 keV² is 6.3 × 10¹¹ V m⁻¹, an atomic-scale field.
Π^μ = −F^(0μ): Πⁱ = Eⁱ = (−0.20, −0.20, −0.30) keV², and Π⁰ = −F⁽⁰⁰⁾ = 0. The 0.80 keV² of ∂ₜA⁰ appears in neither, nor anywhere in ℒ.
E² = 0.04 + 0.04 + 0.09 = 0.170 keV⁴ and B² = 0.0625 keV⁴, so ℒ = ½(E² − B²) = 0.0538 keV⁴ and the energy density is ½(E² + B²) = 0.116 keV⁴. With 1 keV⁴ = 2.085 × 10¹³ J m⁻³, that is 2.42 × 10¹² J m⁻³.
The canonical density Πⁱ∂₀Aᵢ − ℒ adds E⋅∇A⁰ = 0.02 − 0.04 − 0.12 = −0.140 keV⁴, giving −0.024 keV⁴ here. That extra term is ∇⋅(A⁰E) − A⁰∇⋅E: a surface term plus the multiplier term, which Gauss's law sets to zero with no charge. Only the integral is the energy.
The gauge change shifts ∇A⁰ by ∇(∂ₜΛ) to (−0.05, 0.20, 0.40) keV² and ∂ₜA by −∇(∂ₜΛ) to (0.25, 0, −0.10) keV², and ∂ₜA⁰ to 0.80 − 2.00 = −1.20 keV². E = (0.05 − 0.25, −0.20, −0.40 + 0.10) = (−0.20, −0.20, −0.30) keV², unchanged.
So Π^i, B, ℒ and the energy density are unchanged, and Π⁰ is still 0. Dimensions: [Π] = 2 = [E], and [Aᵢ] + [Πⁱ] = 3 = [δ³], so the commutator for Aᵢ is consistent; for A₀ it would read 0 = iδ³.

AnswerE = Πⁱ = (−0.20, −0.20, −0.30) keV² and Π⁰ = 0, for any ∂ₜA⁰ and before and after the gauge change; ℒ = 0.0538 keV⁴; energy density 0.116 keV⁴ = 2.42 × 10¹² J m⁻³.

MediumFor one Fourier mode A^μ = ε^μ e^(−ik⋅x), Maxwell's equation reads K^(μν)ε_ν = j̃^μ with K^(μν) = −k²g^(μν) + k^μ k^ν. Take k^μ = (5, 0, 0, 3) GeV. Show that k is a zero mode of K, Kk = 0, although k² ≠ 0, evaluate Kε for ε^μ = (0, 1, 0, 0) GeV and (1, 0, 0, 0) GeV, decide which of the sources j̃^μ = (3, 0, 0, 5) GeV³ and (1, 0, 0, 0) GeV³ can be solved, and find all its solutions and the one with ε⁰ = 0.

k² = 25 − 9 = 16 GeV² and k_ν = (5, 0, 0, −3) GeV. K^(μν)k_ν = −k² k^μ + k^μ (k⋅k) = 0 for every μ; for μ = 0, −16 × 5 + 5 × 16 = 0. The pure-gauge mode ε ∝ k is invisible to K.
Kε = −k²ε^μ + k^μ (k⋅ε). For (0, 1, 0, 0) GeV, k⋅ε = 0 and Kε = (0, −16, 0, 0) GeV³. For (1, 0, 0, 0) GeV, k⋅ε = 5 GeV² and Kε = (−16 + 25, 0, 0, 15) = (9, 0, 0, 15) GeV³, with k_μ(Kε)^μ = 5 × 9 − 3 × 15 = 0. K has eigenvalue −16 GeV² on the three directions with k⋅ε = 0 and 0 on k, so det K = 0.
Contracting Kε = j̃ with k_μ gives 0 = k⋅j̃. For (3, 0, 0, 5) GeV³, k⋅j̃ = 15 − 15 = 0: conserved. For (1, 0, 0, 0) GeV³, k⋅j̃ = 5 GeV⁴: a charge with no current to feed it, which no ε can produce.
Since k⋅j̃ = 0, ε = −j̃/k² = (−0.1875, 0, 0, −0.3125) GeV has k⋅ε = 0 and solves the equation. Adding c k^μ, with c any pure number, gives another solution, so the solutions form a line: one free constant per mode. ε⁰ = 0 needs c = 0.1875/5 = 0.0375, giving ε = (0, 0, 0, −0.200) GeV; check Kε = (0, 0, 0, 3.2) + 0.6 × (5, 0, 0, 3) = (3, 0, 0, 5) GeV³.
The field cannot tell them apart. E = −∇A⁰ − ∂ₜA gives E_z = i(k⁰ε³ − k³ε⁰) = i(−1.5625 + 0.5625) = −1.00i GeV² for the first and i(5 × (−0.200)) = −1.00i GeV² for the second; Gauss's law, i k³E_z = 3 GeV³, matches j̃⁰ for both.

AnswerK k = 0 and det K = 0; Kε = (0, −16, 0, 0) GeV³ and (9, 0, 0, 15) GeV³. Only the conserved source (3, 0, 0, 5) GeV³ is solvable: ε = (−0.1875, 0, 0, −0.3125) GeV + c(5, 0, 0, 3) GeV, with ε = (0, 0, 0, −0.200) GeV at ε⁰ = 0; both give E_z = −1.00i GeV².

HardA static charge density ρ = ρ₀ cos kz, with ρ₀ = 1.2 keV³ and k = 4.0 keV (a period of 0.31 nm), fills space with no current. Find A^μ and E in Coulomb gauge. In temporal gauge, A⁰ = 0 with A = 0 at t = 0, find A^z(t), the Cartesian z component, and its size after 1.0 ns. Give the gauge function linking the two, compare E, B and the mean energy density in J m⁻³, and say what becomes of Gauss's law in temporal gauge.

Coulomb gauge: ∇⋅A = 0 and nothing changes in time, so Gauss's law is −∇²A⁰ = ρ, and A⁰ = a cos kz needs k²a = ρ₀: a = 1.2/16 = 0.075 keV, with A = 0. Then E_z = −∂_zA⁰ = (ρ₀/k) sin kz = 0.30 keV² sin kz, and ∂_zE_z = ρ₀ cos kz = ρ. Dimensions: [ρ₀/k²] = 3 − 2 = 1, a potential. Restored, 0.30 keV² is 4.6 × 10¹¹ V m⁻¹.
Temporal gauge: with A⁰ = 0, E = −∂ₜA, so the same static E_z needs A^z = −(ρ₀/k) t sin kz, growing linearly for ever. After 1.0 ns, t = 1.0 × 10⁻⁹ s ÷ 6.582 × 10⁻¹⁹ keV s = 1.52 × 10⁹ keV⁻¹, and A^z's amplitude is 0.30 × 1.52 × 10⁹ = 4.6 × 10⁸ keV, six billion times the Coulomb-gauge A⁰.
Gauge function: A⁰ + ∂ₜΛ = 0 needs Λ = −(ρ₀/k²) t cos kz = −(0.075 keV) t cos kz. Then A^z → A^z + ∂^zΛ = 0 − ∂_zΛ = −(ρ₀/k) t sin kz, exactly step 2's potential.
Physics agrees. E_z = −∂_zA⁰ − ∂ₜA^z = (ρ₀/k) sin kz in both gauges, and B = ∇ × (A^z ẑ) = 0 because A^z depends on z alone. The energy density averaged over z is ½⟨E_z²⟩ = ¼(ρ₀/k)² = 0.0225 keV⁴, or 4.7 × 10¹¹ J m⁻³, in both; Coulomb gauge's ½⟨ρA⁰⟩ = ρ₀²/(4k²) is the same number.
In temporal gauge Gauss's law reads −∂ₜ(∇⋅A) = ρ. Here ∂_zA^z = −ρ₀t cos kz, so −∂ₜ(∂_zA^z) = ρ₀ cos kz = ρ: satisfied, but as a condition on ∂ₜA at one instant. Ampère's law, here ∂ₜ²A^z = 0, keeps it true only because ∂ₜρ + ∇⋅j = 0. Set A⁰ = 0 in ℒ before varying and this equation is lost; it must then be imposed on the initial data by hand.

AnswerCoulomb: A⁰ = (0.075 keV) cos kz, A = 0. Temporal: A^z = −(0.30 keV²) t sin kz, 4.6 × 10⁸ keV after 1.0 ns. Λ = −(0.075 keV) t cos kz. Both: E_z = (0.30 keV²) sin kz, B = 0, mean energy 0.0225 keV⁴ = 4.7 × 10¹¹ J m⁻³.