University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.5
Momentum-space Feynman rules and loop counting
Loop diagrams are almost never evaluated by integrating over where their vertices sit. Fourier transform the lines and each vertex integral collapses to a δ function conserving four-momentum; what the δ's leave unfixed is one integral per loop, and a two-line count gives the power of λ in advance. Every amplitude from here on is written this way.
Build the model
Connect the measurement to the mechanism.
A position-space diagram is a set of propagators glued at points that are integrated over all of spacetime, and integrals over gluing points are convolutions, which Fourier transformation turns into products. Write each DF as topic 1.06's plane-wave integral and every vertex integral collapses into (2π)⁴δ⁴ of the four-momenta meeting there. Momentum conservation at a vertex is therefore not an extra assumption but translation invariance in momentum language: the vertex was integrated over every position. The δ functions then fix every internal momentum they can.
One combination of them involves only the external momenta and becomes the overall δ⁴ of the total momentum, and the I − V + 1 momenta left free are the loops, each with one ∫d⁴l/(2π)⁴. In φ⁴ the leg count 4V = 2I + E turns this into V = L + E/2 − 1, so at fixed E each loop costs exactly one more λ and perturbation theory in λ is a loop expansion. The rules stop in three places. They need each vertex's time integral to run over the whole time axis, which a finite window spoils.
They write down loop integrals that diverge and leave them unevaluated, which is the work of Units 5 and 6. And they produce correlation functions, not scattering amplitudes, until Unit 4's LSZ formula amputates the external lines and puts them on shell.
- Simple definition
- Momentum-space Feynman rules evaluate a diagram by giving each line a propagator i/(p² − m² + iε) in its own four-momentum, each vertex −iλ with four-momentum conserved, and each of the L = I − V + 1 momenta conservation leaves free an integral ∫d⁴l/(2π)⁴.
- Example
- The one-loop s-channel bubble of φ⁴ has V = 2 and I = 2, so L = 2 − 2 + 1 = 1: its lines carry l and p₁ + p₂ − l, and it is (−iλ)²(1/2)∫d⁴l/(2π)⁴ times their two propagators, one power of λ beyond the tree's −iλ.
Moves every spacetime point into an exponential, so the vertex integrals can be done first. The factor i/(p² − m² + iε) is even in p, so DF(x − y) = DF(y − x) and each line's orientation is a free choice.
x, y positions in GeV⁻¹ (1 GeV⁻¹ = 0.1973 fm); p the line's four-momentum and m the mass in GeV; DF in GeV². Quoted.
Energy and three-momentum are conserved exactly at every vertex, because the vertex sits at every point equally: translation invariance written in momentum space.
z the vertex position in GeV⁻¹; momenta in GeV; δ⁴ in GeV⁻⁴. Derived; needs z over all of spacetime.
Derived: the rules in one line. External lines carry fixed pᵢ, internal ones depend on the lⱼ, and S is topic 3.02's symmetry factor, unchanged by the transform.
pᵢ external momenta flowing in, lⱼ the L loop momenta, qₖ each line's momentum, all in GeV; λ dimensionless.
Derived: V δ functions, but their arguments sum to the external total, so only V − 1 constrain internal momenta. Trees, L = 0, need no integral.
I internal lines, V ≥ 1 vertices, L independent loop momenta; pure numbers. The +1 is the overall δ⁴.
At fixed E each loop costs exactly one more λ: the four-point function runs λ, λ², λ³ at zero, one and two loops, and the two-point function λ, λ² at one and two.
E external lines; derived from 4V = 2I + E, four line ends per vertex, two per internal line, one per external line.
Shown: p enters and leaves the same vertex, so conservation never touches l. A constant insertion acts like a shift of m², which topic 3.08 resums.
l the loop momentum in GeV; the insertion has mass dimension 2, in GeV²; ½ is topic 3.02's symmetry factor.
Trade each line for a plane-wave integral
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 3.01's rules write a φ⁴ diagram as (1/S)(−iλ)V times an integral over every vertex position z of a product of DF's, one per line. Integrals over gluing points are convolutions, and Fourier transformation turns convolutions into products, so replace every line with topic 1.06's representation, quoted: DF(x − y) = ∫d⁴p/(2π)⁴ i e(−ip⋅(x − y))/(p² − m² + iε). Each line now carries its own four-momentum, and every spacetime point sits only in exponentials. Because p → −p leaves p² alone, DF(x − y) = DF(y − x), so orient each line however you like: p flows from the tail y to the head x, the head collects e(−ip⋅x) and the tail e(+ip⋅y). The orientation is bookkeeping, not a direction of travel: each line still contains both time orders of its ends, and positive p⁰ against the arrow is as allowed as along it.
Each vertex integral is a δ function of momentum
At a vertex z the four line ends bring e(−ip⋅z) for each momentum flowing in and e(+ip⋅z) for each flowing out, so z appears only in e(−i(Σpᵢₙ − Σpₒᵤₜ)⋅z), and ∫d⁴z gives (2π)⁴δ⁴(Σpᵢₙ − Σpₒᵤₜ), derived. Four-momentum, energy included, is conserved exactly at every vertex because each vertex is integrated over every position. A line that leaves z and returns to it, the tadpole's loop, brings e(−il⋅z)e(+il⋅z) = 1 and is not constrained at all. The δ needs the whole time axis. Cut z⁰ to a window of length T and the energy factor becomes 2 sin(q⁰T/2)/q⁰, with q⁰ the energy mismatch: a peak of height T and width 4π/T that becomes 2πδ(q⁰) only as T → ∞, the limit topic 2.05 takes. At T = 20 GeV⁻¹, about 3.9 fm/c, a mismatch of 0.20 GeV still sits at 45% of the peak height. Any real experiment is far past that: one second is 1.5 × 10²⁴ GeV⁻¹. The figure plots the window.
One overall δ, and one integral for every loop
Fourier transform the external points too, G̃(p₁, …, pE) = ∫d⁴x₁…d⁴xE e(−i(p₁⋅x₁ + … + pE⋅xE)) G, and each external line's momentum is fixed at pᵢ, flowing in, with i/(pᵢ² − m² + iε) left on it. A connected diagram then has I internal momenta and V δ functions, and they do not all bite. Add the arguments of all V δ's: each internal momentum enters one vertex, leaves another and cancels, leaving p₁ + … + pE. One combination constrains only external momenta and survives as the overall (2π)⁴δ⁴(Σpᵢ); the other V − 1 fix internal momenta, and L = I − V + 1 stay free, derived. Each takes one ∫d⁴l/(2π)⁴, and the 2π's look after themselves: (2π)(4V − 4I) = (2π)⁴ × [(2π)⁻⁴]L. Trees, L = 0, have every momentum fixed and no integral. At V = 0 there is no vertex and the count does not apply: the two external integrals supply the δ themselves and give (2π)⁴δ⁴(p₁ + p₂) i/(p₁² − m² + iε), the transform of DF itself, the free-field check.
Route momenta along a spanning tree
Conservation says how many momenta are free, not which ones to call free, and a systematic routing answers that. Pick a spanning tree: V − 1 internal lines that reach every vertex without closing a loop. Give each of the other I − V + 1 internal lines, the chords, a loop momentum of its own; conservation then fixes every tree line, working in from the tips of the tree. Each chord closes exactly one loop through the tree, so L is also the number of independent closed loops a graph has, Euler's count. In the sunset, V = 2: one line is the tree, the other two are chords carrying l and k, and the tree line takes p − l − k. Another tree relabels the loop momenta by integer combinations of one another and of p, in the sunset (l, k) → (p − l − k, k) say, a change of variables with unit Jacobian, so every routing gives the same integral where it converges, and topic 6.03's dimensional regularisation keeps that true for divergent ones, quoted. Choose the routing that makes the external momenta enter as few propagators as possible.
Each loop costs exactly one more λ
A φ⁴ vertex has four line ends; an internal line uses two and an external line one, so 4V = 2I + E. With L = I − V + 1 this gives V = L + E/2 − 1 and I = 2L + E/2 − 2, derived. The four-point function has λ at tree level, λ² at one loop, in the s, t and u bubbles and the tadpoles on its legs, and λ³ at two; the two-point function has λ in the tadpole and λ² in all three of its two-loop diagrams. At fixed E the expansion in λ is the expansion in loops. Dimensions agree: with the overall (2π)⁴δ⁴ stripped off, the amputated E-point function has mass dimension 4 − E at every L, since a new loop adds d⁴l (+4), one vertex (0) and two propagators (−4). For vertices with n legs the same algebra gives Σ(n − 2)Vₙ = 2L + E − 2, so cubic couplings pay two powers per loop. φ⁴ offers no measurement, but QED does, where each loop costs e². Schwinger's α/2π = 0.0011614 exceeds the measured aₑ = 0.00115965 by 1.76 × 10⁻⁶, and the two-loop term −0.328(α/π)² = −1.77 × 10⁻⁶ accounts for it, quoted from topic 14.04.
The tadpole's loop never sees the external momentum
Route the order-λ tadpole of the two-point function: p comes in along one external line, leaves along the other, and the loop line carries l from z back to z. Conservation at z reads p + l = p + l and constrains nothing, so the amputated insertion is (−iλ/2)∫d⁴l/(2π)⁴ i/(l² − m² + iε) = −(iλ/2)DF(0), with p nowhere in it, shown. A constant of mass dimension 2 on a line is what a shift of m² looks like; topic 3.08 resums it and topic 6.05 evaluates it. The s-channel bubble is the contrast. P = p₁ + p₂ enters one vertex, so its lines carry l and P − l, and (−iλ)²(1/2)∫d⁴l/(2π)⁴ [i/(l² − m² + iε)][i/((P − l)² − m² + iε)] depends on P, through s = P² alone once Lorentz invariance has acted. The rule: a loop feels an external momentum only if that momentum must pass through the loop's lines on its way from where it enters to where it leaves, whatever routing you choose.
Change one variable at a time
Make the relationship visible.
Drag T from 5 to 25 GeV⁻¹: the peak grows fivefold and the zeros close in fivefold while the dashed envelope stays put, so only faster oscillation suppresses the tails. Park the marker at q⁰ = 0.5 GeV and its weight crosses zero between T = 12 and 13: exact conservation needs T → ∞.
PEAK HEIGHT T/2π1.91 GeV⁻¹
FIRST ZEROS AT ±2π/T0.524 GeV
δT AT THE MARKER1.033 GeV⁻¹
WINDOW cT2.37 fm
Live interpretationPEAK HEIGHT T/2π: 1.91 GeV⁻¹. FIRST ZEROS AT ±2π/T: 0.524 GeV. δT AT THE MARKER: 1.033 GeV⁻¹. WINDOW cT: 2.37 fm
Catch the common trap
Explain before calculating.
A connected diagram contributing to the φ⁴ four-point function has three loops. How many vertices does it have, and so what power of λ does it carry?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA connected diagram for the φ⁴ six-point function has seven internal lines. Find its vertices, loops and power of λ, how many momentum integrals survive the δ functions, the net power of 2π, and the mass dimension of its amputated amplitude.
- Leg count: 4V = 2I + E with I = 7 and E = 6 gives 4V = 20, so V = 5 and the diagram is order λ⁵.
- Loops: L = I − V + 1 = 7 − 5 + 1 = 3. Check with V = L + E/2 − 1 = 3 + 3 − 1 = 5 ✓.
- Integrals: 7 internal momenta against 5 vertex δ functions. Four of the δ's fix internal momenta and the fifth becomes the overall (2π)⁴δ⁴(p₁ + … + p₆), so 7 − 4 = 3 = L integrals ∫d⁴l/(2π)⁴ remain.
- Factors of 2π: (2π)(4V − 4I) = (2π)(20 − 28) = (2π)⁻⁸, which is (2π)⁴ on the overall δ times (2π)⁻⁴ for each of the three loops, (2π)(4 − 12) ✓.
- Dimension: with the overall (2π)⁴δ⁴ stripped off, an amputated E-point function has mass dimension 4 − E = −2. Directly, three loops give +12, seven internal propagators −14 and λ⁵ nothing: −2 ✓. The six-point tree has V = 2, so the three loops cost exactly three more powers of λ.
AnswerV = 5, so order λ⁵; L = 3, leaving three integrals ∫d⁴l/(2π)⁴ and one overall δ⁴; net (2π)⁻⁸; with that δ⁴ removed the amputated amplitude has mass dimension −2, in GeV⁻².
MediumRoute momenta through the φ⁴ sunset, three lines joining two vertices, and write its amputated integral. Then take m = 0.10 GeV, external p = (0.25, 0, 0, 0) GeV and loop momenta l = (0.12, 0.05, 0, 0) GeV and k = (0.08, 0, 0.06, 0) GeV: find the third line's momentum and each line's virtuality q² − m².
- Count: E = 2, V = 2, I = 3, so L = 3 − 2 + 1 = 2 and V = L + E/2 − 1 = 2: two loop integrals at order λ².
- Route: p enters at z₁. Give two lines l and k, from z₁ to z₂; conservation at z₁ fixes the third at p − l − k. At z₂ the lines deliver l + k + (p − l − k) = p, which leaves, so that δ is the overall one.
- Integral, with topic 3.02's S = 6: (1/6)(−iλ)² ∫d⁴l/(2π)⁴ ∫d⁴k/(2π)⁴ [i/(l² − m² + iε)][i/(k² − m² + iε)][i/((p − l − k)² − m² + iε)]. Dimension: +8 from the measures and −6 from the propagators, so the amputated sunset is in GeV², a mass squared ✓.
- Third line: p − l − k = (0.25 − 0.12 − 0.08, −0.05, −0.06, 0) = (0.05, −0.05, −0.06, 0) GeV. The energies 0.12 + 0.08 + 0.05 sum to 0.25 GeV and every spatial component balances: conservation is exact.
- Virtualities, with m² = 0.0100 GeV²: l² = 0.0144 − 0.0025 = 0.0119, so +0.0019 GeV²; k² = 0.0064 − 0.0036 = 0.0028, so −0.0072 GeV²; (p − l − k)² = 0.0025 − 0.0025 − 0.0036 = −0.0036, so −0.0136 GeV². All three lines are off shell, and the third is spacelike.
- Move to any other point of the integration and the virtualities change while the energy and momentum sums do not: conservation holds everywhere in the integrand, the mass shell almost nowhere. With this routing p sits only in the third propagator, so unlike the tadpole the sunset depends on p².
Answer(1/6)(−iλ)²∫∫ of propagators in l, k and p − l − k: two loops at order λ², in GeV². Third line (0.05, −0.05, −0.06, 0) GeV; virtualities +0.0019, −0.0072 and −0.0136 GeV².
HardIn φ⁴, momenta p₁ and p₂ enter at vertex z₁, two lines join z₁ to z₂, two more join z₂ to z₃, and p₃ and p₄ leave at z₃. Count, route and write the amputated amplitude with its symmetry factor, and show that it factorises. Then evaluate s, t and u for p₁ and p₂ head-on in the centre-of-mass frame with energy 0.40 GeV each, m = 0.14 GeV and a 90° scattering angle.
- Count: E = 4, V = 3, I = 4, so L = 4 − 3 + 1 = 2 and V = L + E/2 − 1 = 3: two loops at order λ³, one λ beyond the one-loop bubble ✓.
- Route: P = p₁ + p₂ enters z₁, so the first pair of lines carries l and P − l. They deliver P to z₂, so the second pair carries k and P − k, and at z₃ the conservation of P = p₃ + p₄ is the overall δ⁴(p₁ + p₂ − p₃ − p₄). Two free momenta, l and k ✓.
- Symmetry: with the external points labelled, the automorphisms swap the two z₁–z₂ lines and, independently, the two z₂–z₃ lines, so S = 2 × 2 = 4 by topic 3.02's count.
- Amplitude: (−iλ)³(1/4)B(s)², with B(s) = ∫d⁴l/(2π)⁴ [i/(l² − m² + iε)][i/((P − l)² − m² + iε)]. Only the fixed P crosses z₂, so the k loop never sees l and the integral factorises: it is −iλ × [(−iλ/2)B]², the one-loop bubble's factor taken twice. Dimension: +8 − 8 = 0, like −iλ ✓.
- Kinematics: s = P² = (0.40 + 0.40)² = 0.64 GeV², above the two-particle threshold (0.28 GeV)² = 0.0784 GeV². With |p|² = 0.40² − 0.14² = 0.1404 GeV², t = −2|p|²(1 − cos 90°) = −0.2808 GeV² and u = −2|p|²(1 + cos 90°) = −0.2808 GeV².
- Check: s + t + u = 0.64 − 0.5616 = 0.0784 GeV², the sum of the four masses squared ✓. The t- and u-channel chains are the same function at t and u, so at 90° they are equal. The 2π count, (2π)(4V − 4I) = (2π)⁻⁴, is the overall (2π)⁴ times (2π)⁻⁸ for two loops ✓.
AnswerV = 3, L = 2, order λ³ and S = 4: the amplitude is (−iλ)³(1/4)B(s)², dimensionless and factorised because only P crosses z₂. s = 0.640 GeV², t = u = −0.281 GeV², s + t + u = 0.0784 GeV².