University Quantum Field Theory I · Interacting Fields: the Interaction Picture, the Gell-Mann-Low Formula and Wick's Theorem · 2.4
The S-operator T exp(i∫d⁴x ℒᵢₙₜ) is Lorentz invariant
Every amplitude in this course is computed in whichever frame is easiest and quoted as a function of invariants. This topic earns that licence: it isolates the one frame-dependent step in Dyson's series, the time ordering, shows why microcausality makes it harmless, and marks where derivative couplings need more care.
Build the model
Connect the measurement to the mechanism.
Topic 2.03 built the evolution operator as a time-ordered exponential of HI. Send its endpoints to ±∞ and, for couplings without derivatives, where ℋᵢₙₜ = −ℒᵢₙₜ, the S-operator becomes S = T exp(i∫d⁴x ℒᵢₙₜ(φI(x))). Order n is then (iⁿ/n!) times an integral over n spacetime points of a time-ordered product of a Lorentz-scalar density, and every piece of that is manifestly invariant except one: T sorts the densities by x⁰, and x⁰ belongs to a frame. Invariance therefore stands or falls on whether two observers can write the same product in different orders.
A proper orthochronous transformation never reverses a timelike or lightlike pair, so the dispute is confined to spacelike pairs, and there the free scalar fields commute by UAQM-09.04's microcausality, quoted, and so does every polynomial built from them, while Dirac fields anticommute (UAQM-09.07), so their bilinears commute too. Both orders are then one operator, and U(Λ)SU†(Λ) = S, derived: amplitudes depend only on invariants. The argument has three edges. At coincident points T is not fixed by step functions, and that is where renormalisation will act.
With derivative couplings ℋᵢₙₜ differs from −ℒᵢₙₜ by non-covariant terms that cancel the contact terms of T, leaving a covariant T* product, stated. And reading S between free states as physical amplitudes remains an assumption until Unit 4's LSZ formula.
- Simple definition
- The S-operator S = T exp(i∫d⁴x ℒᵢₙₜ) is Lorentz invariant, U(Λ)SU†(Λ) = S, because its one frame-dependent ingredient, the time ordering, changes only for spacelike-separated vertices, where the interaction densities commute.
- Example
- Vertices with Δz = 2.0 fm and cΔt = 1.0 fm have s² = −3.0 fm². An observer moving at 0.80c along z finds cΔt′ = −1.0 fm, so T writes ℒᵢₙₜ(x)ℒᵢₙₜ(y) where the lab wrote ℒᵢₙₜ(y)ℒᵢₙₜ(x), yet the two products are one operator.
Topic 2.03's T exp(−i∫HI dt) with ℋᵢₙₜ = −ℒᵢₙₜ, valid without derivative couplings: the time and space integrals merge into one over all spacetime.
φI is the free field. ℒᵢₙₜ has mass dimension 4 and d⁴x has −4, so the exponent is a pure number; in SI it is (i/ħc)∫d⁴x ℒᵢₙₜ, J m⁻³ × m⁴ over J m.
A boost cannot mix powers of the coupling, so S is invariant only if every order is. Order 1 has nothing to sort; the question first arises at n = 2.
n = 0, 1, 2, …; T puts later densities to the left. In φ⁴, ℒᵢₙₜ = −(λ/4!)φI⁴ with λ dimensionless.
The sign can flip only if |cΔt| is smaller than |Δz|, a spacelike pair, once β passes cΔt/Δz. Timelike and lightlike pairs keep their order in every frame.
β = v/c along z, γ = (1 − β²)(−1/2); cΔt and Δz in fm, with 1 fm/c = 3.34 × 10⁻²⁴ s.
Densities even in fermion fields commute, so the two orders a pair of frames dispute are one operator, and T ℒᵢₙₜ(x)ℒᵢₙₜ(y) is the same in every frame.
From [φI(x), φI(y)] = 0 (UAQM-09.04) and [ψ(x), ψ̄(y)]₊ = 0 (UAQM-09.07) outside the cone, by Leibniz's rule.
So ⟨Λp₁ …|S|Λk₁ …⟩ = ⟨p₁ …|S|k₁ …⟩: amplitudes are functions of invariants such as s and t, and any frame may be used to compute them.
Uses U(Λ)ℒᵢₙₜ(x)U†(Λ) = ℒᵢₙₜ(Λx), quoted from topic 1.03, and d⁴(Λx) = d⁴x for det Λ = 1.
The δ⁴ sits in the 00 slot in every frame, so it is no tensor. T*, derivatives outside T, drops it; Matthews' theorem, stated, cancels it against ℋᵢₙₜ's extra terms.
From the step functions and the equal-time [φ̇(x), φ(y)] = −iδ³(x − y). Both sides have mass dimension 4.
From U(t, t₀) to an integral over all spacetime
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 2.03 gave U(t, t₀) = T exp(−i∫HI dt′), with HI(t) = ∫d³x ℋᵢₙₜ written in free fields, and showed ℋᵢₙₜ = −ℒᵢₙₜ when ℒᵢₙₜ holds no time derivatives, since the canonical momentum is then unchanged. Send t₀ → −∞ and t → +∞ and the time and space integrals merge: S = U(+∞, −∞) = T exp(i∫d⁴x ℒᵢₙₜ). Expanded, order n is (iⁿ/n!) ∫d⁴x₁ … d⁴xₙ T ℒᵢₙₜ(x₁) … ℒᵢₙₜ(xₙ). Dimensions: ℒᵢₙₜ has mass dimension 4 in all three of topic 2.01's models, −(λ/4!)φ⁴, −gψ†ψφ with [g] = 1 and −gφψ̄ψ, while d⁴x has −4; with ħ and c restored the exponent is (i/ħc)∫d⁴x ℒᵢₙₜ, J m⁻³ × m⁴ over J m, a pure number. Limits: as the coupling goes to zero S → 1, and the first-order term i∫d⁴x ℒᵢₙₜ has nothing to sort, so the question of frames first arises at second order.
Everything except the ordering is already a scalar
Topic 1.03 represented a proper orthochronous Λ on the free Fock space by a unitary U(Λ) taking |p⟩ to |Λp⟩, with U(Λ)φI(x)U†(Λ) = φI(Λx); a Dirac field also picks up a spinor matrix, which cancels in the scalar bilinear ψ̄ψ and becomes a Lorentz matrix on the index of ψ̄γμ ψ, so a density with every index contracted is a scalar. A density built as a scalar from fields at one point therefore obeys U(Λ)ℒᵢₙₜ(x)U†(Λ) = ℒᵢₙₜ(Λx), and d⁴(Λx) = |det Λ| d⁴x = d⁴x. Conjugate the second-order integrand: U(Λ)[T ℒ(x)ℒ(y)]U†(Λ) = θ(x⁰ − y⁰)ℒ(Λx)ℒ(Λy) + θ(y⁰ − x⁰)ℒ(Λy)ℒ(Λx). Now rename x′ = Λx and y′ = Λy. The densities and the measure come out right, but the step functions still compare the old frame's times, where T ℒ(x′)ℒ(y′) would compare x′⁰ with y′⁰. That mismatch is the whole question. Topic 1.05 met it for the c-number DF; here it is an operator statement about composite densities at n points.
Which pairs two frames can put in different orders
A proper orthochronous Λ never changes the sign of the time component of a timelike or lightlike vector, so for those pairs θ(x⁰ − y⁰) = θ(x′⁰ − y′⁰) and nothing is disputed. For a spacelike pair, cΔt′ = γ(cΔt − βΔz) changes sign for any boost faster than cΔt/Δz, which is below 1 precisely because |cΔt| is smaller than |Δz|. Take cΔt = 1.0 fm and Δz = 2.0 fm: s² = (cΔt)² − Δz² = −3.0 fm², and at β = 0.80, where γ = 5/3, cΔt′ = (5/3)(1.0 − 1.6) fm = −1.0 fm. The lab writes ℒ(y)ℒ(x), the moving observer ℒ(x)ℒ(y), each putting the later density on the left; she sees y first by 1.0 fm/c = 3.3 × 10⁻²⁴ s. A timelike pair with cΔt = 2.0 fm and Δz = 1.0 fm would need β = 2.0: no observer. The non-relativistic limit is a check: restoring c, the flip needs v = c²Δt/Δz, which runs off to infinity as c → ∞, so Galilean time order is absolute and the issue is purely relativistic.
Microcausality makes both orders one operator
UAQM-09.04's microcausality, quoted, gives [φI(x), φI(y)] = 0 for spacelike x − y, and likewise with φI† for a complex field. Leibniz's rule [AB, C] = A[B, C] + [A, C]B extends it to any polynomial at x against any at y, so φ⁴(x) commutes with φ⁴(y). Dirac fields anticommute instead, [ψ(x), ψ̄(y)]₊ = 0 outside the cone by UAQM-09.07, quoted, and the identity [AB, CD] = A[B, C]₊D − AC[B, D]₊ + [A, C]₊DB − C[A, D]₊B shows that bilinears such as ψ̄ψ and ψ̄γμ ψ commute. A density odd in fermion fields would anticommute, but none is a Lorentz scalar. For a disputed pair the step functions may weight ℒ(x)ℒ(y) and ℒ(y)ℒ(x) however they like: it is one operator. With n vertices, pass from one frame's order to the other's by swapping neighbours; each swap exchanges a pair the frames order differently, necessarily spacelike, so it costs nothing. Hence U(Λ)SU†(Λ) = S, derived, and amplitudes are functions of invariants alone.
Derivative couplings: T, T* and a contact term
Suppose ℒᵢₙₜ contains ∂₀φ, as in g jμ ∂μ φ for a c-number current j, topic 1.09's source coupled through a derivative, or as in scalar QED. Then π = φ̇ + gj⁰, and the Legendre transform gives ℋᵢₙₜ = −ℒᵢₙₜ + ½g²(j⁰)², a term that singles out the time component. A second non-covariant piece hides in T itself. Differentiating the step functions twice, with the equal-time [φ̇(x), φ(y)] = −iδ³(x − y), gives ⟨0|T φ̇(x)φ̇(y)|0⟩ = ∂₀ˣ∂₀ʸDF − iδ⁴(x − y), so in momentum space the 00 slot reads iEₚ²/(p² − m²) instead of ip₀²/(p² − m²). At p⁰ = 2.0 GeV, |p| = 1.0 GeV and m = 1.0 GeV these are 1.0 and 2.0 in units of i; from a frame moving at 0.6c along p they are 0.53 and 1.53. The difference is −1 in the 00 slot of every frame and zero elsewhere, which no tensor does. The covariant T* product, derivatives outside T, keeps the second. Matthews showed in 1949, stated here, that the extra terms of ℋᵢₙₜ cancel the contact terms order by order, so S = T* exp(i∫d⁴x ℒᵢₙₜ). QED's −eψ̄γμ ψAμ has no derivative, so Unit 9 needs none of this.
What is now earned, and what is still assumed
Invariance of S lets every later unit compute where it is easiest: a φ⁴ amplitude found in the centre-of-mass frame is a function of s, t and u alone and holds in the laboratory. It is also measured. Topic 5.03 will show, quoted here, that a decay rate's only frame dependence is its 1/(2E), so a particle in flight lives γ times longer: muons circulating in CERN's storage ring at γ = 29.3, with τ₀ = 2.197 μs at rest, should live γτ₀ = 64.4 μs, and the lifetime measured in 1977 agreed with that to about 0.1%. Two edges are flagged rather than crossed. At coincident points T is not fixed by step functions and products of fields are singular; Unit 6's counterterms define them. And S has been sandwiched between eigenstates of H₀, free quanta of mass m, while real particles are eigenstates of H; that this still yields physical amplitudes is assumed here and settled by Unit 4's LSZ formula. Weinberg ran the argument backwards, stated: requiring [ℋᵢₙₜ(x), ℋᵢₙₜ(y)] = 0 outside the cone forces fields to pair annihilators with creators, so charged particles need antiparticles.
Change one variable at a time
Make the relationship visible.
At cΔt = 1 fm and Δz = 2 fm, drag β up through 0.50: y drops below the tilted line, cΔt′ changes sign and T swaps ℒ(x) and ℒ(y), harmless because s² is negative. Then raise cΔt above Δz: the β needed passes 1 and the curve never crosses zero.
INTERVAL s²-3.00 fm²
cΔt′ FOR THE OBSERVER0.52 fm
Δt′ FOR THE OBSERVER1.72 × 10⁻²⁴ s
β NEEDED TO EQUALISE THE TIMES0.50
Live interpretationINTERVAL s²: −3.00 fm². cΔt′ FOR THE OBSERVER: 0.52 fm. Δt′ FOR THE OBSERVER: 1.72 × 10⁻²⁴ s. β NEEDED TO EQUALISE THE TIMES: 0.50
Catch the common trap
Explain before calculating.
In the second-order term of S for φ⁴ theory, vertex y lies at cΔt = 0.5 fm and Δz = 2.5 fm from vertex x in the lab, so the lab's T writes ℒᵢₙₜ(y)ℒᵢₙₜ(x). What does an observer moving at 0.60c along z conclude about this term?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA second-order φ⁴ term has vertex x at the origin and vertex y at cΔt = 2.0 fm, Δz = 5.0 fm in the lab. Classify the separation, find the boost that makes the vertices simultaneous, and find cΔt′ and Δt′ for an observer moving at 0.80c along z. Which product does T write in each frame, and is the term the same?
- s² = (cΔt)² − Δz² = (2.0 fm)² − (5.0 fm)² = −21 fm², negative, so the pair is spacelike: neither vertex lies inside the other's light cone.
- cΔt′ = γ(cΔt − βΔz) vanishes at β = cΔt/Δz = 2.0/5.0 = 0.40, below 1, so an observer at 0.40c sees the two vertices simultaneous.
- At β = 0.80, γ = 1/√(1 − 0.64) = 5/3: cΔt′ = (5/3)(2.0 − 0.80 × 5.0) fm = −3.33 fm and Δz′ = (5/3)(5.0 − 0.80 × 2.0) fm = 5.67 fm. Check with the exact −10/3 and 17/3: 11.11 − 32.11 = −21.00 fm², the interval unchanged.
- Δt′ = −3.33 fm ÷ (2.998 × 10²³ fm s⁻¹) = −1.11 × 10⁻²³ s, so y is now earlier. The lab's T writes ℒᵢₙₜ(y)ℒᵢₙₜ(x) and the observer's writes ℒᵢₙₜ(x)ℒᵢₙₜ(y).
- Each density is −(λ/4!)φI⁴, and [φI⁴(x), φI⁴(y)] = 0 at spacelike separation by microcausality and Leibniz's rule, so the two products are one operator. A timelike pair, cΔt = 5.0 fm with Δz = 2.0 fm, would need β = 2.5: no observer reverses it.
AnswerSpacelike, s² = −21 fm²; simultaneous for an observer at 0.40c; at 0.80c, cΔt′ = −3.3 fm and Δt′ = −1.1 × 10⁻²³ s, so T reverses the product, but the densities commute and the term is unchanged.
MediumA third-order φ⁴ term has vertices x₁ = (0, 0), x₂ = (1.0, 3.0) and x₃ = (2.5, 1.0), each written (ct, z) in fm. Find every time order of the three vertices that some observer moving along z can see, with the boosts that produce it, and check that each differs from the lab's only by swaps of spacelike pairs.
- Intervals: x₂ − x₁ gives 1.0² − 3.0² = −8.0 fm², spacelike; x₃ − x₁ gives 2.5² − 1.0² = +5.25 fm², timelike with x₃ later; x₃ − x₂ = (1.5, −2.0) fm gives 2.25 − 4.00 = −1.75 fm², spacelike.
- Lab order, earliest first: x₁, x₂, x₃, so T writes ℒ(x₃)ℒ(x₂)ℒ(x₁). x₁ precedes x₃ in every frame, since γ(2.5 − 1.0β) fm stays positive for every |β| below 1; that rules out three of the 3! = 6 orders.
- Pair (x₁, x₂): γ(1.0 − 3.0β) changes sign at β = 1.0/3.0 = 0.33. Pair (x₂, x₃): γ(1.5 + 2.0β) changes sign at β = −0.75. Both flips at once would need β above 0.33 and below −0.75, which no observer has.
- At β = 0.60, γ = 1.25: ct′ = 0, −1.00 and 2.38 fm for x₁, x₂, x₃, so the order is x₂, x₁, x₃ and T writes ℒ(x₃)ℒ(x₁)ℒ(x₂). At β = −0.80, γ = 5/3: ct′ = 0, 5.67 and 5.50 fm, so the order is x₁, x₃, x₂ and T writes ℒ(x₂)ℒ(x₃)ℒ(x₁).
- Each product differs from the lab's by one swap of neighbours, (x₁, x₂) or (x₂, x₃), both spacelike, so microcausality makes all three one operator. At β = 0.60, x₂ precedes x₁ by 1.00 fm/c = 3.34 × 10⁻²⁴ s.
AnswerThree orders occur: x₁, x₂, x₃ for β between −0.75 and 0.33; x₂, x₁, x₃ above 0.33; x₁, x₃, x₂ below −0.75. Each is one spacelike swap from the lab's, and x₁ always precedes x₃.
HardA free scalar of mass m couples to a c-number current through ℒᵢₙₜ = g jμ ∂μ φ. Find ℋᵢₙₜ, show that its extra piece cancels the contact term of T φ̇φ̇ in the order-g² vacuum amplitude, and, for jμ = uμ f(x) with f a scalar profile, find the factor by which each non-covariant piece grows when the source moves at 0.60c.
- With ℒ = ½∂μ φ ∂μ φ − ½m²φ² + g(j⁰φ̇ + jⁱ∂ᵢφ), π = φ̇ + gj⁰. Then ℋ = πφ̇ − ℒ = ½π² + ½(∇φ)² + ½m²φ² − gj⁰π − gjⁱ∂ᵢφ + ½g²(j⁰)², and with πI = φ̇I, ℋᵢₙₜ = −ℒᵢₙₜ + ½g²(j⁰)². Since [g j] = 2, the extra piece has mass dimension 4, as a density must.
- In ⟨0|S|0⟩ at order g², the extra piece at first order gives −(i/2)g²∫d⁴x (j⁰)², and −ℒᵢₙₜ at second order gives ((−i)²/2) g²∫∫ jμ(x) jν(y) ⟨0|T ∂μ φ(x) ∂ν φ(y)|0⟩.
- Only ∂₀ against ∂₀ makes a contact term: ⟨0|T φ̇(x)φ̇(y)|0⟩ = ∂₀ˣ∂₀ʸDF − iδ⁴(x − y), from differentiating θ(x⁰ − y⁰) with [φ̇, φ] = −iδ³ at equal times. Its share is (−g²/2)(−i)∫d⁴x (j⁰)² = +(i/2)g²∫d⁴x (j⁰)², cancelling step 2's first term exactly.
- What survives is −(g²/2)∫∫ jμ(x) jν(y) ∂μ ∂′ν DF(x − y), the T* result: a full contraction of tensors, the same in every frame. As g → 0 it vanishes and ⟨0|S|0⟩ → 1.
- With u the source's four-velocity, j⁰ = γf in a frame where it moves at β = 0.60, γ = 1.25, so ∫d⁴x (j⁰)² = γ²∫d⁴x f² = 1.5625∫d⁴x f², while ∫d⁴x jμ jμ = ∫d⁴x f² is unchanged. Each non-covariant piece grows by 56%; only their sum, zero, is frame independent.
Answerℋᵢₙₜ = −ℒᵢₙₜ + ½g²(j⁰)². The pieces ∓(i/2)g²∫d⁴x (j⁰)² cancel, leaving the covariant −(g²/2)∫∫ jμ jν ∂μ ∂′ν DF. Each piece alone is γ² = 1.56 times its rest-frame value at 0.60c.