University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.5
Time ordering and the Feynman propagator DF(x − y)
Every Feynman diagram in this course is stitched together from one object: two free fields put in time order and sandwiched between vacua. Here we build it from Advanced QM's vacuum amplitude, settle why an ordering that observers dispute still gives a Lorentz scalar, and read what its tail outside the light cone does and does not mean.
Build the model
Connect the measurement to the mechanism.
Advanced QM handed over the vacuum amplitude D(x − y) = ⟨0|φ(x)φ(y)|0⟩ of UAQM-09.04: a quantum made at y and absorbed at x, whatever the times. An interacting theory asks for something else. Unit 2's Dyson series multiplies interaction densities in time order, so its vacuum expectations contain ⟨0|T φ(x)φ(y)|0⟩, with the later field always on the left.
That object, the Feynman propagator, is θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x): a quantum made at whichever point is earlier and absorbed at the later, carrying positive energy forward in |x⁰ − y⁰|. The step functions look fatal, because observers disagree about which point is earlier. They disagree only when x − y is spacelike, and there D(x − y) = D(y − x), so the disputed choice changes nothing: DF is a Lorentz scalar, even under x ↔ y, and a function of (x − y)² alone.
The same fact makes DF equal to D outside the light cone, non-zero with an e(−mr) tail, which is an amplitude and carries no signal; causality is carried by the commutator. The construction is a free-field one. Interactions change the vacuum and the fields, a Dirac field's time order needs a minus sign (Unit 7), and on the light cone DF is singular and needs the iε of topic 1.06 to be defined at all.
- Simple definition
- The Feynman propagator DF(x − y) is the vacuum expectation of the time-ordered product of two free fields: the amplitude for a quantum created at whichever of x and y is earlier to be absorbed at the later one.
- Example
- For a pion-mass field, two points 0.50 fm apart at equal times have DF = mK₁(mr)/(4π²r) = 9.06 × 10⁻² fm⁻², 89% of the massless 1/(4π²r²), and an observer who sees y first and one who sees x first both get that number.
The later time stands on the left. For a boson field the product is symmetric in x and y, and with n fields every factor is placed in order of decreasing time.
φ a real scalar field, mass dimension 1 (fm⁻¹ through ħc); θ the unit step; x⁰ = ct in fm, 1 fm/c = 3.336 × 10⁻²⁴ s.
A quantum made at the earlier point and absorbed at the later. Swapping x and y swaps the terms, so DF(x − y) = DF(y − x), while D(y − x) = D(x − y)*.
D(x − y) = ⟨0|φ(x)φ(y)|0⟩, quoted from UAQM-09.04; DF has mass dimension 2: fm⁻², or MeV² after multiplying by (ħc)².
Every mode carries e(−iEₚ|t|), whichever point is later: the time order sends positive energy from the earlier point to the later. Topic 1.06 recovers it from poles.
t = x⁰ − y⁰ and r the spatial separation, in fm; Eₚ = √(p² + m²) in fm⁻¹; p⋅r is the three-dimensional product.
Timelike and null pairs keep their order for every observer; spacelike pairs do not, so θ(x⁰ − y⁰) is not a Lorentz scalar.
Boost of speed β (units of c) along x − y, r = |x − y|; t, t′ and r in fm; γ = 1/√(1 − β²).
Boost to equal times and relabel p → −p. The disputed step function then multiplies two equal terms, so DF is invariant and depends on (x − y)² alone.
(x − y)² = t² − r², in fm²; negative means outside the light cone, where the time order is frame-dependent.
Schrödinger's free kernel, with the rest-energy phase and the 1/(2m) of covariant normalisation; for a pion at |t| = 10 fm its size is 0.4% below the exact value.
Valid for m|t| ≫ 1 and r ≪ |t|; m = mc/ħ in fm⁻¹, t and r in fm.
Two orderings, one object
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, with ħc = 197.327 MeV fm turning masses into inverse lengths. The time-ordered product puts the field at the later time on the left: T φ(x)φ(y) = θ(x⁰ − y⁰)φ(x)φ(y) + θ(y⁰ − x⁰)φ(y)φ(x), and a product of more fields is put in order of decreasing time. Take its vacuum expectation with UAQM-09.04's amplitude D(x − y) = ⟨0|φ(x)φ(y)|0⟩, quoted, the same mode integral for the real field of UAQM-09.02, in which φ(y) makes a quantum and φ(x) absorbs it. The result, derived in one line, is DF(x − y) = θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x): the quantum is always made at the earlier point. Two properties follow at once. Swapping x and y swaps the two terms, so DF(x − y) = DF(y − x), unlike D, for which D(y − x) = D(x − y)*. And since φ has mass dimension 1, DF has dimension 2: a number in fm⁻², which (ħc)² = 3.894 × 10⁴ MeV² fm² turns into MeV².
Positive energy runs from the earlier point to the later
Write t = x⁰ − y⁰ and r for the spatial separation. The quoted mode integral is D(x − y) = ∫ d³p e(−iEₚt + ip⋅r)/((2π)³ 2Eₚ), with p⋅r the three-dimensional product, and D(y − x) is the same with t → −t and r → −r; relabelling p → −p in it restores +ip⋅r. So DF(x − y) = ∫ d³p e(−iEₚ|t| + ip⋅r)/((2π)³ 2Eₚ), derived: every mode runs forward in |t| with positive energy, whichever way t points, and topic 1.06 recovers this from the poles of i/(p² − m² + iε). Now the non-relativistic check. Deep inside the cone, with r ≪ |t| and m|t| ≫ 1, the integral is dominated by p ≪ m, so Eₚ ≈ m + p²/(2m) and 1/(2Eₚ) ≈ 1/(2m). The Gaussian integral then gives DF ≈ (e(−im|t|)/(2m))(m/(2πi|t|))³⁄² e(imr²/(2|t|)), derived: Schrödinger's free kernel, times the rest-energy phase and the 1/(2m) of covariant normalisation. Unlike Schrödinger's retarded kernel, it does not vanish for negative t.
A boost can reverse the order only outside the light cone
Under a proper orthochronous Lorentz transformation D(x − y) is unchanged, because d³p/2Eₚ and p⋅(x − y) are invariant (topic 1.03), but θ(x⁰ − y⁰) need not be. Rotations leave t alone, so boosts decide, and only the component of β along the separation matters. With r the distance between the two points, t′ = γ(t − βr). If x − y is timelike or null, r ≤ |t|, so |βr| is strictly smaller than |t| and t′ has the sign of t for every observer: the order is absolute, and the smallest |t′| anyone measures is the proper time √(t² − r²). If x − y is spacelike, |t| is smaller than r, and a boost at β = t/r makes the events simultaneous, while any faster one reverses them. Take t = 0.75 fm and r = 1.50 fm, that is 2.50 × 10⁻²⁴ s and 1.50 × 10⁻¹⁵ m in SI. An observer at β = 0.60 has γ = 1.25 and t′ = 1.25 × (0.75 − 0.90) fm = −0.19 fm. For that observer x comes first, the second term of DF is the one switched on, and the step functions are plainly not Lorentz scalars.
Where the order is disputed, the two terms agree
The disputed case is exactly where D(x − y) = D(y − x), derived here in two steps. Boost to the frame in which a spacelike pair is simultaneous; there D(0, r) = ∫ d³p e(ip⋅r)/((2π)³ 2Eₚ), and relabelling p → −p, which leaves d³p/2Eₚ unchanged, turns it into D(0, −r), so D(0, r) = D(0, −r). Since D is invariant, the equality holds in every frame, the same fact UAQM-09.04 used to close the commutator. So outside the cone DF = [θ(t) + θ(−t)]D = D for every observer, while inside or on the cone the step functions themselves are invariant. Together, DF(Λ(x − y)) = DF(x − y) for every proper orthochronous Λ, and DF is a function of (x − y)² alone. The operator says the same. With U φ(x) U⁻¹ = φ(Λx) from topic 1.03, U Tφ(x)φ(y) U⁻¹ differs from Tφ(Λx)φ(Λy) only when the order flips, and then by the commutator [φ(Λx), φ(Λy)], which vanishes there by microcausality, quoted from UAQM-09.04. Unit 2 needs exactly this for products of interaction densities.
Outside the cone: an amplitude, not a signal
At spacelike interval s = √(−(x − y)²), DF therefore equals D, whose size UAQM-09.04 derived: DF = mK₁(ms)/(4π²s), quoted, K₁ the modified Bessel function. Checks: the dimension is m/s, mass²; as m → 0, K₁(z) → 1/z gives the massless 1/(4π²s²); for ms ≫ 1 it falls as e(−ms) over one Compton length. For the charged pion, measured at 139.57 MeV, m = 0.7073 fm⁻¹ and the range is 1.414 fm, the reach of the one-pion-exchange tail of the nuclear force. At s = 1.00 fm, DF = 1.85 × 10⁻² fm⁻², or 722 MeV², which is 0.73 of the massless 2.53 × 10⁻² fm⁻². None of this is a signal. DF is the amplitude for a quantum made at the earlier point to be found at the later one. Whether an action at y can change a measurement at x is decided by the commutator, zero here, and by the retarded function DR of topic 1.07, which vanishes outside the forward cone.
Why time order: the Dyson series asks for it
Why build an object that observers order differently at all? Because time evolution does. Quantum Mechanics II's Dyson series, quoted, writes the interaction-picture evolution as U = T exp(−i∫d⁴x ℋI(x)), and its second-order term is ((−i)²/2!)∫d⁴x d⁴y T[ℋI(x)ℋI(y)]: a time-ordered product appears because later interactions act after earlier ones, not by taste. Between vacua, with interaction densities built from free fields, those products reduce to products of DF, which Wick's theorem in Unit 2 makes systematic: stated here, shown there. The simplest case is a classical source J with ℋI = −Jφ, whose vacuum amplitude at second order is −½∫d⁴x d⁴y J(x)DF(x − y)J(y), the object topic 1.09 evaluates. Time order is also what makes DF a Green's function: (□ + m²)DF = −iδ⁴(x − y), shown in topic 1.06, the δ coming from differentiating the step functions.
Change one variable at a time
Make the relationship visible.
Drag β past {dt/dx|2}: the tilted line crosses x, x⁰′ − y⁰′ changes sign and the weight moves from D(x − y) to D(y − x), equal out here. Then raise x⁰ − y⁰ above x¹ − y¹: x enters the cone and no β reaches it.
INTERVAL (x − y)²-1.69 fm²
x⁰′ − y⁰′ FOR THE OBSERVER0.31 fm
THE SAME IN SECONDS1.05 × 10⁻²⁴ s
β TO REVERSE (NONE IF |β| ≥ 1)0.50
Live interpretationINTERVAL (x − y)²: −1.69 fm². x⁰′ − y⁰′ FOR THE OBSERVER: 0.31 fm. THE SAME IN SECONDS: 1.05 × 10⁻²⁴ s. β TO REVERSE (NONE IF |β| ≥ 1): 0.50
Catch the common trap
Explain before calculating.
Under a boost the step function θ(x⁰ − y⁰) can jump from 1 to 0. Which statement correctly explains why DF(x − y) = ⟨0|T φ(x)φ(y)|0⟩ is nevertheless the same for every inertial observer?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA two-point function is needed at points with x⁰ − y⁰ = 2.5 × 10⁻²⁴ s and |x − y| = 1.20 fm in the lab, and at a second pair with x⁰ − y⁰ = 5.0 × 10⁻²⁴ s and the same 1.20 fm. Classify each separation, find the slowest boost along x − y that reverses each time order, and say which term of DF each observer uses.
- Restore c: ct = (2.998 × 10⁸ m s⁻¹)(2.5 × 10⁻²⁴ s) = 7.495 × 10⁻¹⁶ m = 0.7495 fm for the first pair, and 1.4990 fm for the second.
- Intervals: (0.7495 fm)² − (1.20 fm)² = 0.5618 − 1.4400 = −0.878 fm², spacelike; (1.4990 fm)² − (1.20 fm)² = 2.2470 − 1.4400 = +0.807 fm², timelike.
- First pair: the order reverses for β beyond 0.7495/1.20 = 0.62. At β = 0.75, γ = 1.512 and t′ = 1.512 × (0.7495 − 0.9000) fm = −0.228 fm, or −7.6 × 10⁻²⁵ s, while the interval stays −0.878 fm².
- The lab therefore uses θ(x⁰ − y⁰)D(x − y) and the moving observer θ(y⁰ − x⁰)D(y − x); at spacelike separation D(x − y) = D(y − x), so both get the same DF.
- Second pair: reversal would need β = 1.4990/1.20 = 1.25, faster than light, so every observer sees x later and uses D(x − y).
AnswerFirst pair spacelike, −0.878 fm², reversed by any boost above 0.62c, with the same DF for both observers; second pair timelike, +0.807 fm², order fixed for all observers.
MediumA pion-mass scalar field (mc² = 139.57 MeV) is probed at one point of space at two times 10.0 fm/c, or 3.34 × 10⁻²³ s, apart. Use the non-relativistic limit of DF to find |DF| in fm⁻² and MeV², compare it with the exact 8.47 × 10⁻⁴ fm⁻² from UAQM-09.04's quoted Bessel form, and say what changes when the order of the two times is reversed.
- Natural units: m = 139.57 MeV ÷ 197.327 MeV fm = 0.7073 fm⁻¹ and |t| = 10.0 fm, so m|t| = 7.07 ≫ 1, and r = 0 ≪ |t|: the non-relativistic form applies.
- At r = 0, |DF| = (1/(2m))(m/(2π|t|))³⁄². Here 1/(2m) = 0.7069 fm and m/(2π|t|) = 0.7073/62.83 = 1.1257 × 10⁻² fm⁻², whose 3/2 power is 1.1944 × 10⁻³ fm⁻³.
- So |DF| = 0.7069 fm × 1.1944 × 10⁻³ fm⁻³ = 8.44 × 10⁻⁴ fm⁻². Dimension check: fm × fm⁻³ = fm⁻², mass², as DF must be; times (ħc)² = 3.894 × 10⁴ MeV² fm² it is 32.9 MeV².
- Against the exact value: 8.44/8.47 = 0.996, a 0.4% shortfall, the size of the leading relativistic correction to the magnitude, 3/(16(m|t|)²) = 0.0037.
- Reversing the order changes nothing: DF(x − y) = DF(y − x), and both carry e(−im|t|) with m|t| = 7.07 rad. The bare amplitude D would instead turn into its complex conjugate.
Answer|DF| ≈ 8.4 × 10⁻⁴ fm⁻², or 33 MeV², 0.4% below the exact 8.47 × 10⁻⁴ fm⁻²; reversing the two times leaves DF unchanged, phase included.
HardFor a pion-mass field (ħ/mc = 1.414 fm) compare F₁ = θ(x⁰ − y⁰)D(x − y), F₂ = θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x) and F₃ = θ(x⁰ − y⁰)[D(x − y) − D(y − x)] for a separation with ct = 0.90 fm and r = 1.50 fm, in the lab and for an observer at 0.80c along x − y, using D = mK₁(ms)/(4π²s) (quoted, K₁(0.8488) = 0.7865). Then say why the timelike pair ct = 1.50 fm, r = 0.90 fm cannot tell them apart.
- Interval: (0.90 fm)² − (1.50 fm)² = 0.81 − 2.25 = −1.44 fm², spacelike, with s = 1.20 fm. The order reverses for β beyond 0.90/1.50 = 0.60.
- At β = 0.80, γ = 5/3: ct′ = (5/3)(0.90 − 0.80 × 1.50) fm = −0.50 fm and r′ = (5/3)(1.50 − 0.80 × 0.90) fm = 1.30 fm. Check: 0.25 − 1.69 = −1.44 fm², unchanged. The observer sees x first.
- Value: ms = 0.7073 × 1.20 = 0.8488, so D = 0.7073 × 0.7865/(4π² × 1.20) fm⁻² = 1.17 × 10⁻² fm⁻², or 457 MeV², the same for D(x − y) and D(y − x) at this spacelike interval.
- F₁ is 1.17 × 10⁻² fm⁻² in the lab and 0 for the observer: not a Lorentz scalar. F₂ switches from D(x − y) to D(y − x) and stays 1.17 × 10⁻² fm⁻²: invariant, and it is DF. F₃ is 0 in both frames, because its bracket vanishes outside the cone: invariant, and it is the retarded function DR of topic 1.07.
- Timelike pair: t′ = γ(1.50 − 0.90β) fm is never below the proper time √(1.50² − 0.90²) fm = 1.20 fm for any |β| below 1, so no observer reverses it and all three candidates are trivially invariant. Only spacelike pairs test the construction.
AnswerF₁ jumps from 1.17 × 10⁻² fm⁻² to 0, so it is not invariant; F₂ = DF = 1.17 × 10⁻² fm⁻² (457 MeV²) for both observers; F₃ = 0 for both. A timelike pair, whose order is absolute, cannot distinguish them.