University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.7

Reading Feynman rules off a Lagrangian

After topics 3.01 and 3.06 you should never need Wick's theorem again to find a vertex. This lesson turns that bookkeeping into a rule you apply by inspection, then takes it into the two places students slip: derivatives on the fields, and terms with only two fields, which is where the mass and field counterterms live.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Every vertex met so far came out of the same three moves. The Gell-Mann-Low numerator brings down i∫d⁴z ℒᵢₙₜ(z), since Hᵢₙₜ = −ℒᵢₙₜ when no field is differentiated; each field at z is contracted with a line; and lines of one species can be attached to the identical copies of that field in every order, each giving the same integral. So a term c φ₁^(n₁) ⋯ φₖ^(nₖ) in ℒ is a vertex i c n₁! ⋯ nₖ!, which is why φ⁴ is written with 1/4! and why −gψ†ψφ, three different fields, needs nothing.

A contracted field becomes a plane wave e^(−ip⋅z), so ∂_μ on a line carrying momentum p into the vertex becomes −ip_μ, and the attachments must then be summed, because they no longer give equal factors. A quadratic term is a vertex with two legs: (1/2)δ_Z(∂φ)² − (1/2)δₘφ² gives i(p²δ_Z − δₘ), and one insertion of it between free propagators reproduces the exact propagator of the shifted quadratic Lagrangian to first order, a check no other vertex allows. The recipe stops in three places.

With time derivatives π is no longer φ̇, Hᵢₙₜ gains non-covariant terms, and reading the vertex straight off ℒ rests on Matthews' theorem and the covariant T* product, quoted, not proved. It gives vertices, not the 1/S a whole diagram keeps, external-state factors or fermion signs. And it never says what a counterterm's coefficient is: Unit 6's renormalisation conditions do.

Simple definition: Reading Feynman rules off a Lagrangian assigns each term of ℒᵢₙₜ a vertex: i times its coefficient, times the number of ways of attaching identical lines, with −ip_μ for every derivative on a line carrying momentum p into the vertex.
Example: For real φ and χ, −(a/2)φ²χ with a = 0.60 GeV attaches its two identical φ lines in 2! ways and its single χ line in one: i(−0.30 GeV)(2!)(1) = −0.60i GeV, a mass, as every three-leg vertex must be.

Vertex of a monomial in ℒᵢₙₜℒᵢₙₜ ⊃ c φ₁^(n₁) ⋯ φₖ^(nₖ) → vertex = i c n₁! n₂! ⋯ nₖ!

Derived: i from the Dyson exponent, nᵢ! from attaching identical lines. It returns −iλ for −(λ/4!)φ⁴ and −ig for −gψ†ψφ.

c in GeV^{(4 − Σnᵢdᵢ)}, dᵢ the mass dimension of φᵢ; nᵢ counts identical copies of φᵢ; φ and φ† are different species.

A derivative on a line∂_μ φ → −ip_μ, with p flowing into the vertex

Derived from the plane wave e^(−ip⋅z) each contraction leaves at the vertex. Using it straight from ℒ needs the T* product: Matthews' theorem, quoted.

p in GeV; an outgoing momentum q counts as p = −q, so ∂_μ on an outgoing line gives +iq_μ.

Ordinary against covariant time ordering⟨0|T φ̇(x)φ̇(y)|0⟩ = ∂₀ˣ∂₀ʸ D_F(x − y) − iδ⁴(x − y)

Derived from [φ̇, φ] = −iδ³ at equal times. This contact term is the non-covariant piece that cancels the extra terms derivative couplings put in Hᵢₙₜ.

φ̇ = ∂₀φ; every term in GeV⁴. T* keeps only the first term, derivatives taken outside the ordering.

A two-derivative coupling−(g/2) φ ∂_μ χ ∂^μ χ → i g k₁⋅k₂

Derived: each of the 2! attachments gives (−ik₁)⋅(−ik₂), cancelling the 1/2. It is symmetric in k₁ and k₂, as two identical χ legs require, and grows as momentum squared where a φχ² vertex would stay constant.

g in GeV⁻¹, the operator having dimension 5; k₁, k₂ the incoming χ momenta in GeV; the vertex is a mass, as three legs require.

Two-point vertex(1/2)δ_Z ∂_μ φ ∂^μ φ − (1/2)δₘ φ² → i(p²δ_Z − δₘ)

Derived. δₘ acts the same at every p² while δ_Z grows with p²; it is the form Unit 6's mass and field counterterms take, their values fixed there.

δ_Z dimensionless, δₘ in GeV², p the momentum through the vertex; each 1/2 is cancelled by 2! attachments.

One insertion against the exact propagatori/((1 + δ_Z)p² − m² − δₘ) = D + D i(p²δ_Z − δₘ) D + O(δ²)

Derived by expanding the exact inverse: the check that fixes both the i and the 2!. Summing every insertion is topic 3.08's job.

D = i/(p² − m² + iε), in GeV⁻²; exact pole at (m² + δₘ)/(1 + δ_Z), residue 1/(1 + δ_Z).

The vertex is the Dyson exponent with its legs attached

Work in ħ = c = 1 with the metric (+, −, −, −). Topic 2.03 showed that without derivatives Hᵢₙₜ = −ℒᵢₙₜ, so the Gell-Mann-Low numerator carries T exp(i∫d⁴z ℒᵢₙₜ). Take one term, c φ₁^(n₁) ⋯ φₖ^(nₖ), at first order: i c ∫d⁴z φ₁(z)^(n₁) ⋯ φₖ(z)^(nₖ). Wick's theorem pairs each field at z with a line. If nᵢ lines of species i meet the vertex, they pair with the nᵢ identical copies of φᵢ in nᵢ! orders, all giving the same integral, so the vertex is i c n₁! ⋯ nₖ!, derived. At order n the 1/n! of the exponential is cancelled by relabelling the n vertices, as in topic 3.01, so every vertex keeps exactly this factor; what is overcounted when a diagram has a symmetry of its own is topic 3.02's 1/S, which belongs to the diagram, not the vertex. For −(λ/4!)φ⁴ the count gives i(−λ/4!)(4!) = −iλ; for −gψ†ψφ, three different fields, it gives −ig, as topic 3.06 found by contracting directly.

Conventions in ℒ, and the dimension check on every vertex

The factorials written into a Lagrangian are conventions chosen so that vertices come out clean, and the rule shows what each one buys. A real cubic is written −(κ/3!)φ³ to give −iκ; a mixed term −(h/4)φ²χ² gives i(−h/4)(2!)(2!) = −ih. Complex fields need care: φ and φ† are different species, so −(λ/4)(φ†φ)² gives −iλ, not the −6iλ that treating all four fields as identical would produce. QED reads the same way: with D_μ = ∂_μ + ieA_μ the Dirac Lagrangian holds −eψ̄γ^μ ψA_μ, three different fields with no derivative on any, so the vertex is −ieγ^μ with no new combinatorics, |e| being √(4πα) = 0.3028; topic 9.02 adds the spinors and polarisations. Then check dimensions. ℒ has mass dimension 4, so a vertex joining n scalar lines, momentum factors included, has dimension 4 − n: −iλ is a pure number with four legs, −iκ a mass with three, and any two-leg vertex must be a mass squared.

Derivatives become momenta, and attachments are summed

In momentum space each contraction ending at z leaves a plane wave e^(−ip⋅z), p being the momentum its line carries into the vertex, so ∂_μ on that field gives −ip_μ, derived. An outgoing momentum q leaves e^(+iq⋅z) and gives +iq_μ: the same rule with p = −q. Once some copies of a field carry derivatives and others do not, different attachments give different factors, and n! is no longer a common multiplier: the vertex is i c times the sum over attachments. Take −gφ ∂_μ φ ∂^μ χ with incoming k₁, k₂ on the φ legs and k₃ on χ. Putting leg 1 on the bare φ and leg 2 under the derivative gives (−ik₂)⋅(−ik₃) = −k₂⋅k₃; the swap gives −k₁⋅k₃. The vertex is i(−g)(−k₁⋅k₃ − k₂⋅k₃) = ig(k₁ + k₂)⋅k₃, and conservation, k₁ + k₂ = −k₃, makes it −igk₃². It is symmetric in the two identical φ legs, as it must be, and has dimension GeV⁻¹ × GeV² = GeV, right for three legs.

Integration by parts is a free check

Two Lagrangians that differ by a total derivative ∂_μK^μ give the same action, so they must give the same vertices. The momentum rule respects this automatically: the outer ∂_μ acts on the whole product at the vertex and brings −i times the sum of the incoming momenta, which conservation sets to zero. So integrating by parts tests any vertex you have read. For −gφ ∂_μ φ ∂^μ χ, write φ ∂_μ φ = (1/2)∂_μ(φ²) and move the derivative: up to a total derivative the term is (g/2)φ²□χ, whose vertex is i(g/2)(2!)(−k₃²) = −igk₃², matching the sum over attachments. Using the free equation □χ = −M²χ is a further step, valid only on shell, where the coupling acts like −(gM²/2)φ²χ with vertex −igM². For a massless χ it then vanishes on shell, yet not off shell: inside a diagram the part −ig(k₃² − M²) cancels the χ propagator it multiplies and leaves a contact term.

Why reading ℒ is allowed: T*, contact terms and Matthews

A time derivative in ℒᵢₙₜ changes the canonical momentum, so Hᵢₙₜ is no longer −ℒᵢₙₜ. The two-point term shows how. With ℒ = (1/2)(1 + δ_Z)(∂φ)² − (1/2)(m² + δₘ)φ², π = (1 + δ_Z)φ̇ and Hᵢₙₜ = −ℒ_ct + δ_Z²φ̇²/(2(1 + δ_Z)), a non-covariant extra, derived. Ordinary T has an extra of its own: [φ̇(x), φ(y)] = −iδ³(x − y) at equal times gives ⟨T φ̇φ̇⟩ = ∂₀∂₀D_F − iδ⁴(x − y). At order δ_Z² the two meet. Two δ_Z vertices joined through the contact term make an effective vertex +iδ_Z²p₀², and the extra term in Hᵢₙₜ is a vertex −iδ_Z²p₀²; they cancel, leaving the covariant answer that topic 2.04's T*, derivatives pulled outside the ordering, gives at once. That this cancellation holds at every order for any derivative coupling is Matthews' theorem of 1949, quoted. QED's −eψ̄γ^μ ψA_μ has no derivative, so topic 9.02 never needs it; scalar QED's vertex −ie(p + p')^μ in topic 8.09 does.

Two-point vertices: counterterms and a measured mixing

A term with two fields is a vertex with one line in and one out. For (1/2)δ_Z ∂_μ φ ∂^μ φ − (1/2)δₘφ², the line brings p in and takes p out, so the derivatives give (−ip)⋅(+ip) = p², the two attachments cancel each 1/2, and the vertex is i(p²δ_Z − δₘ), derived. A quadratic Lagrangian can also be solved exactly, so this rule alone can be checked against an exact answer: expanding i/((1 + δ_Z)p² − m² − δₘ) to first order gives the free D plus one insertion of the vertex between two D's. Two-point vertices can join different states too. For the neutral kaon's complex field K, a term −(b/2)(K² + K†²) gives −ib and turns K⁰ into K̄⁰, splitting the pair into masses √(m² ± b), so Δm ≈ b/m. Ignoring widths and CP violation, the measured Δm_K = 0.5293 × 10¹⁰ ħ s⁻¹ = 3.484 × 10⁻¹² MeV fixes |b| = m_KΔm_K = 1.73 × 10⁻⁹ MeV², with m_K = 497.6 MeV. Unit 6 writes its mass and field counterterms in exactly this form and fixes δ_Z and δₘ by renormalisation conditions.

Change one variable at a time

Make the relationship visible.

Interactive model

Field counterterm δ_Z0.30

Mass counterterm δₘ0.10 m²

Probe momentum p²2.00 m²

Set δ_Z to zero and slide δₘ: the solid line moves parallel and the gap is the same at every p². Then set δₘ to zero and raise δ_Z: the line pivots about p² = 0, the gap grows with p², and the pole slides below m² as the residue 1/(1 + δ_Z) falls.

VERTEX ÷ i AT PROBE0.500 m²

EXACT POLE p²0.846 m²

FIRST-ORDER POLE0.800 m²

RESIDUE 1/(1 + δ_Z)0.769

Live interpretationVERTEX ÷ i AT PROBE: 0.500 m². EXACT POLE p²: 0.846 m². FIRST-ORDER POLE: 0.800 m². RESIDUE 1/(1 + δ_Z): 0.769

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that a vertex is simply i times the coefficient in ℒ, so φ⁴ would give −iλ/4! and the two-point term (i/2)(p²δ_Z − δₘ). Both drop the attachments: the n! ways of joining identical lines to identical fields belong to the vertex, and only the overcount that a whole diagram's symmetry leaves behind, topic 3.02's 1/S, is divided out. The opposite slip comes with derivatives. Multiplying one attachment by n! assumes every attachment gives the same factor, which fails once derivatives sit on some copies and not others. For −gφ ∂_μ φ ∂^μ χ, 2! times the single attachment −k₂⋅k₃ gives 2igk₂⋅k₃, which is not even symmetric in the two identical φ legs; the correct sum, ig(k₁ + k₂)⋅k₃ = −igk₃², is. Symmetry under exchange of identical legs is the quick test that the attachments were summed.

A real scalar φ of mass m couples to a massless real scalar χ through ℒᵢₙₜ = −(g/2)φ ∂_μ χ ∂^μ χ. At first order the decay φ → χχ has iM equal to this vertex with both χ lines outgoing and on shell. What is M?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA complex scalar φ and a real scalar χ interact through ℒᵢₙₜ = −(λ/4)(φ†φ)² − yφ†φχ − (h/2)φ†φχ², with λ = 0.80, y = 1.5 GeV and h = 0.24. Read off the three vertex factors, check each one's mass dimension, and say what the first would have been had φ and φ† been counted as identical fields.

Species and copies: (φ†φ)² holds two φ and two φ†, which are different fields; φ†φχ holds one each of three fields; φ†φχ² holds one φ, one φ† and two identical χ.
(φ†φ)²: i(−λ/4)(2!)(2!) = −iλ = −0.80i, a vertex with two arrows in and two out.
φ†φχ: i(−y)(1)(1)(1) = −iy = −1.5i GeV. φ†φχ²: i(−h/2)(1)(1)(2!) = −ih = −0.24i.
Dimensions: a three-leg vertex carries 4 − 3 = 1, and −1.5i GeV is a mass; four-leg vertices carry 0, and λ and h are pure numbers.
Counting all four fields of (φ†φ)² as identical would give i(−λ/4)(4!) = −6iλ = −4.8i, six times too large: it counts pairings Wick's theorem sets to zero, since only φ contracts with φ† (topic 2.07).

AnswerVertices −0.80i for φφ†φφ†, −1.5i GeV for φφ†χ and −0.24i for φφ†χχ; the identical-field miscount would give −4.8i.

MediumA real scalar of mass m = 0.50 GeV carries the quadratic terms (1/2)δ_Z ∂_μ φ ∂^μ φ − (1/2)δₘφ² with δ_Z = 0.030 and δₘ = 0.012 GeV². Find the two-point vertex at p² = 0.25 GeV² and at p² = 0.50 GeV², check one insertion against the exact propagator at p² = 0.50 GeV², and locate the exact pole and its residue.

The vertex is i(p²δ_Z − δₘ). At p² = m² = 0.25 GeV²: i(0.0075 − 0.012) = −0.0045i GeV². At p² = 0.50 GeV², twice m²: i(0.015 − 0.012) = +0.0030i GeV². It changes sign at p² = δₘ/δ_Z = 0.40 GeV², because only the δ_Z part grows with p².
Exact propagator at p² = 0.50 GeV²: (1.030)(0.50) − 0.25 − 0.012 = 0.253 GeV², so i/0.253 = 3.953i GeV⁻².
Free plus one insertion: D = i/0.25 = 4.000i GeV⁻², and D(0.0030i)D = (4.000i)²(0.0030i) = −0.048i GeV⁻², a total of 3.952i GeV⁻². The 0.0006i left over is second order: 4.000 × (0.0030 × 4.000)² = 0.00058.
Pole: (1 + δ_Z)p² = m² + δₘ gives p² = 0.262/1.030 = 0.25437 GeV², so the pole mass is 0.50435 GeV, 4.35 MeV above m. The first-order estimate m² + δₘ − m²δ_Z = 0.25450 GeV² gives 0.50448 GeV.
Residue 1/(1 + δ_Z) = 0.971. Only δ_Z moves it, because only the p²-dependent part of the vertex changes the slope of the inverse propagator. Limit: δ_Z = δₘ = 0 returns the pole m² with residue 1.

AnswerVertex −0.0045i GeV² at p² = m² and +0.0030i GeV² at p² = 0.50 GeV²; one insertion gives 3.952i GeV⁻² against the exact 3.953i GeV⁻²; pole at 0.5044 GeV with residue 0.971.

HardA real scalar of mass m = 0.80 GeV has ℒᵢₙₜ = −(κ/2)φ ∂_μ φ ∂^μ φ with κ = 0.20 GeV⁻¹. Read off the three-point vertex for incoming momenta k₁, k₂, k₃, simplify it with momentum conservation, confirm it by integrating by parts, and find the coupling κ₃ of a term −(κ₃/3!)φ³ that gives the same vertex with all three legs on shell. How do the two compare at threshold, with one leg at k² = 2.56 GeV², four times m²?

Three identical fields, one bare and two differentiated: each of the 3! attachments puts one leg on the bare φ and the other two, r and s, under the two derivatives, giving (−ikᵣ)⋅(−ikₛ) = −kᵣ⋅kₛ. Each pair of differentiated legs appears twice.
Sum: i(−κ/2)(2)(−1)(k₁⋅k₂ + k₂⋅k₃ + k₃⋅k₁) = iκ(k₁⋅k₂ + k₂⋅k₃ + k₃⋅k₁), symmetric in the three legs, as identical legs require.
Squaring k₁ + k₂ + k₃ = 0 gives k₁⋅k₂ + k₂⋅k₃ + k₃⋅k₁ = −(k₁² + k₂² + k₃²)/2, so the vertex is −(iκ/2)(k₁² + k₂² + k₃²). Dimension: GeV⁻¹ × GeV² = GeV, right for three legs.
By parts: φ ∂_μ φ = (1/2)∂_μ(φ²), so up to a total derivative the term is (κ/4)φ²□φ. □ on a leg of momentum k gives −k²; any of the 3 legs can carry it and the other two attach in 2! ways: i(κ/4)(2)(−Σk²) = −(iκ/2)Σk², the same vertex.
On shell, kᵢ² = m²: the vertex is −(3/2)iκm² = −(3/2)(0.20 GeV⁻¹)(0.64 GeV²)i = −0.192i GeV. A φ³ term gives −iκ₃, so κ₃ = (3/2)κm² = 0.192 GeV, as □φ = −m²φ turns (κ/4)φ²□φ into −(κm²/4)φ³.
At threshold, two legs on shell and one at k² = 2.56 GeV²: the derivative vertex is −(iκ/2)(0.64 + 0.64 + 2.56) GeV² = −0.384i GeV, twice the φ³ value. The difference, −(iκ/2)(k² − m²), cancels the propagator on that line and acts as a contact term.

AnswerVertex −(iκ/2)(k₁² + k₂² + k₃²); on shell −0.192i GeV, matching −(κ₃/3!)φ³ with κ₃ = (3/2)κm² = 0.192 GeV; at threshold the derivative vertex is −0.384i GeV, twice the φ³ value.