University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.3

Photon exchange between static charges: like charges repel

Any quantum theory of light and charge has to give Coulomb's law back, sign and all, and this is where QED does it. You carry one diagram, e⁻μ⁻ → e⁻μ⁻, down to the static limit, find the sign of the force in the g₀₀ of the photon propagator, and see why the same photon pulls an electron and a positron together.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Coulomb's law is not an input to QED but the static limit of one-photon exchange, and this topic recovers it with its sign. In e⁻μ⁻ → e⁻μ⁻ there is a single t-channel diagram, because no vertex turns an electron into a muon. As both particles slow down, the charge densities ūγ⁰u tend to 2m δₛₛ′ while the spatial currents fall with the speeds, so g_μν J₁^μ J₂^ν keeps only J₁⁰J₂⁰; and elastic scattering in the centre-of-mass frame has q⁰ = 0, so the photon's −ig₀₀/q² is +i/|q|², opposite in sign to a scalar's −i/(|q|² + m_φ²).

Matching to Quantum Mechanics II's first Born amplitude turns that sign into V(r) = +α/r: like charges repel, where topic 7.05's scalar attracted. For an electron and a positron the bilinear v̄γ⁰v = v†v is positive, not negative like v̄v, so nothing cancels the −1 from ordering the operators and V = −α/r. The result is the leading term of an expansion in v and α: currents, spin and retardation enter at order v² as the Breit interaction, annihilation adds a contact term for e⁻e⁺, the Born step needs α ≪ v, and bound states such as positronium come from iterating the potential, not from one diagram.

Simple definition: One-photon exchange between two slow charged fermions acts, at first Born order, as the potential V(r) = +α/r between like charges and −α/r between a particle and an antiparticle, the sign fixed by the propagator's g₀₀ and by the ordering of the fermion operators.
Example: An electron and a μ⁻ 1.0 Å apart have V = αħc/r = 14.40 eV Å ÷ 1.0 Å = +14.4 eV, a repulsion; an electron and a μ⁺ at that distance have −14.4 eV, and a massless scalar of the same strength would give −14.4 eV to both pairs.

Electron–muon scattering at tree levelM = e²[ū(p₃)γ^μ u(p₁)][ū(p₄)γ_μ u(p₂)]/t

A single diagram: no vertex turns an electron into a muon, so there is no u-channel and no annihilation. The ordering sign is +, from six inversions.

e dimensionless, e²/4π = α = 1/137.036; t = (p₁ − p₃)² in GeV²; each spinor product in GeV, so M is dimensionless

Charge density of a slow fermionū(p′, s′)γ⁰u(p, s) = 2m δₛₛ′ + O(|p|²/m)

Slow charges carry charge but almost no current, so g_μν J₁^μ J₂^ν reduces to J₁⁰J₂⁰ up to a fraction of order v₁v₂.

m the fermion mass, p and p′ its three-momenta, in GeV; the correction includes the spin-flip term iξ′†σ⋅(p′ × p)ξ/(E + m); the spatial current ū(p′, s′)γⁱu(p, s) is of order |p| = mv

Static limit of one-photon exchangeM → −4mₑm_μ e² δₛ₁ₛ₃ δₛ₂ₛ₄/|q|²

Negative, where topic 7.05's scalar gave +4m₁m₂g²/(|q|² + m_φ²): the photon's −ig₀₀/q² is +i/|q|², the scalar's i/(q² − m_φ²) is −i/(|q|² + m_φ²).

q the three-momentum transfer in GeV; elastic scattering in the centre-of-mass frame has q⁰ = 0, so t = −|q|² exactly

Born matchingṼ(q) = −M/(4mₑm_μ) = +e²/|q|²

The 4mₑm_μ undoes relativistic normalisation, cancelling the two charge densities 2m. A negative M means a positive Ṽ: repulsion.

Ṽ, in GeV⁻², is the Fourier transform of V; from f = M/(8π√s), f = −(mᵣ/2π)Ṽ and √s → mₑ + m_μ

Like charges repelV(r) = +α/r, or +αħc/r with units restored

Coulomb's law, with e²/(4πε₀) = αħc = 2.307 × 10⁻²⁸ J m; topic 7.05's massless scalar gives the same 1/r with the opposite sign.

αħc = 1.440 MeV fm = 14.40 eV Å; the transform of 1/|q|² is 1/(4πr), the m_φ → 0 limit of Yukawa's

Electron and positron: the ordering sign survivesMₜ = −e²[ū(p₃)γ^μ u(p₁)][v̄(p₂)γ_μ v(p₄)]/t

So Mₜ → +4mₑ²e²/|q|² and V = −α/r: opposite charges attract. Annihilation adds only a contact term of relative order v², set aside.

− from seven inversions of d₄b₃(ψ̄γψ)(ψ̄γψ)b†₁d†₂; v̄₂γ⁰v₄ = v₂†v₄ → +2m δₛ₂ₛ₄, a norm, not −2m like v̄v

One diagram, its sign fixed by operator ordering

Work in natural units, ħ = c = 1, with metric (+, −, −, −). Topic 9.02's rules, quoted, give the vertex −ieγ^μ (e = −|e| for electron and muon alike) and the Feynman-gauge propagator −ig_μν/(q² + iε). In e⁻(p₁)μ⁻(p₂) → e⁻(p₃)μ⁻(p₄) the photon must join the electron line to the muon line, since no vertex turns one into the other: one t-channel diagram, with no u-channel partner and no annihilation. Ordering ⟨0|b₄b₃, the two currents and b†₁b†₂|0⟩ into adjacent contracted pairs takes six inversions, topic 7.05's direct count, so the sign is +, and iM = (−ie)²[ū₃γ^μ u₁](−ig_μν/q²)[ū₄γ^ν u₂], with ūᵢ = ū(pᵢ, sᵢ), gives M = e²[ū₃γ^μ u₁][ū₄γ_μ u₂]/t, t = (p₁ − p₃)², derived. Only e² appears, so the sign of the charge drops out. Dimensions: e is a pure number, each spinor product carries one power of mass and t two, so M is dimensionless, as a 2 → 2 amplitude must be. A general covariant gauge adds a q_μ q_ν term, which these on-shell currents annihilate, as topic 9.04 shows.

Slow charges carry charge, not current

With UAQM-04's Dirac-representation spinors, quoted, and |p′| = |p| as in elastic centre-of-mass scattering, the bilinears are exactly ū(p′, s′)γ⁰u(p, s) = (E + m)ξ′†ξ + ξ′†[p′⋅p + iσ⋅(p′ × p)]ξ/(E + m) and ū(p′, s′)γⁱu(p, s) = ξ′†[(p + p′)ⁱ + i((p − p′) × σ)ⁱ]ξ, derived, ξ and ξ′ being two-component spin states. As |p| → 0 the first tends to 2m δₛₛ′, every correction carrying |p|²/m², while the second, a convection current plus the spin's magnetisation current, is of order |p| = mv. So J₁⁰J₂⁰ − J₁⋅J₂ → 4mₑm_μ δₛ₁ₛ₃δₛ₂ₛ₄, the current product suppressed by vₑv_μ and the densities' own corrections by v². Elastic scattering in the centre-of-mass frame also transfers no energy, q⁰ = 0, so t = −|q|² exactly and nothing is retarded. At |p| = 0.10 MeV and 90°, spins along the beam, the current product is 4.6 × 10⁻⁵ of the charge product: the muon, at v_μ ≈ 10⁻³, is a static charge even while the electron moves at 0.19c.

The sign sits in g₀₀

Assembled, M → e²(2mₑ)(2m_μ)g₀₀/t = −4mₑm_μ e²/|q|² for unflipped spins, derived. Topic 7.05's scalar gave +4m₁m₂g²/(|q|² + m_φ²), and line by line the two calculations match: (−ie)² against (−ig)², charge density 2m against scalar density 2m, ordering sign + in both. Only the propagator differs: the photon's −ig₀₀/q² is +i/|q|² at spacelike q, the scalar's i/(q² − m_φ²) is −i/(|q|² + m_φ²). The minus in front of g_μν gives each spatial component, −gᵢᵢ = +1, the sign of a positive-norm scalar, and Lorentz covariance hands the time component the opposite one. The spatial terms keep the scalar's sign, so −J₁⋅J₂/t = +J₁⋅J₂/|q|² makes parallel currents attract, Ampère's force, at order vₑv_μ. Nor is the repulsion a gauge artefact. In Feynman gauge it rides on the timelike photon, whose negative norm (topic 8.02) is that −1; in Coulomb gauge (topic 8.04) the whole J⁰J⁰ term is the instantaneous Coulomb energy, positive between like charges, with no unphysical photon in sight.

Match to the Born amplitude: V = +α/r

Topic 5.06 makes f = M/(8π√s) the amplitude with dσ/dΩ = |f|², and the first Born approximation (UQMII-15.08, quoted) gives f = −(mᵣ/2π)Ṽ(q), Ṽ being the Fourier transform of V and mᵣ = mₑm_μ/(mₑ + m_μ). As √s → mₑ + m_μ the masses collapse to Ṽ(q) = −M/(4mₑm_μ) = +e²/|q|², and the transform of 1/|q|², the m_φ → 0 limit of the Yukawa transform, quoted, is 1/(4πr). So V(r) = e²/(4πr) = +α/r, derived: a positive energy that falls with distance, pushing the pair apart. Dimensions: Ṽ has mass dimension −2 and α/r is an energy. Restoring units, αħc = 1.440 MeV fm = 14.40 eV Å, which in SI is e²/(4πε₀) = 2.307 × 10⁻²⁸ J m, so the Heaviside-Lorentz e² = 4πα is Coulomb's constant in disguise. The cross-section |f|² = 4mᵣ²α²/|q|⁴ is Rutherford's and cannot see the sign; the sign shows wherever the amplitude interferes or binds. The match needs v ≪ 1 for static currents and α ≪ v for the Born step.

Electron and positron: a minus sign nothing cancels

For e⁻(p₁)e⁺(p₂) → e⁻(p₃)e⁺(p₄) take |in⟩ = b†₁d†₂|0⟩ and ⟨out| = ⟨0|d₄b₃. In the t-channel diagram the positron line gives v̄₂γ^ν v₄, read against its arrow, and making the contracted pairs adjacent costs seven inversions, as in topic 7.05, so Mₜ = −e²[ū₃γ^μ u₁][v̄₂γ_μ v₄]/t, derived. Here the photon parts company with the scalar: v̄₂γ⁰v₄ = v₂†v₄ is a norm at low speed and tends to +2m δₛ₂ₛ₄, where v̄v tended to −2m. For the scalar two minus signs cancelled and the pair attracted; for the photon only the ordering sign is left, so Mₜ → +4mₑ²e²/|q|², Ṽ = −e²/|q|² and V = −α/r, derived. In operator language the charge e∫:ψ†ψ: d³x = e Σ(b†b − d†d) gives the positron −e, while the scalar density counted both alike. For e⁻μ⁺ the same steps give −α/r and no other diagram exists. e⁻e⁺ also annihilates, but v̄γ⁰u vanishes in the centre-of-mass frame, and the spatial annihilation current over s ≈ 4m² gives a contact term of relative order |q|²/m² in the spin triplet, set aside.

What one diagram leaves out, and a measured test

V = ±α/r is the first term of an expansion in v. At order v² come the densities' own corrections, the current–current terms, the spin–orbit and spin–spin couplings carried by the spin terms of both bilinears, and, for bound particles where q⁰ no longer vanishes, retardation: together the Breit interaction, named only here, the source of fine and hyperfine structure. Repeated exchange, which builds the Coulomb phase and the bound states themselves, and loops (Unit 12) lie beyond one diagram. Iterating −α/r in the Schrödinger equation, quoted, with mᵣ = mₑ/2 gives positronium the levels −mₑα²/(4n²): the ground state at −6.803 eV and a 1S–2S interval of (3/16)mₑα² = 5.10213 eV, or 1.23369 × 10¹⁵ Hz, against the measured 1.23361 × 10¹⁵ Hz. The gap, 7 parts in 10⁵, is the size of the relative order-α² corrections that the Breit and annihilation terms supply. The e⁻μ⁻ pair, under +α/r, has no bound state at all.

Change one variable at a time

Make the relationship visible.

Interactive model

Muon charge ÷ electron's: +1 μ⁻, −1 μ⁺1

Boson spin: 0 scalar, 1 photon1

Separation r1.0 Å

Start with the photon and a μ⁻: the curve lies above the axis and the arrows point apart. Flip the muon's charge: only the ordering sign changes, and the pair attracts. Then set the spin to 0: the scalar attracts both partners, since its bilinear sign follows the ordering sign and cancels it.

ORDERING SIGN1

BILINEAR SIGN1

PROPAGATOR SIGN-1

V(r)14.40 eV

Live interpretationORDERING SIGN: 1. BILINEAR SIGN: 1. PROPAGATOR SIGN: −1. V(r): 14.40 eV

Catch the common trap

Explain before calculating.

Misconception check

“Scalar exchange attracts, vector exchange repels” is half a rule, and used as a whole one it gets e⁻e⁺ wrong. What the photon propagator fixes is that sources of the same sign repel: −ig₀₀/q² multiplies the product of the two charge densities. Whether two particles' charges share a sign is a separate fact, carried by the spinors and the operator ordering. For e⁻μ⁻ both densities are +2m and the ordering sign is +, so V = +α/r. For e⁻e⁺ the densities are still both +2m, since v̄γ⁰v = v†v is a norm and tends to +2m, but the ordering costs one more inversion, so V = −α/r and positronium binds by 6.8 eV. The mirror error, “an antiparticle's v̄ always flips the sign”, fails for the scalar, whose ψψ̄ pair attracts. Count all three signs, propagator, bilinear and ordering, every time.

In e⁻μ⁺ → e⁻μ⁺ at low speed, with states ordered as b†d†|0⟩ and ⟨0|d b, the t-channel term carries v̄(p₂)γ⁰v(p₄) → +2m_μ just as ū(p₄)γ⁰u(p₂) does in e⁻μ⁻ → e⁻μ⁻, where V = +α/r. Why is the e⁻μ⁺ potential −α/r?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron and a μ⁻ scatter in their centre-of-mass frame with |p| = 20 keV through θ = 60°, neither spin flipping. Evaluate the static tree amplitude M, the momentum-space potential Ṽ(q) in MeV⁻², and V at r = 0.50 Å in eV, then say what a massless scalar coupled with g²/4π = α to both particles would give instead.

Momentum transfer: |q| = 2|p| sin(θ/2) = 2 × 20 keV × 0.5 = 20 keV, so |q|² = 4.0 × 10⁻⁴ MeV². The electron's speed is |p|/Eₑ = 0.039, slow enough for the static limit, and e² = 4πα = 4π/137.036 = 0.091701.
Static amplitude: M = −4mₑm_μ e²/|q|² = −(4 × 0.51100 MeV × 105.658 MeV) × 0.091701 ÷ (4.0 × 10⁻⁴ MeV²) = −215.965 × 0.091701/(4.0 × 10⁻⁴) = −4.95 × 10⁴. MeV² over MeV² leaves a pure number, as a 2 → 2 amplitude must be.
Born matching: Ṽ(q) = −M/(4mₑm_μ) = e²/|q|² = 0.091701 ÷ (4.0 × 10⁻⁴ MeV²) = +229 MeV⁻². Positive, so repulsive; mass dimension −2 is right for ∫d³r e^(−iq⋅r) V(r).
Position space: V = αħc/r = 14.400 eV Å ÷ 0.50 Å = +28.8 eV. In SI, e²/(4πε₀r) = 2.3071 × 10⁻²⁸ J m ÷ 5.0 × 10⁻¹¹ m = 4.614 × 10⁻¹⁸ J = 28.8 eV, the same number.
Scalar: its propagator i/(q² − m_φ²) → −i/|q|² at m_φ = 0 is opposite to the photon's +i/|q|², while vertices, densities 2m and ordering sign are unchanged, so M = +4.95 × 10⁴, Ṽ = −229 MeV⁻² and V(0.50 Å) = −28.8 eV.

AnswerM = −4.95 × 10⁴, Ṽ = +229 MeV⁻² and V(0.50 Å) = +28.8 eV, a repulsion; the massless scalar gives the same magnitudes with the opposite signs, an attraction.

MediumThe same pair now has |p| = 0.10 MeV in the centre-of-mass frame and scatters through θ = 90°, both spins up along the incoming electron's direction and unflipped. Using the exact bilinears, find J₁⁰, J₂⁰ and J₁⋅J₂, check q⋅J₁ = 0, and compare the full tree amplitude with its static value.

Kinematics: with m_μ = 105.6584 MeV, Eₑ = √(0.51100² + 0.10²) MeV = 0.52069 MeV, so vₑ = 0.192; E_μ − m_μ = 4.7 × 10⁻⁵ MeV and v_μ = 9.5 × 10⁻⁴. Take p₁ = |p|ẑ and p₃ = |p|x̂, so |q|² = 2|p|²(1 − cos 90°) = 0.0200 MeV² and t = −0.0200 MeV².
Charge densities: p₃ × p₁ lies along y while ξ†σξ = ẑ, so the spin term drops, and J₁⁰ = (Eₑ + mₑ) + |p|² cos θ/(Eₑ + mₑ) = 1.03169 MeV, which is 1.0095 × 2mₑ. For the muon J₂⁰ = 211.3168 MeV, 2m_μ to 2 parts in 10⁷.
Currents: J₁ = (p₁ + p₃) + i(p₁ − p₃) × ẑ = (0.10, 0.10i, 0.10) MeV and J₂ = −J₁, so J₁⋅J₂ = −0.0100 + 0.0100 − 0.0100 = −0.0100 MeV². Conservation: q = p₁ − p₃ = (−0.10, 0, 0.10) MeV gives q⋅J₁ = −0.0100 + 0.0100 = 0, and q⁰ = 0, so q_μ J₁^μ = 0.
Amplitude: M = e²(J₁⁰J₂⁰ − J₁⋅J₂)/t = 0.091701 × (218.014 + 0.010) MeV² ÷ (−0.0200 MeV²) = −999.7, against the static −4mₑm_μ e²/|q|² = −990.2.
Reading: the current product is 0.010/218.0 = 4.6 × 10⁻⁵ of the charge product, close to vₑv_μ/4 = 4.5 × 10⁻⁵, while the electron's own density, J₁⁰/2mₑ = 1.0095, carries almost all of the 0.95% gap, an order-vₑ² correction of the kind topic 9.09 collects into Mott's factor.

AnswerJ₁⁰ = 1.0317 MeV, J₂⁰ = 211.317 MeV, J₁⋅J₂ = −0.0100 MeV² and q⋅J₁ = 0; M = −999.7 against the static −990.2. The magnetic term is 5 × 10⁻⁵ of the Coulomb term, and the 0.95% gap is the electron's order-v² charge-density correction.

HardAn electron and a positron collide in their centre-of-mass frame with |p| = 20 keV and scatter through θ = 60°. With |in⟩ = b†₁d†₂|0⟩ and ⟨out| = ⟨0|d₄b₃, fix the sign of the t-channel term, evaluate it in the static limit, bound the annihilation term, and test the resulting potential against positronium's measured 1S–2S interval, 1.23361 × 10¹⁵ Hz.

Ordering: number d₄, b₃, ψ̄ and ψ at the first vertex, ψ̄ and ψ at the second, b†₁ and d†₂ as 1 to 8. The t-channel pairs (b₃ψ̄)(ψb†₁)(d₄ψ)(ψ̄d†₂) give the order 2 3 4 7 1 6 5 8, with 1 + 1 + 1 + 3 + 0 + 1 = 7 inversions: sign −1, so Mₜ = −e²[ū₃γ^μ u₁][v̄₂γ_μ v₄]/t.
Static limit: |q|² = 4|p|² sin²30° = 4.0 × 10⁻⁴ MeV² and t = −|q|²; ū₃γ⁰u₁ → 2mₑ and v̄₂γ⁰v₄ = v₂†v₄ → +2mₑ, so Mₜ = −e²(2mₑ)²/t = 4 × 0.26112 MeV² × 0.091701 ÷ (4.0 × 10⁻⁴ MeV²) = +239.45.
Potential: Ṽ = −Mₜ/(4mₑ²) = −e²/|q|² = −229 MeV⁻², so V = −α/r, an attraction. Dropping the ordering sign, with v̄γ⁰v positive, would have made the pair repel.
Annihilation: v̄₂γ⁰u₁ = 0 in this frame and each spatial current is at most about 2√2 mₑ, so with s = 4(mₑ² + |p|²) = 1.0461 MeV², |Mₛ| ≤ 8mₑ²e²/s = 2.0890 × 0.091701/1.0461 = 0.183, which is 7.6 × 10⁻⁴ of Mₜ: a contact term of relative order |q|²/mₑ².
Positronium: the Schrödinger levels for −α/r, quoted, with mᵣ = mₑ/2 are −mₑα²/(4n²), so 1S lies at −6.803 eV and 1S–2S = (3/16)mₑα² = 5.10213 eV. Dividing by h = 4.135668 × 10⁻¹⁵ eV s gives 1.23369 × 10¹⁵ Hz, 7 parts in 10⁵ above the measured value, the size of the order-α² corrections left out.

AnswerSign −1 from seven inversions; Mₜ = +239, Ṽ = −229 MeV⁻² and V = −α/r, an attraction; |Mₛ| ≤ 0.18, under 0.08% of Mₜ; positronium's 1S–2S comes out at 1.23369 × 10¹⁵ Hz against 1.23361 × 10¹⁵ Hz measured.