University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.1
Position-space Feynman rules for φ⁴
This is where Wick's theorem stops being algebra and becomes drawing. We expand the two-point function to first order in λ by hand, once, sort the fifteen contractions into pictures, and read off the rules that let you write every later term without contracting anything.
Build the model
Connect the measurement to the mechanism.
The Gell-Mann-Low formula turns every correlation function of φ⁴ theory into free-field vacuum expectation values of time-ordered products, and Wick's theorem turns each of those into a sum of products of DF. A Feynman diagram is a picture of one such product: a line for each contraction, a dot for each spacetime point. Expanding ⟨Ω|T φ(x)φ(y)|Ω⟩ to first order in −(λ/4!)φ⁴ shows the mechanism in a case small enough to check by hand.
Six fields give fifteen full contractions in two shapes. Twelve join x and y to the vertex z and close the last pair into a loop, the tadpole −(iλ/2)∫d⁴z DF(x − z)DF(0)DF(z − y); three join x to y and pair the four fields at z among themselves, a figure-eight bubble beside the free line, which the denominator removes. Read backwards, the pictures are the rules: DF for every line, −iλ with an integral over all spacetime for every vertex, the 4! ways of attaching lines to a vertex cancelling the 1/4! of ℒ and the n! labellings of vertices cancelling the 1/n! of the exponential.
The rules stop short in two places. Where a diagram has a symmetry, fewer than 4! attachments exist, leaving the tadpole's 1/2, which topic 3.02 counts in general. And because the interaction is not normal ordered, a line may start and end on one vertex, bringing DF(0), which is infinite: the tadpole is a mass shift (λ/2)DF(0) that only Unit 6 makes finite.
- Simple definition
- Position-space Feynman rules turn each term of a correlation function's perturbation series into a picture and back: DF(a − b) for every line joining points a and b, and −iλ with ∫d⁴z over all spacetime for every φ⁴ vertex z, divided by the diagram's symmetry factor.
- Example
- At order λ the two-point numerator has 5!! = 15 contractions: 12 make the tadpole, weight 12/4! = 1/2 on −iλ∫d⁴z DF(x − z)DF(0)DF(z − y), and 3 make the figure-eight beside DF(x − y), weight 3/4! = 1/8.
Derived from topic 2.06's numerator. Six fields have 5!! = 15 full contractions (topic 2.08), and only full contractions survive in the vacuum.
λ is dimensionless and d⁴z in GeV⁻⁴; every field is free. It comes from exp(i∫d⁴z ℒᵢₙₜ), as HI = −∫d³z ℒᵢₙₜ.
Derived: x takes one of the four fields at z, y one of the other three, and the last two join, making the loop DF(0) that starts and ends at z.
DF in GeV², so the term is GeV², like DF(x − y); the 1/2 is 12 contractions ÷ 4!.
Derived. The Gell-Mann-Low denominator carries the same term at order λ, so the ratio drops it; topic 3.03 shows the cancellation at every order.
∫d⁴z DF(0)² = DF(0)² V₄, a constant in GeV⁴ times the spacetime volume V₄ in GeV⁻⁴, so a pure number; the 1/8 is 3 pairings ÷ 4!.
Derived by reading the first-order terms backwards: 4! attachments cancel each 1/4!, and n! vertex labellings cancel the exponential's 1/n!.
S, the symmetry factor, is a pure number: 2 for the tadpole, 8 for the figure-eight, 1 for a tree with labelled external points.
Derived by Fourier transform. It makes the tadpole δm² ∂DF/∂m² with δm² = (λ/2)DF(0): the first-order term of a heavier free line.
Both sides are pure numbers: GeV⁻⁴ × GeV⁴ on the left, GeV² ÷ GeV² on the right; m is the mass in the free propagator.
Derived from the mode sum at x = y. The Λ² growth belongs to the cutoff and its coefficient to this one: a Euclidean four-momentum cutoff gives Λ²/(16π²), and dimensional regularisation, in topic 6.05, gives no Λ² at all.
The mode sum ∫d³p/((2π)³2Eₚ) cut off at |p| = Λ, in GeV; DF(0) in GeV², near Λ²/(8π²) for Λ ≫ m.
Bring down one vertex and count the fields
Work in ħ = c = 1 with the metric (+, −, −, −). Topic 2.06 writes ⟨Ω|T φ(x)φ(y)|Ω⟩ as a ratio whose numerator is ⟨0|T φ(x)φ(y) exp(−i∫dt HI)|0⟩, every field free. With no derivative coupling HI = −∫d³z ℒᵢₙₜ (topic 2.03), so the exponent is i∫d⁴z ℒᵢₙₜ = −(iλ/4!)∫d⁴z φ(z)⁴, with T → ∞(1 − iε) letting z run over all of spacetime. To first order the numerator is DF(x − y) plus −(iλ/4!)∫d⁴z ⟨0|T φ(x)φ(y)φ(z)φ(z)φ(z)φ(z)|0⟩. The four fields at z are four operators that happen to share a point, and Wick's theorem (topic 2.08) treats them as distinct: six fields, 5!! = 5 × 3 × 1 = 15 full contractions, each a product of three DF's, every term with a normal-ordered remainder vanishing in the vacuum. Nothing about the vertex has been assumed yet; the whole calculation is counting which field pairs with which.
Fifteen contractions, two shapes
Sort the fifteen by what x contracts with. If x pairs with y, the four fields at z pair among themselves in 3 ways, the first choosing one of three partners, each giving DF(x − y)DF(z − z)². If x pairs with a field at z it has 4 to choose from; y then takes one of the 3 left, and the last two pair with each other: 12 contractions, each DF(x − z)DF(z − z)DF(z − y). The count 3 + 12 = 15 confirms nothing is missed. Writing DF(z − z) = DF(0), the numerator to order λ is DF(x − y) − (iλ/8)DF(x − y)∫d⁴z DF(0)² − (iλ/2)∫d⁴z DF(x − z)DF(0)DF(z − y), derived, from 3/4! = 1/8 and 12/4! = 1/2. Dimensions check it: [φ] = 1, so DF is GeV², d⁴z is GeV⁻⁴ and λ is a pure number, making the tadpole GeV², like the free term, and the bubble factor a pure number. At λ → 0 only the free propagator is left. φ⁴ has no measured coupling to compare with, so dimensions, limits and the mass-shift reading below are the checks.
The bubble belongs to the denominator
The middle term is the free line times a factor that knows nothing of x and y. Its integrand is constant, so ∫d⁴z DF(0)² = DF(0)²V₄, with V₄ the volume of spacetime: infinite even before DF(0) is. Nothing joins z to the external points; it is a vacuum bubble. The Gell-Mann-Low denominator ⟨0|T exp(−i∫dt HI)|0⟩ holds at first order only the four fields at z, 3 contractions and the same factor: 1 + B, with B = −(iλ/8)∫d⁴z DF(0)². Call the tadpole A. Then [DF(x − y)(1 + B) + A]/(1 + B) = DF(x − y) + A + O(λ²), because A is already of order λ, so ⟨Ω|T φ(x)φ(y)|Ω⟩ = DF(x − y) − (iλ/2)∫d⁴z DF(x − z)DF(0)DF(z − y) + O(λ²), derived. At this order the bubble is not made small; it is removed, because the numerator's bubble and the denominator's are the same term. Why that happens at every order, and what the factor of V₄ means physically, is topic 3.03. The working rule starts here: keep only diagrams in which every vertex is joined to an external point.
Draw the terms and read the rules back
Draw each product: a dot for each spacetime point, a line for each DF. The tadpole is a line from x to z, a line from z to y and a loop from z back to z; the bubble term is the line from x to y beside a figure-eight at z. Now run the dictionary backwards. A line joining a and b is DF(a − b), with no arrow, since DF(a − b) = DF(b − a) for a real field. A vertex is −iλ times ∫d⁴z over all spacetime; external points stay fixed. Why −iλ and not −iλ/4!? Wick's theorem produced every assignment of a diagram's lines to the four fields at z. Were all four lines distinguishable, 4! = 24 assignments would give the same integral and cancel the 1/4!. In the tadpole two of the lines are the two ends of one loop, and exchanging them returns the same contraction, so only 24/2 = 12 exist: the 1/2. The rules read: DF per line, −iλ∫d⁴z per vertex, all divided by a symmetry factor S, with S = 2 for the tadpole and 8 for the figure-eight. Topic 3.02 finds S for any diagram.
At order n the 1/n! goes to relabelling vertices
At order λⁿ the exponential brings (1/n!)(−iλ/4!)ⁿ and n integration points z₁, …, zₙ. They are dummy variables, so permuting their labels turns one contraction into another with an equal integral, and each diagram appears once per labelling. Count two tadpoles in series at order λ²: x joined to one vertex, y to the other, one line between them and a loop on each. x takes one of 8 fields, 2 vertices times 4; y one of the 4 at the other vertex; the link one of 3 × 3 pairs; and each vertex closes its loop with its last two fields. That is 8 × 4 × 9 = 288 contractions, weight 288/(2! × 4! × 4!) = 288/1152 = 1/4. The 2 inside the 8 is the relabelling z₁ ↔ z₂, cancelling the 1/2!; the other 4 × 4 × 9 = 144 attachments fall short of 4! × 4! = 576 by the 2 × 2 of the two loops. The rule, derived: drop the 1/n!, draw each distinct diagram once with −iλ per vertex, and divide by S. Only a diagram that swapping vertices maps onto itself spoils the n!, and that too is topic 3.02's S.
Tadpoles stay, and a tadpole is a mass shift
ℒᵢₙₜ is not normal ordered, so a field at z may contract with another at z, leaving DF(0) = ⟨0|φ(z)²|0⟩ = ∫d³p/((2π)³2Eₚ). Cut off at |p| = Λ, the mode sum gives [Λ√(Λ² + m²) − m² asinh(Λ/m)]/(8π²), derived, near Λ²/(8π²) for Λ ≫ m; topic 1.08's 1/(4π²r²) shows the same divergence as r → 0. The tadpole shifts the mass. By Fourier transform ∫d⁴z DF(x − z)DF(z − y) = i∂DF(x − y)/∂m², so the tadpole is δm²∂DF/∂m² with δm² = (λ/2)DF(0): the first-order term of a free propagator of mass² m² + δm², derived. At equal times and separation r, DF = mK₁(mr)/(4π²r), quoted from topic 1.08, and the tadpole multiplies it by 1 − (δ/2) mr K₀(mr)/K₁(mr), with δ = δm²/m²: correlations fall faster, as a heavier field's must. Normal ordering ℒᵢₙₜ would delete every such diagram, the same as adding a mass term that cancels δm² (topic 2.09); the course keeps them and lets Unit 6's mass counterterm absorb DF(0) (topic 6.05).
Change one variable at a time
Make the relationship visible.
Set λ to 0 and the curves coincide. Raise Λ/m: δm² grows as Λ², and the solid first-order curve falls away from the faint heavier-mass curve at large m r; at λ = 1 and Λ/m = 8 it crosses zero near m r = 5.6, where first order has run out.
DF(0)/m² WITH CUTOFF0.431
SHIFT δm²/m²0.129
HEAVIER MASS m′/m1.063
TADPOLE G/DF AT PROBE0.895
Live interpretationDF(0)/m² WITH CUTOFF: 0.431. SHIFT δm²/m²: 0.129. HEAVIER MASS m′/m: 1.063. TADPOLE G/DF AT PROBE: 0.895
Catch the common trap
Explain before calculating.
In φ⁴ theory with ℒᵢₙₜ = −(λ/4!)φ⁴, the order-λ numerator of ⟨Ω|T φ(x)φ(y)|Ω⟩ contains the diagram in which x and y both join the vertex z and the two remaining fields at z join each other. Which term is it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA real scalar has ℒᵢₙₜ = −(η/6!)φ⁶. Expand the numerator of ⟨Ω|T φ(x)φ(y)|Ω⟩ to first order in η, sort the full contractions into diagrams, give each its weight, and check the count and the mass dimensions.
- First order brings −(iη/6!)∫d⁴z ⟨0|T φ(x)φ(y)φ(z)⁶|0⟩: eight fields, so 7!! = 7 × 5 × 3 × 1 = 105 full contractions.
- x with y: the six fields at z pair among themselves in 5!! = 15 ways, each DF(x − y)DF(0)³. Weight 15/6! = 15/720 = 1/48.
- x with a field at z (6 ways), y with one of the other 5, the remaining four paired in 3 ways: 6 × 5 × 3 = 90 contractions, each DF(x − z)DF(0)²DF(z − y). Weight 90/720 = 1/8. Check: 15 + 90 = 105.
- So the numerator is DF(x − y) − (iη/8)∫d⁴z DF(x − z)DF(0)²DF(z − y) − (iη/48)DF(x − y)∫d⁴z DF(0)³; the last term is a bubble beside the free line and divides out as in φ⁴.
- Dimensions: [η] = 4 − 6 = −2, so η DF(0)²∫d⁴z DF DF is GeV⁻² × GeV⁴ × GeV⁻⁴ × GeV⁴ = GeV², matching DF(x − y), and η∫d⁴z DF(0)³ is GeV⁻² × GeV⁻⁴ × GeV⁶, a pure number.
AnswerA double-loop tadpole −(iη/8)∫d⁴z DF(x − z)DF(0)²DF(z − y) from 90 contractions, and a bubble term of weight 1/48 from 15; 90 + 15 = 105 = 7!!.
MediumA real scalar of mass m = 1.00 GeV has ℒᵢₙₜ = −(κ/3!)φ³ with κ = 0.30 GeV. Find ⟨Ω|φ(x)|Ω⟩ to first order in κ by contracting, evaluate it with DF(0) = 1.235 GeV², the value a cutoff Λ = 10.0 GeV on |p| gives, and check it against the vacuum average of the field equation.
- First order brings −(iκ/3!)∫d⁴z ⟨0|T φ(x)φ(z)³|0⟩. Four fields have 3!! = 3 full contractions, all one diagram: x joins one of the three fields at z and the other two close a loop. Weight 3/3! = 1/2, so the term is −(iκ/2)∫d⁴z DF(x − z)DF(0), as the rules give with vertex −iκ and S = 2.
- The denominator's ⟨0|T φ(z)³|0⟩ has an odd number of fields and vanishes, and at order κ² there are seven fields, so ⟨Ω|φ(x)|Ω⟩ = −(iκ/2)DF(0)∫d⁴z DF(x − z) + O(κ³).
- The integral is the propagator's Fourier transform at zero momentum: ∫d⁴z DF(x − z) = i/(0 − m² + iε) = −i/m². So ⟨Ω|φ(x)|Ω⟩ = −(iκ/2)(−i/m²)DF(0) = −κDF(0)/(2m²), real and the same at every x, as translation invariance requires.
- Dimensions: [κ] = 4 − 3 = 1, so κDF(0)/m² is GeV × GeV² ÷ GeV² = GeV, the dimension of φ. Numbers: −0.30 × 1.235/(2 × 1.00) GeV = −0.185 GeV.
- Check: the field equation is (□ + m²)φ = −(κ/2)φ². In the vacuum □⟨φ⟩ = 0 and ⟨φ²⟩ = DF(0) at leading order, so m²⟨φ⟩ = −(κ/2)DF(0), the same result: the loop is the fluctuation ⟨φ²⟩ and the line at zero momentum is the static response 1/m².
Answer⟨Ω|φ(x)|Ω⟩ = −κDF(0)/(2m²) = −0.185 GeV at this cutoff. A φ³ tadpole shifts the field's vacuum value, which the φ → −φ symmetry of φ⁴ forbids at every order.
HardA φ⁴ scalar of mass m = 139.57 MeV has λ = 0.50, with DF(0) cut off at Λ = 1.00 GeV on |p|. For x and y at equal times, r = 2.00 fm apart, find the first-order ⟨Ω|T φ(x)φ(y)|Ω⟩ as a multiple of the free DF and compare it with the free propagator of the shifted mass. Use DF = mK₁(mr)/(4π²r) at spacelike separation and d(zK₁(z))/dz = −zK₀(z).
- To first order G = ⟨Ω|T φ(x)φ(y)|Ω⟩ = DF + δm² ∂DF/∂m². With z = mr, DF = zK₁(z)/(4π²r²) and ∂/∂m² = (r²/(2z))∂/∂z at fixed r, so ∂DF/∂m² = −K₀(mr)/(8π²) and G/DF = 1 − (δ/2) mr K₀(mr)/K₁(mr), with δ = δm²/m². Limit: r → 0 gives 1, since short distances do not feel the mass.
- m² = 0.019480 GeV² and Λ/m = 7.165: DF(0) = [1.00 × 1.00969 − 0.019480 × asinh 7.165]/78.957 GeV² = (1.00969 − 0.05196)/78.957 = 0.012130 GeV², so δm² = 0.25 × 0.012130 = 0.0030325 GeV² and δ = 0.15567.
- Restore ħc: mr = 139.57 MeV × 2.00 fm/(197.327 MeV fm) = 1.4146, where K₀ = 0.23902 and K₁ = 0.31402, a ratio of 0.76117.
- G/DF = 1 − (0.15567/2) × 1.4146 × 0.76117 = 1 − 0.08381 = 0.9162: the tadpole lowers the equal-time correlation at 2.00 fm by 8.4%.
- Shifted mass: m′ = m√1.15567 = 150.04 MeV, so m′r = 1.5207 and K₁(m′r) = 0.26926; the ratio (m′/m)K₁(m′r)/K₁(mr) = 1.07502 × 0.26926/0.31402 = 0.9218, within 0.006 of the first-order value, the difference being second order in δm².
- For mr ≫ 1, K₀/K₁ → 1 and the correction grows as (δ/2) mr, reaching 1 at mr = 2/δ = 12.85, that is r = 12.85 × 197.327 MeV fm/139.57 MeV = 18.2 fm: there first order fails and the insertions must be summed, which topic 3.08 does.
AnswerG/DF = 0.916 at r = 2.00 fm, against 0.922 for the free propagator of mass 150.0 MeV; first order in the tadpole holds while (δ/2) mr ≪ 1, that is r ≪ 18 fm.