University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.5
Fermion signs between diagrams and scalar attraction
This is where fermion signs stop being bookkeeping and become physics. You fix every sign in ψψ → ψψ and ψψ̄ → ψψ̄ by counting swaps of anticommuting operators, watch the minus between the t- and u-channel diagrams enforce Pauli exclusion, and carry the amplitude down to a potential you can quote in MeV at a distance in fm.
Build the model
Connect the measurement to the mechanism.
Fermions bring two signs that scalar field theory never had, and this topic shows that both are physics rather than convention. The first sits between diagrams. In ψψ → ψψ the t- and u-channel exchanges differ only in which outgoing fermion leaves which vertex, and turning one pairing of anticommuting operators into the other is a single transposition, so the two diagrams enter with opposite signs and the amplitude changes sign when the final fermions are swapped: the Pauli principle, produced by Wick's theorem instead of imposed on a wavefunction.
The second is the sign of the force. As |p|/m → 0, ū(p′, s′)u(p, s) → 2m δₛₛ′, the t-channel amplitude becomes 4m²g²/(|q|² + mφ²), and matching it to Quantum Mechanics II's Born amplitude gives V(r) = −(g²/4π) exp(−mφ r)/r: attraction. For a fermion and an antifermion v̄v = −2m looks as if it should turn this into repulsion, but the order of the b and d operators supplies a second −1 and the two cancel, so that pair attracts too, and so do two antifermions.
The picture is tree level and non-relativistic: it needs |p| ≪ m and, for the Born step, αY m/mφ ≪ 1 with αY = g²/4π. Beyond that, spin-orbit terms of relative order |p|²/m², the ladder of repeated exchanges that builds bound states, and the s-channel pole at s = mφ² take over, and none of them is a single static potential.
- Simple definition
- Two Feynman diagrams that differ only by exchanging two identical external fermions enter with opposite signs, because turning one pairing of anticommuting operators into the other takes a single transposition.
- Example
- For ψψ → ψψ with both spins up along the beam at |p| = 0.3 GeV and 90° in the centre of mass, both diagrams carry momentum transfer √2 × 0.3 = 0.42 GeV, so t = u = −0.18 GeV², the four spinor products are equal, and the two terms cancel: M = 0 exactly.
Only signs relative to the identity part of S are physical. Reorder one external state and every term flips together, the no-scattering term included.
N, a pure count, is the number of inversions that make every contracted pair adjacent: 6 for the t-channel, 5 for the u-channel
Swapping (p₃, s₃) with (p₄, s₄) swaps t with u and one term with the other, so M → −M. For equal spins along the beam at 90°, where t = u, the amplitude vanishes.
g dimensionless; t = (p₁ − p₃)², u = (p₁ − p₄)² and mφ² in GeV²; each spinor product ūᵢuⱼ in GeV, so M is dimensionless
The scalar density weighs each slow fermion as 2m and ignores its spin, so the leading potential is spin-independent.
m the fermion mass and p, p′ the three-momenta, all in GeV; the spin-flip piece −iσ⋅(p′ × p)/(4m²) enters at relative order |p|²/m²
A positive tree amplitude means a negative Ṽ, which is attraction. The 4m² is relativistic normalisation, cancelled by the two factors ūu = 2m.
From f = M/(8π√s) with √s → 2m and f = −(mᵣ/2π)Ṽ with mᵣ = m/2; Ṽ in GeV⁻², q the momentum transfer in GeV
The range is set by the scalar's mass alone. As mφ → 0 it becomes −αY/r, the Coulomb shape but attractive between like particles.
In laboratory units V = −αY ħc e(−r/λ)/r with αY = g²/4π and λ = ħ/(mφc) = 197.327 MeV fm ÷ mφc²
The pair attracts exactly as two fermions do. Two antifermions, with eight inversions and a factor (−2m)², attract as well.
The − is the ordering sign, from seven inversions; with v̄₂v₄ → −2m δₛ₂ₛ₄ it gives Mₜ → +4m²g²/(|q|² + mφ²), the ψψ value
Order the operators against the external states
Work in natural units, ħ = c = 1, with metric (+, −, −, −). Take identical fermions in, |p₁s₁, p₂s₂⟩ = b†(p₁, s₁)b†(p₂, s₂)|0⟩, and out, ⟨p₃s₃, p₄s₄| = ⟨0|b(p₄, s₄)b(p₃, s₃), the adjoint of the ket built in the same order. At order g², with the two φ fields contracted into the scalar propagator, the fermion operators to pair are b₄ b₃ (ψ̄ψ)ₓ (ψ̄ψ)y b†₁ b†₂; number them 1 to 8 from the left. A b against a ψ̄ gives ū and a ψ against a b† gives u (topic 7.03, quoted), and Wick's theorem for fermions (topic 7.02) weights each pairing by (−1)N, N being the inversions in the permutation that makes every contracted pair adjacent. The direct pairing, b₃ψ̄ₓ, ψₓ b†₁, b₄ψ̄y, ψy b†₂, is the order 2 3 4 7 1 5 6 8: six inversions, sign +. The exchange pairing, b₄ with ψ̄ₓ and b₃ with ψ̄y, is 1 3 4 7 2 5 6 8: five inversions, sign −. Swapping the labels x and y repeats each pairing once more, which cancels the 1/2! of the Dyson series.
A relative minus sign makes the amplitude antisymmetric
With vertex −ig and scalar propagator i/(q² − mφ²) (topic 7.04, quoted), the two pairings give M = g²[ū₃u₁ ū₄u₂/(mφ² − t) − ū₄u₁ ū₃u₂/(mφ² − u)], where ūᵢ = ū(pᵢ, sᵢ), t = (p₁ − p₃)² and the Mandelstam u = (p₁ − p₄)². Swap the outgoing labels, (p₃, s₃) ↔ (p₄, s₄): t and u trade places, so do the two terms, and M → −M. The amplitude is antisymmetric in the identical final fermions, derived, and for equal spins along the beam at 90° in the centre of mass, where t = u and the four spinor products are equal, it vanishes. The overall sign follows the ordering of the states: build the in-state as b†₂b†₁ and both terms flip, but so does the identity part of S, ⟨p₃p₄|p₁p₂⟩, so the sign relative to no scattering is fixed once in- and out-states are ordered alike. A field quantised with commutators would put a plus between the diagrams, and UAQM-09.06 showed such a Dirac field has negative norms or no lowest energy. The measured trace of the minus: low-energy pp and nn scattering has no ³S₁ wave, the channel in which np binds as the deuteron.
At low speed ū u weighs each fermion as 2m
For elastic scattering in the centre of mass |p′| = |p|, and with UAQM-04's Dirac-representation spinors, quoted, the bilinear is exactly ū(p′, s′)u(p, s) = ξ′†[(E + m) − (p′⋅p + iσ⋅(p′ × p))/(E + m)]ξ, with ξ and ξ′ the two-component spin states. At p = p′ = 0 it is 2m ξ′†ξ = 2m δₛₛ′, and every correction carries |p|²/m². For m = 1.0 GeV, |p| = 0.2 GeV and θ = 60° it gives 2.0099 GeV without a spin flip, against 2m = 2 GeV, and 0.0172 GeV in size between opposite spins along the beam: the spin-orbit piece is under 1%. An elastic centre-of-mass collision transfers no energy, q⁰ = 0, so t = −|q|² and the direct term becomes Mₜ = 4m²g² δₛ₁ₛ₃ δₛ₂ₛ₄/(|q|² + mφ²), derived: positive, spin-independent, and set by |q| alone. The u-channel term is the same function of |p₁ − p₄|, with the minus sign and the spin labels crossed.
Match to the Born amplitude and read off the potential
Topic 5.06 showed that f = M/(8π√s) is the amplitude with dσ/dΩ = |f|², and the first Born approximation (UQMII-15.08, quoted) gives f = −(mᵣ/2π)Ṽ(q), Ṽ being the Fourier transform of V. With √s → 2m and reduced mass mᵣ = m/2 the direct term gives Ṽ(q) = −Mₜ/(4m²) = −g²/(|q|² + mφ²), and the transform of 1/(|q|² + mφ²), quoted, gives V(r) = −(g²/4π) exp(−mφ r)/r, derived: attraction, as topic 5.06 found for scalars. Dimensions: [ψ] = 3/2 and [φ] = 1 make g dimensionless, so g²/(4πr) is a mass, an energy. Topic 5.06's cubic scalar coupling carried mass dimension and left 1/(4MM′) in V; here the two factors ūu = 2m cancel the 1/(4m²) of relativistic normalisation instead. As mφ → 0 the potential becomes −αY/r with αY = g²/4π, long-ranged like Coulomb but attractive between like particles; restoring units, the range ħ/(mφc) is 0.395 fm for mφ = 500 MeV. The match needs |p| ≪ m and, for Born, αY mᵣ/mφ ≪ 1.
Fermion and antifermion: two minus signs that cancel
For ψ(p₁)ψ̄(p₂) → ψ(p₃)ψ̄(p₄) take |in⟩ = b†₁d†₂|0⟩ and ⟨out| = ⟨0|d₄b₃. In the t-channel b₃ψ̄ₓ and ψₓ b†₁ give ū₃ and u₁ as before, while d₄ψy gives v₄ and ψ̄y d†₂ gives v̄₂, so the second vertex carries v̄₂v₄, read against the antifermion's arrow. The pairing order is 2 3 4 7 1 6 5 8: seven inversions, sign −. At low speed v̄(p, s)v(p′, s′) → −2m δₛₛ′ (UAQM-04, quoted), and (−1) × (2m) × (−2m) = +4m², the two-fermion value, so Mₜ and V(r) are unchanged: the pair attracts, derived. Two antifermions give eight inversions and (−2m)², and attract too. The annihilation diagram ψψ̄ → φ → ψψ̄ carries v̄(p₂)u(p₁), which in the centre of mass is −2η†(σ⋅p)ξ in the Dirac representation: zero for a pair at rest, because a scalar makes a fermion pair only in a P wave (topic 7.07). Away from its pole at s = mφ² it corrects the potential only at order |p|²/m².
Why scalar exchange attracts every pair
Read the result through the source. φ couples to the scalar density ψ̄ψ, which is even under charge conjugation (UAQM-08, quoted): for one slow antifermion ⟨p̄|:ψ̄ψ:|p̄⟩ = −v̄v = +2m, normal ordering d d† into −d†d undoing the sign of v̄v just as the operator ordering did above. Fermion and antifermion carry the same scalar charge, the amplitude goes as g² times positive weights, and a spin-0 boson exchanged between like charges attracts. A vector is different (Unit 8): ψ̄γ⁰ψ counts particles minus antiparticles, v̄γ⁰v = +2m leaves the ordering sign uncancelled, and the force reverses between ψψ and ψψ̄, so like charges repel and opposite charges attract, quoted. The general rule, quoted, is that even-spin exchange attracts like sources and odd-spin exchange repels them, which is why gravity's spin 2 attracts all mass. Two different species with couplings g₁ and g₂ feel −g₁g₂ exp(−mφ r)/(4πr), so a scalar repels only when the two couplings have opposite signs.
Change one variable at a time
Make the relationship visible.
Leave the sign at −1, the value anticommuting fields force: the solid curve is odd about 90° and crosses zero there. Drag |p|/mφ down to 0.1 and it flattens onto zero, since equal-spin fermions have no s-wave; set the sign to +1, the commutator answer, and the 90° value is twice the t term instead.
t TERM AT 90°0.333
SUM AT 90°0.000
SUM AT θ = 0°0.800
SUM AT θ = 180°-0.800
Live interpretationt TERM AT 90°: 0.333. SUM AT 90°: 0.000. SUM AT θ = 0°: 0.800. SUM AT θ = 180°: −0.800
Catch the common trap
Explain before calculating.
In ψψ̄ → ψψ̄ by scalar exchange, with in- and out-states ordered alike as b†d†|0⟩, the t-channel term carries v̄(p₂)v(p₄), which tends to −2m at low speed, while the ψψ direct term carries ū(p₄)u(p₂) → +2m. What decides whether the fermion–antifermion pair attracts?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA heavy Dirac fermion couples to a scalar of mass mφ = 400 MeV through −g φ ψ̄ψ with g = 2.0. Using the low-speed potential derived here, find its range in fm and V(r) at 0.50 fm and 1.0 fm in MeV, and find the separation beyond which the Coulomb repulsion of two unit charges, α ħc/r, outweighs it.
- Dimensions: with [ψ] = 3/2 and [φ] = 1, the term g φ ψ̄ψ has mass dimension 4 only if g is dimensionless, so αY = g²/4π = 4.0/12.566 = 0.318 is a pure number and αY ħc/r is an energy.
- Range: λ = ħc/(mφc²) = 197.327 MeV fm ÷ 400 MeV = 0.493 fm.
- At r = 0.50 fm: αY ħc/r = 0.3183 × 197.327 MeV fm ÷ 0.50 fm = 125.6 MeV, and e(−r/λ) = e(−1.014) = 0.363, so V = −45.6 MeV.
- At r = 1.0 fm: 62.8 MeV × e(−2.027) = 62.8 MeV × 0.1317 = 8.27 MeV, so V = −8.27 MeV. Limit check: with mφ → 0 it would be −62.8 MeV, so the Yukawa factor has removed 87% of it.
- Coulomb: α ħc = 1.440 MeV fm with the measured α = 1/137.036, so at 1.0 fm the scalar wins by 8.27/1.44 = 5.7. They balance where αY e(−r/λ) = α, at r = λ ln(αY/α) = 0.493 fm × ln 43.6 = 1.86 fm; at 2.0 fm the values are −0.545 MeV against +0.720 MeV.
Answerλ = 0.49 fm; V(0.50 fm) = −45.6 MeV and V(1.0 fm) = −8.3 MeV; beyond 1.86 fm the unit-charge Coulomb repulsion is larger. The same V acts between ψ and ψ̄.
MediumIn a Yukawa model with m = 1.0 GeV, mφ = 0.5 GeV and g = 0.5, two identical fermions scatter at centre-of-mass momentum |p| = 0.2 GeV, all spins up along the beam before and after. Find the t- and u-channel amplitudes and their sum at θ = 60° at low speed, then dσ/dΩ in nb sr⁻¹, compare with what a plus sign between the diagrams would give, and test the low-speed answer against the exact spinor products.
- Regime: |p|/m = 0.2, so each spinor product sits within about |p|²/m² = 4% of 2m, and αY mᵣ/mφ = (0.25/4π) × 0.5/0.5 = 0.020 ≪ 1, so the Born-level tree is enough.
- Momentum transfers: |q|² = 4|p|² sin²30° = 0.040 GeV² for the t-channel and |p₁ − p₄|² = 4|p|² cos²30° = 0.120 GeV² for the u-channel; mφ² = 0.25 GeV².
- All spins equal, so both Kronecker deltas are 1: Mₜ = 4m²g²/(|q|² + mφ²) = 1.00/0.29 = 3.448 and Mᵤ = −1.00/0.37 = −2.703, so M = 0.746. All three are dimensionless, as a 2 → 2 amplitude must be.
- f = M/(8π√s) with √s ≈ 2m = 2.0 GeV gives f = 0.746/50.27 GeV = 0.0148 GeV⁻¹, so dσ/dΩ = f² = 2.20 × 10⁻⁴ GeV⁻² sr⁻¹; with 1 GeV⁻² = 0.3894 mb that is 8.6 × 10⁻⁵ mb sr⁻¹ = 86 nb sr⁻¹.
- Comparisons: a plus between the diagrams gives M = 6.151 and (6.151/0.746)² = 68 times the rate; spins ↑↓ → ↑↓ keep only the t-term, (3.448/0.746)² = 21 times. At 120° the sum is −0.746, the antisymmetry M(π − θ) = −M(θ).
- Exact check: for spins along the beam each product is (E + m) − |p|² cos θᵢⱼ/(E + m) with E = 1.0198 GeV, so 2.0099 GeV for the two 60° pairs and 2.0297 GeV for the two 120° pairs. Then Mₜ = 3.483, Mᵤ = −2.784 and M = 0.699, and with √s = 2E = 2.040 GeV, dσ/dΩ = 1.86 × 10⁻⁴ GeV⁻² sr⁻¹ = 72 nb sr⁻¹. Each term moves by only 1% or 3%, but the near-cancellation magnifies that to 7% in M, so the low-speed 86 nb sr⁻¹ is about 18% high.
AnswerAt low speed M = 3.448 − 2.703 = 0.746 and dσ/dΩ ≈ 86 nb sr⁻¹ (8.6 × 10⁻⁵ mb sr⁻¹); the exact spinor products give M = 0.699 and 72 nb sr⁻¹, the near-cancellation magnifying the |p|²/m² corrections to 18%; a plus sign between the diagrams would give 68 times as much.
HardSame model (m = 1.0 GeV, mφ = 0.5 GeV, g = 0.5), now ψ(p₁)ψ̄(p₂) → ψ(p₃)ψ̄(p₄) at |p| = 0.2 GeV and θ = 60°, with |in⟩ = b†₁d†₂|0⟩ and ⟨out| = ⟨0|d₄b₃. Fix the sign of the t-channel term, evaluate it at low speed, give V at 0.40 fm in MeV, bound the annihilation diagram, and say what changes for a vector coupling, where ūγ⁰u and v̄γ⁰v both tend to +2m.
- Number d₄ b₃ ψ̄ₓ ψₓ ψ̄y ψy b†₁ d†₂ as 1 to 8. The t-channel pairs (b₃ψ̄ₓ)(ψₓ b†₁)(d₄ψy)(ψ̄y d†₂) are the order 2 3 4 7 1 6 5 8, with inversions 1 + 1 + 1 + 3 + 0 + 1 + 0 + 0 = 7: sign −1.
- At low speed ū₃u₁ → 2m and v̄₂v₄ → −2m, so Mₜ = −g²[ū₃u₁][v̄₂v₄]/(|q|² + mφ²) = −0.25 × (2.0)(−2.0)/0.29 = +3.448, identical to the ψψ direct term; Ṽ = −Mₜ/(4m²) = −0.862 GeV⁻², an attraction.
- Units restored: αY = 0.25/4π = 0.0199 and λ = 0.197327 GeV fm ÷ 0.5 GeV = 0.395 fm, so V(0.40 fm) = −0.0199 × 197.327 MeV fm × e(−1.014)/0.40 fm = −3.56 MeV.
- Annihilation: the s-channel term carries [v̄₂u₁][ū₃v₄]/(s − mφ²), and in the Dirac representation v̄(−p)u(p) = −2η†(σ⋅p)ξ, at most 2|p| = 0.4 GeV in size. With s = 4(m² + |p|²) = 4.16 GeV², |Mₛ| ≤ 0.25 × 0.4 × 0.4/(4.16 − 0.25) = 0.0102, 0.3% of Mₜ, and it vanishes as |p| → 0.
- Vector contrast: the ordering sign is the same −1, but v̄γ⁰v → +2m, so the ratio of the ψψ̄ to the ψψ t-term is (−1)(+2m)/(2m) = −1, against (−1)(−2m)/(2m) = +1 for the scalar: a vector force reverses between the two pairs, the scalar force does not.
AnswerSign −1 from seven inversions; Mₜ = +3.45, the ψψ value, so the pair attracts with V(0.40 fm) = −3.56 MeV; the annihilation term is under 0.3% of Mₜ; a vector coupling would make the ψψ̄ force opposite to the ψψ one.