University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.2

Gauge fixing, Feynman gauge and photons of negative norm

This is the step Advanced QM only named: quantising all four components of A_μ with Lorentz covariance in plain view. One added term gives A₀ the momentum Maxwell denied it; then g_μν, the only constant tensor covariance offers, hands the timelike photon a negative norm. Watch where that sign lands, and why it cannot be sent anywhere harmless.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 8.01 found that −¼F² gives A₀ no conjugate momentum, so there is nothing to impose a commutator on. The covariant cure changes the Lagrangian: add −(1/(2ξ))(∂⋅A)², a Lorentz scalar that is not gauge invariant. Then π⁰ = −(1/ξ)∂⋅A, the field equation becomes □A^μ − (1 − 1/ξ)∂^μ(∂⋅A) = j^μ, and at ξ = 1, Feynman gauge, each component obeys a massless wave equation, derived.

Imposing [A_μ, π^ν] = iδ_μ^ν δ³ on all four gives [a_μ(k), a_ν†(k′)] = −g_μν(2π)³2|k|δ³(k − k′), derived: three components behave like ordinary scalars and A₀ carries the opposite sign, which no covariant choice can remove, since g_μν and k_μk_ν are the only tensors available. With a_μ|0⟩ = 0 the minus signs in H and in the commutator cancel, so every energy eigenvalue is a sum of positive |k|, but a₀†|0⟩ has negative norm, derived. Declaring a₀† the annihilator instead makes every norm positive and gives each timelike quantum energy −|k| with no floor, UAQM-09.06's dilemma again, and a vector field cannot escape it through anticommutators.

So the four-photon Fock space has an indefinite inner product and is not a Hilbert space. That is where the model stops: covariant bookkeeping, not yet physics, until topic 8.03 finds a subspace of non-negative norm in which timelike and longitudinal photons do nothing, as blackbody radiation, counting two photon states and not four, demands.

Simple definition: Feynman-gauge quantisation adds −½(∂⋅A)² to Maxwell's Lagrangian so that all four components of A_μ have conjugate momenta, then imposes covariant commutators [a_μ(k), a_ν†(k′)] = −g_μν(2π)³2|k|δ³(k − k′), which give the timelike photon a negative norm.
Example: A timelike photon of energy 2.0 eV, from a packet scaled so that |⟨ψ|ψ⟩| = 1, has H eigenvalue +2.0 eV but norm −1, so ⟨ψ|H|ψ⟩ = −2.0 eV; a transverse photon of the same energy has norm +1 and ⟨ψ|H|ψ⟩ = +2.0 eV.

Gauge-fixed Maxwell LagrangianL = −¼F_μνF^μν − (1/(2ξ))(∂_μA^μ)² − j_μA^μ

Lorentz invariant, but gauge invariant only under A_μ → A_μ + ∂_μΛ with □Λ = 0. As ξ → ∞ the added term vanishes and Maxwell's Lagrangian returns.

A^μ in GeV (mass dimension 1), ∂⋅A in GeV², j^μ in GeV³, L in GeV⁴; ξ is a real, dimensionless gauge parameter.

The momentum A₀ was missingπ⁰ = ∂L/∂(∂₀A₀) = −(1/ξ) ∂_μA^μ

Non-zero, so [A₀(x), π⁰(y)] = iδ³(x − y) can be imposed at last. It tends to 0 as ξ → ∞, which is topic 8.01's constraint π⁰ = 0 again.

π⁰ in GeV² (mass dimension 2), conjugate to A₀ in GeV; ξ dimensionless. Derived, from ∂(∂_νA^ν)/∂(∂₀A_μ) = g^(μ0).

Field equation at general ξ□A^μ − (1 − 1/ξ) ∂^μ(∂_νA^ν) = j^μ

At ξ = 1, Feynman gauge, it reads □A^μ = j^μ: four uncoupled massless wave equations. At ξ ≠ 1 the free field obeys □²A^μ = 0 and admits t e^(−ik⋅x) modes.

□ = ∂ₜ² − ∇², in GeV²; every term in GeV³. Derived from the Euler-Lagrange equation.

Equal-time commutator in Feynman gauge[A_μ(t, x), Ȧ_ν(t, y)] = −i g_μν δ³(x − y)

+iδᵢⱼ δ³ for the spatial components, a scalar field's sign; −iδ³ for A₀. Covariance offers g_μν and k_μk_ν, never δ_μν, so one sign must differ.

A_μ in GeV, Ȧ_ν in GeV², δ³ in GeV³. From [A_μ, π^ν] = iδ_μ^ν δ³, as π^ν = −Ȧ^ν + gradients. Derived.

Photon creation and annihilation operators[a_μ(k), a_ν†(k′)] = −g_μν (2π)³ 2|k| δ³(k − k′)

a₁, a₂ and a₃ are ordinary oscillators, but ⟨0|a₀(k)a₀†(k′)|0⟩ = −(2π)³2|k|δ³(k − k′): with a₀|0⟩ = 0 a timelike photon has negative norm.

a_μ in GeV⁻¹; A_μ = ∫ d³k/((2π)³2|k|) [a_μ e^(−ik⋅x) + h.c.], k⁰ = |k|, topic 1.03's normalisation. Derived.

Feynman-gauge HamiltonianH = ∫ d³k/((2π)³2|k|) |k| (−g^μν) a_μ†(k) a_ν(k)

[H, a_ν†(k)] = |k| a_ν†(k) for all four ν, the −g in H and the −g in the commutator multiplying to +1: every energy eigenvalue is a sum of positive |k|.

H in GeV, normal ordered so that H|0⟩ = 0; the μ = ν = 0 term carries −g⁰⁰ = −1. Derived.

A Lorentz scalar that gives A₀ a momentum

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout, with Heaviside-Lorentz charge so that α = e²/(4π). Topic 8.01, quoted, found that −¼F_μνF^μν holds no ∂₀A₀, because F is antisymmetric, so π⁰ vanishes identically and A₀ cannot be quantised. Add a Lorentz scalar that does hold it: L = −¼F_μνF^μν − (1/(2ξ))(∂_μA^μ)² − j_μA^μ, with ξ a real number. As ∂(∂_νA^ν)/∂(∂₀A_μ) = g^(μ0), only A₀'s momentum changes: π⁰ = −(1/ξ)∂_μA^μ, while πⁱ = −F⁽⁰ⁱ⁾ = Eⁱ as before, derived. Dimensions: A has mass dimension 1, so (∂⋅A)² has 4, a Lagrangian density's, and ξ is a pure number. The new term is not gauge invariant: A_μ → A_μ + ∂_μΛ shifts ∂⋅A by □Λ, so only Λ with □Λ = 0 survives. This is a new Lagrangian, not a gauge condition laid on Maxwell's. Limits: as ξ → ∞ the term and π⁰ vanish, topic 8.01 again; as ξ → 0 any non-zero ∂⋅A costs unbounded action, Landau gauge, which topic 8.04 reaches through the propagator.

The field equation, and why ξ = 1

From ∂L/∂(∂_μA_ν) = −F^μν − (1/ξ)g^μν ∂⋅A and ∂L/∂A_ν = −j^ν, the Euler-Lagrange equation is ∂_μF^μν + (1/ξ)∂^ν(∂⋅A) = j^ν; with ∂_μF^μν = □A^ν − ∂^ν(∂⋅A) it becomes □A^ν − (1 − 1/ξ)∂^ν(∂⋅A) = j^ν, derived. Its divergence is (1/ξ)□(∂⋅A) = ∂⋅j, so with a conserved current ∂⋅A is a free massless field: it need not vanish, but no charge sources it. At ξ = 1 the equation is □A^μ = j^μ, four uncoupled wave equations, and L = −½∂_μA_ν∂^μA^ν up to a total derivative, shown: three massless scalars with the usual sign and A₀ with the opposite one. At ξ ≠ 1 the free equation implies □²A^μ = 0, whose solutions include t e^(−ik⋅x), so the modes are not plain waves; Feynman gauge avoids them. Static check: a charge e at rest gives −∇²A⁰ = eδ³(r), so A⁰ = e/(4πr) with ∂⋅A = 0, and a second charge e at 1 fm has energy αħc/r = 1.44 MeV: Coulomb's law, with the gauge-fixing term silent.

Covariant commutators leave one sign odd

At ξ = 1, π⁰ = −Ȧ₀ + ∂ᵢAᵢ and πⁱ = Ȧᵢ − ∂ᵢA₀: each π^μ is −Ȧ^μ plus spatial gradients of A, which commute with A at equal times. So [A_μ(x), π^ν(y)] = iδ_μ^ν δ³(x − y) becomes [A_μ(x), Ȧ_ν(y)] = −ig_μν δ³(x − y), derived: +iδᵢⱼ δ³ for the spatial components, a scalar field's sign, and −iδ³ for A₀. Expand A_μ = ∫ d³k/((2π)³2|k|) [a_μ(k)e^(−ik⋅x) + a_μ†(k)e^(ik⋅x)] and try [a_μ(k), a_ν†(k′)] = C_μν(2π)³2|k|δ³(k − k′): each exponential contributes ½iC_μν δ³(x − y), so C_μν = −g_μν, derived. The odd sign is not a convention. A Lorentz-covariant coefficient must be built from g_μν and k_μk_ν, and nothing covariant equals δ_μν. Even C = −g + b k̂k̂, with k̂_μ = (1, 0, 0, −1) for k along z, fails for every b: its timelike-longitudinal block has C₀₀ = b − 1, C₀₃ = −b and C₃₃ = b + 1, determinant −1, so one direction of negative norm always remains, shown.

Positive energies, negative norms

Up to surface terms the Feynman-gauge energy density is ½Σᵢ[(Ȧⁱ)² + (∇Aⁱ)²] − ½[(Ȧ⁰)² + (∇A⁰)²], classically indefinite, and in modes H = ∫ d³k/((2π)³2|k|) |k|(−g^μν)a_μ†a_ν, derived. Choose a_μ(k)|0⟩ = 0 for all four μ. Then [H, a_ν†(q)] = |q|g^μρ g_ρν a_μ†(q) = |q|a_ν†(q): the −g in H and the −g in the commutator multiply to +δ, so every state built on |0⟩ has energy Σ|k| ≥ 0, derived, and momentum Σk. The sign has gone into the norm: ⟨0|a₀(k)a₀†(k′)|0⟩ = −(2π)³2|k|δ³(k − k′), and for a packet a₀†[f] = ∫ d³k/((2π)³2|k|) f(k)a₀†(k), ⟨0|a₀[f]a₀†[f]|0⟩ = −∫ d³k/((2π)³2|k|) |f(k)|², negative, derived. A state of n timelike photons has norm (−1)ⁿ times a positive number. The classical negative energy survives only as ⟨ψ|H|ψ⟩ = E⟨ψ|ψ⟩: a positive E times a negative norm.

Swap a₀ and a₀†, and the floor goes instead

The commutators do not say which of a₀(k) and a₀†(k) annihilates the vacuum. Write b(k) = a₀†(k), b†(k) = a₀(k), and choose b|0′⟩ = 0. Then [b(k), b†(k′)] = +(2π)³2|k|δ³(k − k′) and every norm is positive, but [H, b†(q)] = −|q|b†(q): each timelike quantum carries energy −|q| and momentum −q, shown. One mode holds any number of them, so n quanta of 1 eV lie n eV below |0′⟩ with nothing to bound n, and |0′⟩ is degenerate with a transverse photon of momentum k plus a b quantum of momentum −k, at zero total energy and momentum: any interaction linking them lets the vacuum decay. This is UAQM-09.06's dilemma for the Dirac field under commutators, quoted: negative norm, or energy unbounded below. The Dirac field escaped through anticommutators. A vector field cannot: by UAQM-09.09's sign rule, quoted, integer spin under anticommutators is causal only with negative norms, which is no escape. The covariant photon keeps the positive energies and must live with negative norms.

An indefinite space, and a subspace still to find

With a₀|0⟩ = 0 the Fock space has an inner product that is not positive, so it is not a Hilbert space, stated. A non-zero vector can have zero norm: (a₁†[f] + a₀†[f])|0⟩, with ∫ d³k/((2π)³2|k|) |f|² = 1, has norm 1 − 1 = 0, yet a₁[f] sends it to |0⟩. The Born rule fails with it: in the completeness relation 1 = Σ|n⟩⟨n|/⟨n|n⟩ the weights of a state can exceed 1 or turn negative. Physics must live on a subspace of non-negative norm where the bad photons do nothing, and a count says how much it keeps. Four photon states per momentum, all of positive energy, would double the Stefan-Boltzmann constant; radiometry measured σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴ to about one part in 10⁴, the two-polarisation value π²k_B⁴/(60ħ³c²), quoted. So timelike and longitudinal photons must drop out of every physical quantity, which topic 8.03's condition achieves by admitting them only in combinations of zero norm.

Change one variable at a time

Make the relationship visible.

Interactive model

Component μ of a_μ(k)0

Vacuum reading v (+1: a₀|0⟩ = 0)1

Quanta n in the mode1

Start at μ = 0, v = +1: the energy climbs while the norm flips sign with every quantum. Set v = −1: every norm turns +1 and the staircase falls with no floor. Then move μ to 1, 2 or 3: v changes nothing, since only g₀₀ has the odd sign.

NORM FACTOR PER QUANTUM-1

ENERGY PER QUANTUM1 |k|

NORM ⟨n|n⟩ / n!-1

ENERGY Eₙ1 |k|

Live interpretationNORM FACTOR PER QUANTUM: −1. ENERGY PER QUANTUM: 1 |k|. NORM ⟨n|n⟩ / n!: −1. ENERGY Eₙ: 1 |k|

Catch the common trap

Explain before calculating.

Misconception check

Seeing −a₀†a₀ in the Hamiltonian, or the classical energy density −½[(Ȧ⁰)² + (∇A⁰)²], students conclude that a timelike photon has negative energy and that this is what covariant quantisation gets wrong. With a₀(k)|0⟩ = 0 it does not: the commutator [a₀, a₀†] is negative too, so −a₀†a₀ counts timelike photons with +1 and [H, a₀†(k)] = +|k|a₀†(k). Every energy eigenvalue is a sum of positive |k|. What is negative is the norm, ⟨0|a₀a₀†|0⟩, and ⟨ψ|H|ψ⟩ = E⟨ψ|ψ⟩ is negative only because ⟨ψ|ψ⟩ is. Negative eigenvalues belong to the other reading, a₀†|0⟩ = 0, where every norm is positive and the energy has no floor. The sign of g₀₀ can be moved but not removed, and those two readings are the only places it can go.

In Feynman gauge with a_μ(k)|0⟩ = 0 for all four μ, a timelike photon |ψ⟩ = a₀†[f]|0⟩ is built from a packet f sharply peaked at |k| = 2.0 eV and scaled so that |⟨ψ|ψ⟩| = 1. Which statement about its norm and energy is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWith no source, test the plane wave A^μ(x) = (c₀, 0, 0, c₃) cos φ, φ = |k|(t − z), in the gauge-fixed theory, taking |k| = 0.30 GeV, c₀ = 0.020 GeV and c₃ = 0.050 GeV. Find ∂_μA^μ and the momentum π⁰ at ξ = 1, 3 and ∞, and decide whether the wave solves the field equation at ξ = 3 and at ξ = 1. Which waves of this form solve it at every ξ?

∂₀A⁰ = −c₀|k| sin φ and ∂₃A³ = ∂_z(c₃ cos φ) = +c₃|k| sin φ, so ∂_μA^μ = |k|(c₃ − c₀) sin φ = 0.30 × 0.030 sin φ = 9.0 × 10⁻³ sin φ GeV². Dimensions: GeV × GeV = GeV², a field's derivative.
π⁰ = −(1/ξ)∂⋅A = −9.0 × 10⁻³ sin φ GeV² at ξ = 1 and −3.0 × 10⁻³ sin φ GeV² at ξ = 3, tending to 0 as ξ → ∞: Maxwell's missing momentum.
k^μ = |k|(1, 0, 0, 1) is null, so □A^μ = −k²A^μ = 0 for every component: on its own, the wave already solves the ξ = 1 equation □A^μ = 0.
∂^μ(∂⋅A) = |k|²(c₃ − c₀) cos φ (1, 0, 0, 1), an amplitude of 0.090 × 0.030 = 2.7 × 10⁻³ GeV³ in the 0 and 3 components. At ξ = 3 the left side is −(1 − 1/3) × 2.7 × 10⁻³ cos φ = −1.8 × 10⁻³ cos φ GeV³, not zero: no solution.
At ξ = 1 the coefficient 1 − 1/ξ vanishes, so it is a solution. With c₃ = c₀, A^μ ∝ k^μ, ∂⋅A = 0 and the wave solves the equation at every ξ; so does a transverse A¹ = c₁ cos φ, which adds nothing to ∂⋅A.

Answer∂⋅A = 9.0 × 10⁻³ sin φ GeV²; π⁰ = −9.0 × 10⁻³ sin φ GeV² at ξ = 1, −3.0 × 10⁻³ sin φ GeV² at ξ = 3, 0 as ξ → ∞. A solution at ξ = 1 only (residual −1.8 × 10⁻³ cos φ GeV³ at ξ = 3); c₃ = c₀ and transverse waves work at every ξ.

MediumUse a packet f with ∫ d³k/((2π)³2|k|) |f|² = 1, sharply peaked at |k| = 1.5 eV, so that [a_μ[f], a_ν†[f]] = −g_μν. With a_μ|0⟩ = 0, take |ψ⟩ = (0.90 a₁†[f] + 0.50 a₀†[f])|0⟩. Find ⟨ψ|ψ⟩, the energy, ⟨ψ|H|ψ⟩, the would-be Born weights of the transverse and timelike photons, and when the norm vanishes. Then build the analogous state on the swapped vacuum, a₀†|0′⟩ = 0, and repeat.

[a₁[f], a₁†[f]] = −g₁₁ = +1 and [a₀[f], a₀†[f]] = −g₀₀ = −1, and different components commute, so ⟨ψ|ψ⟩ = 0.90² − 0.50² = 0.81 − 0.25 = 0.56.
Both terms hold one photon of 1.5 eV, so H|ψ⟩ = 1.5 eV |ψ⟩ whatever the norms, and ⟨ψ|H|ψ⟩ = 1.5 eV × 0.56 = 0.84 eV.
Completeness here reads 1 = Σ|n⟩⟨n|/⟨n|n⟩. With |γ₁⟩ = a₁†[f]|0⟩ and |γ₀⟩ = a₀†[f]|0⟩: the transverse weight is |⟨γ₁|ψ⟩|²/⟨ψ|ψ⟩ = 0.81/0.56 = 1.45; ⟨γ₀|ψ⟩ = 0.50 × (−1), so the timelike weight is 0.25/((−1) × 0.56) = −0.45. They sum to 1.00, and neither is a probability.
The norm 0.81 − c₀², with c₀ the coefficient of a₀†[f], vanishes at |c₀| = 0.90, equal amplitudes, and is negative beyond: (0.50 a₁†[f] + 0.90 a₀†[f])|0⟩ has norm −0.56 yet still energy 1.5 eV.
Swapped: a₀†|0′⟩ = 0, so the analogue is (0.90 a₁†[f] + 0.50 b†[f])|0′⟩ with b†[f] = ∫ d³k/((2π)³2|k|) f(k)a₀(k). Its norm is 0.81 + 0.25 = 1.06, and the weights 0.81/1.06 = 0.764 and 0.25/1.06 = 0.236 are honest probabilities.
But the b quantum carries −1.5 eV: H returns +1.5 eV with probability 0.764 and −1.5 eV with 0.236, so ⟨H⟩ = 1.5 × (0.764 − 0.236) = 0.79 eV. Unnormalised, ⟨ψ|H|ψ⟩ is 0.84 eV in both readings.

AnswerWith a₀|0⟩ = 0: ⟨ψ|ψ⟩ = 0.56, E = 1.5 eV, ⟨ψ|H|ψ⟩ = 0.84 eV, weights 1.45 and −0.45, zero norm at equal amplitudes. Swapped vacuum: norm 1.06, probabilities 0.76 and 0.24, but the 0.24 branch has E = −1.5 eV, and ⟨H⟩ = 0.79 eV.

HardA classical timelike wave has energy density −½[(Ȧ⁰)² + (∇A⁰)²] ≤ 0. Model one box mode of |k| = 0.80 eV, with [a₀, a₀†] = −1 and H = −|k|a₀†a₀, by the coherent state |α⟩ = exp(αa₀†)|0⟩ with |α|² = 2.5. Find ⟨α|α⟩, ⟨α|H|α⟩ and their ratio from the sums over n. Then do the same in the swapped reading, b = a₀† with b|0′⟩ = 0 and |β|² = 2.5, and find the probability that H is measured at −4.8 eV or below.

(a₀†)ⁿ|0⟩ has H eigenvalue +n|k| and, as each a₀ pulled through gives −1, norm (−1)ⁿn!. So ⟨α|α⟩ = Σ(−|α|²)ⁿ/n!: terms 1, −2.5, 3.125, −2.604, 1.628, −0.814, 0.339, … with partial sums 1, −1.5, 1.625, −0.979, 0.648, −0.165, 0.174, converging to e^(−2.5) = 0.0821.
⟨α|H|α⟩ = |k|Σ n(−|α|²)ⁿ/n! = −|k||α|²e^(−|α|²) = −0.80 × 2.5 × 0.0821 = −0.164 eV, assembled from the positive eigenvalues 0.80, 1.60, 2.40 eV, … weighted by alternating norms.
Ratio: ⟨α|H|α⟩/⟨α|α⟩ = −|k||α|² = −2.00 eV. Directly, [a₀, a₀†] = −1 gives a₀|α⟩ = −α|α⟩, so ⟨α|a₀†a₀|α⟩ = |α|²⟨α|α⟩. The units are eV, an energy, as they should be.
Swapped: [b, b†] = +1 and, normal ordered on |0′⟩, H = −|k|b†b. Then |β⟩ = exp(βb†)|0′⟩ has ⟨β|β⟩ = e^(2.5) = 12.18 and ⟨H⟩/⟨β|β⟩ = −|k||β|² = −2.00 eV, now from eigenvalues −n × 0.80 eV with Poisson weights of mean 2.5.
−4.8 eV or below means n ≥ 6: P = 1 − e^(−2.5)(1 + 2.5 + 3.125 + 2.604 + 1.628 + 0.814) = 1 − 0.958 = 0.042, and nothing bounds n.
For contrast a transverse mode gives norm e^(2.5) and ⟨H⟩/⟨⟩ = +2.00 eV. Both readings reproduce the timelike wave's classical negative energy: one hides its sign in alternating norms, the other in negative eigenvalues.

AnswerWith a₀|0⟩ = 0: ⟨α|α⟩ = 0.0821, ⟨α|H|α⟩ = −0.164 eV, ratio −2.00 eV, though every eigenvalue is ≥ 0. Swapped: ⟨β|β⟩ = 12.2, ratio −2.00 eV from eigenvalues −n × 0.80 eV, and P(E ≤ −4.8 eV) = 0.042.