University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.4
Connected diagrams and the four-point function at order λ
Most of the φ⁴ four-point function is two quanta ignoring each other. Here we expand it to order λ, sort all 105 contractions, and strip out the products of two-point functions that describe only propagation; what survives is one vertex, the connected function every φ⁴ scattering amplitude starts from.
Build the model
Connect the measurement to the mechanism.
In natural units ħ = c = 1 with metric (+, −, −, −), the four-point function ⟨Ω|Tφ(x₁)φ(x₂)φ(x₃)φ(x₄)|Ω⟩ is the smallest correlation function in which the φ⁴ interaction can join two quanta to each other rather than merely dress one. Topic 2.06's Gell-Mann-Low formula and topic 2.08's Wick theorem compute it: at order λ the numerator holds eight fields and 7!! = 105 contractions, in three families. Nine carry the figure-eight bubble and cancel against the denominator.
Seventy-two hang a tadpole on one line of a free pairing. Twenty-four join each external point to a different leg of the vertex and cancel the 1/4! exactly. The first two families say nothing new: they are products of two-point functions, two quanta propagating independently, each with its own order-λ mass shift.
What carries the interaction is what survives once every product of lower functions is subtracted, the connected four-point function. It vanishes for the free field, which Wick's theorem makes Gaussian, and at order λ it is the single term −iλ∫d⁴z DF(x₁ − z)DF(x₂ − z)DF(x₃ − z)DF(x₄ − z): local, because all four lines meet at one point, and integrated over all spacetime, so one diagram holds every time ordering. It is not yet an amplitude: it is first order only, it keeps four external propagators, and that only connected diagrams reach scattering is stated here and proved with LSZ in Unit 4.
- Simple definition
- The connected four-point function is the four-point correlation function minus every product of lower correlation functions; in φ⁴ that means minus the three products of two-point functions, which leaves exactly the diagrams joining all four external points into one piece.
- Example
- At order λ, 105 contractions split into 24 connected, 72 tadpole-on-a-line and 9 bubble contractions. 24/4! = 1, so Gc⁽⁴⁾ = −iλ∫d⁴z DF(x₁ − z)DF(x₂ − z)DF(x₃ − z)DF(x₄ − z), with symmetry factor 1.
Derived: the 4! ways of putting four distinct external fields on the vertex's four legs all give this term and cancel the 1/4!, so its symmetry factor is 1.
λ dimensionless; each DF in GeV²; d⁴z in GeV⁻⁴; so Gc⁽⁴⁾ is in GeV⁴, as four fields of dimension 1 require
Derived by direct count. External fields enter the vertex in even numbers only, 4, 2 or 0, and the families summing to 105 is the check that no contraction was lost.
eight fields, four external and four at z; each count times −iλ/4! gives the weight −iλ, −iλ/2 or −iλ/8
Quoted. Multiply two of these and keep order λ: a free pairing plus a tadpole on either of its lines, so three pairings give exactly the nine non-vertex terms.
G⁽²⁾ and DF in GeV²; DF(0) is the tadpole loop, in GeV², divergent and left to topic 6.05
A definition: the fourth cumulant. It is zero for a free field, whose correlators Wick's theorem builds from pairs alone, so it measures the interaction and nothing else.
i stands for xᵢ; all terms in GeV⁴; G⁽¹⁾ and G⁽³⁾ vanish in φ⁴ because of φ → −φ, so no products with them
Derived in one line, since each factor depends on one difference. The vertex term carries a single δ⁴(p₁ + p₂ + p₃ + p₄), so only it moves momentum between the pairs.
transform with ∫d⁴xⱼ e(−ipⱼ⋅xⱼ), all pⱼ incoming, in GeV; each δ⁴ in GeV⁻⁴
Derived from topic 1.06's p⁰ contour, e(−im|t − z⁰|)/(2m) per line. With k lines arriving from earlier the phase mΣᵢ|tᵢ − z⁰| has slope (2k − 4)m: flat only for 2 in, 2 out.
V the volume of space; tᵢ = xᵢ⁰ and z⁰ in GeV⁻¹, or fm via ħc = 0.1973 GeV fm; m in GeV
Eight fields, 105 contractions, three families
At order λ the Gell-Mann-Low numerator is (−iλ/4!)∫d⁴z ⟨0|Tφ₁φ₂φ₃φ₄φ(z)⁴|0⟩, with φᵢ = φ(xᵢ). Wick's theorem, topic 2.08, quoted, turns the eight fields into a sum over full contractions, 7!! = 105 of them. Sort them by how many external fields contract into the vertex. That number is even, because the legs left over must pair among themselves. Four: φ₁ has 4 legs to choose from, φ₂ then 3, φ₃ 2 and φ₄ 1, so 4! = 24 contractions. Two: choose the pair in 6 ways, attach it in 4 × 3 = 12, close the last two legs into a loop and contract the other two externals together, 72 in all. None: 3 pairings of the externals times 3 pairings of the legs, 9. The total, 24 + 72 + 9 = 105, is the check that nothing is lost. Multiplying by the prefactor −iλ/4! = −iλ/24 gives the weights, all derived: −iλ for the connected term, −iλ/2 for each of six tadpole terms and −iλ/8 for each of three bubble terms.
The bubbles cancel; the tadpoles sit on single lines
The nine bubble contractions are the free four-point function times (−iλ/8)∫d⁴z DF(0)², the figure-eight with symmetry factor 8. The denominator through order λ is 1 + (−iλ/8)∫d⁴z DF(0)², so dividing removes them, as topic 3.03 showed to all orders. The 72 tadpole contractions make six terms such as −(iλ/2)DF(0)∫d⁴z DF(x₁ − z)DF(z − x₂) × DF(x₃ − x₄): a loop hung on the line from x₁ to x₂, with the line from x₃ to x₄ untouched. The first factor is exactly the order-λ part of topic 3.01's two-point function. So through order λ the four-point function has ten terms: the three free pairings D₁₂D₃₄ + D₁₃D₂₄ + D₁₄D₂₃, writing Dᵢⱼ = DF(xᵢ − xⱼ), the six tadpole terms and the vertex. Each has mass dimension 4, since DF carries 2, d⁴z −4 and λ none. Set λ = 0 and topic 2.09's three pairings return.
Subtract the products, and the vertex is what is left
Multiply G⁽²⁾(x₁, x₂)G⁽²⁾(x₃, x₄), each through order λ: D₁₂D₃₄, plus a tadpole on either line, plus terms of order λ². The three pairings give 3 + 6 = 9 terms, exactly the nine non-vertex terms. So defining Gc⁽⁴⁾ as G⁽⁴⁾ minus the three products leaves −iλ∫d⁴z DF(x₁ − z)DF(x₂ − z)DF(x₃ − z)DF(x₄ − z) + O(λ²), shown. Only two-point products appear because φ → −φ makes G⁽¹⁾ and G⁽³⁾ vanish; a field without that symmetry, such as the scalar Yukawa model's φ, needs products with them as well. This is the cumulant of probability theory. A Gaussian variable has ⟨x⁴⟩ = 3⟨x²⟩², and the free field is Gaussian, so Gc⁽⁴⁾ = 0 at λ = 0. That the subtraction leaves exactly the connected diagrams at every order is stated, its generating-functional proof only named. At λ² there are seven: topic 3.02's three one-loop bubbles, and four with a tadpole on one external leg of the vertex.
One vertex, anywhere in spacetime
The connected term is local: −(λ/4!)φ⁴ multiplies four fields at one point, so all four lines end at the same z. That point is not put anywhere; it is integrated over all of spacetime, and the propagators decide where it matters. At spacelike separation r, DF is UAQM-09.04's mK₁(mr)/(4π²r), quoted, dying as e(−mr) beyond the reduced Compton wavelength ħ/(mc), 0.66 fm for m = 0.30 GeV. At timelike separation it oscillates and falls only as a power, so a vertex far from every external point still counts if it sits between them in time, as the next section shows. Three checks hold for the term as written. Dimensions: four DF give 8, d⁴z gives −4, total 4. Symmetry: the product of four DF is unchanged by all 24 permutations of the xᵢ, as identical bosons require, while each disconnected product is unchanged by only 8 and the three together restore the rest. Invariance: DF(x) depends on x only through x², and d⁴z is invariant, so Gc⁽⁴⁾ is unchanged when one Lorentz transformation and one translation act on all four points.
Every time ordering lives in one term
Each DF(x − z) is θ(x⁰ − z⁰)⟨0|φ(x)φ(z)|0⟩ + θ(z⁰ − x⁰)⟨0|φ(z)φ(x)|0⟩: a quantum created at whichever end is earlier and absorbed at the later. Fix t₁ ≤ t₂ ≤ t₃ ≤ t₄ and let z⁰ run. In each of the five intervals k lines arrive at z from earlier points and 4 − k leave for later ones: 0 → 4, 1 → 3, 2 → 2, 3 → 1, 4 → 0. Integrate each external field over all space, so that only times matter; topic 1.06's p⁰ contour makes each line e(−im|tᵢ − z⁰|)/(2m), and the z⁰ integrand is e(−iΦ) with Φ = mΣ|tᵢ − z⁰|. Its slope, (2k − 4)m, is the energy arriving at the vertex minus the energy leaving it. Only the 2 → 2 interval is flat, so it contributes its whole length t₃ − t₂, growing with the time available, while each other interval oscillates and gives at most 2/ΔE for a mismatch ΔE. Old-fashioned perturbation theory, quoted, draws these as five diagrams with energy denominators; the Feynman diagram is one Lorentz-invariant term.
Why only the connected part can scatter, stated
Translate x₁ and x₂ together, leaving x₃ and x₄ alone: D₁₂D₃₄ does not change, so the pair (1, 2) exchanges nothing with the pair (3, 4). Fourier transforming says it with delta functions: the product carries δ⁴(p₁ + p₂)δ⁴(p₃ + p₄), two separate conservation laws, while the vertex term carries the single δ⁴(p₁ + p₂ + p₃ + p₄), so only it can turn incoming momenta into different outgoing ones. A tadpole term is no better; its loop rides on one line and shifts that quantum's mass, topic 3.08's subject. Hence the claim, stated here and shown with LSZ in Unit 4: only connected diagrams with every external point attached contribute to scattering, the rest giving the 1 in S = 1 + iT, or a lower-point process beside spectators. Two cautions. The connected function is not yet an amplitude: it still carries four external propagators, which topic 4.05 amputates to leave ℳ = −λ, quoted. And pulling the pairs far apart in space leaves D₁₂D₃₄ fixed while the connected part dies away exponentially, the cluster property.
Change one variable at a time
Make the relationship visible.
Slide z⁰ from −5.75 to +5.75 fm and the arrows turn one at a time, 0 → 4, 1 → 3, 2 → 2, 3 → 1, 4 → 0: five processes, one diagram. Raise m and only the energy-violating stretches oscillate faster; widen the gap and the flat stretch grows while the rest stays fixed.
LINES ARRIVING AT z2
ENERGY MISMATCH AT z0.00 GeV
OTHER FOUR ORDERINGS, |SUM|0.375 fm
2 → 2 STRETCH ÷ REST10.7
Live interpretationLINES ARRIVING AT z: 2. ENERGY MISMATCH AT z: 0.00 GeV. OTHER FOUR ORDERINGS, |SUM|: 0.375 fm. 2 → 2 STRETCH ÷ REST: 10.7
Catch the common trap
Explain before calculating.
In φ⁴ theory, what is the connected four-point function Gc⁽⁴⁾(x₁, x₂, x₃, x₄) through order λ, once the vacuum bubbles have cancelled against the denominator?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyShrink spacetime to one point: every DF becomes 1, every ∫d⁴z becomes 1 and −iλ becomes −λ, so averages are taken with the weight e(−φ²/2 − λφ⁴/4!) and Wick's counting is unchanged. Find ⟨φ²⟩ and ⟨φ⁴⟩ to order λ, form ⟨φ⁴⟩c = ⟨φ⁴⟩ − 3⟨φ²⟩², and compare with −0.04255, the value numerical integration gives at λ = 0.05.
- Gaussian moments count full contractions: ⟨φ²ⁿ⟩₀ = (2n − 1)!!, so ⟨φ²⟩₀ = 1, ⟨φ⁴⟩₀ = 3, ⟨φ⁶⟩₀ = 15 and ⟨φ⁸⟩₀ = 105, the same 105 as the four-point function at order λ.
- To order λ the numerator is ⟨φ⁴(1 − λφ⁴/24)⟩₀ = 3 − 105λ/24 = 3 − 4.375λ, and the denominator is 1 − 3λ/24 = 1 − λ/8, the figure-eight's 1/8.
- Divide: ⟨φ⁴⟩ = (3 − 4.375λ)(1 + 0.125λ) = 3 − 4λ + O(λ²). The 9 bubble contractions, −0.375λ, are cancelled by the denominator's +0.375λ, leaving the 24 connected ones, −λ, and the 72 tadpole ones, −3λ.
- Likewise ⟨φ²⟩ = (1 − 15λ/24)(1 + λ/8) = 1 − λ/2: of the 15 contractions, 12 make the tadpole with weight 12/24 = 1/2 and 3 make the bubble that cancels.
- Connected: 3 − 4λ − 3(1 − λ/2)² = 3 − 4λ − 3 + 3λ = −λ + O(λ²). The tadpole terms cancel against the cross terms of 3⟨φ²⟩², and the vertex survives with weight 1. At λ = 0 the cumulant vanishes, as it must for a Gaussian.
- At λ = 0.05 this gives −0.0500 against −0.04255, an error of 0.0075 that is order λ²: the seven connected diagrams there, each worth 1/2, add 3.5λ² = +0.00875 and give −0.04125. As a check, the exact (⟨φ⁴⟩c + λ)/λ² is 3.488 at λ = 0.001, tending to 3.5.
Answer⟨φ⁴⟩c = −λ + O(λ²), the vertex with symmetry factor 1; at λ = 0.05 that is −0.0500 against the exact −0.04255, and −0.04125 once the seven λ² diagrams are added.
MediumExpand the φ⁴ six-point function ⟨Ω|Tφ(x₁) … φ(x₆)|Ω⟩ to order λ. Count its contractions and sort them into families, show that the connected six-point function has no order-λ part, and find how many diagrams make up its first non-zero order and the weight of each.
- Ten fields, six external and four at z, give 9!! = 945 full contractions. External fields enter the vertex in even numbers: 4, 2 or 0.
- Four in: choose them in 15 ways and attach them in 4! = 24, the other two contracting together. That is 360 contractions in 15 terms of weight −iλ, each the order-λ vertex on four points times DF joining the remaining two.
- Two in: 15 choices, 4 × 3 = 12 attachments, a loop from the last two legs, and the other four externals paired in 3 ways: 540 contractions in 45 tadpole terms of weight −iλ/2. None in: 15 pairings × 3 = 45 bubble contractions, which the denominator removes. Check: 360 + 540 + 45 = 945.
- With G⁽¹⁾ = G⁽³⁾ = 0, G⁽⁶⁾ = Gc⁽⁶⁾ + Σ Gc⁽⁴⁾G⁽²⁾ + Σ G⁽²⁾G⁽²⁾G⁽²⁾, each sum over 15 ways of splitting the points. At order λ the first sum supplies the 15 vertex terms and the second the 15 free pairings with their 15 × 3 = 45 tadpole terms.
- Every one of the 900 contractions that survive the denominator is accounted for, so Gc⁽⁶⁾ = 0 + O(λ²): one four-leg vertex cannot join six points. The first connected diagrams have two vertices joined by one line and three external points on each, 20/2 = 10 of them.
- Weight: 2 ways to give the triples to the vertices, 4 × 3 × 2 = 24 for one triple on its vertex, 4 for the joining line's far leg and 3! = 6 for the other triple, so 2 × 24 × 4 × 6 = 1152 contractions against (1/2!)(1/4!)² = 1/1152. Each diagram is (−iλ)²∫d⁴z₁d⁴z₂ times seven DF, weight 1, symmetry factor 1, dimension 7 × 2 − 8 = 6 as six fields need.
Answer945 = 360 + 540 + 45 contractions; all 900 surviving ones are products of lower functions, so Gc⁽⁶⁾ = O(λ²), where it begins with 10 tree diagrams of weight (−iλ)² = −λ², symmetry factor 1.
HardIn φ⁴ with m = 0.30 GeV, integrate each external field over all space and place them at ct₁ = 0, ct₂ = 0.80 fm, ct₃ = 12.80 fm and ct₄ = 13.60 fm. Split the order-λ connected term's vertex-time integral into its five time orderings, find what each contributes, and compare the energy-conserving one with the other four together. What changes if t₃ and t₄ are both delayed by a further 12.0 fm/c?
- At zero momentum each line is e(−im|t − z⁰|)/(2m), so the term is −iλV(2m)⁻⁴ I with I = ∫dz⁰ e(−iΦ) and Φ = mΣ|tᵢ − z⁰|. Restoring ħ and c, m/ħc = 0.30/0.197327 = 1.5203 fm⁻¹, and I is measured as a length cz⁰ in fm.
- The five intervals: before t₁, 0 lines arrive and 4 leave, 0 → 4, energy mismatch 4mc² = 1.20 GeV; t₁ to t₂, 1 → 3, 0.60 GeV; t₂ to t₃, 2 → 2, balanced; t₃ to t₄, 3 → 1, 0.60 GeV; after t₄, 4 → 0, 1.20 GeV.
- On the 2 → 2 stretch Φ is a constant Φ₀, so that stretch contributes e(−iΦ₀) × 12.0 fm in full. Factor out e(−iΦ₀) and measure the other four against it.
- Each side stretch, length a = 0.80 fm with ma = 1.2163: (1 − e(−2ima))/(2im), modulus |sin ma|/m = 0.9378 × 0.6578 fm = 0.617 fm. Each tail, damped by topic 2.05's iε: e(−2ima)/(4im), modulus ħc/(4mc²) = 0.164 fm, the reach ħ/ΔE of a 1.20 GeV fluctuation.
- Added with their phases the four give (1 − e(−2ima)/2)/(im) = 0.214 − 0.907i fm, modulus √(5/4 − cos 2ma)/m = 0.932 fm, less than the 1.56 fm sum of moduli because they partly cancel. The 2 → 2 stretch is 12.0/0.932 = 12.9 times larger and gives 98% of |I| = 12.25 fm.
- Delaying t₃ and t₄ by 12.0 fm/c doubles the flat stretch to 24.0 fm and leaves the rest, which depends only on m and a, at 0.932 fm: the ratio becomes 25.7. The energy-conserving ordering grows with the time available, the seed of a rate; the others stay capped near ħ/ΔE.
AnswerThe 2 → 2 ordering contributes 12.0 fm; the other four together 0.214 − 0.907i fm, modulus 0.932 fm, a ratio of 12.9. The delay doubles the first and leaves the rest, raising the ratio to 25.7.