University Physics IV · Particle Physics · 14.4
Feynman Diagrams & Perturbative Amplitudes
Every diagram you draw is one term of an expansion in the coupling. Learn to read the order straight off the picture, to treat an internal line as a propagator rather than a particle, and to know the point at which adding terms stops helping.
Build the model
Connect the measurement to the mechanism.
Quantum field theory almost never hands you an exact amplitude, so it is computed as a series: expand the interaction in powers of the coupling and keep the first few terms. A Feynman diagram is one such term, drawn instead of written out. Its vertices are factors of the coupling, its internal lines are propagators, and its external lines are the particles you actually prepared and detected.
Two consequences run through everything that follows. First, the picture is arithmetic and not history: the internal momenta are integrated over, no single diagram is measurable on its own, and diagrams of the same order interfere, so amplitudes must be summed before they are squared. Second, the whole scheme lives or dies on the coupling being small.
One more loop costs about α/π ≈ 2.3×10⁻³ in QED, which is why the electron's magnetic moment is the most precisely verified prediction in physics — but the coefficients grow factorially, so the series has zero radius of convergence and only ever approaches the truth to within roughly e(−1/λ) before it turns and diverges. In QED that floor sits near 10⁻¹⁸⁷ and will never be met. In QCD below a few GeV, where αₛ approaches one, the series stops paying after a couple of terms and the diagrams have to be abandoned for the lattice.
- Simple definition
- A Feynman diagram is shorthand for one term in the perturbative expansion of a transition amplitude: each vertex supplies a factor of the coupling, each internal line a propagator, and only the sum over all diagrams of a given order means anything physically.
- Example
- Electron–positron annihilation to a muon pair has one two-vertex diagram, so its amplitude carries e² ∝ α and its cross-section α² ≈ (1/137)² = 5.3×10⁻⁵; the four-vertex diagrams shift that by only about α/π ≈ 0.2%.
Counting vertices on a sketch gives the power of α before any integral is attempted: two vertices means σ ∝ α² ≈ 5.3×10⁻⁵.
n = number of vertices; α = e²/4πε₀ħc = 1/137.036 at low q². Each vertex is worth √α.
It carries all the q-dependence of the result. A massless mediator's 1/q² squares to the 1/q⁴ behind Rutherford's sin⁻⁴(θ/2).
q = four-momentum flowing along the line, in GeV/c; m = mediator mass. Off shell: q² is fixed by the external kinematics.
The weak interaction is weak at low q because of the propagator, not the coupling: αw is 4.6 times α.
ħ = c = 1 in the second relation. MW = 80.4 GeV, GF = 1.166×10⁻⁵ GeV⁻² ⇒ g² = 0.426, αw = g²/4π = 1/29.5.
Range is not separate physics — it is the Fourier transform of the same denominator you put in the amplitude.
MW = 80.4 GeV/c² gives R = 2.5×10⁻³ fm; M → 0 restores the infinite-range 1/r Coulomb potential.
One diagram lands within 0.15% of experiment; three loops agree with it to eight significant figures.
α/π = 2.32282×10⁻³. Diagrams contributing at each order: 1, 7, 72, 891, 12672.
Zero radius of convergence: past n* every further diagram makes the answer worse, not better.
λ is the coupling per order: α/π = 2.3×10⁻³ in QED, so n* ≈ 430; αₛ/π ≈ 0.16 at 1 GeV, so n* ≈ 6.
A diagram is a term in a series, not a photograph of an event
Perturbation theory expands the transition amplitude in powers of the coupling: M = M₁ + M₂ + M₃ + …, the subscript counting powers of e. Writing those terms out algebraically is punishing, so Feynman's rules let you draw them instead. Each topologically distinct way of joining the given incoming and outgoing particles with vertices and internal lines is one term, and the rules translate the drawing back into an integral. What the drawing does not encode is where or when anything happened. The internal momenta are integrated over — every value, including ones no real particle could carry — so the position of a vertex on the page is a bookkeeping convenience. Deform the drawing however you like and it is the same term; change which external legs meet which vertex and it is a different term that must also be included. Every diagram at a given order contributes, and they are added, not chosen between.
External lines, vertices, propagators — and what is legal
Every diagram is built from three pieces. External lines are the real, on-shell particles you prepared and detected, each carrying a spinor or a polarisation vector. Vertices are the interaction points; each supplies a factor of the coupling — √(4πα) in QED — together with exact energy–momentum conservation at that point. Internal lines join two vertices and supply a propagator. The rules then say what is legal. QED has exactly one vertex, with two fermion legs and one photon leg, so charge conservation, lepton-number conservation and the continuity of fermion lines are automatic: an arrow runs unbroken along the fermion line, drawn against the reading direction for an antiparticle. There is no three-photon vertex, because the photon is uncharged. QCD differs precisely here — gluons carry colour, so three- and four-gluon vertices are in the rule book, and that single change is what eventually produces confinement.
Read the order straight off the picture
Count the vertices. n vertices means n factors of e in the amplitude, so M ∝ α(n/2), and probabilities go as |M|². Electron–positron annihilation to a muon pair has one tree diagram with two vertices, so M₀ ∝ α and σ ∝ α² ≈ 5.3×10⁻⁵. The next diagrams have four vertices — a photon exchanged between the outgoing muons, a vertex correction, a vacuum-polarisation loop — and each gives M₁ ∝ α². The trap is to square that and expect a change of order α² ≈ 5×10⁻⁵. Amplitudes add before they are squared: |M₀ + M₁|² = |M₀|² + 2Re(M₀*M₁) + |M₁|², and the interference term is only one power of α down on the tree. A four-vertex diagram therefore shifts the cross-section by of order α, and once the 1/π that a loop integral characteristically supplies is folded in, the working number for one more loop is α/π ≈ 0.2%. Tree level before loops, and never square a correction on its own.
The internal line is off shell, and that is the whole point
An internal line is not a particle: it is the factor 1/(q² − m²c²), with q fixed by conservation at the vertices it joins. Because q is whatever the external kinematics forces it to be, q² generally does not equal m²c² — the line is off the mass shell. The photon exchanged in electron scattering has q² < 0, a negative invariant mass squared, which no detectable photon can have. That denominator carries all the q-dependence of the answer. For a massless mediator it gives 1/q², and squaring it produces the 1/q⁴ that is Rutherford's sin⁻⁴(θ/2). For a heavy mediator at small transfer, q² is negligible beside M²c² and the propagator freezes to the constant −1/M²c²: exchange becomes a contact interaction, which is exactly Fermi's four-fermion theory of beta decay. That freezing, not a small coupling, is why the weak interaction is weak below the W mass — αw ≈ 1/29.5 is 4.6 times α.
Loops diverge, and force you to say which coupling you mean
Tree diagrams are finite; loops are not. A loop closes an internal momentum that the external kinematics does not fix, so the rules instruct you to integrate over it from zero to infinity, and the integral diverges. Renormalisation is the observation that the divergences appear only in a fixed set of places — the electron's mass, its charge, the field normalisations — and those quantities were never predicted anyway; they are measured. Write the answer in terms of the measured charge rather than the bare one, and every remaining prediction comes out finite. The price is that the coupling now needs a scale attached to it. Vacuum-polarisation loops screen the bare charge, so α runs: α = 1/137.036 measured at large distance, but α(MZ) ≈ 1/128. Quote a cross-section to a per cent and you must say which α you used, because at LEP energies the difference is about 7%.
Where the series stops helping
Nothing guarantees that adding diagrams converges, and it does not. The number of diagrams at order n grows roughly like n!, and Dyson gave the reason convergence is impossible: a convergent series would converge in a disc around α = 0, and so would also describe α < 0 — a world in which like charges attract and the vacuum is unstable against making pairs. That physics is not analytic at zero, so the radius of convergence is zero. What you get instead is an asymptotic expansion: the terms shrink for a while, reach a minimum near n* ≈ 1/λ where λ is the cost of one order, then grow without bound, and the best accuracy available is the size of that smallest term, about e(−1/λ). For QED with λ = α/π ≈ 2.3×10⁻³ the floor sits near 10⁻¹⁸⁷ and no experiment will ever meet it. For QCD, αₛ/π ≈ 0.038 at the Z still leaves many usable orders; by 1 GeV αₛ ≈ 0.5 and the turnaround arrives by the sixth term; near ΛQCD ≈ 0.2 GeV there is no small number to expand in at all. That is why 99% of the proton's mass is computed on a lattice and not from diagrams.
Change one variable at a time
Make the relationship visible.
Slide λ from 0.03 to 0.6 and watch the valley march left: at λ = 0.03 the terms keep shrinking out to order 33, at λ = 0.12 they turn round by order 8, and at λ = 0.6 the very first term is already the best the series will ever manage.
OPTIMAL ORDER n* = 1/λ8
RATIO OF SUCCESSIVE TERMS0.96
log10 TERM AT ORDER n-2.8
log10 SMALLEST TERM-2.8
Live interpretationOPTIMAL ORDER n* = 1/λ: 8. RATIO OF SUCCESSIVE TERMS: 0.96. log10 TERM AT ORDER n: −2.8. log10 SMALLEST TERM: −2.8
Catch the common trap
Explain before calculating.
Electron–positron annihilation to a muon pair has one tree diagram with two vertices, giving σ ∝ α². You now also include a four-vertex diagram, in which a photon is exchanged between the outgoing muons. By roughly what fraction does the predicted cross-section change?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCompton scattering, γ + e⁻ → γ + e⁻, has two tree diagrams, each with two QED vertices. (a) What power of α does the tree-level cross-section carry, numerically? (b) A four-vertex diagram is now included. By roughly what fraction does the predicted cross-section change?
- Each QED vertex contributes one factor of the charge e to the amplitude, and α = e²/4πε₀ħc, so a vertex is worth √α. Two vertices give M₀ ∝ e² ∝ α.
- A rate is |M|², so σ ∝ α² = (1/137.036)² = 5.33×10⁻⁵. The classical limit confirms the counting: σT = (8π/3)rₑ² with rₑ = αħ/mec, so the Thomson cross-section does carry α².
- The extra diagram has four vertices, so M₁ ∝ e⁴ ∝ α². Do not square it alone: |M₀ + M₁|² = |M₀|² + 2Re(M₀*M₁) + |M₁|², and the interference term 2Re(M₀*M₁) ∝ α³ is the leading correction.
- Fractional shift = α³/α² = α = 7.30×10⁻³. A loop integral characteristically supplies a further 1/π, so the working estimate for one more loop is α/π = 2.32×10⁻³.
Answerσₜᵣₑₑ ∝ α² = 5.33×10⁻⁵; the four-vertex diagram changes it by of order α/π ≈ 0.23%, not by α² ≈ 5×10⁻⁵. Squaring a correction on its own is the standard way to come out a factor of 137 too small.
MediumWeak and electromagnetic processes both run through two vertices and one exchanged boson, yet beta decay is slow. Take α = 1/137.036, αw = g²/4π = 1/29.5 and MW = 80.4 GeV/c². Compare the W-exchange amplitude with the photon-exchange amplitude at a momentum transfer q = 0.10 GeV/c, typical of nuclear beta decay, and then at q = 80.4 GeV/c.
- Same vertex count, so the couplings enter identically: Mγ ∝ 4πα/q² and MW ∝ 4παw/(q² + MW²c²). Everything else cancels in the ratio, leaving MW/Mγ = (αw/α) × q²/(q² + MW²c²).
- At q = 0.10 GeV/c, q² = 0.010 GeV² is utterly negligible beside MW² = 6.46×10³ GeV², so the propagator freezes to the constant 1/MW² — a contact interaction, which is Fermi's four-fermion theory.
- Ratio = (αw/α)(q²/MW²) = (137.0/29.5) × (0.010/6460) = 4.65 × 1.55×10⁻⁶ = 7.20×10⁻⁶. The amplitude is suppressed by six orders of magnitude, and the rate by twelve.
- At q = 80.4 GeV/c the denominator is q² + MW² = 2MW², so the ratio becomes 4.65 × ½ = 2.3. The propagator suppression has evaporated and the weak amplitude is now the larger of the two.
AnswerAbout 7.2×10⁻⁶ at q = 0.10 GeV/c, rising to about 2.3 at q ≈ MW. The weak interaction is weak at nuclear energies only because the propagator freezes to 1/MW²; its coupling αw ≈ 1/29.5 is 4.6 times the electromagnetic one.
HardQED gives the electron's anomalous magnetic moment as aₑ = ½(α/π) − 0.328479(α/π)² + 1.181241(α/π)³ − …, the three terms coming from 1, 7 and 72 diagrams. Take α⁻¹ = 137.035999 and the measured aₑ = 1.15965218×10⁻³. Evaluate the partial sums through three loops, say what each further loop buys, and estimate where the series would stop improving.
- One more loop costs one more power of α/π. Here α/π = 7.2973526×10⁻³ / 3.1415927 = 2.3228195×10⁻³, so (α/π)² = 5.395490×10⁻⁶ and (α/π)³ = 1.253275×10⁻⁸.
- One loop, Schwinger's single diagram: aₑ = ½ × 2.3228195×10⁻³ = 1.1614097×10⁻³. Against the measured value that is high by 1.758×10⁻⁶, a relative error of 1.52×10⁻³ — 0.15% from one diagram.
- Two loops, 7 diagrams: add −0.328479 × 5.395490×10⁻⁶ = −1.77231×10⁻⁶, giving 1.1596374×10⁻³. The relative error falls to 1.27×10⁻⁵.
- Three loops, 72 diagrams: add +1.181241 × 1.253275×10⁻⁸ = +1.48042×10⁻⁸, giving 1.15965223×10⁻³ against a measured 1.15965218×10⁻³ — a relative error of 4.5×10⁻⁸, so all eight quoted figures agree.
- Each loop buys a factor of a few hundred, of order π/α times a coefficient near one. But the coefficients themselves grow like n!, so the terms turn round near n* ≈ π/α = 430, and the smallest term — the accuracy floor — is about e⁻⁴³⁰ ≈ 10⁻¹⁸⁷.
Answer1.1614097×10⁻³, then 1.1596374×10⁻³, then 1.15965223×10⁻³, against a measured 1.15965218×10⁻³. Three loops match to eight figures, and the asymptotic floor near 10⁻¹⁸⁷ is why QED can be pushed to five loops without ever meeting it.