University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.8
Wick's theorem for time-ordered products, by induction
Topic 2.07 split two fields into a normal-ordered piece and a propagator; a term of the Dyson series can hold dozens. This theorem reduces all of them to those same two pieces. Prove it once, by induction, and learn to count the pairings, so you know how many terms a vacuum value holds before Unit 3 draws them.
Build the model
Connect the measurement to the mechanism.
Wick's theorem says a time-ordered product of free fields is fixed by its two-field pieces: T(φ₁⋯φₙ) equals the normal-ordered product plus every way of replacing pairs of fields by Feynman propagators DF(xᵢ − xⱼ), the unpaired fields left normal-ordered. It rests on two facts about a free field. It is linear in ladder operators, φ = φ⁺ + φ⁻, so carrying an annihilation part past a creation part costs only the c-number [φ⁺(x), φ⁻(y)] = D(x − y); and time ordering puts the latest field on the left, so that c-number is exactly the contraction.
Induction does the rest: put the latest field first, expand the other n − 1 by hypothesis, and push its φ⁺ to the right, where each commutator it leaves is one new contraction with a field still unpaired. In the vacuum every normal-ordered remainder dies, so ⟨0|T(φ₁⋯φ₂ₙ)|0⟩ is a sum of (2n − 1)!! products of n propagators, and an odd number of fields gives zero: the free vacuum is Gaussian, and this is Isserlis' theorem for its moments. The theorem ends where free fields end: it reduces the interaction-picture fields that the Gell-Mann-Low formula supplies, never the interacting theory's own Heisenberg fields.
- Simple definition
- Wick's theorem writes a time-ordered product of free fields as a sum over every way of choosing disjoint pairs among them: each term is the product of the chosen pairs' Feynman propagators times the normal-ordered product of the fields left unpaired.
- Example
- T(φ₁⋯φ₆) expands into 76 terms: 1 with no contraction, 15 with one, 45 with two and 15 with three. Only the last 15, each a product of three propagators of mass dimension 2, survive in ⟨0|T(φ₁⋯φ₆)|0⟩, and 15 = 5!! = 5 × 3 × 1.
Derived here, by induction on n. Every interaction-picture correlator becomes c-number propagators times normal-ordered operators, and the vacuum kills the latter.
φᵢ = φ(xᵢ), free field, mass dimension 1 (GeV); Dᵢⱼ = DF(xᵢ − xⱼ), GeV²; :⋯:ᵢⱼ omits φᵢ and φⱼ; sums run over disjoint pairs
Derived. Each creation part φₖ⁻ that φ₁⁺ crosses leaves one c-number; when x₁⁰ is the latest time, D(x₁ − xₖ) is exactly the contraction DF(x₁ − xₖ).
φ⁺ annihilation part, φ⁻ creation part; D(x − y) = [φ⁺(x), φ⁻(y)], GeV², quoted (UAQM-09.04); :⋯:ₖ omits φₖ
Derived: ⟨0|:⋯:|0⟩ = 0 whenever an operator is left, since φ⁺|0⟩ = 0 and ⟨0|φ⁻ = 0. These products are what every Feynman diagram evaluates.
n propagators of GeV² in each term, mass dimension 2n as 2n fields require; an odd number of fields gives 0
Derived: φ₁ has 2n − 1 possible partners, and what is left is the same problem with two fewer fields. It outruns any power of n, growing as √2 (2n/e)ⁿ.
dimensionless count: 1, 3, 15, 105, 945 for 2, 4, 6, 8, 10 fields; 654 729 075 for 20
Derived: in a(n − 1) terms φ₁ stays in the normal-ordered product; in the rest it pairs with one of n − 1 partners and the other n − 2 fields expand freely.
a(n): all terms in the expansion of n fields, dimensionless; 2, 4, 10, 26, 76, 232, 764 for n = 2 to 8
Stated, not proved; its vacuum value is topic 1.09's exp(−½∫∫jDF j). Matching order j²ⁿ needs (2n − 1)!!/(2n)! = 1/(2ⁿ n!), the same count again.
j(x) a real c-number source of mass dimension 3 (GeV³), so both exponents are dimensionless
What is claimed, and why the latest field may go first
Natural units, ħ = c = 1, and the metric (+, −, −, −) hold throughout. Write φᵢ for the free, interaction-picture field φ(xᵢ), split as φ = φ⁺ + φ⁻: φ⁺ holds the annihilation operators and kills |0⟩, φ⁻ holds the creation operators. Normal ordering puts every φ⁻ left of every φ⁺, and topic 2.07, quoted here, showed T(φ₁φ₂) = :φ₁φ₂: + DF(x₁ − x₂). Wick's theorem claims the same structure for any n: a sum over every set of disjoint pairs, one propagator per pair, the unpaired fields normal-ordered. For three fields, T(φ₁φ₂φ₃) = :φ₁φ₂φ₃: + D₁₂φ₃ + D₁₃φ₂ + D₂₃φ₁, four terms, with Dᵢⱼ = DF(xᵢ − xⱼ). Both sides are symmetric under relabelling the points: the left because T sorts bosonic fields whatever order they are listed in, the right because it sums over all pairings and DF(x − y) = DF(y − x). So nothing is lost by calling the latest point x₁. Ties are harmless: fields at equal times are spacelike separated and commute (UAQM-09.04), so T is unambiguous there too.
The lemma: one c-number for every creation part crossed
The proof needs one identity. A normal-ordered product :φ₂⋯φₙ: is a sum of terms with creation parts on the left and annihilation parts on the right. Multiplying by φ₁⁻ from the left keeps it normal-ordered. Multiplying by φ₁⁺ does not, since φ₁⁺ now stands left of creation parts, so push it right. It commutes with every annihilation part, and each time it crosses a φₖ⁻ it leaves behind [φ₁⁺, φₖ⁻] = D(x₁ − xₖ), the c-number vacuum amplitude ⟨0|φ(x₁)φ(xₖ)|0⟩ of UAQM-09.04, times the product with φₖ removed, which is still normal-ordered. Adding both parts, φ₁ :φ₂⋯φₙ: = :φ₁φ₂⋯φₙ: + Σₖ D(x₁ − xₖ) :φ₂⋯φₙ:ₖ at any times. It becomes a statement about contractions only when x₁⁰ is later than every xₖ⁰, for then DF(x₁ − xₖ) = θ(x₁⁰ − xₖ⁰)D(x₁ − xₖ) + θ(xₖ⁰ − x₁⁰)D(xₖ − x₁) reduces to D(x₁ − xₖ). The commutator being a number, not an operator, is the whole secret.
The induction step, and why each term appears exactly once
The case n = 1 is trivial, T(φ₁) = φ₁ = :φ₁:, and n = 2 is topic 2.07. Assume the theorem for n − 1 fields and let x₁ be the latest point. Time ordering puts φ₁ leftmost, so T(φ₁⋯φₙ) = φ₁ T(φ₂⋯φₙ), and the hypothesis expands T(φ₂⋯φₙ) into terms of the form (propagators) × :unpaired fields:. The propagators are c-numbers and let φ₁ through; the lemma then splits each term two ways: φ₁ joins the normal-ordered product, or φ₁ is contracted with one of that term's unpaired fields. Now run it backwards. A term of the n-field expansion either leaves φ₁ unpaired, and then it came from the (n − 1)-field term with the same pairs, or pairs φ₁ with some φₖ, and then it came from the term with the remaining pairs, in which φₖ was unpaired. Each arises once and only once, which completes the induction. The count follows the proof, a(n) = a(n − 1) + (n − 1)a(n − 2): 4, 10, 26 and 76 terms for three, four, five and six fields.
In the vacuum only full contractions survive
Sandwich the theorem between ⟨0| and |0⟩. A normal-ordered product that still holds an operator has a φ⁺ on its right, which kills |0⟩, or a φ⁻ on its left, which ⟨0| kills, so its vacuum value is zero, and only terms with every field contracted survive. An odd number of fields always leaves one unpaired, so ⟨0|T(φ₁⋯φ₂ₙ₊₁)|0⟩ = 0, the free theory's φ → −φ symmetry at work. For 2n fields, count the full pairings: φ₁ picks a partner in 2n − 1 ways, the first field still unpaired picks in 2n − 3, and so on, giving (2n − 1)!! = (2n)!/(2ⁿ n!). Four fields give 3 products, the free four-point function topic 2.09 evaluates; six give 15; eight give 105, the count for the first-order φ⁴ four-point numerator; twenty give 654 729 075. Terms with exactly k contractions number C(n, 2k)(2k − 1)!!, which fields pair times how; these are the terms that survive between the vacuum and an (n − 2k)-particle state, the use topic 2.09 makes of them.
Three checks: dimensions, a single mode, the generating form
Dimensions first. φ has mass dimension 1 and DF mass dimension 2, so each term of ⟨0|T(φ₁⋯φ₂ₙ)|0⟩, a product of n propagators, has dimension 2n, as 2n fields must. Next the smallest free field there is, one mode in 0 + 1 dimensions: x(t) = (a e(−iωt) + a† e(iωt))/√(2mω), whose contraction is e(−iω|t|)/(2mω). At equal times Wick predicts ⟨0|x⁴|0⟩ = 3⟨x²⟩², and ladder algebra agrees, its two routes from |0⟩ back to |0⟩ weighing 2 and 1; for x⁶ the five routes weigh 6, 4, 2, 2 and 1, total 15 = 5!!. These are the moments of the Gaussian ground state, Isserlis' 1918 theorem, of which Wick's 1950 theorem is the operator form; balanced homodyne detection of the optical vacuum records exactly that Gaussian quadrature noise. Last, the generating form, stated: at order j²ⁿ its vacuum value exp(−½∫∫jDF j) requires (2n − 1)!!/(2n)! = 1/(2ⁿ n!), which is (2n)! = 2ⁿ n! (2n − 1)!! once more.
Where it stops: anticommuting and interacting fields
Two facts did the work: [φ⁺, φ⁻] is a c-number, and T of bosonic fields does not care in which order they are listed. Dirac fields keep the first, with an anticommutator in place of the commutator, but their T-product changes sign at every exchange of two fields. UAQM-10.05's four-operator rule, quoted, already shows the outcome: the same sum over pairings, each term signed by the parity of the permutation that brings its paired fields together, with ⟨0|Tψ(x)ψ̄(y)|0⟩ = SF(x − y) as the contraction; topic 7.02 recovers it from this proof with the signs kept. The first fact is the real boundary. In φ⁴ theory the Heisenberg field U(t, t₀)†φ(x)U(t, t₀) picks up, already at order λ, terms cubic in the free field, so it has no split into creation and annihilation parts whose commutator is a number, and the lemma has nothing to act on. The cost shows in the first-order four-point numerator: of its 105 contractions, the 24 that tie each external point to a different field at z carry 24/4! = 1 and give −iλ∫d⁴z DF(x₁ − z)DF(x₂ − z)DF(x₃ − z)DF(x₄ − z), a connected term of mass dimension 4 that no product of two-point functions contains.
Change one variable at a time
Make the relationship visible.
Step n from 5 to 6: a bar of 15 full contractions appears at k = 3, entirely above the dashes, because a full contraction must pair φ₁. Step on to 7 and the dot goes, since an odd number of fields always leaves one unpaired.
TERMS IN T(φ₁ … φₙ)76 terms
φ₁ UNPAIRED a(n − 1)26 terms
φ₁ PAIRED (n − 1)a(n − 2)50 terms
IN THE VACUUM (n − 1)!!, 0 IF n ODD15 terms
Live interpretationTERMS IN T(φ₁ … φₙ): 76 terms. φ₁ UNPAIRED a(n − 1): 26 terms. φ₁ PAIRED (n − 1)a(n − 2): 50 terms. IN THE VACUUM (n − 1)!!, 0 IF n ODD: 15 terms
Catch the common trap
Explain before calculating.
To prove Wick's theorem for T(φ₁φ₂φ₃) with x₁⁰ = 3.0 GeV⁻¹, x₂⁰ = 1.0 GeV⁻¹ and x₃⁰ = 2.0 GeV⁻¹, the product is written φ₁ T(φ₂φ₃) = φ₁[:φ₂φ₃: + DF(x₂ − x₃)]. What is the next step, and what does it produce?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA free real scalar field is taken at five distinct points. How many terms does Wick's expansion of T(φ₁⋯φ₅) contain, how many survive in ⟨0|T(φ₁⋯φ₅)|0⟩, and how many contribute to ⟨p|T(φ₁⋯φ₅)|0⟩ for a one-particle state |p⟩? Check the mass dimension of that matrix element.
- Terms with k contractions number C(5, 2k)(2k − 1)!!: k = 0 gives 1, k = 1 gives C(5, 2) × 1 = 10, k = 2 gives C(5, 4) × 3 = 15. A third pair would need six fields. Total 26.
- Check with the induction's recursion: a(5) = a(4) + 4a(3) = 10 + 4 × 4 = 26. In 10 terms φ₁ stays unpaired; in 16 it pairs with one of four partners and the other three fields expand in 4 ways each.
- Vacuum: five is odd, so every term keeps at least one unpaired field, and ⟨0|:⋯:|0⟩ = 0 while any operator is left. So ⟨0|T(φ₁⋯φ₅)|0⟩ = 0.
- One particle: ⟨p|:⋯:|0⟩ survives only if the creation parts make exactly one quantum, so exactly one field is unpaired. Those are the 15 terms with k = 2: the unpaired field in 5 ways, the other four paired in 3. With ⟨p|φ(x)|0⟩ = e(ip⋅x), each is DF DF e(ip⋅xₖ).
- Dimensions: with ⟨p|q⟩ = 2Eₚ(2π)³δ³(p − q), |p⟩ has mass dimension −1, so the matrix element has 5 − 1 = 4. Each term, two propagators of dimension 2 times a dimensionless phase, has 4 as well.
Answer26 terms (1 + 10 + 15); the vacuum value is 0; 15 terms reach ⟨p|, each two propagators times e(ip⋅xₖ), of mass dimension 4.
MediumThe free field in 0 + 1 dimensions is one oscillator, x(t) = (a e(−iωt) + a† e(iωt))/√(2mω), with m = 1.0 GeV and ω = 0.50 GeV. Use Wick's theorem for G(t) = ⟨0|T x(t)x(t)x(0)x(0)|0⟩ with t positive, check it by ladder algebra, and evaluate G at t = π/(2ω) in fm⁴.
- The contraction is ⟨0|T x(t)x(0)|0⟩ = e(−iω|t|)/(2mω), whose equal-time value is ⟨x²⟩ = 1/(2mω) = 1/(2 × 1.0 × 0.50) GeV⁻² = 1.0 GeV⁻².
- Four fields have 3 pairings. Pairing x(t) with x(t) and x(0) with x(0) gives ⟨x²⟩²; each of the two cross pairings gives [e(−iωt)/(2mω)]². So G = ⟨x²⟩²(1 + 2e(−2iωt)), and at t = 0 this is 3⟨x²⟩², the Gaussian fourth moment.
- Ladder check: x(0)²|0⟩ = ⟨x²⟩(|0⟩ + √2|2⟩) and ⟨0|x(t)² = ⟨x²⟩(⟨0| + √2 e(−2iωt)⟨2|), so G = ⟨x²⟩²(1 + 2e(−2iωt)). The 1 is the self-pairings through |0⟩; the 2e(−2iωt) is the two cross pairings through the two-quantum state.
- At t = π/(2ω) = π/(1.0 GeV) = 3.14 GeV⁻¹, e(−2iωt) = e(−iπ) = −1, so G = −⟨x²⟩² = −1.0 GeV⁻⁴: the cross pairings outweigh the self-pairing and flip the sign.
- Restore units: 1 GeV⁻¹ = 0.1973 fm, so ⟨x²⟩ = 0.03894 fm² and G = −1.52 × 10⁻³ fm⁴, at t = 3.14 GeV⁻¹ × 6.582 × 10⁻²⁵ GeV s = 2.07 × 10⁻²⁴ s. Four powers of length, as four factors of x need.
AnswerG(t) = ⟨x²⟩²(1 + 2e(−2iωt)) with ⟨x²⟩ = 1.0 GeV⁻² = 0.0389 fm²; at t = π/(2ω) = 2.07 × 10⁻²⁴ s, G = −1.52 × 10⁻³ fm⁴, against 3⟨x²⟩² = 4.55 × 10⁻³ fm⁴ at t = 0.
HardAt first order in φ⁴ theory the Gell-Mann-Low numerator for the two-point function contains ⟨0|T φ(x₁)φ(x₂)φ(z)⁴|0⟩, multiplied by −iλ/4! and integrated over z. Count its full contractions by how the external points attach, find the coefficient each class carries after the 1/4!, and check the mass dimension of the order-λ term.
- Six fields, so Wick's theorem gives 5!! = 15 full contractions. Sort them by where φ(x₁) goes: to the other external point, or to one of the four fields at z.
- x₁ to z: 4 choices of field. Then x₂, with x₁ taken, pairs with one of the 3 remaining z fields, and the last two z fields pair with each other in 1 way: 4 × 3 × 1 = 12 terms, each DF(x₁ − z)DF(x₂ − z)DF(0), with the divergent DF(0) kept as a symbol.
- x₁ to x₂: the four z fields pair among themselves in 3!! = 3 ways, each DF(x₁ − x₂)DF(0)². Total 12 + 3 = 15, the full count, so no class is missing.
- Coefficients: 12/4! = 1/2 and 3/4! = 1/8, so the order-λ numerator is −iλ∫d⁴z [½DF(x₁ − z)DF(x₂ − z)DF(0) + ⅛DF(x₁ − x₂)DF(0)²]. The second class is DF(x₁ − x₂) times the first-order vacuum value −(iλ/8)∫d⁴z DF(0)²; Unit 3 reads these numbers as diagrams.
- Dimensions: λ is dimensionless, d⁴z carries −4 and three propagators carry 6, net 2, the dimension of the zeroth-order term DF(x₁ − x₂). Free limit: λ → 0 leaves DF(x₁ − x₂) alone, topic 2.07's result.
Answer15 = 12 + 3 full contractions; coefficients ½ (both external points joined at z) and ⅛ (x₁ joined to x₂, z closed on itself); the order-λ term has mass dimension 2, like DF.