University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.4

The photon propagator in covariant gauges and Coulomb gauge

Every internal photon in QED is this one object, so it is worth building twice. You will see why Maxwell's operator cannot be inverted as it stands, invert it once gauge-fixed for any ξ, and watch the gauge choice vanish between conserved currents, leaving Coulomb's instantaneous force plus two transverse photons.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Every internal photon line in QED stands for one object, the time-ordered product ⟨0|T A_μ(x) A_ν(y)|0⟩, and this topic builds it two ways that must agree. In Feynman gauge, ξ = 1, topic 8.02 left four free massless fields whose operators obey [a_μ, a_ν†] ∝ −g_μν, so the propagator is −g_μν times topic 1.05's D_F at m = 0, or −i g_μν/(k² + iε): three components propagate like a massless scalar and A₀ with the opposite sign, the negative norm resurfacing as a residue. The second route explains why gauge fixing was unavoidable.

Maxwell's quadratic form −k² g^μν + k^μ k^ν annihilates k_ν, the pure-gauge direction, at every k, so it has no inverse; the term −(1/(2ξ))(∂⋅A)² gives that direction the coefficient −k²/ξ, and inverting on transverse and longitudinal projectors gives −i[g_μν − (1 − ξ)k_μ k_ν/k²]/(k² + iε) in every covariant gauge. This family is not a family of physics. When both ends of the line are conserved currents, which means on-shell charged particles, the k_μ k_ν term dies, and the timelike and longitudinal photons combine into a term with no energy dependence, Coulomb's instantaneous 1/|k|², leaving Coulomb gauge's transverse photons.

The agreement stops where conservation stops: off-shell Green's functions keep their ξ.

Simple definition: The photon propagator is the Fourier transform of ⟨0|T A_μ(x) A_ν(y)|0⟩, −i[g_μν − (1 − ξ)k_μ k_ν/k²]/(k² + iε) in the covariant gauge labelled ξ; ξ changes only the part along k, which a conserved current never sees.
Example: A photon with ω = 0.06 GeV and |k| = 0.10 GeV has k² = −0.0064 GeV², so Feynman gauge's D₀₀ is 156i GeV⁻²; add the longitudinal photon and two conserved currents feel exactly 1/|k|² = 100 GeV⁻², Coulomb's static value.

Feynman gauge, ξ = 1, from the modes⟨0|T A_μ(x) A_ν(y)|0⟩ = ∫ d⁴k/(2π)⁴ e^(−ik⋅z) (−i g_μν)/(k² + iε), z = x − y

Derived from topic 8.02's [a_μ, a_ν†] ∝ −g_μν. A₁, A₂ and A₃ propagate like massless scalars; A₀'s opposite sign is the timelike photon's negative norm.

A_μ of mass dimension 1; k in GeV and z in GeV⁻¹; ħ = c = 1 with metric (+, −, −, −); topic 1.05's D_F at m = 0

No Green's function before gauge fixingK^μν D_νλ = iδ^μ_λ has no solution: k_μ K^μν = 0 turns it into 0 = ik_λ

Derived. The null direction k_ν is the gauge mode A_μ = ∂_μΛ, and it is null at every k, off shell too, so no iε can rescue the inversion.

K^μν = −k²g^μν + k^μ k^ν in GeV², from −(1/4)F² = (1/2)A_μ(g^μν ∂² − ∂^μ ∂^ν)A_ν; D in GeV⁻²

The gauge-fixed operator on projectorsK_ξ^μν = −k² P_T^μν − (k²/ξ) P_L^μν, with P_L^μν = k^μ k^ν/k² and P_T^μν = g^μν − P_L^μν

Derived. The projectors do not mix, so inverting K_ξ means inverting two numbers, −k² on P_T and −k²/ξ on P_L.

ξ dimensionless, from adding −(1/(2ξ))(∂⋅A)²; P_T and P_L idempotent, orthogonal and summing to g^μν; K_ξ in GeV²

Propagator in a general covariant gaugeD_μν(k) = −i[g_μν − (1 − ξ) k_μ k_ν/(k² + iε)]/(k² + iε)

Derived as a Green's function; that canonical quantisation at ξ ≠ 1 gives the same time-ordered product is stated. ξ sets the part along k and nothing else.

D_μν in GeV⁻²; ξ = 1 Feynman, ξ = 0 Landau, ξ = 3 Fried-Yennie; it gives k^μ D_μν = −iξ k_ν/(k² + iε)

ξ drops out between conserved currentsk_μ J^μ = 0 ⇒ J^μ D_μν(k) J′^ν = −i J⋅J′/(k² + iε) for every ξ

Shown. Conservation needs on-shell charged lines; off-shell quantities, such as the electron's Z₂ of topic 12.05, keep their ξ.

J, J′ the currents at the line's two ends, in GeV for one-particle matrix elements such as (p + p′)^μ; one conserved end suffices

Coulomb gauge recovered−J⋅J′/(k² + iε) = J⁰J′⁰/|k|² + J_T⋅J′_T/(k² + iε)

Shown. The first term has no ω, so it acts at one instant: UAQM-11.01's Coulomb energy. The second is Coulomb gauge's transverse photon with its light-cone pole.

k^μ = (ω, k) in GeV; J_T⋅J′_T the ordinary three-vector product of the parts across k; uses ωJ⁰ = |k| J_L

Feynman gauge: four massless fields, one of opposite sign

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 8.02, quoted, left the ξ = 1 theory as four free massless fields, A_μ(x) = ∫ d³k/((2π)³ 2|k|) [a_μ(k) e^(−ik⋅x) + a_μ†(k) e^(ik⋅x)], with [a_μ(k), a_ν†(k′)] = −g_μν (2π)³ 2|k| δ³(k − k′) and all four a_μ annihilating the vacuum. Only the commutator survives in ⟨0|A_μ(x) A_ν(y)|0⟩, which is therefore −g_μν times the massless form of UAQM-09.04's vacuum amplitude. Time ordering follows topic 1.05 term by term: ⟨0|T A_μ(x) A_ν(y)|0⟩ = −g_μν D_F(x − y), or −i g_μν/(k² + iε) in momentum space, derived. Each Aᵢ propagates exactly like a massless scalar, +i/(k² + iε); A₀ carries −i/(k² + iε), the timelike photon's negative norm reappearing as a residue of the wrong sign. Dimensions agree: [A] = 1, so the Fourier transform over d⁴x has dimension 2 − 4 = −2, that of 1/k². In position space D_F(x) = −1/(4π²(x² − iε)), quoted; at spacelike x it is 1/(4π²r²), the m → 0 limit of topic 1.05's mK₁(mr)/(4π²r), and it is not confined to the light cone: an amplitude, not a signal.

Without gauge fixing there is nothing to invert

Integrating by parts, −(1/4)F_μν F^μν = (1/2)A_μ(g^μν ∂² − ∂^μ ∂^ν)A_ν, so the action is (1/2)∫ Ã_μ(−k) K^μν(k) Ã_ν(k) d⁴k/(2π)⁴ with K^μν = −k² g^μν + k^μ k^ν, derived. For a scalar the same steps give k² − m², and the propagator is i times its inverse. Here K^μν k_ν = −k² k^μ + k^μ k² = 0 at every k: the direction Ã_ν ∝ k_ν, which is A_μ = ∂_μΛ, costs no action. A Green's function would need K^μν D_νλ = i δ^μ_λ, and contracting with k_μ kills the left side and leaves 0 = i k_λ, a contradiction. This is not the on-shell zero of a massive scalar, which iε steps around at k² = m²; it holds off shell, for every k, and no prescription repairs it. It is topic 8.01's undetermined evolution seen in momentum space. On a transverse vector, (0, 1, 0, 0) with k along z, K multiplies by −k², 16 GeV² at k^μ = (3, 0, 0, 5) GeV; on k itself it gives zero.

Gauge fixing gives the longitudinal direction a coefficient

Topic 8.02's term −(1/(2ξ))(∂⋅A)² adds −(1/ξ)k^μ k^ν to K, so K_ξ^μν = −k² g^μν + (1 − 1/ξ)k^μ k^ν. For k² ≠ 0 define P_L^μν = k^μ k^ν/k² and P_T^μν = g^μν − P_L^μν. Contracting indices, P_L P_L = P_L, P_T P_T = P_T, P_T P_L = 0 and P_T + P_L = g: P_L projects onto the line of k, P_T onto the three directions orthogonal to it. Then K_ξ = −k² P_T − (k²/ξ)P_L, and since the projectors do not mix the inverse is −P_T/k² − ξ P_L/k². Multiplying by i, D_μν(k) = −i[g_μν − (1 − ξ)k_μ k_ν/k²]/(k² + iε), derived as a Green's function, each 1/k² taking the +iε of topic 1.06. ξ = 1 is Feynman gauge; ξ = 0 is Landau gauge, where k^μ D_μν = 0; ξ = 3, the Fried-Yennie gauge, is named. In general k^μ D_μν = −iξ k_ν/(k² + iε): ξ sets the longitudinal part and touches nothing else. At ξ ≠ 1 the field equation couples components and the modes are no longer pure plane waves; that canonical quantisation still gives this time-ordered product is stated, not derived.

Between conserved currents ξ has nowhere to act

An internal photon ends on two currents, so every amplitude contains J^μ D_μν(k) J′^ν. Its ξ-dependent piece is i(1 − ξ)(k⋅J)(k⋅J′)/(k² + iε)², which vanishes as soon as one end satisfies k_μ J^μ = 0, shown: a property of the currents, not of the propagator. A charged scalar's transition current is (p + p′)^μ, the matrix element of UAQM-09.03's Noether current, quoted, and with k = p′ − p the photon sees k⋅(p + p′) = p′² − p², zero only when p′² = p², as when both lines sit on the same mass shell. One-photon exchange between external lines, which LSZ puts on shell, is therefore the same in every covariant gauge. Inside loops the charged lines are off shell and ξ cancels only in the sum of diagrams: topic 8.06 argues it in outline, topic 9.04 checks it along a fermion line and topic 12.08 carries it to all orders. Off-shell functions keep their ξ: the electron's field-strength renormalisation Z₂ has a one-loop pole proportional to ξ that is absent in Landau gauge, quoted from topic 12.05, while the measured pole mass is gauge independent. Choose ξ for convenience, never for physics.

Timelike plus longitudinal photons make Coulomb's term

Take k^μ = (ω, 0, 0, |k|) and split each current into J⁰, a part J_L along k and a part J_T across it. Conservation, ωJ⁰ = |k| J_L, ties J_L to the charge density. In Feynman gauge −i J^μ D_μν J′^ν = −J⋅J′/(k² + iε), to which the timelike photon contributes −J⁰J′⁰/(k² + iε) and the longitudinal one J_L J′_L/(k² + iε) = (ω²/|k|²)J⁰J′⁰/(k² + iε). Their sum is J⁰J′⁰(ω² − |k|²)/(|k|² k²) = J⁰J′⁰/|k|², shown. Each unphysical photon carries the light-cone pole, but the sum has none and no ω, so it is δ(t − t′) in time and, through the transform of 1/|k|², ρ(x)ρ′(y)/(4π|x − y|) in space: UAQM-11.01's instantaneous Coulomb energy, quoted. What remains, J_T⋅J′_T/(k² + iε), is exchange of the two transverse photons through Coulomb gauge's i(δᵢⱼ − k̂ᵢk̂ⱼ)/(k² + iε). Coulomb gauge, with D₀₀ = i/|k|² and D₀ᵢ = 0, therefore gives exactly the covariant exchange between conserved currents, though its propagator is not a Lorentz tensor.

Checks: static charges, slow charges and the photon's mass

At ω = 0 the k_μ k_ν term has no 00 entry, so D₀₀ = i/|k|² in every covariant gauge and in Coulomb gauge. The transform of 1/|k|² is 1/(4πr), so with e² = 4πα an electron and a proton a distance r apart have energy −αħc/r, the sign topic 9.03 fixes. At the Bohr radius, 0.05292 nm, that is −1.43996 eV nm ÷ 0.05292 nm = −27.21 eV; the ground state, at half of it by the virial theorem, lies at −13.61 eV, or 13.598 eV deep with the proton's recoil, the measured ionisation energy. For slow charges J_T ≈ ρv, so transverse exchange is down on Coulomb by about vv′; for atomic electrons, with v of order α, that is α² = 5.3 × 10⁻⁵, the order of the magnetic Breit interaction, named only. For two charged scalars in their centre-of-mass frame the ratio is exactly v₁v₂ cos²(θ/2), worked below, so near light speed the transverse photons are as large as Coulomb. The pole at k² = 0 is the photon's masslessness: a mass m_γ would turn 1/r into e^{(−m_γ r)}/r, and the measured bound m_γ ≤ 1 × 10⁻¹⁸ eV, quoted, puts the range above ħc/m_γ = 2 × 10¹¹ m, beyond the Earth-Sun distance.

Change one variable at a time

Make the relationship visible.

Interactive model

Gauge parameter ξ1.00

Transverse overlap J_T⋅J′_T ÷ J⁰J′⁰0.3

Marker at ω/|k|0.50

Drag ξ from 0, Landau, through 1, Feynman, to 3: the two dashed curves swing apart by ±(1 − ξ)u²/(1 − u²)², with u = ω/|k|, yet their sum never leaves the Coulomb line. Set the transverse overlap to 0 and the total falls onto that line too: static charges feel Coulomb alone.

00 PART1.333 J⁰J′⁰/|k|²

03 + 30 + 33 PARTS-0.333 J⁰J′⁰/|k|²

THEIR SUM: COULOMB1.000 J⁰J′⁰/|k|²

TOTAL WITH TRANSVERSE0.600 J⁰J′⁰/|k|²

Live interpretation00 PART: 1.333 J⁰J′⁰/|k|². 03 + 30 + 33 PARTS: −0.333 J⁰J′⁰/|k|². THEIR SUM: COULOMB: 1.000 J⁰J′⁰/|k|². TOTAL WITH TRANSVERSE: 0.600 J⁰J′⁰/|k|²

Catch the common trap

Explain before calculating.

Misconception check

Landau gauge is often called the transverse gauge, and it is tempting to conclude that its propagator carries only the two physical photons while Feynman gauge drags in the unphysical ones. Four-dimensional transversality, k^μ D_μν = 0, is not the transversality of a photon. P_T^μν projects onto the three directions orthogonal to k, its trace g^μ_μ − 1 is 3, not 2, and for the spacelike k of a scattering exchange one of the three is timelike, so Landau gauge's D₀₀ is Feynman gauge's −i/k² times 1 − k₀²/k², a factor above 1 whenever k₀ ≠ 0: the timelike photon has grown, not gone. No covariant propagator holds only the two physical photons, because picking them out needs a frame; Coulomb gauge pays exactly that price, with D₀ᵢ = 0 and spatially transverse Dᵢⱼ. What all gauges share is their effect between conserved currents, not their content.

In a general covariant gauge the photon propagator is D_μν(k) = −i[g_μν − (1 − ξ)k_μ k_ν/k²]/(k² + iε). What is k^μ D_μν, and what does it tell you?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA t-channel photon carries k^μ = (3, 0, 0, 5) GeV. Find the Landau-gauge components D₀₀, D₀₃ and D₃₃, check that k^μ D_μν = 0, and compare D₀₀ with Feynman gauge's.

Lower the index: k_μ = (3, 0, 0, −5) GeV, and k² = 3² − 5² = −16 GeV², spacelike as a scattering exchange must be. Every entry of D_μν will carry GeV⁻², the dimension of 1/k².
Landau gauge, ξ = 0: D_μν = −i[g_μν − k_μ k_ν/k²]/k². For 00 the bracket is 1 − 9/(−16) = 25/16, so D₀₀ = −i(25/16)/(−16) = 25i/256 = 0.0977i GeV⁻².
For 03 the bracket is 0 − (3)(−5)/(−16) = −15/16, so D₀₃ = −15i/256 = −0.0586i GeV⁻². For 33 it is −1 − 25/(−16) = 9/16, so D₃₃ = 9i/256 = 0.0352i GeV⁻².
Contract with k^μ = (3, 0, 0, 5): k^μ D_μ0 = 3(25i/256) + 5(−15i/256) = 0 and k^μ D_μ3 = 3(−15i/256) + 5(9i/256) = 0, as ξ = 0 requires.
Feynman gauge gives D₀₀ = −i/k² = i/16 = 0.0625i GeV⁻² and D₀₃ = 0, with k^μ D_μ0 = 3i/16 = −ik₀/k², not zero. Making the propagator four-dimensionally transverse has enlarged its timelike entry by 25/16 rather than removed it.

AnswerLandau gauge: D₀₀ = 0.0977i GeV⁻², D₀₃ = −0.0586i GeV⁻², D₃₃ = 0.0352i GeV⁻², with k^μ D_μν = 0; Feynman gauge's D₀₀ = 0.0625i GeV⁻² is smaller by the factor 1.5625.

MediumA photon of k^μ = (0.6, 0, 0, 1.0) GeV joins transition currents J^μ = (1.0, 0.3, 0, 0.6) GeV and J′^μ = (2.0, 0.5, 0.4, 1.2) GeV. Check both are conserved, split −i J^μ D_μν J′^ν in Feynman gauge into timelike, longitudinal and transverse parts, and show that Landau gauge regroups the first two but not the total.

k_μ J^μ = 0.6 × 1.0 − 1.0 × 0.6 = 0 and k_μ J′^μ = 0.6 × 2.0 − 1.0 × 1.2 = 0, so both currents are conserved; k² = 0.36 − 1.00 = −0.64 GeV² and |k|² = 1.00 GeV².
Feynman gauge gives −i J^μ D_μν J′^ν = −J⋅J′/k², term by term: timelike −J⁰J′⁰/k² = −2.00/(−0.64) = 3.125; longitudinal J³J′³/k² = 0.72/(−0.64) = −1.125; transverse (J¹J′¹ + J²J′²)/k² = 0.15/(−0.64) = −0.234. Total 1.766, dimensionless as GeV² × GeV⁻² must be.
Timelike plus longitudinal is 3.125 − 1.125 = 2.000 = J⁰J′⁰/|k|² = 2.00/1.00: the instantaneous Coulomb term, with no ω left in it.
Landau gauge uses −J^μ[g_μν − k_μ k_ν/k²]J′^ν/k². The 00 entry gives 2.00 × 1.5625/0.64 = 4.883, the 33 entry 0.72 × 0.5625/0.64 = 0.633 and the 03 and 30 entries 2.40 × (−0.9375)/0.64 = −3.516; these again sum to 2.000, and the transverse part stays −0.234.
So the total is 1.766 in Feynman gauge, in Landau gauge and in Coulomb gauge, where it reads 2.000 − 0.234, as (k⋅J)(k⋅J′) = 0 requires. The transverse photons, the only physical ones, carry −13% of it; currents with J_T = 0, such as static charges, would leave Coulomb alone.

Answer−i J^μ D_μν J′^ν = 1.766 in every gauge: Feynman's 3.125 − 1.125 and Landau's 4.883 + 0.633 − 3.516 both regroup into the Coulomb term 2.000, and the transverse part adds −0.234.

HardCharged scalars 1 (mass 0.60 GeV) and 2 (mass 1.50 GeV), each of charge e, scatter in their centre-of-mass frame at p = 0.80 GeV through θ = 90°. With currents J = p₁ + p₃ and J′ = p₂ + p₄, split one-photon exchange into Coulomb and transverse parts, show their ratio is v₁v₂ cos²(θ/2), and find dσ/dΩ in nb sr⁻¹ with and without the transverse photons.

E₁ = √(0.36 + 0.64) = 1.00 GeV and E₂ = √(2.25 + 0.64) = 1.70 GeV, so √s = 2.70 GeV, v₁ = 0.800 and v₂ = 0.471. Elastic scattering in this frame changes no energy, so ω = 0, |q|² = 4p² sin²(θ/2) = 1.28 GeV² and q² = −1.28 GeV².
J = (2E₁, p sin θ, 0, p(1 + cos θ)) and J′ = (2E₂, −p sin θ, 0, −p(1 + cos θ)), while q = (0, p sin θ, 0, p(cos θ − 1)). Then q⋅J = 0 on shell, so J has no part along q and every ξ gives the same exchange.
Coulomb: J⁰J′⁰/|q|² = (2.00)(3.40)/1.28 = 5.3125. Transverse: J_T⋅J′_T = −p²[sin²θ + (1 + cos θ)²] = −4p² cos²(θ/2) = −1.28 GeV², and over q² = −1.28 GeV² it gives +1.000. The total, 6.3125, equals −J⋅J′/q² = 8.08/1.28.
Their ratio is p² cos²(θ/2)/(E₁E₂) = v₁v₂ cos²(θ/2) = 0.188. It vanishes for slow charges, the static limit, and tends to cos²(θ/2) as both speeds approach 1.
With the vertex −ie(p + p′)^μ quoted from topic 8.09, M = e² J⋅J′/q² = −4π × 6.3125/137.036 = −0.5789, negative as for like charges repelling. Topic 5.04's dσ/dΩ = |M|²/(64π²s) = 0.3351/(631.65 × 7.29) GeV⁻² = 7.277 × 10⁻⁵ GeV⁻², or 28.3 nb sr⁻¹ with 1 GeV⁻² = 0.3894 mb.
Coulomb alone gives (5.3125/6.3125)² = 0.708 of that, 20.1 nb sr⁻¹: at these speeds, dropping the transverse photons loses 29% of the cross-section.

AnswerCoulomb 5.31 and transverse 1.00, in the ratio v₁v₂ cos²(θ/2) = 0.188; dσ/dΩ = 28.3 nb sr⁻¹ from one-photon exchange, against 20.1 nb sr⁻¹ from the Coulomb term alone.