University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.9
Wick's theorem at work: the four-point function and ℳ = −g
This is where Wick's theorem stops being a theorem and starts producing answers. Pair four fields into the free four-point function and see why it holds no scattering; contract three fields against external particles to get the course's first amplitude, ℳ = −g; and see what normal ordering the interaction would quietly throw away.
Build the model
Connect the measurement to the mechanism.
Wick's theorem rewrites any time-ordered product of free fields as normal-ordered pieces plus contractions, and this topic spends it. In a vacuum expectation value every normal-ordered piece dies, so ⟨0|T φ₁φ₂φ₃φ₄|0⟩ is the sum over the three ways of pairing four points, each a product of two Feynman propagators. That sum is the whole answer: no term ties all four points together, which is the exact sense in which free particles never scatter, and the two cross pairings are only the direct and exchange propagation of two identical bosons.
Between particle states one rule changes: a field contracted with an incoming or outgoing particle leaves that particle's plane wave, e(−ik⋅x) or e(+ik⋅x), not a propagator. Contract all three fields of the scalar Yukawa interaction that way, integrate over the vertex, and momentum conservation appears as (2π)⁴δ⁴(k − p − q) multiplying −ig: ℳ = −g, a mass, as the coupling's dimension demands. Contractions inside one vertex are what normal ordering would forbid; in φ⁴ they are a divergent mass shift and a constant energy density.
Two limits frame the result. ℳ is computed between free states built on |0⟩, so calling it the decay amplitude of real, interacting particles waits for Unit 4's LSZ formula. And ℳ only says how strongly: the δ function says whether, and for ψ → ψφ, built from the very same interaction, it has no support at all.
- Simple definition
- Using Wick's theorem means replacing every pair of fields contracted with each other by a Feynman propagator, every field contracted with an incoming or outgoing particle by that particle's plane wave, and discarding every term that leaves an operator uncontracted.
- Example
- With x₁, x₂, x₄ and x₃ taken in order round a 1.0 fm square at one instant, a massless field's three pairings give 9.73, 9.73 and 2.43 × 10⁻⁷ GeV⁴; contracting gψ†ψφ with φ(k) in and ψ(p), ψ̄(q) out gives ℳ = −g, that is −1.5 GeV for g = 1.5 GeV.
Derived from topic 2.08: only the 3!! = 3 full contractions survive in the vacuum. No term links all four points, so the connected part is zero: free particles do not scatter.
φᵢ = φI(xᵢ), a free real field of mass dimension 1 (GeV); DF in GeV²; xᵢ in GeV⁻¹, where 1 GeV⁻¹ is 0.1973 fm
Derived. An incoming particle leaves the plane wave e(−ik⋅x) and an outgoing one, by conjugation, e(+ik⋅x): on shell, with no propagator and nothing left to integrate.
φI⁺ holds the annihilators; |k⟩ = aₖ†|0⟩, of dimension −1, with ⟨q|k⟩ = 2Eₖ(2π)³δ³(k − q); k and Eₖ in GeV
Derived: each field of gψ†ψφ is contracted with one external particle, and ∫d⁴x e(i(p + q − k)⋅x) supplies the δ⁴ of momentum conservation.
φ(k) in, ψ(p) and ψ̄(q) out; four-momenta in GeV; g in GeV; δ⁴ in GeV⁻⁴
Derived at order g, and the same for every crossing of the vertex. Using it as the amplitude of real, interacting particles is assumed here and justified by LSZ in Unit 4.
ℳ is defined by ⟨f|iT|i⟩ = (2π)⁴δ⁴(Pₒᵤₜ − Pᵢₙ) iℳ; three particles give it mass dimension 4 − 3 = 1, in GeV
Derived from the δ⁴, which demands 2√(|p|² + M²) = m. Below threshold the δ⁴ has no support, whatever ℳ is.
m and M the φ and ψ masses in GeV (GeV/c² with c restored); |p| each daughter's momentum in GeV (GeV/c with c restored)
Derived: six ways to contract one pair, three to contract all four. Ordering ℋᵢₙₜ deletes this mass shift and a constant energy density; the course keeps them for Unit 6's counterterms.
λ dimensionless; DF(0) = ∫d³k/((2π)³2Eₖ) in GeV², about Λ²/(8π²) for a cutoff Λ on |k|; δm² in GeV²
Three pairings, and nothing else
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 2.08's theorem, quoted, writes T(φ₁φ₂φ₃φ₄), with φᵢ = φI(xᵢ), as :φ₁φ₂φ₃φ₄:, plus six terms with one contraction and a normal-ordered pair left over, plus three terms with two contractions. Now take the vacuum expectation value. Every term of a normal-ordered product puts creators on the left, where ⟨0| kills them, and annihilators on the right, where |0⟩ does, so only the full contractions survive: G₄ = DF(x₁ − x₂)DF(x₃ − x₄) + DF(x₁ − x₃)DF(x₂ − x₄) + DF(x₁ − x₄)DF(x₂ − x₃), derived. Three checks. Dimensions: [φ] = 1 makes G₄ a fourth power of mass, and each DF carries two. Symmetry: inside T the free fields commute, so G₄ must be symmetric in its four points, and a permutation only shuffles the three pairings. Counting: 3 = 3!!, while six fields give 15 terms and an odd number give none.
What the three terms describe
Let x₁ and x₂ be later than x₃ and x₄ and insert a complete set of free states between the pairs: G₄ = Σₙ ⟨0|T φ₁φ₂|n⟩⟨n|T φ₃φ₄|0⟩. Two free fields acting on |0⟩ reach only the vacuum and two-quantum states. The vacuum gives D₁₂D₃₄: nothing crosses from the early pair to the late one. The two-quantum states give D₁₃D₂₄ + D₁₄D₂₃, shown: two identical quanta made at x₃ and x₄ and absorbed at x₁ and x₂, in both assignments, with a plus sign because they are bosons. Each quantum travels as if the other were absent, so the connected four-point function, G₄ minus its products of two-point functions, is zero. The first connected term is of order λ (topic 3.04, quoted). The limit to check is cluster decomposition. A massless field at equal times has DF = 1/(4π²r²) (topic 1.08), and moving two 1.0 fm pairs from 1.0 fm to 4.0 fm apart raises the D₁₂D₃₄ share from 44.4% to 99.3%; with a mass the cross terms die as e(−2mR).
External particles leave plane waves
Wick's theorem is an operator identity, so it holds between particle states too; the difference is that a normal-ordered leftover now survives when each of its operators has a particle to act on. With topic 1.03's [aₚ, aₖ†] = (2π)³2Eₖ δ³(p − k) and |k⟩ = aₖ†|0⟩, the annihilation part of the field gives φI⁺(x)|k⟩ = ∫d³p/((2π)³2Eₚ) e(−ip⋅x) aₚ aₖ†|0⟩ = e(−ik⋅x)|0⟩, derived, and ⟨k|φI⁻(x) = ⟨0|e(ik⋅x). That is what contracting a field with an external particle means: an incoming particle contributes e(−ik⋅x), an outgoing one e(+ik⋅x), each on shell, with no propagator. Dimensions agree: [φ] = 1 and [|k⟩] = −1 leave the dimensionless |0⟩. For the complex ψ of the scalar Yukawa model, ψ holds a and b†, so it absorbs an incoming ψ or emits an outgoing ψ̄, while ψ† emits ψ or absorbs ψ̄. A φ field cannot touch a ψ or ψ̄ at all.
The first amplitude: ℳ = −g for φ → ψψ̄
With ℋᵢₙₜ = gψ†ψφ (topic 2.01), topic 2.04's S-operator is S = 1 − ig∫d⁴x ψ†ψφ + O(g²). Take φ(k) in and ψ(p), ψ̄(q) out. The 1 gives ⟨p, q|k⟩ = 0, since the particle contents differ. At first order Wick's theorem splits ψ†ψφ into :ψ†ψφ: and a self-contraction DM(0) = ⟨0|ψ†ψ|0⟩ times φ; φ cannot turn |k⟩ into a ψψ̄ pair, so that term vanishes. In the first, ψ† must emit ψ(p), ψ must emit ψ̄(q) and φ must absorb φ(k), the one assignment that leaves nothing acting on the vacuum: ⟨p, q|iT|k⟩ = −ig∫d⁴x e(i(p + q − k)⋅x) = −ig(2π)⁴δ⁴(k − p − q), derived. Matched to ⟨f|iT|i⟩ = (2π)⁴δ⁴(Pₒᵤₜ − Pᵢₙ) iℳ, this is ℳ = −g. Checks: three particles need [ℳ] = 1 and topic 2.01 gave [g] = 1; g → 0 returns zero. The states are built on |0⟩, so ℳ as the decay amplitude of real particles is assumed until Unit 4's LSZ formula, whose √Z per leg is 1 + O(g²).
The δ function decides whether anything happens
ℳ is a coefficient; the process happens only where its δ⁴ has support. For φ → ψψ̄ in the φ rest frame, k = (m, 0) and the daughters carry p and −p, so 2√(|p|² + M²) = m and |p| = √(m²/4 − M²), derived: the decay is open only for m ≥ 2M. With m = 3.0 GeV and M = 1.2 GeV each daughter leaves with |p| = 0.90 GeV at speed 0.60, and at m = 2M exactly both sit at rest. The rate needs phase space and Unit 5; topic 5.03's Γ = (g²/(16πm))√(1 − 4M²/m²) is quoted. Crossed, the vertex gives ψψ̄ → φ the same ℳ = −g, but with support only where s = (p + q)² = m², a single energy until topic 5.09 gives φ a width. And the self-contraction term that vanished above gives ⟨0|iT|k⟩ = −igDM(0)(2π)⁴δ⁴(k): no support for a φ of non-zero mass, so it cannot make a φ vanish into the vacuum, though it is not harmless, as the next section shows.
What normal ordering ℋᵢₙₜ would remove
Contractions inside one vertex are exactly what normal ordering forbids. At a single point the contraction is DF(0) = ∫d³k/((2π)³2Eₖ), positive and divergent, and Wick's theorem gives φ⁴ = :φ⁴: + 6DF(0):φ²: + 3DF(0)²: six ways to choose one pair, three to pair all four. So (λ/4!)φ⁴ = (λ/4!):φ⁴: + ½(λ/2)DF(0):φ²: + (λ/8)DF(0)², derived, and ordering ℋᵢₙₜ would delete a mass shift δm² = (λ/2)DF(0) and a constant energy density. Cut |k| at ten times the mass and DF(0)/m² = 1.235, near Λ²/(8π²m²) = 1.267, so at λ = 1 the deleted shift is δm²/m² = 0.62. In the scalar Yukawa model ψ†ψφ = :ψ†ψφ: + DM(0)φ, so ordering deletes a term linear in φ, which at first order would move ⟨φ⟩ to −gDM(0)/m². The course keeps ℋᵢₙₜ unordered. Ordering reaches only contractions inside one vertex; loops joining two vertices shift the mass as well, and Unit 6's counterterms, fixed by the measured mass, remove both in one scheme.
Change one variable at a time
Make the relationship visible.
Start at the square, R = 1.0 fm and θ = 0°, where D₁₂D₃₄ and D₁₃D₂₄ tie at 44.4%. Pull pair B out to R = 4.0 fm and D₁₂D₃₄ takes 99.3%: the pairs decouple. Back at R = 1.0 fm, turn θ to 180°, which swaps x₃ and x₄: the other two bars trade places and G₄ ends where it began.
D₁₂D₃₄ SHARE44.4 %
D₁₃D₂₄ SHARE44.4 %
D₁₄D₂₃ SHARE11.1 %
G₄ AT THESE POINTS21.89 10⁻⁷ GeV⁴
Live interpretationD₁₂D₃₄ SHARE: 44.4 %. D₁₃D₂₄ SHARE: 44.4 %. D₁₄D₂₃ SHARE: 11.1 %. G₄ AT THESE POINTS: 21.89 10⁻⁷ GeV⁴
Catch the common trap
Explain before calculating.
For a free complex scalar field ψ, what does Wick's theorem give for ⟨0|T ψ(x₁)ψ(x₂)ψ†(x₃)ψ†(x₄)|0⟩? Write DF for topic 1.08's propagator ⟨0|T ψ(x)ψ†(y)|0⟩.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFour points sit at one instant on the corners of a 1.0 fm × 2.0 fm rectangle: x₁ and x₂ 1.0 fm apart on one short side, x₃ and x₄ on the other, with x₃ facing x₁. For a massless free scalar, where DF = (ħc)²/(4π²r²) at equal times (topic 1.08), evaluate the three terms of the free four-point function in GeV⁴ and the share of each.
- Distances: r₁₂ = r₃₄ = 1.0 fm, r₁₃ = r₂₄ = 2.0 fm, and the diagonals r₁₄ = r₂₃ = √5 fm = 2.236 fm. Every separation is spacelike, so every DF is real and positive.
- Restore ħc: DF(r) = (0.197327 GeV fm)²/(4π²r²) = 9.863 × 10⁻⁴ GeV² × (1 fm/r)², which is 9.863 × 10⁻⁴, 2.466 × 10⁻⁴ and 1.973 × 10⁻⁴ GeV² at 1.0, 2.0 and 2.236 fm.
- Terms: D₁₂D₃₄ = 9.728 × 10⁻⁷ GeV⁴, D₁₃D₂₄ = 6.080 × 10⁻⁸ GeV⁴ and D₁₄D₂₃ = 3.891 × 10⁻⁸ GeV⁴, so G₄ = 1.073 × 10⁻⁶ GeV⁴.
- Shares: against the first term the others are (1/(2 × 2))² = 1/16 and (1/(√5 × √5))² = 1/25, so the shares are 1, 0.0625 and 0.04 over 1.1025: 90.7%, 5.7% and 3.6%.
- Checks: GeV⁴ is [φ]⁴, as it must be. Swapping x₃ and x₄ swaps the last two terms and leaves G₄ unchanged. Pull the pairs apart and the first share tends to 1, cluster decomposition; a massless field has no scale, so halving every distance multiplies G₄ by 2⁴ = 16 but leaves every share alone.
AnswerG₄ = 1.07 × 10⁻⁶ GeV⁴, of which D₁₂D₃₄ supplies 90.7%, D₁₃D₂₄ 5.7% and D₁₄D₂₃ 3.6%.
MediumA real scalar χ of mass mχ = 2.6 GeV couples to a real scalar φ of mass m = 1.0 GeV through ℒᵢₙₜ = −(κ/2)χφ², with κ = 0.80 GeV. At first order, contract the fields with the states to find ⟨k₁, k₂|iT|P⟩ for χ(P) → φ(k₁)φ(k₂) and read off ℳ. Check its dimension, find each φ's momentum and speed in the χ rest frame, and say whether φ → χφ can occur.
- ℋᵢₙₜ = (κ/2)χφ², so the first-order term is −i(κ/2)∫d⁴x ⟨k₁, k₂|χφφ|P⟩. Contracting the two φ fields with each other leaves DF(0)⟨k₁, k₂|χ|P⟩, which is zero: a χ field cannot turn one χ into two φ.
- In the surviving term χ absorbs χ(P), giving e(−iP⋅x), and the two φ fields emit φ(k₁) and φ(k₂). Either φ field may take k₁, so there are two assignments, each giving e(i(k₁ + k₂)⋅x).
- Hence ⟨k₁, k₂|iT|P⟩ = −i(κ/2) × 2 × (2π)⁴δ⁴(P − k₁ − k₂), and ℳ = −κ = −0.80 GeV. The ½ in the Lagrangian is there to cancel the two ways identical fields meet identical particles; gψ†ψφ needs no such factor because its three fields are distinct.
- Dimension: χφ² has dimension 3, so [κ] = 4 − 3 = 1, and three external particles need [ℳ] = 1. They agree. As κ → 0 the decay switches off.
- Kinematics: |k| = √(mχ²/4 − m²) = √(1.69 − 1.00) GeV = 0.831 GeV, with E = 1.30 GeV and speed β = 0.831/1.30 = 0.639. Unit 5 will add a further ½ to the width for identical daughters (stated). φ → χφ uses the same vertex but needs m ≥ mχ + m: its δ⁴ has no support, so φ stays stable.
Answerℳ = −κ = −0.80 GeV, of mass dimension 1; each φ leaves with |k| = 0.831 GeV at speed 0.639; φ → χφ is kinematically forbidden.
HardIn φ⁴ theory with m = 0.50 GeV and λ = 0.40, cut the spatial-momentum integral at Λ = 4.0 GeV. Find DF(0), the mass shift δm² that normal ordering ℋᵢₙₜ would delete, and the mass the unordered theory reads at first order. Then find the constant energy density ordering would delete, in GeV fm⁻³ and J m⁻³, and compare it with the measured dark-energy density, 5.2 × 10⁻¹⁰ J m⁻³.
- The radial integral gives DF(0) = (1/(8π²))[Λ√(Λ² + m²) − m² ln((Λ + √(Λ² + m²))/m)], derived. Here Λ√(Λ² + m²) = 4.0 × 4.0311 = 16.125 GeV² and m² ln(…) = 0.25 × 2.7765 = 0.694 GeV², so DF(0) = 15.430/78.957 = 0.1954 GeV², 3.6% below the leading Λ²/(8π²) = 0.2026 GeV².
- Wick at one point: (λ/4!)φ⁴ = (λ/4!):φ⁴: + ½(λ/2)DF(0):φ²: + (λ/8)DF(0)². The middle term is ½δm²:φ²: with δm² = (λ/2)DF(0) = 0.20 × 0.1954 = 0.0391 GeV², 16% of m² = 0.25 GeV².
- In the unordered theory the φ² term of ℋ is ½(m² + δm²)φ², so at first order the pole sits at 0.289 GeV², a mass of 0.538 GeV, 7.5% above m, and the shift is set by Λ: doubling Λ roughly quadruples δm². Checks: [λ] = 0 and [DF(0)] = 2 give δm² dimension 2, and λ → 0 removes it.
- The constant is (λ/8)DF(0)² = 0.05 × 0.03819 GeV⁴ = 1.91 × 10⁻³ GeV⁴, an energy density since [ℋ] = 4. Dividing by (ħc)³ = 7.684 × 10⁻³ GeV³ fm³ gives 0.249 GeV fm⁻³, more than the 0.15 GeV fm⁻³ inside a nucleus (quoted); at 1.602 × 10³⁵ J m⁻³ per GeV fm⁻³ that is 3.98 × 10³⁴ J m⁻³.
- Against the measured 5.2 × 10⁻¹⁰ J m⁻³ (Planck, quoted) the deleted constant is 8 × 10⁴³ times too large. Only gravity weighs a constant energy density, so no prediction of this course feels it and topic 3.03 drops it; that its size cannot be argued away is the cosmological-constant problem, beyond this course.
AnswerDF(0) = 0.195 GeV² and δm² = 0.0391 GeV², so the unordered mass reads 0.538 GeV; the constant is 1.91 × 10⁻³ GeV⁴ = 0.249 GeV fm⁻³ = 4.0 × 10³⁴ J m⁻³, about 8 × 10⁴³ times the measured dark-energy density.