University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.1

The QED Lagrangian from local phase invariance

QED's interaction is not guessed: demand that the electron field's phase may be reset independently at every spacetime point, and a vector field, its coupling to the Noether current and Maxwell's sourced equations are forced. This lesson writes that Lagrangian and sorts what the symmetry decides from what dimension and experiment decide.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The free Dirac Lagrangian ψ̄(iγ^μ ∂_μ − m)ψ is unchanged when every value of ψ is multiplied by the same phase e^(iα), and Noether's theorem turns that into the conserved current ψ̄γ^μ ψ. Let the phase vary from point to point and the mass term still survives, but the derivative does not: ∂_μ compares ψ at neighbouring points whose phase conventions now differ, and the Lagrangian changes by −(∂_μ α)ψ̄γ^μ ψ. Restoring invariance takes a field that records how the convention changes across spacetime, a vector A_μ that shifts by −(1/e)∂_μ α and enters through D_μ = ∂_μ + ieA_μ.

The coupling that results, −eψ̄γ^μ ψ A_μ, is linear in A, attached to exactly the Noether current and free of any derivative of ψ. Adding the gauge-invariant kinetic term −(1/4)F_μν F^μν makes A a dynamical photon, and the field equations become the Dirac equation in a field and Maxwell's equations with the electron as their source. The same symmetry forbids a photon mass.

It does not fix the size of e, does not make the muon's charge equal the electron's, and does not exclude a gauge-invariant Pauli term: the first two are inputs that experiment fixes and renormalisation preserves, and the third is excluded by its dimension. And the redundancy that makes the construction work also leaves A impossible to quantise as it stands, so Unit 8's gauge-fixing term is added before any Feynman rule is read.

Simple definition: The QED Lagrangian, L = ψ̄(iγ^μ D_μ − m)ψ − (1/4)F_μν F^μν with D_μ = ∂_μ + ieA_μ, is the free Dirac Lagrangian made invariant under ψ → e^(iα)ψ, for any function α(x), by a vector field that shifts as A_μ → A_μ − (1/e)∂_μ α.
Example: Take α = ωt with ħω = 1.00 eV. Every energy the electron's wave displays drops by 1.00 eV, while A₀ → A₀ − ω/e lowers its potential energy eA₀ by the same 1.00 eV everywhere: energy differences, and so all measurements, are unchanged.

Local phase and the compensating shiftψ(x) → e^(iα)ψ(x), A_μ(x) → A_μ(x) − (1/e)∂_μ α(x)

Derived as the shift that cancels the −(∂_μ α)ψ̄γ^μ ψ the derivative leaves behind. Constant α is the global symmetry and moves nothing; the sign of e decides which way A moves.

α(x) real, in rad; ψ of mass dimension 3/2, A_μ of dimension 1 (GeV); e dimensionless, −|e| = −0.3028 for the electron

The covariant derivativeD_μ ψ = (∂_μ + ieA_μ)ψ → e^(iα) D_μ ψ

Derived. D_μ ψ transforms like ψ itself, so any term that was invariant under a global phase stays invariant under a local one once every ∂ acting on ψ becomes D.

D_μ ψ in GeV⁵⁄²; iD_μ = i∂_μ − eA_μ is UAQM-04.09's p_μ − eA_μ, quoted

The QED LagrangianL = ψ̄(iγ^μ D_μ − m)ψ − (1/4)F_μν F^μν with F_μν = ∂_μ A_ν − ∂_ν A_μ

Derived. Expanded, the interaction is −eψ̄γ^μ ψ A_μ: one power of A and no derivative of ψ, which is why topic 9.02 finds a single vertex and no seagull.

L in GeV⁴; F_μν in GeV², unchanged by the shift; m = 0.511 MeV; α = e²/(4π) = 1/137.036 in Heaviside-Lorentz units

Dirac equation in the field(iγ^μ ∂_μ − eγ^μ A_μ − m)ψ = 0

Derived. Its wave-equation reading is UAQM-04.09's minimal coupling, quoted; the difference here is that A_μ is itself a quantum field, not a given background.

From varying ψ̄; every term in GeV⁵⁄²; ψ in GeV³⁄², A_μ and m in GeV; e = −|e| for the electron

Maxwell's equations with the electron as source∂_μ F^μν = j^ν with j^ν = e ψ̄γ^ν ψ

Derived. ∂_ν ∂_μ F^μν = 0 identically, so the equation is consistent only if ∂_ν j^ν = 0, which the Dirac equation in the field guarantees.

From varying A_ν; both sides in GeV³; ν = 0 is Gauss's law ∇⋅E = eψ†ψ

A photon mass term is not invariant(1/2)m_γ² A_μ A^μ → (1/2)m_γ² A_μ A^μ − (m_γ²/e)A^μ ∂_μ α + (m_γ²/(2e²))∂_μ α ∂^μ α

Derived: nothing else in L cancels the change, so the symmetry forbids a photon mass. The Pauli term passes the symmetry and is excluded only by its dimension.

m_γ in GeV; each term of dimension 4; the change vanishes only for constant α; measured m_γ below 1 × 10⁻¹⁸ eV, quoted

A global phase is free; a local one breaks the derivative

Work in natural units, ħ = c = 1, with the metric (+, −, −, −) and Heaviside-Lorentz charge, so α = e²/(4π) = 1/137.036 and the electron's charge is e = −|e| = −0.3028. The free Dirac Lagrangian L₀ = ψ̄(iγ^μ ∂_μ − m)ψ of UAQM-09.05 is unchanged by ψ → e^(iα)ψ for constant α, and Noether's theorem (UAQM-09.01), quoted, gives the conserved current ψ̄γ^μ ψ of UAQM-04.04. Now let α depend on x. The mass term still cancels phase against phase, but ∂_μ(e^(iα)ψ) = e^(iα)[∂_μ ψ + i(∂_μ α)ψ], so L₀ → L₀ − (∂_μ α)ψ̄γ^μ ψ, derived: the Noether current times the gradient of the parameter, which is Noether's theorem read backwards. The reason is geometric. ∂_μ ψ subtracts ψ at x from ψ at x + dx, and once the phase convention may differ between the two points the difference measures the convention as well as the field. A phase turning by 1 rad per femtometre adds a spurious ħc/(1 fm) = 197 MeV to every momentum that −i∂ reads.

A field that carries the phase convention

Ask for a derivative that transforms like ψ, D_μ ψ → e^(iα)D_μ ψ, and try D_μ = ∂_μ + ieA_μ with a new vector field. Then D′_μ ψ′ = e^(iα)[∂_μ ψ + i(∂_μ α)ψ + ieA′_μ ψ], equal to e^(iα)D_μ ψ only if A′_μ = A_μ − (1/e)∂_μ α, derived. Dimensions agree: [∂_μ α] = 1 and e is dimensionless, so [A] = 1, as for Unit 8's photon. Replacing ∂ by D in L₀ adds one term, −eψ̄γ^μ ψ A_μ, whose shift, +(∂_μ α)ψ̄γ^μ ψ, cancels the change found above exactly. UAQM-04.09 checked that a given equation survives a gauge transformation; here the logic runs the other way, and three features of the coupling come out forced: it is linear in A, because the Dirac Lagrangian has one derivative and so one D; it puts no derivative on ψ; and it couples to exactly the Noether current. A scalar's |D_μ φ|² has two D's and so the seagull e²A_μ A^μ |φ|² of topic 8.09; the Dirac field has none.

The photon's own term, and why it has no mass

For A_μ to be a field rather than a fixed background it needs a kinetic term that is invariant on its own. The commutator [D_μ, D_ν]ψ = ieF_μν ψ, with F_μν = ∂_μ A_ν − ∂_ν A_μ, identifies the invariant object: under the shift F changes by −(1/e)(∂_μ ∂_ν α − ∂_ν ∂_μ α) = 0, derived. Of the two Lorentz scalars of dimension four built from it, F_μν F̃^μν is a total derivative that alters no field equation, so −(1/4)F_μν F^μν = (E² − B²)/2 is the one that matters, its −1/4 fixed by Maxwell's energy density (E² + B²)/2. A photon mass cannot be added: (1/2)m_γ² A_μ A^μ changes by −(m_γ²/e)A^μ ∂_μ α + (m_γ²/(2e²))(∂α)², and nothing else in L cancels it, so local phase invariance predicts a strictly massless photon, derived. The measured bound, m_γ below 1 × 10⁻¹⁸ eV (quoted), puts its reduced Compton wavelength ħ/(m_γ c) at no less than 1.97 × 10¹¹ m, about 1.3 AU.

The field equations: Dirac in a field, Maxwell with a source

Varying ψ̄ in L = ψ̄(iγ^μ D_μ − m)ψ − (1/4)F² gives (iγ^μ D_μ − m)ψ = 0. Varying A_ν, with ∂L/∂(∂_μ A_ν) = −F^μν and ∂L/∂A_ν = −eψ̄γ^ν ψ, gives ∂_μ F^μν = eψ̄γ^ν ψ, both derived. The ν = 0 component is Gauss's law ∇⋅E = eψ†ψ, so one electron carries charge e = −|e|. The equation checks itself: ∂_ν ∂_μ F^μν = 0 because F is antisymmetric, so it is consistent only if ∂_ν(ψ̄γ^ν ψ) = 0, and subtracting the Dirac equation from its adjoint shows that it is, the two eA terms cancelling. Coupling A to a non-conserved current would be a contradiction. Limits: at e = 0 the free theories of Advanced QM and Unit 8 return, and for a static proton A⁰ = |e|/(4πr), so the electron's potential energy is eA⁰ = −α/r. The ground state has ⟨1/r⟩ = 1/a₀, so ⟨eA⁰⟩ = −αħc/a₀ = −27.21 eV, twice hydrogen's measured 13.6 eV binding energy, as the virial theorem requires.

Before quantising: add Unit 8's gauge fixing

The redundancy that made the construction work also blocks quantisation. A pure-gauge configuration A_μ = ∂_μ λ costs no action, so the kinetic operator of −(1/4)F² annihilates it and, as topic 8.04 showed, has no inverse and no propagator. QED is therefore quantised with topic 8.02's gauge-fixing term added, quoted: L_QED = ψ̄(iγ^μ D_μ − m)ψ − (1/4)F_μν F^μν − (1/(2ξ))(∂_μ A^μ)². The new term contains only A, so it leaves the Dirac equation and the coupling alone and turns the photon equation into ∂²A^ν − (1 − 1/ξ)∂^ν(∂_μ A^μ) = eψ̄γ^ν ψ, derived, which in Feynman gauge, ξ = 1, is simply ∂²A^ν = j^ν. The term is deliberately not gauge-invariant; what survives is current conservation, which Unit 8 showed makes ξ drop out between physical states. For QED nothing more is needed, and this Lagrangian, with its single interaction −eψ̄γ^μ ψ A_μ, is what topic 9.02 turns into Feynman rules.

What the symmetry leaves open: the Pauli term and the muon

Local phase invariance fixes the form of the coupling, not everything else. The Pauli term c ψ̄σ^μν ψ F_μν is invariant, since F is and the phases of ψ̄ and ψ cancel. What removes it is dimension: 3/2 + 3/2 + 2 = 5, so [c] = −1, which power counting (topic 6.01, stated) excludes from a renormalisable Lagrangian. Experiment agrees: the electron's measured anomaly aₑ = 0.00115965 is what loops produce, α/(2π) = 0.0011614 at one loop (topic 14.03, quoted), 0.15% away. Nor does the symmetry fix charges. A muon field transforming as Ψ → e^(iqα)Ψ, with the same α(x), keeps its Lagrangian invariant through D_μ Ψ = (∂_μ + iqeA_μ)Ψ for any real q, so giving the muon the electron's charge, q = 1, is an input. Muonium spectroscopy finds the two charges equal within a few parts in 10⁹ (quoted), and topic 12.07, stated, shows renormalisation keeps them equal: the vertex and self-energy corrections cancel in the charge, leaving only the photon's own, common to every species.

Change one variable at a time

Make the relationship visible.

Interactive model

Phase amplitude a1.50 rad

Phase periods across the window n1.0

Marker position x0.8 λ

Start at n = 0: the phase is global, the wave only slides and no gauge field appears. Raise n and a: crests bunch and spread, −i∂ₓ swings by ±an/3 in units of k, eAˣ swings with it exactly, and −iDₓ never leaves k.

LOCAL PHASE AT MARKER-0.16 rad

CANONICAL WAVENUMBER0.503 k

GAUGE FIELD × CHARGE-0.497 k

MECHANICAL WAVENUMBER1.000 k

Live interpretationLOCAL PHASE AT MARKER: −0.16 rad. CANONICAL WAVENUMBER: 0.503 k. GAUGE FIELD × CHARGE: −0.497 k. MECHANICAL WAVENUMBER: 1.000 k

Catch the common trap

Explain before calculating.

Misconception check

Local phase invariance does not, on its own, give minimal coupling or g = 2. The symmetry demands only that ψ enter through D_μ and A through F_μν, and anything built from those is allowed, including the Pauli term c ψ̄σ^μν ψ F_μν, which would shift the electron's g away from 2 by 8m/|e| times its coefficient c. What keeps it out of the QED Lagrangian is a separate principle: its dimension is 5, so [c] = −1 and the term is not renormalisable. Dirac's g = 2 is therefore gauge invariance plus renormalisability, and the measured g − 2 = 0.0023193 is not a failure of the gauge principle: loops generate an effective Pauli term at order α/π. Whoever says gauge invariance gives g = 2 cannot explain why a gauge-invariant term is absent from the Lagrangian yet visible in the data.

Each term below is added to ψ̄(iγ^μ D_μ − m)ψ − (1/4)F_μν F^μν. Which one is unchanged by ψ → e^(iα)ψ, A_μ → A_μ − (1/e)∂_μ α for every function α(x), yet is still left out of the QED Lagrangian because of its mass dimension?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron moves along x with momentum 3.000 keV and no field present. Apply the local phase α(t, x) = ωt − κx with ħω = 2.00 eV and κ = 10.00 nm⁻¹. Find what i∂ₜ and −i∂ₓ now read, the new eA₀ and eAˣ, and show that the mechanical energy E − eA₀ and momentum p − eAˣ are unchanged.

With ħ = c = 1, ψ ∝ e^(−iEt) e^(ipx) with E = √(m² + p²) = √(510.999² + 3.000²) keV = 511.008 keV. The phase e^(iα) multiplies it by e^(iωt) e^(−iκx), so i∂ₜ now reads E − ω = E − 2.00 eV and −i∂ₓ reads p − κ.
Restoring ħ and c, κ corresponds to ħcκ = 197.327 eV nm × 10.00 nm⁻¹ = 1.973 keV, so −i∂ₓ reads 3.000 − 1.973 = 1.027 keV.
The field shifts by A′_μ = −(1/e)∂_μ α, with ∂₀α = ω and ∂ₓα = −κ. So eA′₀ = −ω = −2.00 eV, and eA′ₓ = +κ, which for the upper-index component is eA′ˣ = −κ = −1.973 keV.
Mechanical values: (E − ω) − (−ω) = E = 511.008 keV, and 1.027 − (−1.973) = 3.000 keV. Both are unchanged, so (E − eA₀)² − (p − eAˣ)² = m² still holds.
No field has been created: F′_μν = −(1/e)(∂_μ ∂_ν α − ∂_ν ∂_μ α) = 0. The new potentials are pure gauge, and their size, A′₀ = 2.00 eV/0.3028 = 6.60 eV in natural units, measures nothing.

Answeri∂ₜ reads E − 2.00 eV and −i∂ₓ reads 1.027 keV; eA₀ = −2.00 eV and eAˣ = −1.973 keV; the mechanical values 511.008 keV and 3.000 keV are unchanged, and F = 0.

MediumSuppose Proca's mass term (1/2)m_γ² A_μ A^μ from topic 8.08 were added to the QED Lagrangian. Show that it breaks local phase invariance, find the static potential of a point charge Q, and for m_γ at the quoted bound 1 × 10⁻¹⁸ eV find the range and the fractional shortfall of the potential against Coulomb's at Earth's radius, 6.371 × 10⁶ m, and at 1 AU, 1.496 × 10¹¹ m.

Under the shift, (1/2)m_γ²(A_μ − (1/e)∂_μ α)(A^μ − (1/e)∂^μ α) = (1/2)m_γ² A_μ A^μ − (m_γ²/e)A^μ ∂_μ α + (m_γ²/(2e²))(∂α)². Nothing else in L produces such terms, so invariance is lost for every non-constant α.
Varying A_ν now gives ∂_μ F^μν + m_γ² A^ν = j^ν. Applying ∂_ν kills the first term, so m_γ² ∂_ν A^ν = ∂_ν j^ν = 0: the Lorenz condition is now forced by the mass, not chosen as a gauge.
For a static charge, j⁰ = Qδ³(r) and ∂_μ F^μ0 = ∇⋅E = −∇²A⁰, so (−∇² + m_γ²)A⁰ = Qδ³(r). The solution, quoted from topic 1.09, is the Yukawa form A⁰ = Q exp(−m_γ r)/(4πr), which returns Coulomb's Q/(4πr) at m_γ = 0.
The range is 1/m_γ, or ħc/(m_γ c²) = 197.327 MeV fm ÷ 1 × 10⁻¹⁸ eV = 1.973 × 10¹¹ m, which is 1.32 AU.
The fractional shortfall is 1 − exp(−m_γ r). At Earth's radius m_γ r = 6.371 × 10⁶ m ÷ 1.973 × 10¹¹ m = 3.23 × 10⁻⁵, so the loss is 3.23 × 10⁻⁵. At 1 AU, m_γ r = 0.758 and the loss is 1 − e^(−0.758) = 0.531.

AnswerRange ħc/(m_γ c²) = 1.97 × 10¹¹ m (1.32 AU); the potential falls short of Coulomb's by 3.2 × 10⁻⁵ at Earth's radius and by 53% at 1 AU, so a bound this tight needs fields measured on solar-system scales.

HardAdd the Pauli term c ψ̄σ^μν ψ F_μν to the electron's Lagrangian. Find [c], reduce the term to a magnetic energy in the non-relativistic limit using σⁱʲ = εⁱʲᵏ Σᵏ, and find the c that reproduces the measured aₑ = 1.15965 × 10⁻³, the anomaly frequency it gives in 1.000 T, and the c the muon's a_μ = 1.16592 × 10⁻³ would need. What does the comparison say?

[ψ] = 3/2 twice and [F] = 2 give dimension 5, so [c] = −1. F and ψ̄σ^μν ψ are each invariant, so the term passes the symmetry and fails only power counting.
With B = ∇ × A, Fᵢⱼ = −εⁱʲᵏ Bᵏ, so σⁱʲ Fᵢⱼ = −2Σ⋅B; the σ⁰ⁱ terms mix upper and lower components and drop at leading order. On upper components ψ̄ ≈ ψ†, so L gains −2cψ†Σ⋅Bψ and the energy gains +2cΣ⋅B = 4c S⋅B: a moment μ = −4cS.
The anomalous moment of g = 2(1 + a) is a(e/m)S, so −4c = ae/m and c = −(a/4)(e/m) = (a/4)|e|/m for the electron. With |e| = 0.30282 and mₑ = 5.1100 × 10⁻⁴ GeV, c = 1.15965 × 10⁻³ × 0.30282/(2.0440 × 10⁻³ GeV) = 0.1718 GeV⁻¹, so 1/c = 5.82 GeV.
The extra spin-flip energy 2aμ_B B equals a times the cyclotron quantum, so fₐ = a|e|B/(2πmₑ) = 1.15965 × 10⁻³ × 27.9925 GHz = 32.46 MHz at 1.000 T, the frequency a Penning-trap g − 2 measurement counts.
For the muon, m_μ = 0.105658 GeV: c = 1.16592 × 10⁻³ × 0.30282/(0.422632 GeV) = 8.354 × 10⁻⁴ GeV⁻¹, so 1/c = 1.20 TeV. The ratio of the two c's is 205.7 = (m_μ/mₑ)(aₑ/a_μ).
Both anomalies sit near α/(2π) = 1.1614 × 10⁻³, so c ≈ α|e|/(8πm) tracks each particle's own mass. A fundamental Pauli term would carry one scale of new physics; this pattern is each particle's own loops (topic 14.03).

Answer[c] = −1. The electron needs c = 0.172 GeV⁻¹ (1/c = 5.82 GeV), giving a 32.46 MHz anomaly frequency at 1 T; the muon needs 8.35 × 10⁻⁴ GeV⁻¹ (1/c = 1.20 TeV). c scales as 1/m, the mark of loops.