University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.1

The Dirac propagator: S_F(p) = i(p̸ + m)/(p² − m² + iε)

Every fermion line in every diagram from here to the Lamb shift is this one function. Learn where its minus sign comes from, why that sign is what keeps it the same in every frame, and how its two poles carry an electron into the future and a positron into the past.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The model is UAQM-09.07's free Dirac field, quantised with anticommutators, and the question is what perturbation theory needs from it. Dyson's series multiplies fields in time order, so the first object is ⟨0|Tψ(x)ψ̄(y)|0⟩, and for anticommuting fields the ordering must carry a minus sign each time it swaps two of them. That sign is not decoration.

Outside the light cone the fields anticommute, so only the signed product is the same in every frame; and in the calculation the ordering's minus cancels the one hidden in the positron spin sum Σvv̄ = p̸ − m, so the two time orders join into one covariant function, (i∂̸ + m)D_F(x − y), which is i(p̸ + m)/(p² − m² + iε) = i/(p̸ − m + iε) in momentum space. It is a Green's function of the Dirac operator, and the iε decides which one: closing the p⁰ contour sends positive frequencies, electrons, forward in time and negative frequencies, positrons, backward, while the retarded function built from the anticommutator sends both forward, as a classical wave equation would. The model stops at free fields.

In Yukawa theory, where every particle is massive, interactions keep a pole but move it to the physical mass, give it a residue Z₂ below 1 and add a multiparticle cut, and reading off that residue is topic 7.03's LSZ for Dirac fields, after Wick's theorem for fermions.

Simple definition: The Dirac propagator S_F(x − y) = ⟨0|Tψ(x)ψ̄(y)|0⟩ is the vacuum amplitude for fermion number to enter at y and leave at x, with the fields time-ordered under a minus sign; in momentum space it is i(p̸ + m)/(p² − m² + iε).
Example: For an electron at rest, S_F(t, 𝐩 = 0)γ⁰ is θ(t)e^(−imt) on the upper two spinor components and −θ(−t)e^(imt) on the lower two. With m = 0.511 MeV the phase turns once every 2πħ/m = 8.09 × 10⁻²¹ s, forward in time for the electron and backward for the positron.

Fermionic time ordering (a definition)T ψ(x)ψ̄(y) = θ(x⁰ − y⁰) ψ(x)ψ̄(y) − θ(y⁰ − x⁰) ψ̄(y)ψ(x)

The sign is forced: [ψ(x), ψ̄(y)]₊ = 0 at spacelike separation, so only the signed product is the same in every frame.

θ is the unit step; ψ has mass dimension 3/2, m^(−3/2) in SI; one minus sign for each swap of two fermion fields

Position space (derived)S_F(x − y) = ⟨0|T ψ(x)ψ̄(y)|0⟩ = (i∂̸ₓ + m) D_F(x − y)

The ordering's minus cancels the minus in Σvv̄ = p̸ − m, so both time orders join into one covariant function.

S_F is a 4 × 4 spinor matrix of mass dimension 3; D_F is topic 1.05's scalar propagator, of dimension 2; m in MeV

Momentum space (derived)S_F(p) = i(p̸ + m)/(p² − m² + iε) = i/(p̸ − m + iε)

(p̸ − m)(p̸ + m) = p² − m² makes it the inverse of p̸ − m; at high energy it falls as 1/|p|, one power slower than a scalar line.

p̸ = γ^μp_μ; p and m in MeV, so S_F(p) is in MeV⁻¹; ε → 0⁺ at the end

Green's function of the Dirac operator (derived)(i∂̸ₓ − m) S_F(x − y) = iδ⁴(x − y)

From (i∂̸ − m)(i∂̸ + m) = −(∂² + m²) and topic 1.06's (∂² + m²)D_F = −iδ⁴. The equation alone does not pick S_F; the iε does.

δ⁴ has mass dimension 4, one more than S_F, as one derivative requires

The two poles in time (derived)S_F(t, 𝐩)γ⁰ = θ(t) Λ₊ e^(−iEₚt) − θ(−t) Λ₋ e^(iEₚt)

Electrons run forward on the +Eₚ pole and positrons backward on −Eₚ, with the ordering's minus; the jump at t = 0 is Λ₊ + Λ₋ = 1.

Λ± = (Eₚ ± H)/(2Eₚ), H = α⋅𝐩 + βm; Eₚ = √(𝐩² + m²) in MeV; t in MeV⁻¹, and ħ/MeV = 6.582 × 10⁻²² s

The retarded function, for contrast (derived)S_R(p) = i(p̸ + m)/((p⁰ + iε)² − 𝐩² − m²)

Same equation, causal boundary conditions: both frequencies run forward and nothing precedes the source. S_F − S_R solves the free Dirac equation.

Both poles lie below the real p⁰ axis; in position space S_R = θ(x⁰ − y⁰)⟨0|[ψ(x), ψ̄(y)]₊|0⟩ = (i∂̸ + m)D_R

The ordering needs a sign, or it depends on the frame

Units are natural, ħ = c = 1, with metric (+, −, −, −) and p⋅x = p⁰x⁰ − 𝐩⋅𝐱; ħc = 197.327 MeV fm converts back. Dyson's series orders fields in time (topic 2.03), so fermions first need Tψ(x)ψ̄(y), defined with a minus sign whenever the ordering reverses the fields as written. The sign is forced, shown here. For spacelike x − y a boost can reverse which event is earlier: for two events r = 386 fm apart at equal times, a frame moving at β = 0.6 along the separation finds x⁰′ − y⁰′ = −γβr = −290 fm/c, or −9.66 × 10⁻²² s. There UAQM-09.07's anticommutator [ψ(x), ψ̄(y)]₊ vanishes (quoted), so ψ̄(y)ψ(x) = −ψ(x)ψ̄(y) and the signed product is ψ(x)ψ̄(y) whichever order a frame sees. Unsigned, it would be sgn(x⁰ − y⁰)ψ(x)ψ̄(y), and a boost would flip a nonzero amplitude: ⟨0|ψ(x)ψ̄(y)|0⟩ survives outside the light cone, its scalar part mD at r = 1/m = 386 fm being m³K₁(1)/(4π²) = 0.0152 × m³ (topic 1.08's K₁ form, quoted), dying off as e^(−mr). Timelike pairs keep their order in every frame, so the spacelike region decides.

Two time orders, two spin sums, one function

Quote UAQM-09.07's field in topic 1.03's normalisation, ψ(x) = ∫dp̃ Σₛ[b(p, s)u(p, s)e^(−ip⋅x) + d†(p, s)v(p, s)e^(ip⋅x)] with dp̃ = d³p/((2π)³2Eₚ) and [b(p, r), b†(k, s)]₊ = 2Eₚ(2π)³δ³(𝐩 − 𝐤)δᵣₛ, likewise for d, and UAQM-04.08's spin sums Σuū = p̸ + m and Σvv̄ = p̸ − m. If x⁰ is later, only b b† survives in the vacuum: ⟨0|ψ(x)ψ̄(y)|0⟩ = ∫dp̃ (p̸ + m)e^{(−ip⋅(x−y))} = (i∂̸ₓ + m)D(x − y), since i∂̸ₓ turns e^(−ip⋅x) into p̸e^(−ip⋅x), with D = ⟨0|φ(x)φ(y)|0⟩ from topic 1.05. If y⁰ is later, only d d† survives: ⟨0|ψ̄(y)ψ(x)|0⟩ = ∫dp̃ (p̸ − m)e^{(ip⋅(x−y))} = −(i∂̸ₓ + m)D(y − x), because i∂̸ₓ now returns −p̸. The ordering's minus cancels this one: S_F = θ(x⁰ − y⁰)(i∂̸ + m)D(x − y) + θ(y⁰ − x⁰)(i∂̸ + m)D(y − x). Moving ∂̸ outside the steps adds iγ⁰δ(x⁰ − y⁰)[D(x − y) − D(y − x)], zero at equal times, so S_F(x − y) = (i∂̸ₓ + m)D_F(x − y), derived. Dimensions: ψ carries 3/2, so S_F carries 3, as D_F's 2 plus one derivative does.

Momentum space: the inverse of p̸ − m

Insert topic 1.06's D_F(x − y) = ∫d⁴p/(2π)⁴ i e^{(−ip⋅(x−y))}/(p² − m² + iε), quoted. On e^(−ip⋅x), i∂̸ₓ becomes p̸, so S_F(p) = i(p̸ + m)/(p² − m² + iε), derived. Because p̸p̸ = p², (p̸ − m)(p̸ + m) = p² − m², and S_F(p) = i/(p̸ − m + iε): the inverse of the Dirac operator, which is (i∂̸ₓ − m)S_F(x − y) = iδ⁴(x − y) read in momentum space. Its mass dimension is −1. Two limits test it. At high energy S_F → ip̸/p², falling as 1/|p| where a scalar line falls as 1/p², so in topic 6.01's power counting a fermion line costs one power of momentum and a scalar line two. Near the mass shell write p⁰ = m + E with E and |𝐩| small against m: the numerator becomes 2m(1 + γ⁰)/2 and p² − m² ≈ 2m(E − 𝐩²/(2m)), so S_F → [(1 + γ⁰)/2] i/(E − 𝐩²/(2m) + iε): i times the free Green's function 1/(E − H₀ + iε) of Quantum Mechanics II's scattering theory, on the upper two components only. The positron pole, 2m away, survives only as corrections of order E/m, which is why non-relativistic theory never needed it.

Close the contour: electrons forward, positrons backward

Do the p⁰ integral at fixed 𝐩. The iε puts the poles at p⁰ = Eₚ − iε′ and −Eₚ + iε′. For positive t = x⁰ − y⁰, e^(−ip⁰t) dies in the lower half-plane: close below and keep +Eₚ, where p̸ + m = Σuū. For negative t close above and keep −Eₚ, where p̸ + m = −(Eₚγ⁰ + γ⋅𝐩 − m) = −Σvv̄ at momentum −𝐩. So S_F(t, 𝐩)γ⁰ = θ(t)Λ₊e^(−iEₚt) − θ(−t)Λ₋e^(iEₚt), derived, with Λ± = (Eₚ ± H)/(2Eₚ). Read it as amplitudes: if x⁰ is later, ψ̄(y) creates an electron and ψ(x) removes it; if y⁰ is later, ψ(x) creates a positron and ψ̄(y) removes it. The fermion-number arrow runs from y to x either way and every quantum carries +Eₚ; backward in time describes the negative-frequency wave, not a particle. At t = 0 the function jumps by Λ₊ + Λ₋ = 1, the equal-time [ψ, ψ†]₊ = δ³ that supplies the iδ⁴. At |𝐩| = 1.00 MeV a share (Eₚ − m)/(2Eₚ) = 0.272 of the upper component runs backward. One pole serves both charges, so the masses must agree: measured equal to 8 parts in 10⁹ (quoted).

The retarded function, and why perturbation theory takes S_F

The Dirac operator has other Green's functions. The retarded one is built from the anticommutator, S_R(x − y) = θ(x⁰ − y⁰)⟨0|[ψ(x), ψ̄(y)]₊|0⟩ = (i∂̸ₓ + m)D_R(x − y), with topic 1.07's D_R; the anticommutator is a c-number, so the vacuum average changes nothing. Both poles now lie below the axis, S_R(p) = i(p̸ + m)/((p⁰ + iε)² − 𝐩² − m²), and S_R(t, 𝐩)γ⁰ = θ(t)[Λ₊e^(−iEₚt) + Λ₋e^(iEₚt)]: both frequencies forward and nothing before the source, the one-particle Dirac wave whose negative-energy half hole theory had to fill. It obeys the same equation, and S_F − S_R = (i∂̸ₓ + m)D(y − x) solves the free one, derived. Perturbation theory takes S_F because Dyson's series is time-ordered, so Wick's theorem (topic 7.02) contracts ψ with ψ̄ into exactly this function; S_R belongs to problems where a source launches a wave and we follow it forward, as in linear response. Beyond free fields: in interacting Yukawa theory, where every particle is massive, ⟨Ω|Tψ(x)ψ̄(y)|Ω⟩ keeps a pole, now at the physical mass with residue Z₂ ≤ 1, and gains a multiparticle continuum; that residue is topic 7.03's.

Change one variable at a time

Make the relationship visible.

Interactive model

Electron momentum |𝐩|1.00 MeV

Ordering sign σ (−1: fermions)−1

At σ = −1 and p = 0 the upper component is pure electron: nothing before t = 0, and S_F matches S_R. Raise p to 2 MeV: a positron share (Eₚ − m)/(2Eₚ) runs into negative t in S_F, while S_R sends both frequencies forward. Then set σ = +1: the jump falls to m/Eₚ and moves with p.

Eₚ = √(𝐩² + m²)1.123 MeV

PERIOD 2πħ/Eₚ3.68 zs

WEIGHT FOR NEGATIVE t, σa₋-0.272

JUMP OF Re AT t = 01.000

Live interpretationEₚ = √(𝐩² + m²): 1.123 MeV. PERIOD 2πħ/Eₚ: 3.68 zs. WEIGHT FOR NEGATIVE t, σa₋: −0.272. JUMP OF Re AT t = 0: 1.000

Catch the common trap

Explain before calculating.

Misconception check

The usual slip is to let the ordering's minus sign survive into the answer, writing S_F = θ(x⁰ − y⁰)(i∂̸ + m)D(x − y) − θ(y⁰ − x⁰)(i∂̸ + m)D(y − x) because fermions are antisymmetric. That forgets a second minus. When ψ̄(y) stands on the left, the surviving term is the positron spin sum Σvv̄ = p̸ − m multiplying e^{(+ip⋅(x−y))}, on which i∂̸ₓ gives −p̸, so ⟨0|ψ̄(y)ψ(x)|0⟩ = −(i∂̸ₓ + m)D(y − x). The two minus signs cancel and the time orders join with a plus into (i∂̸ + m)D_F. Keep only one of them and you have the vacuum function of the unsigned ordering, i(p̸ + m)p⁰/(Eₚ(p² − m² + iε)) − iγ⁰/Eₚ in momentum space: it depends on p⁰ and Eₚ separately, so it changes from frame to frame, and it is not a Green's function of i∂̸ − m. The antisymmetry is real, but it is spent on making the function covariant, not on a visible relative sign.

Why must the time-ordered product of two Dirac fields carry a minus sign when it reverses their order?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyInside a diagram an electron line carries p⁰ = 0.5200 MeV and three-momentum 0.0600 MeV along z. Find how far it is off shell, write S_F(p) as coefficients of 1, γ⁰ and γ³, check that (p̸ − m)S_F(p) = i, and compare its upper-left element in the Dirac representation with the Schrödinger propagator i/(E − 𝐩²/(2m)), where E = p⁰ − m. Take m = 0.511 MeV.

p² = 0.5200² − 0.0600² = 0.2704 − 0.0036 = 0.2668 MeV², so p² − m² = 0.2668 − 0.261121 = 0.005679 MeV². It is not zero, so the line is virtual: p⁰ sits 0.00549 MeV above the on-shell energy Eₚ = 0.51451 MeV.
With p̸ = γ⁰p⁰ − γ³p_z, S_F(p) = i(p⁰γ⁰ − p_zγ³ + m)/(p² − m²). The coefficient of 1 is i × 0.511/0.005679 = 89.98i MeV⁻¹, of γ⁰ it is i × 0.5200/0.005679 = 91.57i MeV⁻¹, and of γ³ it is −i × 0.0600/0.005679 = −10.57i MeV⁻¹. Mass dimension −1, as a fermion line needs.
Check: γ^μγ^ν + γ^νγ^μ = 2g^μν makes p̸p̸ = p², so (p̸ − m)(p̸ + m) = p² − m² and (p̸ − m)S_F(p) = i(p² − m²)/(p² − m²) = i, the momentum-space form of (i∂̸ − m)S_F = iδ⁴.
In the Dirac representation γ⁰ = diag(1, 1, −1, −1) and γ³ has no diagonal entries, so the upper-left element is i(p⁰ + m)/(p² − m²) = 1.031i/0.005679 = 181.5i MeV⁻¹, while the lower-right one is i(m − p⁰)/(p² − m²) = −1.585i MeV⁻¹, over a hundred times smaller.
Non-relativistic limit: E = 0.0090 MeV and 𝐩²/(2m) = 0.0036/1.022 = 0.003523 MeV, so i/(E − 𝐩²/(2m)) = i/0.005477 = 182.6i MeV⁻¹. The relativistic element is 0.56% lower, a correction of order E/m that vanishes as the line comes to rest.

Answerp² − m² = 0.005679 MeV²; S_F(p) = (89.98 + 91.57γ⁰ − 10.57γ³)i MeV⁻¹; its upper-left element, 181.5i MeV⁻¹, agrees with the Schrödinger value 182.6i MeV⁻¹ to 0.56%.

MediumAn electron line has three-momentum 0.400 MeV along z. Do the p⁰ integral of S_F(p)e^(−ip⁰t) by closing the contour, find the upper spin-up element G(t) = [S_F(t, 𝐩)γ⁰]₁₁ in the Dirac representation for positive and negative t, evaluate it at t = +3.00 zs and t = −3.00 zs, and check its jump at t = 0. Use ħ = 0.65821 MeV zs, where 1 zs = 10⁻²¹ s, and m = 0.511 MeV.

Eₚ = √(0.400² + 0.511²) MeV = √0.42112 MeV = 0.64894 MeV. The iε puts the poles of i(p̸ + m)/((p⁰)² − Eₚ² + iε) at p⁰ = Eₚ − iε′ and −Eₚ + iε′.
Positive t: e^(−ip⁰t) decays for negative Im p⁰, so close below, clockwise, around +Eₚ. The residue there is i(p̸ + m)/(2Eₚ) = iΛ₊γ⁰, and −2πi/(2π) times it gives Λ₊γ⁰e^(−iEₚt). Negative t: close above, anticlockwise, around −Eₚ, where p̸ + m = −2EₚΛ₋γ⁰, the residue is iΛ₋γ⁰, and +2πi/(2π) times it gives −Λ₋γ⁰e^(iEₚt).
Multiply by γ⁰ and take the upper-left entry. α_z has no diagonal entries and β = diag(1, 1, −1, −1), so (Λ₊)₁₁ = (Eₚ + m)/(2Eₚ) = 1.15994/1.29788 = 0.8937 and (Λ₋)₁₁ = (Eₚ − m)/(2Eₚ) = 0.1063. Hence G(t) = 0.8937e^(−iEₚt) for positive t and −0.1063e^(iEₚt) for negative t.
Restore ħ: Eₚ|t|/ħ = 0.64894 × 3.00/0.65821 = 2.9577 rad, with cosine −0.98315 and sine 0.18283. At t = +3.00 zs, G = 0.8937(−0.98315 − 0.18283i) = −0.8787 − 0.1634i. At t = −3.00 zs, e^(iEₚt/ħ) = −0.98315 − 0.18283i, so G = −0.1063(−0.98315 − 0.18283i) = +0.1045 + 0.0194i.
Checks: the jump at t = 0 is 0.8937 + 0.1063 = 1.000, the equal-time anticommutator, at any momentum. G is dimensionless, as S_F(t, 𝐩) carries mass dimension 3 − 3 = 0. At rest the backward share would be 0, and for |𝐩| ≫ m it tends to ½. The retarded function instead gives 0 at −3.00 zs and cos − i(m/Eₚ)sin = −0.9831 − 0.1440i at +3.00 zs.

AnswerG(t) = 0.894e^{(−iEₚt/ħ)} for positive t and −0.106e^(iEₚt/ħ) for negative t; G(+3.00 zs) = −0.879 − 0.163i and G(−3.00 zs) = +0.104 + 0.019i, with a jump of exactly 1 at t = 0.

HardSuppose Dirac fields were time-ordered without the sign, T₊ψ(x)ψ̄(y) = θ(x⁰ − y⁰)ψ(x)ψ̄(y) + θ(y⁰ − x⁰)ψ̄(y)ψ(x). Find its vacuum function S₊(p) in momentum space. Then evaluate the scalar part, ¼ tr, of S₊ and of S_F for an internal electron with p = (1.000, 0, 0, 0) MeV in frame S and in a frame S′ moving at 0.600c along +z, and say which is a Lorentz scalar.

The two vacuum pieces are those of S_F; only the sign joining them changes, so S₊(t, 𝐩)γ⁰ = θ(t)Λ₊e^(−iEₚt) + θ(−t)Λ₋e^(iEₚt). Transforming with ∫dt e^(ip⁰t), θ(t)e^(−iEₚt) gives i/(p⁰ − Eₚ + iε) and θ(−t)e^(iEₚt) gives −i/(p⁰ + Eₚ − iε).
With Λ± = (Eₚ ± H)/(2Eₚ): S₊γ⁰ = iΛ₊/(p⁰ − Eₚ) − iΛ₋/(p⁰ + Eₚ) = i(Eₚ² + p⁰H)/(Eₚ(p² − m²)), so S₊(p) = i(p̸ + m)p⁰/(Eₚ(p² − m² + iε)) − iγ⁰/Eₚ. With the sign, the same steps give i(p⁰ + H)/(p² − m²), which is S_F(p)γ⁰.
Scalar parts: tr γ^μ = 0 and tr 1 = 4, so ¼ tr S_F = im/(p² − m²) and ¼ tr S₊ = imp⁰/(Eₚ(p² − m²)). A covariant function obeys S_F(Λp) = M S_F(p) M⁻¹, with M the boost's spinor matrix, and a trace cannot change under that.
Frame S: p² = 1.000 MeV², p² − m² = 0.73888 MeV² and Eₚ = m = 0.511 MeV. Then ¼ tr S_F = 0.511i/0.73888 = 0.6916i MeV⁻¹ and ¼ tr S₊ = 0.6916i × 1.000/0.511 = 1.3534i MeV⁻¹.
Frame S′: γ = 1.25, so p′⁰ = 1.250 MeV and p′_z = −0.750 MeV, with p′² still 1.000 MeV². Eₚ′ = √(0.5625 + 0.261121) MeV = 0.90754 MeV. ¼ tr S_F = 0.6916i MeV⁻¹ again, but ¼ tr S₊ = 0.6916i × 1.250/0.90754 = 0.9526i MeV⁻¹.
Checks: every value is in MeV⁻¹. Near p⁰ = +Eₚ the factor p⁰/Eₚ → 1, so S₊ and S_F share the electron pole's residue; at −Eₚ it → −1 and the residues are opposite, which is exactly the dropped sign. The frames disagree by a factor 1.421 for S₊ and not at all for S_F.

AnswerS₊(p) = i(p̸ + m)p⁰/(Eₚ(p² − m² + iε)) − iγ⁰/Eₚ. ¼ tr S_F = 0.692i MeV⁻¹ in both frames, a scalar; ¼ tr S₊ is 1.353i MeV⁻¹ in S and 0.953i MeV⁻¹ in S′, so without the sign the propagator depends on the frame.