University Quantum Field Theory I · Natural Units, Relativistic States and the Feynman Propagator · 1.6

D_F as a momentum integral: i/(p² − m² + iε) and its poles

Every internal line in Unit 3's Feynman diagrams carries i/(p² − m² + iε), so it pays to see once why its iε is topic 1.05's time ordering and not a convergence trick. Here the p⁰ integral is done by residues both ways, the four ways past two poles are set side by side, and the Feynman choice is shown to permit Unit 6's Wick rotation.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 1.05 built D_F from two step functions, θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x): correct, but no Feynman rule can be read off a step function. This topic packs both terms into one covariant integral, ∫ d⁴p/(2π)⁴ i e^{(−ip⋅(x−y))}/(p² − m² + iε), and shows by doing the p⁰ integral that the iε is the time ordering. The integrand has poles at p⁰ = ±Eₚ, the two signs of energy on the mass shell, and a real-axis integral is undefined until told how to pass them.

The iε puts the positive-energy pole just below the axis and the negative-energy pole just above, as a mass squared m² − iε would. Jordan's lemma fixes the closing: below when x⁰ is later than y⁰, catching the positive-energy pole and giving D(x − y); above otherwise, giving D(y − x), both derived. Because the Klein-Gordon operator cancels the denominator, (□ + m²)D_F = −iδ⁴(x − y) for every placement, so the four ways past two poles are four Green's functions, retarded, advanced, Feynman and anti-Feynman, differing only on the mass shell.

The Feynman pair, in the second and fourth quadrants, leaves the first and third empty, which lets Unit 6 rotate p⁰ onto the imaginary axis. The model is a free field: interactions turn the pole into one at the physical mass with residue Z plus a multiparticle cut (Unit 4, quoted), and an unstable particle's pole sits a finite distance off the axis, iε becoming imΓ (topic 5.09).

Simple definition: The iε prescription is the instruction, carried by i/(p² − m² + iε), to pass the positive-energy pole p⁰ = Eₚ on its upper side and the negative-energy pole p⁰ = −Eₚ on its lower side, which turns the p⁰ integral into the time-ordered vacuum amplitude.
Example: For a mode with Eₚ = 500 MeV and x⁰ − y⁰ = +1.0 fm, closing below picks the pole at 500 MeV and gives e^{(−iEₚ t)}/(2Eₚ): modulus 1.0 × 10⁻³ MeV⁻¹, phase −Eₚ t/ħc = −2.53 rad. At x⁰ − y⁰ = −1.0 fm, closing above gives the same number.

Feynman propagator as a momentum integral, derivedD_F(x − y) = ∫ d⁴p/(2π)⁴ i e^{(−ip⋅(x−y))} / (p² − m² + iε)

One covariant expression for topic 1.05's two step-function terms. Its integrand is the factor i/(p² − m² + iε) that every internal line carries in Unit 3.

ħ = c = 1, metric (+, −, −, −); p and m in MeV; D_F in MeV², or fm⁻² via ħc = 197.327 MeV fm; ε → 0⁺ after integrating.

Where the two poles sit, derivedp⁰ = Eₚ − iε/(2Eₚ) and p⁰ = −Eₚ + iε/(2Eₚ), with Eₚ = √(|𝐩|² + m²)

Only the side of the real axis each pole is on survives ε → 0, not the size of the shift: positive energy just below, negative energy just above.

𝐩 is the three-momentum, Eₚ in MeV. The first pole lies in the fourth quadrant, the second in the second.

The p⁰ integral, closed by the sign of t, derived∫ dp⁰/(2π) i e^(−ip⁰t) / ((p⁰)² − Eₚ² + iε) = e^{(−iEₚ|t|)} / (2Eₚ)

Close below, clockwise, for t positive; above, anticlockwise, for t negative. Either way the wave is positive-frequency, and the d³p integral makes the two cases D(x − y) and D(y − x).

t = x⁰ − y⁰ in MeV⁻¹, or fm through ħc, with 1 fm ↔ 3.336 × 10⁻²⁴ s. The result is in MeV⁻¹.

Green's function, derived(□ + m²) D_F(x − y) = −i δ⁴(x − y)

On e^{(−ip⋅(x−y))} the operator gives m² − p², which cancels the denominator. The iε never enters, so all four pole placements obey this same equation.

□ = ∂²/∂t² − ∇². Both sides carry MeV⁴: D_F is MeV², □ adds two powers, and δ⁴ is MeV⁴.

Off-shell and on-shell parts, derived1/(p² − m² + iε) = P[1/(p² − m²)] − iπ δ(p² − m²)

Every placement shares the principal-value part and differs only on the mass shell, so any two of the four functions differ by a solution of the free equation.

P is the principal value; all terms in MeV⁻². The retarded function has −iπ sgn(p⁰) δ(p² − m²), the anti-Feynman +iπ δ(p² − m²).

Wick rotation of the propagator, derivedWith p⁰ = ip₄: d⁴p i/(p² − m² + iε) becomes d⁴p_E/(p_E² + m²)

Turning the contour anticlockwise through quadrants I and III passes no Feynman pole, and what is left is positive and needs no iε.

p₄ real, p_E² = p₄² + |𝐩|² ≥ 0 in MeV², d⁴p = i d⁴p_E. Valid for integrals with no e^(−ip⁰t), such as Unit 6's loops.

Two poles on the mass shell

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, so p⋅x = p⁰x⁰ − 𝐩⋅𝐱 with 𝐩 the three-momentum, p² = (p⁰)² − |𝐩|², and a time in MeV⁻¹ becomes fm through ħc = 197.327 MeV fm. Split the candidate D_F into ∫ d³p/(2π)³ e^{(i𝐩⋅(𝐱−𝐲))} times the p⁰ integral ∫ dp⁰/(2π) i e^(−ip⁰t)/((p⁰)² − Eₚ² + iε), with t = x⁰ − y⁰ and Eₚ = √(|𝐩|² + m²). Without the iε the integrand blows up at p⁰ = ±Eₚ, on the contour itself, and the integral has no meaning until it is told which side of each pole to pass. With it, (p⁰)² = Eₚ² − iε, whose roots to first order are Eₚ − iε', in the fourth quadrant, and −Eₚ + iε', in the second, where ε' = ε/(2Eₚ). The integrand factorises as i e^(−ip⁰t)/[(p⁰ − Eₚ + iε')(p⁰ + Eₚ − iε')], with residues i e^{(−iEₚ t)}/(2Eₚ) and −i e^{(iEₚ t)}/(2Eₚ), derived.

Close below for x⁰ later than y⁰, above otherwise

On a large arc p⁰ = R e^(iθ), |e^(−ip⁰t)| = e^{(tR sin θ)}. For t positive this dies only where sin θ is negative, so the contour closes in the lower half-plane, clockwise, and Jordan's lemma removes the arc. Only the positive-energy pole is inside: the integral is (−2πi/2π) × i e^{(−iEₚ t)}/(2Eₚ) = e^{(−iEₚ t)}/(2Eₚ). Restoring the 𝐩 integral gives ∫ d³p e^{(−ip⋅(x−y))}/((2π)³ 2Eₚ) with p⁰ = Eₚ, which is UAQM-09.04's vacuum amplitude D(x − y): a quantum made at y and absorbed at x. For t negative the arc must go above, anticlockwise, around −Eₚ: (2πi/2π) × (−i e^{(iEₚ t)}/(2Eₚ)) = e^{(−iEₚ|t|)}/(2Eₚ), two minus signs cancelling. Relabelling 𝐩 → −𝐩, which leaves d³p/Eₚ alone, makes that D(y − x). So the integral equals θ(t)D(x − y) + θ(−t)D(y − x), topic 1.05's time-ordered product, derived. At t = 0 the integrand falls as (p⁰)⁻², either closing works, and both give 1/(2Eₚ).

The iε is the time ordering, and a complex mass

Read the result backwards. Whichever way t points, the pole that gets caught is the one whose wave, e^{(−iEₚ|t|)}, carries positive energy from the earlier event to the later: the placement of the iε is the time ordering, not a regulator. Shifting the mass gives the same placement, since p² − m² + iε = p² − (m² − iε) and Eₚ becomes Eₚ − iε/(2Eₚ), so e^{(−iEₚ t)} gains a factor e^{(−εt/(2Eₚ))} that decays into the future: an infinitesimal width. For an unstable particle topic 5.09 makes it finite, m² − imΓ, quoted. The π⁺, with measured τ = 26.033 ns, has Γ = ħ/τ = 2.53 × 10⁻¹⁴ MeV and Γ/m = 1.8 × 10⁻¹⁶, so an infinitesimal iε describes it on any collision time scale; the Z, with Γ/m = 0.027, needs the finite form. The non-relativistic limit is a second check: near p⁰ = m + ω with ω and |𝐩| much smaller than m, p² − m² + iε ≈ 2m(ω − |𝐩|²/2m + iε/2m), so the propagator is 1/(2m) times Schrödinger's i/(ω − |𝐩|²/2m + iε), which is purely retarded. The other pole has fled to ω ≈ −2m.

A Green's function, whatever the placement

Act with □ + m² under the integral. Each plane wave gives (□ + m²)e^{(−ip⋅(x−y))} = (m² − p²)e^{(−ip⋅(x−y))}, so (□ + m²)D_F = −i∫ d⁴p/(2π)⁴ e^{(−ip⋅(x−y))} (p² − m²)/(p² − m² + iε). As ε → 0 the ratio tends to 1 and the integral is δ⁴(x − y): (□ + m²)D_F = −iδ⁴(x − y), derived. Dimensions agree, since D_F carries mass², □ two more, and δ⁴ mass⁴. Topic 1.07 reaches the same δ for the retarded function by differentiating step functions against the equal-time commutator; here it is algebra, and the iε plays no part. That is the point: the equation does not fix the function. For one mode it reads G'' + Eₚ²G = −iδ(t), whose whole content is a jump of −i in the slope of G at t = 0. The Feynman mode function e^{(−iEₚ|t|)}/(2Eₚ) has slope −i/2 just after t = 0 and +i/2 just before. Any free solution can be added without changing the jump, and choosing one is a boundary condition, which a momentum integral states through the placement of its poles.

Four ways past two poles

Each pole can be passed on either side, giving four functions with one integrand and one equation, shown by closing contours exactly as above. For one mode, with t = x⁰ − y⁰: both poles below gives the retarded G_R = −iθ(t) sin(Eₚ t)/Eₚ, zero for negative t because the upper closing then encloses nothing; both above gives the advanced G_A = iθ(−t) sin(Eₚ t)/Eₚ; positive-energy pole below and negative above gives the Feynman G_F = e^{(−iEₚ|t|)}/(2Eₚ); and the reverse, p² − m² − iε, gives the anti-Feynman −e^(iEₚ|t|)/(2Eₚ) = −G_F*, minus an anti-time-ordered product. In momentum space they differ only on the mass shell: 1/(s + iη) = s/(s² + η²) − iη/(s² + η²) → P(1/s) − iπδ(s), so the Feynman function carries −iπδ(p² − m²) on both energy shells, the retarded −iπ sgn(p⁰)δ(p² − m²). The difference of any two therefore lives on p² = m² and solves the free equation. Which one a calculation needs is physics: Unit 2's amplitudes are time-ordered and need G_F; a classical field driven by a source needs G_R, which topic 1.07 takes up.

Turning the contour onto the imaginary axis

The Feynman placement has one more consequence. Its poles lie in the second and fourth quadrants, so the first and third are empty. Turn the integration line anticlockwise about the origin until it lies along the imaginary axis: it sweeps only quadrants I and III, crosses no pole, and by Cauchy's theorem the integral is unchanged, provided the integrand falls faster than 1/|p⁰| on the arcs. That proviso excludes e^(−ip⁰t) with t ≠ 0, which grows in one of the swept quadrants, so the rotation is made on integrals without a time exponential, such as loops, or with time rotated as well. Set p⁰ = ip₄ with p₄ real: p² = −(p₄² + |𝐩|²) = −p_E², d⁴p = i d⁴p_E, and d⁴p i/(p² − m² + iε) becomes d⁴p_E/(p_E² + m²), positive with no pole in sight. One-mode check: by residues ∫ dp⁰/(2π) i/((p⁰)² − Eₚ² + iε) = 1/(2Eₚ); rotated, ∫ dp₄/(2π) 1/(p₄² + Eₚ²) = (1/2π)(π/Eₚ), the same. Anti-Feynman poles fill I and III and need the clockwise turn; the retarded pair blocks both turns. Topic 6.02 takes this step for a loop.

Change one variable at a time

Make the relationship visible.

Interactive model

Pole near +Eₚ: 1 below, −1 above1

Pole near −Eₚ: 1 below, −1 above−1

Time Eₚ(x⁰ − y⁰)1.5

Start at the Feynman setting, +Eₚ below and −Eₚ above: the modulus stays 1 at every t. Move −Eₚ below for the retarded function, zero for negative t, or both above for the advanced. Every setting keeps the same kink in Im at t = 0, a slope jump of −2: the −iδ of a Green's function.

POLES INSIDE CONTOUR1

RE PART 2Eₚ G0.07

IM PART 2Eₚ G-1.00

MODULUS |2Eₚ G|1.00

Live interpretationPOLES INSIDE CONTOUR: 1. RE PART 2Eₚ G: 0.07. IM PART 2Eₚ G: −1.00. MODULUS |2Eₚ G|: 1.00

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to treat the iε as a convergence factor: harmless, set to zero as soon as convenient, its sign a matter of taste. It is none of these. Set ε = 0 first and the poles sit on the contour; the natural repair, the principal value, gives for one mode −i sin(Eₚ|t|)/(2Eₚ), the average of the retarded and advanced functions. That misses the on-shell part cos(Eₚ t)/(2Eₚ) of the Feynman e^{(−iEₚ|t|)}/(2Eₚ) entirely and is no time-ordered product at all. Flip the sign and the integral becomes the anti-Feynman function −D_F*, an anti-time-ordered amplitude whose Wick rotation turns the wrong way. The size of ε never matters, but its sign and its placement are the physics: they are the boundary condition that the Green's-function equation leaves open.

In D_F(x − y) = ∫ d⁴p/(2π)⁴ i e^{(−ip⋅(x−y))}/(p² − m² + iε), the event x is 2.0 fm earlier than y, so t = x⁰ − y⁰ = −2.0 fm. How is the p⁰ integral done, and what does the whole integral give?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA π⁺ mode has |𝐩| = 300 MeV, with m = 139.570 MeV. Locate the two poles of i/(p² − m² + iε) in the complex p⁰ plane to first order in ε, then evaluate the one-mode Feynman integral G_F(t) = ∫ dp⁰/(2π) i e^(−ip⁰t)/((p⁰)² − Eₚ² + iε) at t = +3.0 fm and at t = −3.0 fm.

Eₚ = √(300² + 139.570²) MeV = √(1.0948 × 10⁵) MeV = 330.88 MeV, so 2Eₚ = 661.75 MeV. The poles are at p⁰ = 330.88 MeV − iε/(661.75 MeV), just below the positive real axis, and p⁰ = −330.88 MeV + iε/(661.75 MeV), just above the negative one.
For t = +3.0 fm, |e^(−ip⁰t)| = e^{(t Im p⁰)} dies in the lower half-plane, so close below, clockwise, around the pole at +Eₚ: G_F = (−2πi/2π) × i e^{(−iEₚ t)}/(2Eₚ) = e^{(−iEₚ t)}/(2Eₚ).
Restore ħc for the phase: Eₚ t/ħc = 330.88 × 3.0/197.327 = 5.030 rad, 0.80 of a cycle. In seconds, t = 1.00 × 10⁻²³ s against a period 2πħ/Eₚ = 1.25 × 10⁻²³ s.
So G_F(+3.0 fm) = (cos 5.030 − i sin 5.030)/(661.75 MeV) = (0.3127 + 0.9499i)/(661.75 MeV) = (4.72 + 14.35i) × 10⁻⁴ MeV⁻¹.
For t = −3.0 fm, close above, anticlockwise, around −Eₚ: (2πi/2π) × (−i e^{(iEₚ t)}/(2Eₚ)) = e^{(−iEₚ|t|)}/(2Eₚ), the same number. Check: the modulus is 1/(2Eₚ) = 1.51 × 10⁻³ MeV⁻¹, or 0.298 fm, both ways, while the retarded function would be zero at t = −3.0 fm.

AnswerPoles at ±330.88 MeV ∓ iε/(661.75 MeV); G_F(±3.0 fm) = (4.72 + 14.35i) × 10⁻⁴ MeV⁻¹ for both signs of t, modulus 1/(2Eₚ) = 1.51 × 10⁻³ MeV⁻¹.

MediumFor a π⁺ mode with |𝐩| = 400 MeV, evaluate J = ∫ dp⁰/(2π) i/((p⁰)² − Eₚ² + iε)², the t = 0 integral met when two propagators carry the same momentum. Do it by residues, again by rotating the contour onto the imaginary axis, and check both against the Eₚ² derivative of the one-propagator result 1/(2Eₚ).

Eₚ = √(400² + 139.570²) MeV = 423.65 MeV. The integrand falls as (p⁰)⁻⁴ and has no exponential, so either half-plane will do; close below, clockwise, around the double pole at Eₚ − iε/(2Eₚ).
At a double pole the residue is the derivative of what multiplies (p⁰ − Eₚ)⁻²: d/dp⁰ [i(p⁰ + Eₚ)⁻²] at p⁰ = Eₚ is −2i/(2Eₚ)³ = −i/(4Eₚ³). Clockwise, J = (−2πi/2π)(−i/(4Eₚ³)) = −1/(4Eₚ³).
Rotation: the poles sit in quadrants II and IV, so the line may turn anticlockwise onto the imaginary axis. With p⁰ = ip₄, (p⁰)² − Eₚ² = −(p₄² + Eₚ²) and dp⁰ = i dp₄, so J = ∫ (i dp₄/2π) i/(p₄² + Eₚ²)² = −(1/2π) × π/(2Eₚ³) = −1/(4Eₚ³), using ∫ dx/(x² + a²)² = π/(2a³) over the whole real line.
Cross-check: ∂/∂(Eₚ²) of i/((p⁰)² − Eₚ²) is i/((p⁰)² − Eₚ²)², so J = ∂/∂(Eₚ²) of 1/(2Eₚ) = −1/(4Eₚ³). Dimensions: dp⁰ in MeV over (MeV²)² leaves MeV⁻³.
Numbers: J = −1/(4 × 423.65³) MeV⁻³ = −3.29 × 10⁻⁹ MeV⁻³, or −0.0253 fm³ after multiplying by (ħc)³. Integrating numerically along the real axis with ε = 500 MeV² gives −(3.288 + 0.014i) × 10⁻⁹ MeV⁻³: the real part already agrees, and the imaginary part, −3ε/(8Eₚ⁵) = −1.4 × 10⁻¹¹ MeV⁻³, vanishes as ε → 0⁺.

AnswerJ = −1/(4Eₚ³) = −3.29 × 10⁻⁹ MeV⁻³ (−0.0253 fm³), the same by residues, by the Wick rotation and by differentiating 1/(2Eₚ).

HardTopic 5.09 gives an unstable particle's propagator the pole p² = m² − imΓ, the iε made finite, with Γ = ħ/τ; quote it here. For a π⁺ beam of momentum |𝐩| = 500 MeV, with m = 139.570 MeV and measured τ = 26.033 ns, locate the positive-energy pole in the p⁰ plane, close the contour for t = x⁰ − y⁰ positive, and find the lab-frame lifetime, the mean decay length and the fraction surviving 10.0 m.

Γ = ħ/τ = (6.5821 × 10⁻²² MeV s)/(2.6033 × 10⁻⁸ s) = 2.528 × 10⁻¹⁴ MeV, so Γ/m = 1.81 × 10⁻¹⁶: tiny but finite, and with the sign of the iε.
Eₚ = √(500² + 139.570²) MeV = 519.11 MeV. The poles solve (p⁰)² = Eₚ² − imΓ, so p⁰ = ±(Eₚ − imΓ/(2Eₚ)); the positive-energy pole is at 519.11 MeV − i(3.40 × 10⁻¹⁵ MeV), still in the fourth quadrant, the other still in the second.
For t positive, close below as before. The residue now gives e^{(−iEₚ t)} e^{(−mΓt/(2Eₚ))}/(2Eₚ), and the probability, its modulus squared, falls as e^{(−(m/Eₚ)Γt)}: the lab-frame decay rate is (m/Eₚ)Γ.
So τ_lab = (Eₚ/m)τ = 3.7194 × 26.033 ns = 96.83 ns. The time-dilation factor γ = Eₚ/m comes out of the invariant pole position m² − imΓ, with no separate relativity argument. Check: |𝐩| → 0 gives Eₚ = m and τ_lab = τ.
Decay length: (|𝐩|/Eₚ) × (Eₚ/m)cτ = (|𝐩|/m)cτ = 3.5824 × 7.8045 m = 27.96 m, where cτ = 7.8045 m follows from the measured τ. The fraction surviving 10.0 m is e^{(−10.0/27.96)} = 0.699.

AnswerPole at 519.11 MeV − i(3.40 × 10⁻¹⁵ MeV); τ_lab = 96.8 ns, mean decay length 28.0 m, and 69.9% of the beam survives 10.0 m.