University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.4
Tree-level Ward identity: kμ Mμ = 0 on a fermion line
This identity is what lets every QED calculation from here on sum photon polarisations with −gμν and pick any gauge for internal photons. Here you prove it by hand: contract a photon with its own momentum, watch the diagrams along one fermion line cancel in neighbouring pairs, and find the exact step where on-shell spinors are needed.
Build the model
Connect the measurement to the mechanism.
Topic 8.06 argued in outline that an amplitude with an external photon has the form εμ Mμ and that current conservation forces kμ Mμ = 0 once the charged particles are on shell, and topic 8.05 showed that Lorentz invariance demands it. This topic checks the claim where it can be checked completely: at tree level, with topic 9.02's rules, along one open fermion line. Put the photon into every slot along the line and contract its vertex −ieγμ with kμ.
The proof rests on one exact line of algebra, k̸ = (p̸ + k̸ − m) − (p̸ − m): the photon's slashed momentum is the difference of the inverse propagators on either side of it. Each contracted insertion therefore collapses into the difference of two strings with the photon removed, neighbouring slots cancel, and the sum telescopes to two end terms, which die only because the external spinors obey the Dirac equation. Nothing about the other vertices is used, so the result holds with the other photons' indices free, and the same algebra applied to an internal photon removes the ξ-dependent part of its propagator.
The proof stops where its inputs stop: a single diagram does not satisfy the identity, an off-shell end leaves a contact term, and a closed fermion loop needs a shift of loop momentum that waits for topic 12.06.
- Simple definition
- At tree level, replacing a photon's polarisation εμ by its momentum kμ and summing over every place it can attach along one open fermion line with on-shell ends gives exactly zero, whatever else attaches to the line.
- Example
- One vertex, both ends on shell: qμ ū(p + q)γμ u(p) = ū(p + q)[(p̸ + q̸ − m) − (p̸ − m)]u(p) = 0. With two other vertices, a 1.00 MeV electron emitting a 0.24 MeV photon gives three diagrams of 10.61, 2.51 and −13.12, common factor removed: sum 0.00.
Derived from k̸ = (p̸ + k̸ − m) − (p̸ − m). The photon is gone: one string with k not yet absorbed, minus one with it absorbed.
SF(p) = i(p̸ − m)⁻¹, mass dimension −1 (GeV⁻¹); k the photon momentum entering the line, GeV; e = −|e|, dimensionless
Derived. The adjacent propagator is cancelled outright, leaving e u(p) on shell. The outgoing end gives ū(p′)(−iek̸)SF(p′ − k) = −e ū(p′).
u(p) of mass dimension ½ with ūu = 2m; the last term vanishes by the Dirac equation (p̸ − m)u(p) = 0, UAQM-04.06
Derived for any vertices V₁, …, Vₙ, so the other photon indices stay free: kμ M(μν…) = 0. No single slot vanishes.
Bⱼ: the line with the photon removed, its j propagators nearest u(p) without k and the rest with k added; n other vertices
Derived. Each diagram fails by e times the photon-free amplitude, so only the pair is gauge invariant, which is why topic 10.04's two Compton diagrams come together.
1: photon next to u(p); 2: next to ū(p′); Γ the other vertex, bringing momentum q; p′ = p + q + k; common factors dropped
Derived, the n = 0 case. It kills topic 8.04's (1 − ξ)qμqν/q² term in topics 9.05 to 9.09, so their amplitudes hold in every covariant gauge.
current in GeV; likewise qμ v̄(p₂)γμ u(p₁) = 0 for q = p₁ + p₂, by v̄(p₂)(p̸₂ + m) = 0
What is claimed, and with which rules
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 9.02's rules give the vertex −ieγμ with e = −|e|, the propagator SF(p) = i(p̸ + m)/(p² − m² + iε), which off its pole is i(p̸ − m)⁻¹, and external spinors that obey UAQM-04.06's Dirac equations, quoted: (p̸ − m)u(p) = 0, ū(p′)(p̸′ − m) = 0, and (p̸ + m)v(p) = 0 for an antifermion. An open fermion line is a spinor chain with ū or v̄ at one end and u or v at the other, through n other vertices V₁, …, Vₙ: external photons, internal photons going to other lines, or an external field, each with its momentum fixed. A further photon of momentum k can sit in any of the n + 1 slots between them, and its amplitude is εμ Mμ with Mμ summed over the slots. The claim is kμ Mμ = 0. Topic 8.05 showed Lorentz invariance requires it and topic 8.06 traced it to current conservation in outline; here it is derived for every tree on such a line.
The photon's k̸ is a difference of inverse propagators
Let the photon carry k into the line where the fermion momentum changes from p to p + k. Contracting the vertex with kμ gives −iek̸, and k̸ = (p̸ + k̸ − m) − (p̸ − m) holds identically. Between two propagators, SF(p + k)(−iek̸)SF(p) = (i)(−ie)(i)[(p̸ − m)⁻¹ − (p̸ + k̸ − m)⁻¹] = e[SF(p) − SF(p + k)], derived: the photon has gone, leaving the line with a single propagator, taken either before or after k entered. Both sides have mass dimension −1. The scalar shadow of the step is the denominators, (p + k)² − m² − (p² − m²) = 2p⋅k + k². As k → 0 the right side becomes −e kμ ∂SF/∂pμ, the tree-level form of the derivative identity Ward published in 1950, which topic 12.07 carries to all orders. For an emitted photon replace k by −k: every sign below flips together, and nothing else changes.
At the ends, the Dirac equation does the work
The two outer slots have a propagator on one side only. Next to the incoming spinor the same split gives SF(p + k)(−iek̸)u(p) = e u(p) − e(p̸ + k̸ − m)⁻¹(p̸ − m)u(p), and the second term vanishes because (p̸ − m)u(p) = 0: the slot leaves +e times the line with the photon removed. Next to the outgoing spinor, ū(p′)(−iek̸)SF(p′ − k) = −e ū(p′), because ū(p′)(p̸′ − m) = 0. An antifermion end works alike, since momentum −p runs along its arrow and (p̸ + m)v(p) = 0 is the same statement. What did the work is the mass shell of the external particles, not k² = 0 and not k⋅ε = 0. In the figure, push the incoming electron 10 per cent below its shell, p² = 0.9 m²: the three slots become 7.22, 2.06 and −11.91, and their sum, −2.635 before rounding, is −2.64 instead of 0. In a Green's function such leftovers are contact terms without the external pole that LSZ keeps; that story is topic 12.06's.
Summing the slots: the line telescopes
Number the other vertices V₁, …, Vₙ from the incoming end, and let Bⱼ be the line with the photon removed in which the j propagators nearest u(p) keep their own momenta and the other n − 1 − j carry k as well. Slot 0, next to u(p), gives +eB₀. Slot j, between Vⱼ and Vⱼ₊₁, gives e(Bⱼ − Bⱼ₋₁). Slot n, next to ū(p′), gives −eBₙ₋₁. The sum is e[B₀ + (Bₙ₋₁ − B₀) − Bₙ₋₁] = 0, derived. The strings Bⱼ do not conserve momentum, since k was taken out; they are bookkeeping, and they cancel. Two consequences are immediate. Nothing about the Vⱼ was used, so kμ M(μν…) = 0 holds with every other photon index free, which is what topic 8.07 needed to replace all the polarisation sums by −gμν at once. And no slot vanishes alone: in the figure's default, a 1.00 MeV electron emitting a 0.24 MeV photon around a static kick and a second vertex, the slots give 10.61, 2.51 and −13.12, common factor removed.
Why ξ drops out of this unit's amplitudes
Topic 8.04's propagator in a covariant gauge is −i[gμν − (1 − ξ)qμqν/q²]/(q² + iε), quoted. Between two open lines A and B its ξ part gives i(1 − ξ)(qμJAμ)(qνJBν)/(q² + iε)², where JAμ is line A with the internal photon's index left open, summed over every slot that photon can take on A. That is the contraction just done, with q for k, so it vanishes when A's ends are on shell; one conserved line suffices. In topics 9.05 to 9.09 each line has one vertex and the check is short: q = p₁ + p₂ gives v̄(p₂)q̸u(p₁) = v̄(p₂)[(p̸₁ − m) + (p̸₂ + m)]u(p₁) = 0, and q = p₃ − p₁ gives ū(p₃)q̸u(p₁) = 0. The dropped term is not small. In the e⁺e⁻ centre-of-mass frame at √s = 10.58 GeV the propagator's 00 entry is −iξ/s: −8.93i × 10⁻³ GeV⁻² in Feynman gauge and zero in Landau gauge. The amplitude cannot tell, since qμJμ = √s J⁰ = 0. Every amplitude of this unit is the same in every covariant gauge, shown, so Feynman gauge is a convenience. No experiment tests this directly; it guarantees that topic 9.06's measured rates meet one prediction, not a family.
Where the tree-level proof stops
Each input to the proof marks a boundary. Tree level: a closed fermion loop has no ends, and its telescoped strings cancel around the loop only after the loop momentum is shifted by k, a step that dimensional regularisation permits and a cutoff does not; topic 12.06 takes it and derives −iqμΓμ(p + q, p) = S⁻¹(p + q) − S⁻¹(p) for the full vertex, of which this lesson is the on-shell, tree-level case. On-shell ends: in Green's functions the end terms survive, and quantities with off-shell legs, such as topic 12.05's self-energy, keep their ξ. A vertex linear in A and free of momenta: a charged scalar's vertex −ie(p + p′)μ also telescopes, since k⋅(2p + k) = [(p + k)² − m²] − [p² − m²], but with two photons its A² seagull must join the sum (topic 8.09). One mass along the line: the split of k̸ used the same m on both sides of each vertex; one joining an electron to a muon would leave a term (mμ − mₑ)ūμ uₑ behind, and QED has none (topic 9.02).
Change one variable at a time
Make the relationship visible.
Drag ω and φ: all three bars move and swap sign, yet the sum stays on the axis. Then push the incoming leg off its shell: the sum bar grows, and the readouts show why, as (p − k)² − m² and −2p⋅k, equal on shell, part by p² − m².
INCOMING p² − m²0.0000 MeV²
(p − k)² − m²-0.0674 MeV²
−2p⋅k-0.0674 MeV²
SECOND VERTEX SUPPLIES Q₂-0.207 MeV
Live interpretationINCOMING p² − m²: 0.0000 MeV². (p − k)² − m²: −0.0674 MeV². −2p⋅k: −0.0674 MeV². SECOND VERTEX SUPPLIES Q₂: −0.207 MeV
Catch the common trap
Explain before calculating.
An electron line has one other vertex Γ and emits a photon of momentum k, so there are two diagrams: the photon next to u(p) or next to ū(p′), with both electrons on shell. What happens when the photon's ε* is replaced by k?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn e⁺e⁻ → μ⁺μ⁻ at √s = 10.58 GeV the photon propagator is taken in a general covariant gauge. Working in the centre-of-mass frame with both masses kept, show that its (1 − ξ) part cannot contribute, and compare the size of what drops out with what is kept.
- With q = p₁ + p₂ = (√s, 0, 0, 0), q² = s = 10.58² = 111.9 GeV², and qμJμ = √s J⁰ for the electron current Jμ = v̄(p₂)γμ u(p₁). It is enough to show J⁰ = 0.
- Quote UAQM-04.06's Dirac-representation spinors: u(p₁) = (√(E + m)χ, σ⋅p⃗₁χ/√(E + m)) and v(p₂) = (σ⋅p⃗₂η/√(E + m), √(E + m)η), E the beam energy. Then J⁰ = v†(p₂)u(p₁) = η†σ⋅(p⃗₁ + p⃗₂)χ, the √(E + m) factors cancelling.
- Back to back, p⃗₂ = −p⃗₁, so J⁰ = 0 for all four spin pairs with mₑ kept, and the muon current's J′⁰ vanishes the same way. This is the formula card's v̄(p₂)q̸u(p₁) = 0 seen in one frame.
- Size of what drops: the 00 component of gμν − (1 − ξ)qμqν/q² is 1 − (1 − ξ) = ξ, so the propagator's 00 entry is −iξ/s, which is −i × 8.93 × 10⁻³ GeV⁻² in Feynman gauge and exactly zero in Landau gauge. In Landau gauge the gauge term is as large as the kept one, and in every gauge it multiplies J⁰J′⁰ = 0.
- Only the spatial currents meet, through gᵢⱼ, whatever ξ is, so topic 9.06's σ = 4πα²/(3s) = 1.993 × 10⁻⁶ GeV⁻² = 0.776 nb holds in every covariant gauge; the muon speed β = 0.9998 changes it by less than 10⁻⁷.
AnswerJ⁰ = η†σ⋅(p⃗₁ + p⃗₂)χ = 0, so the (1 − ξ) term drops although in Landau gauge its 00 entry, 8.93 × 10⁻³ GeV⁻² in size, matches the kept term; M, and σ = 0.776 nb at 10.58 GeV, are the same in every covariant gauge.
MediumA 1.00 MeV electron (m = 0.511 MeV) moving along +z is kicked by a static potential and emits a 0.30 MeV photon along +z, with spins up along z throughout. Replace the photon's ε* by k and compute each of the two diagrams directly, using UAQM-04.06's Weyl-representation spinors, then show that they cancel.
- Use light-cone components q^± = q⁰ ± qz. Incoming: P = √(1 − 0.2611) = 0.8596 MeV, so p⁺ = 1.8596 and p⁻ = 0.1404 MeV. Outgoing: E′ = 0.70 MeV and P′ = √(0.49 − 0.2611) = 0.4784 MeV, so p′⁺ = 1.1784 and p′⁻ = 0.2216 MeV. Photon: k⁺ = 0.60 MeV, k⁻ = 0. The potential supplies Q = P′ + ω − P = −0.081 MeV of z-momentum and no energy.
- For spin up along z, u = (√(p⋅σ)χ, √(p⋅σ̄)χ) reduces to two numbers, (√p⁻, √p⁺) = (0.3747, 1.3637), and ū(p′) to (√p′⁺, √p′⁻) = (1.0855, 0.4707), in √MeV. On these, q̸ acts as the matrix with rows (0, q⁻) and (q⁺, 0), and γ⁰ as the one with rows (0, 1) and (1, 0).
- Photon next to u(p): k̸u = (0, k⁺√p⁻) = (0, 0.2248), and p̸ − k̸ + m turns it into 0.2248 × ((p − k)⁻, m) = (0.0316, 0.1149). The denominator is (p − k)² − m² = 1.2596 × 0.1404 − 0.2611 = −0.0843 MeV², exactly −2p⋅k, so SF(p − k)k̸u(p), its i stripped, is (−0.3747, −1.3637) = −u(p).
- Photon next to ū(p′): ū(p′)k̸ = (k⁺√p′⁻, 0) = (0.2824, 0); times p̸′ + k̸ + m it gives 0.2824 × (m, p′⁻) = (0.1443, 0.0626), and over (p′ + k)² − m² = 1.7784 × 0.2216 − 0.2611 = +0.1330 MeV² = 2p′⋅k it becomes (1.0855, 0.4707) = ū(p′).
- So the diagrams are −ū(p′)γ⁰u(p) and +ū(p′)γ⁰u(p) times the common factor (−ie)²iÃ(Q), with ū(p′)γ⁰u(p) = √(p′⁺p⁺) + √(p′⁻p⁻) = 1.4803 + 0.1764 = 1.657 MeV. Both denominators were cancelled exactly because p² = m² and p′² = m².
Answer−1.657 MeV and +1.657 MeV times the common factor: each diagram is ∓ the photon-free Coulomb amplitude ū(p′)γ⁰u(p), neither vanishes, and their sum is zero.
HardGive the medium problem's 1.00 MeV electron two other vertices: a static kick of 0.15 MeV along z (vertex γ⁰), then a vertex ε̸₂ with ε₂ = (0, 0, 0, 1), a longitudinal polarisation no free photon has. A 0.40 MeV photon is emitted along +z at slot 0, 1 or 2. Compute the three contracted diagrams and their sum, using the reduced two-component matrices.
- Kinematics: p⁺ = 1.8596, p⁻ = 0.1404 MeV as before. After the kick p₁ has energy 1.00 MeV and z-momentum 1.0096 MeV, so p₁⁺ = 2.0096, p₁⁻ = −0.0096 MeV and D₁ = p₁² − m² = −0.2804 MeV². With k also removed, (p₁ − k)⁺ = 1.2096 MeV and D₁ₖ = −0.2727 MeV². Finally E′ = 0.60, P′ = 0.3145 MeV, p′⁺ = 0.9145, p′⁻ = 0.2855 MeV, and the second vertex supplies Q₂ = −0.295 MeV of z-momentum.
- ε̸₂ = −γ³ has rows (0, −1) and (1, 0), so ρ = ū(p′)ε̸₂ = (√p′⁻, −√p′⁺) = (0.5344, −0.9563), and γ⁰u(p) = (√p⁺, √p⁻) = (1.3637, 0.3747). A propagator numerator q̸ + m acting on γ⁰u gives (m√p⁺ + q⁻√p⁻, q⁺√p⁺ + m√p⁻): (0.6932, 2.9319) at q = p₁ and (0.6932, 1.8409) at q = p₁ − k.
- Slots 0 and 2 use the medium problem's end results, SF(p − k)k̸u = −u and ū(p′)k̸SF(p′ + k) = ū(p′), i's stripped. Slot 2 = ρ⋅(0.6932, 2.9319)/D₁ = −2.4332/(−0.2804) = 8.678. Slot 0 = −ρ⋅(0.6932, 1.8409)/D₁ₖ = 1.3900/(−0.2727) = −5.097.
- Slot 1 directly: k̸ sends (0.6932, 2.9319) to (0, k⁺ × 0.6932) = (0, 0.5546); the numerator at p₁ − k gives 0.5546 × ((p₁ − k)⁻, m) = (−0.0053, 0.2834); ρ⋅ of that is −0.2738, and dividing by D₁D₁ₖ = 0.07646 MeV⁴ gives −3.581.
- Sum: −5.097 − 3.581 + 8.678 = 0.000, while the largest slot is 8.678. The middle slot equals 5.097 − 8.678, the difference of the photon-free strings with and without k, as the insertion rule says, and the unphysical ε₂ changed nothing.
AnswerSlots 0, 1 and 2 give −5.097, −3.581 and +8.678 times their common factor and sum to zero to four figures, although ε₂ is a polarisation no free photon has.