University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.2

QED Feynman rules: the photon vertex and propagator

Every QED amplitude in this course is assembled from the pieces built here. The fermion lines you can already write; what is new is the photon. See where its one vertex comes from, why there is no second, and learn to follow its Lorentz index to wherever it ends: a polarisation vector, another line, or the same line again.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

QED's interaction is a single term, ℒᵢₙₜ = −eψ̄γ^μψA_μ, and its Feynman rules are not postulated: they are read off with the machinery of Units 3, 7 and 8. Because the term differentiates no field, Hᵢₙₜ = −ℒᵢₙₜ (topic 2.03); because the Dirac Lagrangian is first order in D_μ, the coupling is linear in A_μ and, unlike topic 8.09's charged scalar, has no seagull. Contracting three different fields at a point gives one vertex, −ieγ^μ; contracting A_μ with A_ν gives −ig_μν/(q² + iε) in Feynman gauge; contracting A_μ with an external photon gives ε_μ coming in and ε*_μ going out.

The fermion side, spinors, S_F and the loop's −1, comes unchanged from topic 7.04, so the new bookkeeping is a Lorentz index: every vertex opens a μ on a fermion chain, and the rest of the diagram must close it, on a polarisation vector, through g_μν onto another vertex, or back on the same line. The muon brings its own copy of the term and none joins it to the electron, so the two lepton numbers are conserved vertex by vertex. The rules hold at every order in |e| = √(4πα) = 0.303, but they are not yet the whole theory: loops diverge, and their counterterm vertices join the list in Unit 12; bound states, where each further Coulomb exchange costs α/v rather than α, need ladders summed to all orders; and near 91 GeV the photon is joined by a Z boson that QED does not contain.

Simple definition: QED's Feynman rules turn each term of the perturbation series into vertices −ieγ^μ, propagators and external factors, every photon index μ opened at a vertex and closed on a polarisation vector, a photon propagator or the same fermion line.
Example: In e⁻μ⁻ → e⁻μ⁻ the two vertices give (−ie)² = −e² and the propagator −ig_μν/q², so iM = ie²[ū₃γ^μu₁][ū₄γ_μu₂]/q²: one diagram, with e² = 4πα = 0.0917, dimensionless as a four-point amplitude must be.

Interaction Hamiltonian, read off ℒHᵢₙₜ = −ℒᵢₙₜ = e ψ̄γ^μψ A_μ

Derived: no field is differentiated, so the canonical momenta are the free ones and topic 2.03 applies. Linear in A_μ: no A² term, so no seagull.

e = −|e|, the electron's charge, a pure number; ψ in GeV³⁄², A_μ in GeV, Hᵢₙₜ in GeV⁴

The QED vertex−ieγ^μ, with (2π)⁴δ⁴(Σpᵢₙ − Σpₒᵤₜ)

Derived: ψ̄, ψ and A_μ are different fields, so no 1/k! survives. The muon has the same vertex, and none joins ψ̄ₑ to ψ_μ: each lepton number is conserved.

γ^μ a 4 × 4 matrix: row index on the arrow leaving, column index on the arrow arriving; μ = 0 to 3 on the photon

Photon propagator, Feynman gaugeD_μν(q) = −ig_μν/(q² + iε)

Quoted from topic 8.04; a contraction of A_μ with A_ν puts it in place. Its spatial parts, +i/q², match a scalar's; its time part, −i/q², has the opposite sign.

q the photon's four-momentum in GeV, routed either way; D_μν in GeV⁻²; gauge parameter ξ = 1

External photonsincoming: ε_μ(k, λ)outgoing: ε*_μ(k, λ)

Quoted from topic 8.06's LSZ step. At a vertex the index is absorbed as −ieε̸, with ε̸ = ε⁰γ⁰ − ε⃗⋅γ⃗; the conjugate matters for helicity states.

k in GeV with k² = 0; ε dimensionless, k⋅ε = 0, ε⋅ε* = −1; for k along z, ε^μ(±) = ∓(0, 1, ±i, 0)/√2

One photon exchanged between two lines(−ie)² J₁^μ (−ig_μν/q²) J₂^ν = ie² J₁⋅J₂/q²

Derived: g_μν lowers one index, so the two currents meet as a Minkowski product, and the result is dimensionless, as a four-point amplitude must be.

J₁^μ = ū(p₃)γ^μu(p₁), J₂ alike, in GeV; q² in GeV²; J₁⋅J₂ = J₁⁰J₂⁰ − J⃗₁⋅J⃗₂; e² = 4πα = 0.0917

Forward exchange numeratorJ₁⋅J₂ → 4p₁⋅p₂ = 4(E₁E₂ − p⃗₁⋅p⃗₂) = 4m₁m₂γᵣₑₗ

Derived: the time part 4E₁E₂ and the space part −4p⃗₁⋅p⃗₂ each change from frame to frame; only their sum is invariant. The figure plots all three.

as q → 0, ū(p, s′)γ^μu(p, s) = 2p^μδₛ′s; γᵣₑₗ the Lorentz factor of one particle in the other's rest frame

No derivative and no A², so Hᵢₙₜ = −ℒᵢₙₜ

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 9.01 supplies ℒ = ψ̄(iγ^μD_μ − m)ψ − ¼F_μνF^μν − (1/(2ξ))(∂⋅A)² with D_μ = ∂_μ + ieA_μ, e the electron's charge −|e|, so ℒᵢₙₜ = −eψ̄γ^μψA_μ. Two features of this term fix everything after it. It differentiates no field, so the momenta conjugate to ψ and A_μ are the free ones and topic 2.03's argument gives Hᵢₙₜ = −ℒᵢₙₜ = eψ̄γ^μψA_μ, derived. And the Dirac Lagrangian is first order in D_μ, so the coupling enters linearly: there is no e²A_μA^μ term and no two-photon contact vertex, where topic 8.09's |D_μφ|² produced the seagull 2ie²g^μν. Dimensions agree: [ψ] = 3/2 and [A] = 1 make [ψ̄γ^μψA_μ] = 4, so e is a pure number, |e| = √(4π/137.036) = 0.3028. The fermion side is quoted unchanged from topic 7.04: S_F along the arrow, u, ū, v and v̄ at the ends, chains read against the arrow, relative signs between diagrams that exchange fermions, and −1 per closed loop.

The vertex −ieγ^μ and its three indices

At order n the Dyson series brings (−i)ⁿ/n! times n copies of ∫d⁴x Hᵢₙₜ, and the n! cancels against relabelling the vertices, as in topic 3.01. At one vertex ψ̄, ψ and A_μ are three different fields, so each line can attach in only one way and no 1/k! survives: the vertex is −i times the coefficient in Hᵢₙₜ, −ieγ^μ, with (2π)⁴δ⁴ of the momenta meeting there, derived, the same as topic 3.07's i times the coefficient in ℒ. It carries three indices: a spinor row index for the arrow leaving, a column index for the arrow arriving, and the Lorentz index μ, which belongs to the photon line and stays open until that line's other end decides what it meets. The muon enters through its own copy, −eψ̄_μγ^νψ_μA_ν with ψ_μ the muon field and the same charge, so its vertex is the same −ieγ^μ. No term joins ψ̄ₑ to ψ_μ, so arrows keep their flavour through every vertex, and electron and muon numbers are each conserved, derived: μ⁻ → e⁻γ has no diagram at any order, a prediction the first worked example sets against experiment.

The photon propagator, and the sign of its time part

Contracting A_μ(x) with A_ν(y) between two vertices gives ⟨0|TA_μ(x)A_ν(y)|0⟩, which topic 8.04 found to be −g_μν times the massless D_F in Feynman gauge, quoted; in momentum space each internal photon is −ig_μν/(q² + iε), with q routed either way because it is even in q. Its spatial components, −igᵢᵢ = +i, match a massless scalar's i/(q² + iε); its time component, −ig₀₀ = −i, has the opposite sign, the trace of topic 8.02's negative-norm timelike photon. Nothing unphysical leaks out: topic 9.03 traces like-charge repulsion to exactly this sign, and topic 8.07 shows the timelike and longitudinal pieces cancel in physical sums. In a general covariant gauge the propagator is −i[g_μν − (1 − ξ)q_μq_ν/q²]/(q² + iε); the extra piece meets each vertex as q_μ times a current, which vanishes along a line with on-shell ends, topic 9.04's result, so these rules are stated at ξ = 1 and lose nothing. Dimensions: [D_μν] = −2, like any boson propagator, and its pole sits at q² = 0, where measurement puts the photon mass below 1 × 10⁻¹⁸ eV, quoted.

External photons: ε_μ coming in, ε*_μ going out

Topic 8.06's LSZ step, quoted, puts ε_μ(k, λ) where an incoming photon of momentum k and polarisation λ meets a vertex, and ε*_μ(k, λ) for an outgoing one: A_μ annihilates with ε_μe^(−ik⋅x) and creates with ε*_μe^(ik⋅x). For k = ω(1, 0, 0, 1) the linear vectors have upper components (0, 1, 0, 0) and (0, 0, 1, 0), and the helicity vectors are ε(±) = ∓(0, 1, ±i, 0)/√2; each obeys k⋅ε = 0 and ε⋅ε* = −1, negative because ε is spacelike in (+, −, −, −). At the vertex the index is absorbed into the chain as −ieε̸, with ε̸ = ε^μγ_μ = ε⁰γ⁰ − ε⃗⋅γ⃗, so a real ε⃗ along x gives ε̸ = −γ¹ and ε̸² = ε⋅ε = −1, a sign easily lost. The conjugate matters only for complex ε: an outgoing photon of helicity +1 carries ε*(+), and writing ε(+) there computes the other helicity. And one vertex with every line on shell is empty: (p + k)² = m² needs p⋅k = 0, but p⋅k = ω(E − |p⃗|cos θ) is positive for any real photon, 0.511 MeV × ω for an electron at rest, so every QED process has at least two vertices and starts at order e².

Following μ: the three ways a photon index closes

Every vertex opens one Lorentz index on a fermion chain, and a complete amplitude leaves none open, so each μ closes in one of three ways. On an external photon it becomes ε̸ or ε̸* inside the chain. Through a propagator to another line, g_μν lowers one index and the two chains meet as a Minkowski product: for e⁻μ⁻ → e⁻μ⁻, iM = (−ie)²[ū(p₃)γ^μu(p₁)](−ig_μν/q²)[ū(p₄)γ^νu(p₂)] = ie²Jₑ⋅J_μ/q², with q = p₁ − p₃ and J⋅J′ = J⁰J′⁰ − J⃗⋅J⃗′, derived. Through a propagator back to the same line, the index is contracted inside one chain: the self-energy's numerator holds γ^μ(p̸ − k̸ + m)γ_μ = −2(p̸ − k̸) + 4m by topic 7.06's identities, quoted; on a closed loop it sits inside the trace. Along one line the matrices stand in the order the photons meet the line, read against the arrow, so ū(p′)ε̸′*S_F(p + k)ε̸u(p) and ū(p′)ε̸S_F(p − k′)ε̸′*u(p) are different diagrams of one process. Dimensions check every closure: [u] = 1/2 makes a current GeV¹, and J⋅J′/q² is dimensionless, as a four-point amplitude must be.

The forward contraction is invariant only as a whole

As q → 0 a current with no spin flip tends to ū(p, s)γ^μu(p, s) = 2p^μ, UAQM-05.07's Gordon decomposition at q = 0, quoted, so forward exchange between any two charged fermions has numerator J₁⋅J₂ = 4p₁⋅p₂ = 4(E₁E₂ − p⃗₁⋅p⃗₂), derived. The time part 4E₁E₂ is carried by g₀₀ = +1 and the space part −4p⃗₁⋅p⃗₂ by gᵢᵢ = −1; neither is invariant, only their sum, 4m₁m₂γᵣₑₗ, with γᵣₑₗ the Lorentz factor of one particle seen from the other. Two charges moving in parallel at β = 0.8 have γ = 5/3: in units of 4m₁m₂ the time part is γ² = 2.78, the space part −γ²β² = −1.78, and the sum exactly 1, the value they see at rest together, the magnetic attraction of parallel currents cancelling the growth of the electric term. Head-on at the same speeds the space part adds, and the sum is γ²(1 + β²) = 4.56. Keeping only the time part, the electric J⁰J⁰ term, is right in a frame where one particle is at rest and wrong in every other. The figure follows the three pieces.

Change one variable at a time

Make the relationship visible.

Interactive model

Speed β₁ of lepton 10.60

Speed β₂ of lepton 20.60

Angle α between the velocities60 °

Set both speeds to 0.80 and α to 0°: the time part climbs to 2.78, the space part falls to −1.78, and the sum is exactly 1, what the two charges see at rest together. Swing α to 180° and the space part adds instead, taking the sum to 4.56.

TIME PART γ₁γ₂1.563

SPACE PART −γ₁γ₂β₁β₂cos α-0.281

SUM = γᵣₑₗ1.281

RELATIVE SPEED βᵣₑₗ0.625

Live interpretationTIME PART γ₁γ₂: 1.563. SPACE PART −γ₁γ₂β₁β₂cos α: −0.281. SUM = γᵣₑₗ: 1.281. RELATIVE SPEED βᵣₑₗ: 0.625

Catch the common trap

Explain before calculating.

Misconception check

Carrying topic 8.09's scalar rules over to the electron: a momentum-dependent vertex −ie(p + p′)^μ, and a seagull joining two photons at a point. Neither is QED. The seagull came from |D_μφ|², quadratic in D_μ; ψ̄iγ^μD_μψ is linear in D_μ, so ℒ holds no A_μA^μ term, and a Dirac contact term ψ̄ψA_μA^μ would have mass dimension 5, needing a coupling of dimension −1, like the Pauli term. Nor is momentum dependence lost: the vertex is the constant matrix γ^μ, and once it is sandwiched, UAQM-05.07's Gordon decomposition, ū(p′)γ^μu(p) = ū(p′)[(p + p′)^μ + iσ^μνq_ν]u(p)/(2m) with q = p′ − p, quoted, returns the scalar's convection current plus a spin term with g = 2. A seagull would also break what works: topic 10.04 finds Compton's two Dirac diagrams obey k_μM^μν = 0 alone, and a contact term would add a multiple of k⋅ε′*, non-zero in general away from the forward direction.

At lowest order in Feynman gauge, with q = p₁ − p₃, which expression is iM for e⁻(p₁) μ⁻(p₂) → e⁻(p₃) μ⁻(p₄)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUsing only the QED Feynman rules, write iM at lowest order for e⁻(p₁) μ⁻(p₂) → e⁻(p₃) μ⁻(p₄), count its diagrams, check its mass dimension, and find the lowest order in e at which μ⁻ → e⁻γ could occur.

Every vertex is ψ̄_ℓγ^μψ_ℓA_μ for a single flavour ℓ, so the electron's arrow runs unbroken from p₁ to p₃ and the muon's from p₂ to p₄. The only way to join the two lines at order e² is one photon between them, carrying q = p₁ − p₃, so that t = q².
Read each line against its arrow, as topic 7.04 prescribes: ū(p₃)(−ieγ^μ)u(p₁) and ū(p₄)(−ieγ^ν)u(p₂), joined by −ig_μν/(q² + iε). The constants give (−ie)²(−i) = ie², and g_μν lowers the muon's index: iM = ie²[ū(p₃)γ^μu(p₁)][ū(p₄)γ_μu(p₂)]/t, with e² = 4πα = 0.0917.
There is no second diagram with p₃ and p₄ exchanged: it would need a vertex holding ψ̄ₑ and ψ_μ, which ℒᵢₙₜ does not contain. e⁻e⁻ → e⁻e⁻, built from the same vertex, has both, as Unit 10 shows.
Dimensions: [u] = 1/2, so each current carries GeV¹, their product GeV², and dividing by t leaves M dimensionless, as 4 − n requires for n = 4 external lines.
μ⁻ → e⁻γ needs a line that enters as a muon and leaves as an electron. No vertex changes flavour, so no diagram exists at any order in e and QED predicts a rate of exactly zero. The MEG experiment's 2016 bound on the charge-conjugate decay μ⁺ → e⁺γ, a branching ratio below 4.2 × 10⁻¹³ at 90% confidence, quoted, is consistent with it.

AnswerOne diagram: M = e²[ū(p₃)γ^μu(p₁)][ū(p₄)γ_μu(p₂)]/t with e² = 0.0917, dimensionless. μ⁻ → e⁻γ has no diagram at any order in QED.

MediumAn electron of energy 1.000 GeV meets a muon of momentum 0.500 GeV head-on along z. For forward single-photon exchange, split Jₑ⋅J_μ into what the photon's time and space components contribute, repeat in the muon's rest frame, and find M at t = −1.0 × 10⁻⁴ GeV².

As t → 0 each current loses its spin flip and tends to 2p^μ, UAQM-05.07's Gordon decomposition at q = 0, quoted, so Jₑ⋅J_μ = 4pₑ⋅p_μ = 4EₑE_μ − 4p⃗ₑ⋅p⃗_μ: the first term carried by g₀₀ = +1, the second by gᵢᵢ = −1.
Kinematics, with m_μ = 0.10566 GeV: E_μ = √(0.500² + 0.10566²) = 0.5110 GeV. The electron's |p⃗| = √(1.000² − 0.000511²) GeV = 0.99999987 GeV, equal to 1.000 GeV to six figures, and head-on p⃗ₑ⋅p⃗_μ = −0.500 GeV².
Time part 4EₑE_μ = 4(1.000)(0.5110) = 2.044 GeV²; space part −4p⃗ₑ⋅p⃗_μ = +2.000 GeV²; total 4.044 GeV². The magnetic part is almost as large as the electric one: keeping J⁰J⁰ alone would come out low by a factor 1.98.
In the muon's rest frame J⃗_μ = 0, so the time part is everything. There the electron has E′ = pₑ⋅p_μ/m_μ = 1.011/0.10566 = 9.569 GeV, and 4m_μE′ = 4.044 GeV²: the same total, split differently, as an invariant must be. γᵣₑₗ = E′/mₑ = 1.87 × 10⁴.
Then M = e²Jₑ⋅J_μ/t = 0.0917 × 4.044/(−1.0 × 10⁻⁴) = −3.71 × 10³, dimensionless. The forward form is safe here: the full result differs by a fraction of order |t|/(2pₑ⋅p_μ) = 5 × 10⁻⁵.

AnswerJₑ⋅J_μ = 2.044 + 2.000 = 4.044 GeV² head-on, and 4.044 + 0 GeV² in the muon's rest frame; M ≈ −3.71 × 10³ at t = −1.0 × 10⁻⁴ GeV².

HardUse the forward currents to derive dσ/dt for e⁻μ⁻ → e⁻μ⁻ at small |t| in terms of the relative speed βᵣₑₗ, check its static limit against Rutherford's formula, and evaluate it for 150 GeV muons striking electrons at rest, integrated over 1.0 × 10⁻³ GeV² ≤ |t| ≤ 1.0 × 10⁻² GeV², in microbarns.

In the centre-of-mass frame both leptons turn through the same angle, and as t → 0 each current loses its spin flip and tends to 2p^μ. The rules then give M = e²(2p₁)⋅(2p₂)/t = 4e²(p₁⋅p₂)/t for each of the four spin states, so the spin average is |M|² = 16e⁴(p₁⋅p₂)²/t², an invariant valid in every frame.
Unit 5's dσ/dΩ = |M|²/(64π²s) for elastic scattering, with t = −2p*²(1 − cos θ), gives dσ/dt = |M|²/(64πsp*²) = |M|²/(16πλ), where λ = 4sp*² = 4[(p₁⋅p₂)² − mₑ²m_μ²]. With p₁⋅p₂ = mₑₘ_μγᵣₑₗ, dσ/dt = 4πα²γᵣₑₗ²/((γᵣₑₗ² − 1)t²) = 4πα²/(βᵣₑₗ²t²), derived: GeV⁻⁴, as dσ/dt must be, and free of spin and frame.
Static limit: with the muon as a fixed target and the electron slow, of speed v and momentum p, t = −4p²sin²(θ/2) and dt = (p²/π)dΩ, so dσ/dΩ = α²/(4p²v²sin⁴(θ/2)): Rutherford's formula for Z = 1, Quantum Mechanics II's Born result (UQMII-15.09), quoted.
For 150 GeV muons on electrons at rest, p₁⋅p₂ = mₑE_μ = 0.07665 GeV² and γᵣₑₗ = E_μ/m_μ = 1420, so βᵣₑₗ² = 1 − 5.0 × 10⁻⁷ and dσ/dt = 4πα²/t², with 4πα² = 6.692 × 10⁻⁴.
Integrate: σ = 4πα²(1/|t|ₘᵢₙ − 1/|t|ₘₐₓ) = 6.692 × 10⁻⁴ × 900 GeV⁻² = 0.602 GeV⁻². With 1 GeV⁻² = 0.3894 mb, σ = 0.235 mb = 235 μb.
Check against the full spin-averaged result, which topic 9.09 derives and is quoted here: it multiplies the forward |M|² by 1 + t/(2p₁⋅p₂) + t(mₑ² + m_μ²)/(4(p₁⋅p₂)²) + t²/(8(p₁⋅p₂)²), which lowers the integral by 1.8%, to 230 μb. The forward rules hold to a few percent while |t| ≪ 2p₁⋅p₂ = 0.153 GeV².

Answerdσ/dt = 4πα²/(βᵣₑₗ²t²) at small |t|, Rutherford's formula in the static limit; for 150 GeV muons on electrons at rest σ ≈ 235 μb over 10⁻³ to 10⁻² GeV², 1.8% above the full result, 230 μb.