University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.3
Interaction-picture fields and the time-ordered U(t, t₀)
Every Feynman diagram in this course is one term of a single expansion, and this is the topic that builds it. Give the free Hamiltonian to the fields, give the interaction to one operator U(t, t₀), and learn exactly what that operator is: why it is built from free fields, and why it must be time-ordered.
Build the model
Connect the measurement to the mechanism.
Nobody can solve the Heisenberg equations of an interacting field, but the free ones are solved, so the interaction picture splits the work. Write H = H₀ + Hᵢₙₜ, both built from the same φ(t₀, x) and π(t₀, x). Let operators evolve with H₀ alone: the result, φI, is the free field, mode operators and all.
Everything H₀ cannot do goes into one unitary operator, U(t, t₀) = e(iH₀(t−t₀)) e(−iH(t−t₀)), which turns the free field into the true one, φ = U†φI U. Differentiating U leaves a Schrödinger equation driven only by HI(t), the interaction written in free fields, so U can be expanded in objects whose contractions are known. The expansion is a time-ordered exponential because HI at two different times does not commute with itself, and the ordering is not cosmetic: for a field mode under a static source it carries the whole shift of the vacuum energy.
The construction has edges. ℋᵢₙₜ = −ℒᵢₙₜ needs an interaction free of time derivatives. Free and interacting dynamics must share one Hilbert space, which Haag's theorem denies in infinite volume, so the formalism is a regulated device whose ratios survive the limit. And as t₀ → −∞ the vacuum's phase e(−iΔE(t−t₀)) winds without bound, so U alone has no limit; topics 2.05 and 2.06 tame it.
- Simple definition
- The interaction picture evolves operators with the free Hamiltonian H₀ and puts the rest of the dynamics into U(t, t₀) = e(iH₀(t−t₀)) e(−iH(t−t₀)), which equals the time-ordered exponential T exp(−i∫HI dt′) of the interaction written in free fields.
- Example
- Drive one field mode of frequency ω with a static source of strength κ = 0.5. After one period, ωt = 2π, the time-ordered U returns the mode to its vacuum with amplitude e(2πiκ²) = e(iπ/2) = i; the unordered exp(−i∫HI dt) would return 1.
Derived to obey (∂² + m²)φI = 0: it is topic 1.02's free field, each aₚ turned into aₚ e(−iEₚ(t−t₀)), so its contractions are known.
H₀ the free Hamiltonian, mass dimension 1 (GeV); φ and φI mass dimension 1; t₀ where all pictures agree
Every question about the interacting field becomes one about free fields and a single operator. U is not e(−iHᵢₙₜ(t−t₀)), because [H₀, Hᵢₙₜ] ≠ 0.
U dimensionless and unitary, U(t₀, t₀) = 1; H = H₀ + Hᵢₙₜ in GeV. Derived, exact at any coupling
H₀ cancels out: only the interaction drives U, and it does so as a polynomial in free fields, so every term of U can be evaluated.
HI(t) = e(iH₀(t−t₀)) Hᵢₙₜ e(−iH₀(t−t₀)), mass dimension 1; ℋᵢₙₜ mass dimension 4. Derived
The T stays because free fields at separations inside or on the light cone do not commute (a massless one only on it), so neither do HI(t₁) and HI(t₂). At n = 1 it is idle: U ≈ 1 − i∫HI dt′.
every tᵢ runs from t₀ to t; T puts later times to the left; n! ordered copies of one simplex fill the cube
Evolution can be cut at any time. Setting t₃ = t₁ gives U(t₁, t₂)⁻¹ = U(t₁, t₂)† = U(t₂, t₁), and topic 2.06 uses the cut to merge Heisenberg fields.
U(t, t′) = e(iH₀(t−t₀)) e(−iH(t−t′)) e(−iH₀(t′−t₀)); dimensionless; any order of t₁, t₂, t₃
Then U = T exp(i∫d⁴x ℒᵢₙₜ(φI)) over t₀ ≤ x⁰ ≤ t, the form topic 2.04 proves Lorentz invariant; derivative couplings are handled there.
valid when ℒᵢₙₜ holds no ∂₀φ, so π = ∂ℒ/∂(∂₀φ) is unchanged; φ⁴: ℋᵢₙₜ = +λφ⁴/4!, mass dimension 4. Shown
Hand the free Hamiltonian to the field
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Split H = H₀ + Hᵢₙₜ, both built from φ(t₀, x) and π(t₀, x) at a reference time t₀ where every picture agrees, with H₀ the free Hamiltonian of topic 1.02. The interaction-picture field is what the field would do if only H₀ acted: φI(t, x) = e(iH₀(t−t₀)) φ(t₀, x) e(−iH₀(t−t₀)). Since H₀ = ∫ d³p/(2π)³ Eₚ aₚ† aₚ gives [H₀, aₚ] = −Eₚ aₚ, conjugation turns each aₚ into aₚ e(−iEₚ(t−t₀)), so φI = ∫ d³p/(2π)³ (aₚ e(−ip⋅x) + aₚ† e(ip⋅x))/√(2Eₚ) with x⁰ = t − t₀. That is the free field, solving (∂² + m²)φI = 0 exactly, derived. Its aₚ are the free modes of φ(t₀), not the particle operators of the interacting theory, whose states and masses differ; relating the two is the job of Unit 4's LSZ formula.
Peel the free evolution off the Heisenberg field
The Heisenberg field is φ(t, x) = e(iH(t−t₀)) φ(t₀, x) e(−iH(t−t₀)). Insert 1 = e(−iH₀(t−t₀)) e(iH₀(t−t₀)) on each side of φ(t₀, x). The middle three factors assemble into φI(t, x), the outer ones into U(t, t₀) = e(iH₀(t−t₀)) e(−iH(t−t₀)) and its adjoint, so φ = U†φI U, derived and exact. U is unitary as a product of unitaries, dimensionless, and equal to 1 at t = t₀. It is not e(−iHᵢₙₜ(t−t₀)): exponents add only for commuting operators, and in φ⁴ theory H₀ contains π² while Hᵢₙₜ contains φ⁴, with [φ(x), π(y)] = iδ³(x − y). The Baker-Campbell-Hausdorff formula puts the first difference at +½(t − t₀)²[H₀, Hᵢₙₜ] in the exponent, second order in the elapsed time. Free limit: at zero coupling H = H₀, U = 1 and φ = φI.
Differentiate U and only the interaction is left
Differentiate U: i ∂U/∂t = e(iH₀(t−t₀)) (H − H₀) e(−iH(t−t₀)). Write H − H₀ = Hᵢₙₜ and slip 1 = e(−iH₀(t−t₀)) e(iH₀(t−t₀)) in after it: i ∂U/∂t = HI(t)U with U(t₀, t₀) = 1, derived, where HI(t) = e(iH₀(t−t₀)) Hᵢₙₜ e(−iH₀(t−t₀)). Conjugation respects products, so Hᵢₙₜ = ∫ d³x ℋᵢₙₜ(φ(t₀, x)) becomes HI(t) = ∫ d³x ℋᵢₙₜ(φI(t, x)), the same polynomial in free fields. If ℒᵢₙₜ holds no time derivative, π = ∂ℒ/∂(∂₀φ) is untouched, so ℋ = πφ̇ − ℒ₀ − ℒᵢₙₜ = ℋ₀ − ℒᵢₙₜ and ℋᵢₙₜ = −ℒᵢₙₜ, shown. In φ⁴ theory ℒᵢₙₜ = −(λ/4!)φ⁴ gives HI = (λ/4!) ∫ d³x φI⁴, a potential bounded below only for positive λ; in the scalar Yukawa model HI = g ∫ d³x ψI†ψI φI. Dimensions: [λ] = 0, [φI⁴] = 4 and [d³x] = −3, so HI is an energy and ∫HI dt is dimensionless, as an exponent must be.
Order the kicks: the Dyson series for fields
Quantum Mechanics II solved this equation (UQMII-12.02), quoted here. Iterating U = 1 − i∫dt₁ HI(t₁)U(t₁, t₀) nests the integrals as t ≥ t₁ ≥ t₂ ≥ … ≥ t₀ with later times to the left, and the time-ordering symbol T lets each run over the whole cube [t₀, t]ⁿ, which holds n! copies of that simplex. Hence U(t, t₀) = Σₙ ((−i)ⁿ/n!) ∫dt₁…dtₙ T[HI(t₁)…HI(tₙ)], written T exp(−i∫HI dt′). In a field theory the T cannot be dropped, because two different time slices always contain timelike- and lightlike-separated pairs of points, where free fields fail to commute (a massless field only on the light cone). Checks: at n = 1 the T is idle and U ≈ 1 − i∫HI dt′; over a short interval δt, U ≈ 1 − iHI(t₀)δt; at zero coupling every term beyond n = 0 vanishes; and each term is dimensionless because each ∫HI dt is. For φ⁴ the commutators are operators and the series never closes, which is why later units evaluate it term by term with Wick's theorem.
What H₀ and the T each do: one driven mode
Couple one mode of frequency ω to a static source switched on at t₀ = 0: Hᵢₙₜ = ωκ(a + a†), κ dimensionless, so HI(t) = ωκ(a e(−iωt) + a† e(iωt)). Here [HI(t₁), HI(t₂)] = −2iω²κ² sin(ω(t₁ − t₂)) is a number, so the series closes: U = e(Ω) e(−i∫HI dt′) with Ω = iκ²(ωt − sin ωt), derived by differentiating, where Ω′ cancels the commutator left over from the plain exponential. So ⟨0|U(t, 0)|0⟩ = e(−2κ² sin²(ωt/2)) e(iκ²(ωt − sin ωt)); its modulus squared is topic 1.09's Poisson persistence e(−⟨N⟩), quoted. Two ablations isolate the ingredients. Freeze HI at Hᵢₙₜ and P falls as e(−κ²ω²t²): without H₀ turning HI, a static source pumps quanta without limit. Drop the T and the modulus survives but the phase is lost, and its secular part κ²ωt is −ΔE t for ΔE = −κ²ω, the shift found by completing the square in H. At κ = 0.3 and ωt = 4.0 the ordering supplies 0.428 rad. Summed over the modes of a field coupled to two static point sources, these shifts build topic 1.09's Yukawa energy, quoted.
Cut the evolution anywhere: the composition law
Define U(t, t′) = e(iH₀(t−t₀)) e(−iH(t−t′)) e(−iH₀(t′−t₀)), which is U(t, t₀) at t′ = t₀. It obeys i ∂U/∂t = HI(t)U with U(t′, t′) = 1, so for t ≥ t′ it is the T exp of HI from t′ to t. In a product the inner factors e(−iH₀(t₂−t₀)) e(iH₀(t₂−t₀)) cancel: U(t₁, t₂)U(t₂, t₃) = U(t₁, t₃), shown, for any order of the three times, and t₃ = t₁ gives U(t₂, t₁) = U(t₁, t₂)†. In the series the law is visible directly: every time in [t₂, t₁] is later than every time in [t₃, t₂], so the T exp splits with the later factor on the left. It lets two Heisenberg fields with x⁰ ≥ y⁰ collapse: U†(x⁰, t₀)φI(x)U(x⁰, t₀)U†(y⁰, t₀)φI(y)U(y⁰, t₀) = U(t₀, x⁰)φI(x)U(x⁰, y⁰)φI(y)U(y⁰, t₀), the string topic 2.06 turns into one time-ordered product. The law holds for operators, not vacuum amplitudes: ⟨0|U(t₁, t₃)|0⟩ needs every intermediate state.
Change one variable at a time
Make the relationship visible.
Drag ωt up from 0: the top curves agree until ωt ≈ 1, since U and e(−iHᵢₙₜ t) first differ at order t²[H₀, Hᵢₙₜ]. Then the frozen guess drains the vacuum while the true P returns to 1 each period. Raise κ and the bottom curve winds faster, at the rate |ΔE| = κ²ω.
P WITH THE TRUE U1.000
P IF U = e(−iHᵢₙₜ t)0.000
PHASE FROM T1.571 rad
VACUUM SHIFT ΔE/ω-0.250
Live interpretationP WITH THE TRUE U: 1.000. P IF U = e(−iHᵢₙₜ t): 0.000. PHASE FROM T: 1.571 rad. VACUUM SHIFT ΔE/ω: −0.250
Catch the common trap
Explain before calculating.
In φ⁴ theory, ℒ = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴. To first order in λ, which expression is the interaction-picture evolution operator U(t, t₀)? Each integral covers t₀ ≤ x⁰ ≤ t.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn φ⁴ theory for a scalar with the charged pion's mass, m = 139.57 MeV, one term of HI(t) annihilates two quanta at rest and creates a pair with momenta ±p, |p| = 100 MeV. Show that the term carries a factor e(iΔE(t−t₀)), and find ΔE and the period of the oscillation in fm/c and in seconds.
- HI(t) = e(iH₀(t−t₀)) Hᵢₙₜ e(−iH₀(t−t₀)) is Hᵢₙₜ with every ladder operator conjugated. From [H₀, aₖ] = −Eₖ aₖ, each aₖ becomes aₖ e(−iEₖ(t−t₀)) and each aₖ† becomes aₖ† e(iEₖ(t−t₀)).
- The term a†(p) a†(−p) a(0) a(0) therefore picks up e(i(Eₚ + Eₚ − m − m)(t−t₀)): every created quantum adds its energy to the phase and every annihilated one subtracts it, so ΔE = 2(Eₚ − m).
- Eₚ = √(p² + m²) = √(100² + 139.57²) MeV = 171.70 MeV, so ΔE = 2 × (171.70 − 139.57) MeV = 64.25 MeV.
- Restore ħ: the period is 2πħ/ΔE = 2π × 197.327 MeV fm ÷ 64.25 MeV = 19.30 fm/c, and dividing 19.30 × 10⁻¹⁵ m by 2.998 × 10⁸ m s⁻¹ gives 6.44 × 10⁻²³ s.
- Checks: ΔE(t − t₀) is a pure number, as a phase must be; as |p| → 0 the term conserves energy, ΔE → 0, and stops oscillating. Over t − t₀ much longer than 19 fm/c its share of ∫HI dt averages away, which is where energy conservation in the S-matrix will come from.
AnswerThe term carries e(iΔE(t−t₀)) with ΔE = 2(Eₚ − m) = 64.3 MeV, so it oscillates with period 2πħ/ΔE = 19.3 fm/c ≈ 6.44 × 10⁻²³ s.
MediumA field mode of frequency ω is driven by a static source of strength κ = 0.30 switched on at t = 0, so that U(t, t′) = e(iκ²(ωΔt − sin ωΔt)) D(α), with Δt = t − t′, α = −κ(e(iωt) − e(iωt′)) and D(α) = exp(αa† − α*a). Build U(2π/ω, 0) from two half periods with the composition law, and compare ⟨0|U(2π/ω, 0)|0⟩ with the product of the two half-period vacuum amplitudes.
- First half, 0 to π/ω: α₁ = −κ(e(iπ) − 1) = 2κ = 0.60, and the phase is κ²(π − sin π) = 0.09π = 0.283 rad. So U(π/ω, 0)|0⟩ = e(0.283i)|0.60⟩, a coherent state with mean quanta |α₁|² = 0.36.
- Second half, π/ω to 2π/ω: α₂ = −κ(e(2πi) − e(iπ)) = −2κ = −0.60, and the phase is again 0.283 rad, because the interval has the same length.
- Compose: U(2π/ω, π/ω)U(π/ω, 0) = e(0.18πi) D(−0.60)D(0.60). Displacements multiply as D(β)D(α) = e((βα* − β*α)/2) D(α + β), and that phase vanishes for real α and β, so the product is e(0.565i) D(0) = e(0.565i).
- Cross-check against the closed form for the full period: phase κ²(2π − sin 2π) = 0.18π = 0.565 rad and α = −κ(e(2πi) − 1) = 0. The mode is back in its vacuum with certainty: ⟨0|U(2π/ω, 0)|0⟩ = e(0.565i).
- Each half-period vacuum amplitude is e(−|α|²/2) e(0.283i) = e(−0.18) e(0.283i), of modulus 0.835, so their product has modulus 0.835² = 0.698 = e(−0.36). The missing 0.302 e(0.565i) is the sum over n ≥ 1 of ⟨0|U(2π/ω, π/ω)|n⟩⟨n|U(π/ω, 0)|0⟩: quanta made in the first half are absorbed in the second.
Answer⟨0|U(2π/ω, 0)|0⟩ = e(0.565i), modulus 1, while the product of half-period vacuum amplitudes is 0.698 e(0.565i). The law composes operators; vacuum amplitudes compose only after summing intermediate states, which supply 0.302 e(0.565i).
HardA static point source, ℒᵢₙₜ = j(x)φ(x) with j = g δ³(x) and g = 1.0, is switched on at t₀ in a free scalar field of mass m = 139.57 MeV. Using φ = U†φI U with U = T exp(−i∫HI dt), find the Heisenberg field to all orders in g, show that it obeys the classical field equation, and evaluate ⟨0|φ|0⟩ at r = 2.00 fm from the source long after the switch-on.
- No time derivatives, so ℋᵢₙₜ = −ℒᵢₙₜ and HI(t) = −∫d³y j(y)φI(t, y). The commutator of HI with itself at two times is ∫∫ j j [φI, φI], a number, so exactly as for one mode U = e(Ω) e(−iA), with Ω a number and A = ∫HI dt′ from t₀ to x⁰.
- Then φ(x) = U†φI(x)U = e(iA) φI(x) e(−iA) = φI(x) + i[A, φI(x)], exactly: [A, φI] is a number, so every nested commutator after it vanishes.
- i[A, φI(x)] = i∫d⁴y j(y)[φI(x), φI(y)] over t₀ ≤ y⁰ ≤ x⁰, and there the commutator is the retarded propagator DR(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩ of topic 1.07. So φ(x) = φI(x) + i∫d⁴y DR(x − y)j(y), and (∂² + m²)DR = −iδ⁴ gives (∂² + m²)φ = j, the Euler-Lagrange equation of ℒ₀ + jφ.
- ⟨0|φI|0⟩ = 0, so ⟨0|φ(x)|0⟩ = g∫dy⁰ iDR(x⁰ − y⁰, x) from t₀ to x⁰. Long after the switch-on this integrates iDR over every delay, and integrating (∂² + m²)iDR = δ⁴ over time leaves (−∇² + m²)G = δ³, so ⟨0|φ⟩ → g e(−mr)/(4πr), the Yukawa profile of topic 1.09, quoted.
- Numbers: 1/m = ħc/mc² = 197.327 MeV fm ÷ 139.57 MeV = 1.414 fm, so ⟨0|φ⟩ = e(−2.00/1.414)/(4π × 2.00 fm) = 0.2430/25.13 fm⁻¹ = 9.67 × 10⁻³ fm⁻¹, or 9.67 × 10⁻³ × 197.327 MeV = 1.91 MeV.
- Checks: mass dimension 1, as a scalar field's must be; as m → 0 the profile becomes the Coulomb-like g/(4πr); and the range ħ/mc = 1.41 fm is the measured range of the nuclear force.
Answerφ(x) = φI(x) + i∫d⁴y DR(x − y)j(y), exact, obeying (∂² + m²)φ = j. Long after the switch-on, ⟨0|φ|0⟩ = g e(−mr)/(4πr) = 9.67 × 10⁻³ fm⁻¹ ≈ 1.91 MeV at r = 2.00 fm.