University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.7

The retarded propagator and where causality lives

Two propagators solve the same equation with the same δ-function source, yet one leaks outside the light cone and the other cannot. Build the retarded one from the commutator, set it against D_F pole by pole, and locate causality where it actually lives: in commutators, never in amplitudes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The commutator of a free field with itself is a c-number, the Pauli-Jordan function iΔ(x − y) that UAQM-09.04 derived and showed to vanish at spacelike separation. Multiply it by θ(x⁰ − y⁰) and you have the retarded propagator D_R. The step function is harmless to Lorentz invariance precisely because the commutator has already switched itself off wherever the time order of two events depends on the frame, and it buys one thing: D_R is zero everywhere outside the forward light cone of y. Differentiate twice in time and the step meets the equal-time commutators, [φ, φ] = 0 and [π, φ] = −iδ³, leaving (□ + m²)D_R = −iδ⁴(x − y): a Green's function with exactly the source D_F has.

Two Green's functions with one source differ by a free solution, and here it is explicit, D_F − D_R = D(y − x). In momentum space the whole difference is pole placement: D_F straddles the real p⁰ axis, D_R has both poles below it. The two then divide the labour. Every response to a disturbance is a retarded commutator, so the classical field a source radiates is i∫D_R j; amplitudes come from the time-ordered Dyson series and are built from D_F, which leaks a Compton length outside the cone.

Nothing travels through that leak, because causality is a property of commutators, not amplitudes. The argument is a free-field one: for interacting fields local commutativity is imposed, not derived.

Simple definition: The retarded propagator D_R(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩ is the field's commutator kept only when x is later than y: a Green's function of the Klein-Gordon operator that gives the response at x to a disturbance at y and vanishes outside the forward light cone.
Example: For the pion field, take x 3.0 fm from y and 2.0 fm/c later, so σ = √(3.0² − 2.0²) fm = 2.24 fm outside the cone: D_R(x − y) = 0 exactly, while D_F(x − y) = D(x − y) = 1.98 × 10⁻³ fm⁻², 39% of the massless 1/(4π²σ²).

Retarded propagator, definedD_R(x − y) = θ(x⁰ − y⁰) ⟨0|[φ(x), φ(y)]|0⟩ = iθ(x⁰ − y⁰) Δ(x − y)

The step costs no Lorentz invariance because Δ already vanishes wherever time order depends on the frame; the product is zero outside the forward light cone.

ħ = c = 1, metric (+, −, −, −); Δ is UAQM-09.04's real Pauli-Jordan function; D_R has mass dimension 2, read in fm⁻².

Green's-function equation, derived(□ₓ + m²) D_R(x − y) = −iδ⁴(x − y)

∂₀² hits the step twice: [φ, φ] = 0 at equal times kills one δ term, [π, φ] = −iδ³ leaves the other. D_F has exactly the same source.

□ = ∂²/∂t² − ∇²; m in fm⁻¹ through ħc = 197.327 MeV fm; δ⁴ in fm⁻⁴.

Momentum space: both poles below, derivedD_R(z) = ∫ d⁴p/(2π)⁴ i e^(−ip⋅z) / ((p⁰ + iε)² − |p|² − m²), z = x − y

x⁰ earlier than y⁰: close above and enclose nothing. x⁰ later: close below round both poles, giving iD_R(t, p) = θ(t) sin(Eₚ t)/Eₚ mode by mode.

Poles at p⁰ = ±Eₚ − iε, Eₚ = √(|p|² + m²) in MeV; ε → 0⁺ only at the end.

Feynman minus retarded, derivedD_F(x − y) − D_R(x − y) = D(y − x) = ⟨0|φ(y)φ(x)|0⟩

One source, two boundary conditions: D_R is zero before the source, D_F sends positive frequencies forward and negative ones back.

In momentum space the difference is 2πδ(p² − m²)θ(−p⁰): on the mass shell only, so it solves the free equation.

Classical field of a source, derived⟨φ(x)⟩ = i ∫ d⁴y D_R(x − y) j(y)

A static source j = gδ³(x) gives g e^(−mr)/(4πr), the Yukawa field, with range 1.414 fm for the measured pion mass.

For (□ + m²)φ = j, with j in fm⁻³ and φ in fm⁻¹; iD_R is real. Exact for a free field.

Response to any local disturbance, quoted from QM IIδ⟨O(x)⟩ = −i ∫ d⁴y θ(x⁰ − y⁰) ⟨[O(x), ℋᵢₙₜ(y)]⟩

Every response is a retarded commutator, so microcausality makes it vanish exactly at spacelike separation. D_F, an amplitude, never enters.

First order in ℋᵢₙₜ, the interaction Hamiltonian density (fm⁻⁴); O any local observable; QM II's interaction picture.

Keep the commutator, but only for later times

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. For the real Klein-Gordon field of UAQM-09.02, D(x − y) = ⟨0|φ(x)φ(y)|0⟩, and UAQM-09.04's argument, quoted and run for the real field, gives [φ(x), φ(y)] = D(x − y) − D(y − x) ≡ iΔ(x − y): a c-number, i times the real Δ, the same in every state, zero whenever (x − y)² is negative. Keep it only when x is later: D_R(x − y) = θ(x⁰ − y⁰)⟨0|[φ(x), φ(y)]|0⟩. A step in time normally breaks Lorentz invariance, since a boost can reverse the order of two spacelike-separated events. Here it cannot: wherever the order is frame-dependent Δ is already zero, and inside the cone the sign of x⁰ − y⁰ is invariant. The product θ(x⁰ − y⁰)D(x − y) alone fails this test, because D survives outside the cone. So D_R is invariant and lives on and inside the forward cone of y, where UAQM-09.04's Δ, quoted, gives iD_R = δ((x − y)²)/(2π) − mJ₁(mλ)/(4πλ) with λ = √((x − y)²).

Where the δ-function comes from

Write t = x⁰ − y⁰ and C = ⟨[φ(x), φ(y)]⟩. One time derivative of θ(t)C gives δ(t)C + θ(t)∂₀C, and the δ term dies because [φ(x), φ(y)] = 0 at equal times. The second gives δ(t)∂₀C + θ(t)∂₀²C, and this δ survives: at equal times ∂₀C = ⟨[π(x), φ(y)]⟩ = −iδ³(x − y), the canonical commutator of UAQM-09.02. Spatial derivatives and m² pass through θ untouched and C solves the free equation, so (□ + m²)D_R(x − y) = −iδ⁴(x − y), derived. Dimensions: D_R carries mass², so □D_R carries mass⁴, as δ⁴ does. UQFTI-01.05 found the same −iδ⁴ for D_F from the same two commutators. Massless check: only the cone term survives, iD_R = θ(t)δ(t − r)/(4πr), the retarded Green's function behind every retarded potential of electrodynamics. A massive field adds a wake −mJ₁(mλ)/(4πλ) inside the cone; its first zero, at mλ = 3.832, lies at λ = 5.42 fm for the pion.

Both poles below the axis

In momentum space, with z = x − y, D_R(z) = ∫ d⁴p/(2π)⁴ i e^(−ip⋅z) / ((p⁰ + iε)² − Eₚ²) and Eₚ = √(|p|² + m²). The poles sit at p⁰ = ±Eₚ − iε, both below the real axis, where D_F's i/(p² − m² + iε) in UQFTI-01.06 put one below and one above. For t = x⁰ − y⁰ negative, e^(−ip⁰t) decays in the upper half plane: close there, enclose nothing, and D_R = 0. For t positive, close below, clockwise, and collect both residues: mode by mode D_R(t, p) = (e^{(−iEₚ t)} − e^{(iEₚ t)})/2Eₚ = −i sin(Eₚ t)/Eₚ, which reassembles into θ(t)[D(x − y) − D(y − x)]. The two prescriptions differ by one sign: (p⁰ + iε)² − Eₚ² → p² − m² + iε sgn(p⁰). Each mode is a kicked oscillator, iD_R(t, p) = θ(t) sin(Eₚ t)/Eₚ, leaving t = 0 with unit slope whatever p is, because δ³ kicks every mode equally. For the pion at rest the period 2π/m is 8.88 fm/c, or 2.96 × 10⁻²³ s.

One source, two boundary conditions

Two Green's functions with the same source differ by a solution of the free equation, and here that solution can be written down. From D_F = θ(t)D(x − y) + θ(−t)D(y − x) and D_R = θ(t)[D(x − y) − D(y − x)], D_F − D_R = [θ(t) + θ(−t)]D(y − x) = D(y − x), derived. D(y − x) = ⟨0|φ(y)φ(x)|0⟩ = ∫ d³p e^(ip⋅z)/((2π)³2Eₚ), with z = x − y and p⁰ = Eₚ, is a superposition of on-shell waves, so □ + m² annihilates it; in momentum space D̃_F − D̃_R = 2πδ(p² − m²)θ(−p⁰), on the mass shell and nowhere else. The advanced propagator D_A = −θ(−t)⟨[φ(x), φ(y)]⟩ has the same source too, and D_R − D_A = iΔ, the commutator itself. What tells the family apart is the boundary condition: D_R vanishes before the source, D_A after it, and D_F carries positive frequencies into the future and negative frequencies into the past. Before the source, the free wave D(y − x) is the whole of D_F.

The classical field is retarded

Add a classical source, ℒ → ℒ + jφ, so the field equation becomes (□ + m²)φ = j. Quantum Mechanics II's interaction picture, to first order in ℋᵢₙₜ = −jφ, gives for any observable δ⟨O(x)⟩ = −i∫d⁴y θ(x⁰ − y⁰)⟨[O(x), ℋᵢₙₜ(y)]⟩; for O = φ that is ⟨φ(x)⟩ = i∫d⁴y D_R(x − y)j(y), exact here because the commutator is a c-number. Check: (□ + m²)iD_R = δ⁴, so ⟨φ⟩ solves the field equation, and iD_R is real, as a classical field must be. A static point source j = gδ³(x) needs only p⁰ = 0, where D̃_R = −i/(|p|² + m²), and the Fourier integral gives ⟨φ⟩ = g e^(−mr)/(4πr), the Yukawa field, derived; m → 0 returns the Coulomb form g/(4πr). With the measured pion mass, 139.570 MeV, the range ħ/mc = 1.414 fm is the reach of the long-distance nuclear force from which Yukawa predicted the pion in 1935. At p⁰ = 0, D̃_R and D̃_F coincide, which is why UQFTI-01.09 can draw the same Yukawa energy from D_F.

Causality lives in commutators, not amplitudes

D_F does not vanish outside the cone. At spacelike separation σ = √(−(x − y)²) a boost can reverse the time order, but there D(x − y) = D(y − x), so D_F = D = mK₁(mσ)/(4π²σ), quoted from UAQM-09.04: 1.98 × 10⁻³ fm⁻² for the pion at σ = 2.24 fm, 39% of the massless value, falling as e^(−mσ). That is an amplitude, and the Dyson series of Unit 2 builds S-matrix elements from such time-ordered products; no measurement is described by an amplitude alone. Everything a disturbance at y can do to a reading at x is a retarded commutator, and microcausality makes every commutator of local observables vanish at spacelike separation: exactly zero, not e^(−mσ) small. The leak in D_F is a correlation the vacuum already holds between the two points, and correlations carry no signal; the commutator removes it by subtracting the reverse-ordered D(y − x). All of this is proved for a free field. For interacting fields the response formula still holds, but local commutativity becomes an axiom rather than a theorem about modes.

Change one variable at a time

Make the relationship visible.

Interactive model

Field mass m140 MeV

Mode momentum |p|0 MeV

Probe time t3.0 fm/c

Drag t below zero: the solid curve sits at exactly zero while the dashed one keeps oscillating. Then raise m or |p|: amplitude 1/Eₚ and period both shrink, yet the solid curve always leaves t = 0 along the slope-1 line, because [π, φ] = −iδ³ kicks every mode equally.

MODE ENERGY Eₚ140 MeV

i D_R AT t1.196 fm

Re i D_F AT t0.598 fm

Re i D_F − i D_R-0.598 fm

Live interpretationMODE ENERGY Eₚ: 140 MeV. i D_R AT t: 1.196 fm. Re i D_F AT t: 0.598 fm. Re i D_F − i D_R: −0.598 fm

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that D_F ≠ 0 at spacelike separation means virtual quanta outrun light, and that the causal choice would be to build amplitudes from D_R. Both halves are wrong. D_F is an amplitude, the vacuum expectation of a time-ordered product that the Dyson series hands you; no detector reads it. What an action at y does to a measurement at x is the retarded commutator −i∫d⁴y θ(x⁰ − y⁰)⟨[O(x), ℋᵢₙₜ(y)]⟩, and microcausality makes that exactly zero at spacelike separation, not merely e^(−mσ) small. The leak in D_F is a correlation the vacuum already carries, and correlations send no signals. Nor would D_R make amplitudes more causal; it would make them wrong. Put D_R in place of D_F in the vacuum persistence amplitude exp(−½∫∫ j D_F j) and the exponent turns purely imaginary, so |⟨0|0⟩|² = 1 and no source could ever radiate a quantum.

A classical source j(y) is nonzero only within 0.2 fm of the origin and for |y⁰| ≤ 0.2 fm/c. A detector reads ⟨φ⟩ at r = 3.0 fm and t = 2.0 fm/c. The free scalar field of mass m starts in its vacuum. What does the detector find?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFor the pion field (m = 139.57 MeV, so m = 0.7073 fm⁻¹), put a source event y at the origin and take three events at r = 3.0 fm: event A at t = 4.0 fm/c, event B at t = 2.0 fm/c and event C at t = −4.0 fm/c. Say where D_R(x − y), the commutator and D_F(x − y) vanish, then evaluate the commutator at A and C and D_F at B from the forms UAQM-09.04 derived and quoted.

Intervals (x − y)² = t² − r²: A gives 16 − 9 = +7 fm², timelike with x later; B gives 4 − 9 = −5 fm², spacelike; C gives +7 fm², timelike with x earlier.
D_R is θ(x⁰ − y⁰) times the commutator. At B the commutator vanishes by microcausality, so D_R = 0; at C the step vanishes, so D_R = 0 although the commutator does not. Only A has D_R ≠ 0. D_F vanishes at none of the three.
At A, λ = √7 fm = 2.646 fm and mλ = 1.871, off the cone, so only the Bessel term of Δ survives: Δ = mJ₁(mλ)/(4πλ) = 0.7073 × 0.5817/(4π × 2.646) fm⁻² = 0.0124 fm⁻². The commutator is iΔ = 0.0124i fm⁻², and D_R(A) equals it.
At C, Δ carries sgn(x⁰) = −1: the commutator is −0.0124i fm⁻², the same size, nonzero, and removed from D_R only by the step function.
At B, σ = √5 fm = 2.236 fm and mσ = 1.582. With K₁(1.582) = 0.2470, D_F = D = mK₁(mσ)/(4π²σ) = 0.7073 × 0.2470/(4π² × 2.236) fm⁻² = 1.98 × 10⁻³ fm⁻², against 1/(4π²σ²) = 5.07 × 10⁻³ fm⁻² for a massless field: real, 39% of it, and not zero.

AnswerD_R ≠ 0 only at A, where it equals the commutator, 0.0124i fm⁻². The commutator also survives at C (−0.0124i fm⁻²) but vanishes at B, while D_F survives everywhere: 1.98 × 10⁻³ fm⁻² at the spacelike event B.

MediumTake the pion field (m = 139.57 MeV) and the single spatial mode |p| = 300 MeV. Do the p⁰ integral D_R(t, p) = ∫ dp⁰/(2π) i e^(−ip⁰t)/((p⁰ + iε)² − Eₚ²) for both signs of t, evaluate iD_R at t = ±1.00 fm/c, and check that D_F − D_R at t = +1.00 fm/c is the free mode e^{(iEₚ t)}/2Eₚ.

Eₚ = √(300² + 139.57²) MeV = 330.88 MeV. D_R(t, p) has mass dimension −1, read in fm through ħc/Eₚ = 197.327/330.88 fm = 0.5964 fm, and at t = 1.00 fm/c the phase is Eₚ t/ħc = 330.88 × 1.00/197.327 = 1.6768 rad.
The poles sit at p⁰ = ±Eₚ − iε, both below the axis. For negative t, e^(−ip⁰t) decays where Im p⁰ is positive: close above, enclose nothing, so D_R(−1.00 fm/c, p) = 0.
For positive t, close below, clockwise, with factor −2πi/(2π) = −i. The residues of i e^(−ip⁰t)/((p⁰ − Eₚ)(p⁰ + Eₚ)) are i e^{(−iEₚ t)}/2Eₚ and −i e^{(iEₚ t)}/2Eₚ, so D_R = (e^{(−iEₚ t)} − e^{(iEₚ t)})/2Eₚ = −i sin(Eₚ t)/Eₚ and iD_R = sin(Eₚ t)/Eₚ.
At t = 1.00 fm/c, iD_R = 0.5964 fm × sin(1.6768) = 0.5964 × 0.9944 fm = 0.593 fm. Its slope just after t = 0 is cos 0 = 1, for this mode and every other: the coefficient of the δ-function.
D_F(t, p) = e^{(−iEₚ|t|)}/2Eₚ = 0.2982 fm × (cos 1.6768 − i sin 1.6768) = (−0.0315 − 0.2965i) fm. Subtracting D_R = −0.593i fm gives (−0.0315 + 0.2965i) fm, and e^{(iEₚ t)}/2Eₚ = 0.2982 fm × (−0.1058 + 0.9944i) = (−0.0315 + 0.2965i) fm: equal.

AnsweriD_R = 0.593 fm at t = +1.00 fm/c and exactly 0 at t = −1.00 fm/c; D_F − D_R = (−0.0315 + 0.2965i) fm, the free mode e^{(iEₚ t)}/2Eₚ, as D(y − x) requires.

HardA point source is switched on suddenly and held, j(x) = gθ(t)δ³(x), in the pion field. Using ⟨φ⟩ = i∫D_R j and the inside-cone form iD_R = δ(τ − r)/(4πr) − mJ₁(mλ)/(4πλ), λ = √(τ² − r²), quoted from UAQM-09.04, find the field at r = 1.00 fm before the light cone arrives, just after, and at late times, and compare the last with the static Yukawa field.

[j] = 3 and [δ³] = 3, so g is dimensionless and ⟨φ⟩ comes out in fm⁻¹. With τ = t − t′ the source history 0 ≤ t′ ≤ t gives ⟨φ(t, r)⟩ = g∫₀^t dτ iD_R(τ, r): the field now is the retarded kernel summed over everything the source has done.
Before arrival, t below r/c = 1.00 fm/c = 3.34 × 10⁻²⁴ s, every τ in the integral is less than r, outside the forward cone, where iD_R = 0. The field is exactly zero, not exponentially small.
Just after arrival only the δ on the cone has contributed: ⟨φ⟩ = g/(4πr) = 0.0796g fm⁻¹ = 15.7g MeV, the massless Coulomb value whatever m is. The wavefront is carried by the highest frequencies, for which the mass is invisible.
At late times the wake adds −(gm/4π)∫ from r to ∞ of J₁(mλ)/λ dτ. With dτ = λ dλ/√(λ² + r²) this is −(gm/4π) times the integral over λ from 0 to ∞ of J₁(mλ)/√(λ² + r²) dλ, and that quoted integral equals (1 − e^(−mr))/(mr), 0.7168 at mr = 0.7073, checked numerically. So ⟨φ⟩ → g/(4πr) − g(1 − e^(−mr))/(4πr) = g e^(−mr)/(4πr).
e^(−0.7073) = 0.4930, so the late field is 0.0392g fm⁻¹ = 7.74g MeV, the Yukawa field a static source gives through D̃_R at p⁰ = 0. Integrated numerically, the field undershoots to 0.0341g fm⁻¹ near t = 5.5 fm/c and rings down onto it. As m → 0 the wake disappears and the field stays at the Coulomb value.

Answer⟨φ⟩ = 0 until t = 1.00 fm/c (3.34 × 10⁻²⁴ s); it jumps to g/(4πr) = 0.0796g fm⁻¹ on the light cone and settles to the Yukawa value g e^(−mr)/(4πr) = 0.0392g fm⁻¹ = 7.74g MeV.