University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.3
LSZ for Dirac fields: external spinors u, v, ū and v̄
Every fermion amplitude in this course begins and ends on a spinor, and this lesson earns them. Rerun topic 4.04's reduction with the Dirac operator and each external propagator collapses to its numerator; at the pole that numerator is a sum over two spins, so one spinor stays in the amplitude and its partner names the state.
Build the model
Connect the measurement to the mechanism.
Topic 4.04 made an S-matrix element the residue of a correlation function at its external poles, with √Z per leg. A Dirac field changes two things. Its pole is a matrix: near p² = m² the exact propagator is iZ₂(p̸ + m)/(p² − m²), and on shell p̸ + m is Σₛ u ū, a sum of outer products over the two spin states.
And its reduction operator is first order: a projection with ūγ⁰ pulls b out of ψ, and the projection's time derivative becomes ū(p, s)(i∂̸ − m)ψ, a square root of □ + m². Meeting the pole, (p̸ − m)(p̸ + m) = p² − m² cancels the denominator and leaves one spinor per leg: √Z₂ u for an incoming fermion, √Z₂ ū for an outgoing one, √Z₂ v̄ for an incoming antifermion and √Z₂ v for an outgoing one. Which spinor goes where is decided by which operator in the field does the job, and contracting the interaction-picture field with the external state reproduces the same four factors with Z₂ = 1, the tree-level shortcut.
Z₂ is the one-particle weight in ρ₁, one of the two spectral functions of a spin-½ field, and the canonical anticommutator fixes ∫ρ₁ dμ² = 1 (quoted), so 0 ≤ Z₂ ≤ 1. The construction needs an isolated pole, so it fails for a fermion that decays and for an electron whose eγ continuum, the photon being massless, starts at the pole itself; QED's Z₂ waits for topic 12.05.
- Simple definition
- LSZ for Dirac fields turns each external fermion line of an amputated correlation function into √Z₂ times a plane-wave spinor: u in and ū out for a fermion, v̄ in and v out for an antifermion, Z₂ being the residue of the full propagator at the physical mass.
- Example
- An incoming fermion at rest with spin up contributes √Z₂ u = √Z₂ √(2m)(1, 0, 0, 0) in the Dirac representation, 2m being 1.022 MeV for an electron. At tree level Z₂ = 1; were Z₂ = 0.94, the four legs of ψψ̄ → ψψ̄ would multiply the amputated diagrams by 0.94² = 0.884.
Derived from b̃(t) = ∫ d³x e(ip⋅x) ūγ⁰ψ. The Dirac operator does the work of topic 4.04's □ + m², and the right side vanishes for a free field.
b̃ = √(2E) b has mass dimension −1, ψ has 3/2 and ū ½; m is the physical mass and Z₂ is dimensionless
Follows from the one-particle term of the stated spectral form. The residue is a sum of outer products of spinors, which is why amputation leaves exactly one spinor per leg.
S′ has mass dimension −1; Z₂ is fixed by ⟨Ω|ψ(0)|p, s⟩ = √Z₂ u(p, s); the remainder is regular at p² = m²
Derived: each reduction factor cancels a pole of residue Z₂ and brings 1/√Z₂. One overall sign per process comes from the order of the anticommuting operators (topic 7.05).
each spinor has mass dimension ½; p is the on-shell momentum and s the spin of the external state
Shown: the tree-level shortcut, LSZ at Z₂ = 1. ψ annihilates fermions with u and creates antifermions with v; ψ̄ creates fermions with ū and annihilates antifermions with v̄.
ψI⁺ is the b part of ψI; |p, s⟩ = √(2E) b†(p, s)|0⟩, normalised to 2E(2π)³δ³(p − q)
Derived by expanding p̸ − m₀ − Σ(p̸) about its zero. At tree level Σ = 0 and Z₂ = 1; positivity of the spectral weight demands dΣ/dp̸ ≤ 0 at the pole.
Σ(p̸) = A(p²)p̸ + B(p²) has mass dimension 1, m₀ is the Lagrangian mass; dΣ/dp̸ = A + 2m²A′ + 2mB′ is dimensionless, a prime meaning d/dp², so A′ has mass dimension −2 and B′ −1
Stated, not derived. One particle gives ρ₁ = Z₂δ(μ² − m²) and ρ₂ = mZ₂δ(μ² − m²), so the sum rule holds 0 ≤ Z₂ ≤ 1.
μ² from 0 to ∞; ρ₁ has mass dimension −2 and ρ₂ −1; ρ₁ ≥ 0, |ρ₂| ≤ μρ₁, ∫ρ₁ dμ² = 1; parity conserved
A projection that isolates b
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 4.04 pulled aᵢₙ and aₒᵤₜ out of a scalar field with the Klein-Gordon product; the Dirac field is first order in time, so a spatial projection will do, once its spinor kills the antiparticle half. Define b̃(p, s; t) = ∫ d³x e(ip⋅x) ū(p, s)γ⁰ψ(x). For UAQM-09.07's free field, ψ = ∫ d³k/((2π)³√(2Eₖ)) Σᵣ [b u e(−ik⋅x) + d† v e(ik⋅x)], the x integral sets k = p in the b term and k = −p in the d† term, and u†(p, s)u(p, r) = 2Eδᵣₛ with u†(p, s)v(−p, r) = 0, quoted from UAQM-04, leave b̃ = √(2E) b(p, s): the operator that removes |p, s⟩ = √(2E) b†|0⟩. The γ⁰ is essential. ū(p)v(−p) is not zero for a moving particle, and without γ⁰ the d†(−p) term would leak in with phase e(2iEt). With interactions b̃ depends on t, and as t → ∓∞ it tends to √Z₂ b̃ᵢₙ and √Z₂ b̃ₒᵤₜ, weak limits for topic 4.04's reason.
The Dirac operator replaces □ + m²
Write the difference of the two limits as the time integral of a derivative. ∂₀ acting on e(ip⋅x)ūγ⁰ψ gives ūγ⁰∂₀ψ + ip₀ūγ⁰ψ under the exponential. Now expand ū(i∂̸ − m)ψ, integrate its spatial derivatives by parts against e(ip⋅x), and use ū(p̸ − m) = 0 to replace ū pⱼγj by ū(m − p₀γ⁰): the integrands agree, so √Z₂[b̃ₒᵤₜ − b̃ᵢₙ] = −i ∫ d⁴x e(ip⋅x) ū(p, s)(i∂̸ − m)ψ(x), derived. For a free field the right side vanishes and nothing scatters, as it must, and since (i∂̸ + m)(i∂̸ − m) = −(□ + m²) the new operator is a square root of topic 4.04's. Trading each out operator in ⟨β out| and each in operator in |α in⟩ for such an integral, the T product keeping them in time order, puts one factor on every external fermion field of ⟨Ω|T ψ ⋯ ψ̄ ⋯|Ω⟩. An incoming fermion takes the Hermitian conjugate, −(i/√Z₂) ∫ d⁴x e(−ip⋅x) ψ̄(x)(−i∂̸← − m)u(p, s), the arrow meaning that the derivative acts on ψ̄.
Amputation leaves one spinor per leg
In momentum space an outgoing fermion's leg ends in the full propagator, iZ₂(p̸ + m)/(p² − m²) near its pole. Its reduction factor brings −(i/√Z₂)ū(p, s)(p̸ − m), and because p̸ commutes with itself, (p̸ − m)(p̸ + m) = p² − m² exactly: the pole cancels and (−i)(i)(Z₂/√Z₂) leaves √Z₂ ū(p, s), derived. An incoming fermion gives √Z₂ u(p, s) the same way. Read at the pole, the residue Z₂Σᵣ u ū splits into a spinor that names the external state and a spinor that stays in the amplitude. Antifermions use d̃† = ∫ d³x e(−ip⋅x) v̄(p, s)γ⁰ψ, and the algebra repeats with p → −p, because v̄(p̸ + m) = 0. Their legs carry momentum −p along the arrow, where the pole term is −iZ₂(p̸ − m)/(p² − m²) = −iZ₂Σᵣ v v̄/(p² − m²). The results are √Z₂ v̄(p, s) for an incoming antifermion and √Z₂ v(p, s) for an outgoing one, derived, each up to a sign that every diagram of the process shares and the order of the anticommuting operators fixes, as topic 7.05 does.
The same factors from one contraction
At tree level there is a shortcut, and it must agree. Take |p, s⟩ = √(2E) b†(p, s)|0⟩, where the anticommutator of b(k, r) with b†(p, s) is (2π)³δ³(k − p)δᵣₛ. The b part of ψI removes the particle: ψI⁺(x)|p, s⟩ = ∫ d³k/((2π)³√(2Eₖ)) Σᵣ u(k, r)e(−ik⋅x) √(2E) (2π)³δ³(k − p)δᵣₛ |0⟩ = e(−ip⋅x)u(p, s)|0⟩, shown, the √(2E) of the state cancelling the 1/√(2E) of the mode. The partners follow identically, with |p̄, s⟩ = √(2E) d†(p, s)|0⟩ the antifermion: ⟨p, s|ψ̄I(x) = ⟨0|ū(p, s)e(ip⋅x), ψ̄I(x)|p̄, s⟩ = v̄(p, s)e(−ip⋅x)|0⟩ and ⟨p̄, s|ψI(x) = ⟨0|v(p, s)e(ip⋅x). The rule is which operator does the job: ψ annihilates fermions with u and creates antifermions with v, while ψ̄ creates fermions with ū and annihilates antifermions with v̄. These are LSZ's factors at Z₂ = 1, and the exponentials become momentum conservation at the vertex. Beyond tree level the shortcut misses Z₂ and the self-energies on external legs, the defect topic 4.09 found for scalars.
Z₂ lives in the spin-½ spectral functions
Inserting a complete set of states between ψ(x) and ψ̄(0) gives, in a parity-conserving theory, S′(p) = ∫ dμ² i[ρ₁(μ²)p̸ + ρ₂(μ²)]/(p² − μ² + iε) with μ² from 0 to ∞, stated here and not derived. There are two functions because a numerator may hold p̸ and 1 in any proportion. Positivity of norms gives ρ₁ ≥ 0 and |ρ₂| ≤ μρ₁, and the equal-time anticommutator of ψ with ψ†, δ³(x − y) whenever the coupling has no derivatives, gives ∫ρ₁ dμ² = 1; both are quoted. The one-particle state, with ⟨Ω|ψ(0)|p, s⟩ = √Z₂ u(p, s), puts Z₂δ(μ² − m²) into ρ₁ and mZ₂δ(μ² − m²) into ρ₂, saturating the bound, so near p² = m² the numerator is Z₂(p̸ + m), a multiple of the projector onto the u's. Everything else is a continuum, starting in Yukawa theory at the lighter of m + mφ and 3m. Hence 0 ≤ Z₂ ≤ 1, with Z₂ = 1 only if ψ creates one fermion and nothing more, as a free field does.
Z₂ from the self-energy, and the checks
Summing self-energy insertions gives S′ = i/(p̸ − m₀ − Σ(p̸)) with Σ = A(p²)p̸ + B(p²). The pole sits where p̸ − m₀ − Σ has the eigenvalue zero on the mass shell, m = m₀ + Σ(p̸ = m); expanding about it, p̸ − m₀ − Σ ≈ (p̸ − m)(1 − dΣ/dp̸), so 1/Z₂ = 1 − dΣ/dp̸ at p̸ = m, derived, with dΣ/dp̸ = A + 2m²A′ + 2mB′ because Σ also depends on p̸ through p² = p̸². At tree level Σ = 0 and Z₂ = 1, the shortcut's value. Dimensions: u has mass dimension ½, since ū u = 2m, so an n-particle amplitude keeps topic 4.01's dimension 4 − n; a Yukawa vertex between ū and v has dimension 1, right for three particles. Non-relativistic limit: ū(p′, s′)u(p, s) → 2mδₛₛ′ (quoted), so each line supplies the 2m that relativistic normalisation needs when an amplitude is matched to a Born amplitude, as in topic 5.06, and keeps the spin; topic 7.05 builds on it. No experiment measures Z₂; what is measured, such as topic 9.06's 1 + cos²θ in e⁺e⁻ → μ⁺μ⁻, comes from these spinors traced together.
Change one variable at a time
Make the relationship visible.
Swing κ from −1 to +1: both curves move everywhere except at p⁰ = ±m, where the one-particle state pins them to 2Z₂ and 0. Lower Z₂ and the 2Z₂ pins drop with it while the zeros stay. Shrink mφ to 0.2m and the cut closes in on the pole, the gap LSZ cannot do without.
√Z₂ PER EXTERNAL LEG0.894
Z₂² FOR ψψ → ψψ0.640
γ⁰ = +1 VALUE AT p⁰ = 0.9m1.530 × m
γ⁰ = −1 VALUE AT p⁰ = 0.9m0.070 × m
Live interpretation√Z₂ PER EXTERNAL LEG: 0.894. Z₂² FOR ψψ → ψψ: 0.640. γ⁰ = +1 VALUE AT p⁰ = 0.9m: 1.530 × m. γ⁰ = −1 VALUE AT p⁰ = 0.9m: 0.070 × m
Catch the common trap
Explain before calculating.
In Yukawa theory a fermion ψ(p₁, s₁) and an antifermion ψ̄(p₂, s₂) annihilate into two scalars, ψψ̄ → φφ. Which external factors does LSZ attach to the fermion line, and where do they stand in its spinor chain?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA fermion's spectral function ρ₁ puts 6.0% of its total weight in the ψφ continuum and the rest in the one-particle pole. Find Z₂. For ψ(p₁, s₁) ψ̄(p₂, s₂) → ψ(p₃, s₃) ψ̄(p₄, s₄), write the factor LSZ attaches to each external leg and the product multiplying the amputated diagrams; repeat for ψψ̄ → φφ, taking the scalar's residue Zφ = 0.97.
- The canonical anticommutator fixes ∫ρ₁ dμ² = 1 (quoted), and the pole contributes Z₂δ(μ² − m²), so Z₂ = 1 − 0.060 = 0.940, inside the range 0 to 1 that positivity allows.
- Assign each spinor by the operator that handles the state. Incoming ψ is removed by ψ: √Z₂ u(p₁, s₁). Incoming ψ̄ is removed by ψ̄: √Z₂ v̄(p₂, s₂). Outgoing ψ is created by ψ̄: √Z₂ ū(p₃, s₃). Outgoing ψ̄ is created by ψ: √Z₂ v(p₄, s₄).
- Four fermion legs give (√Z₂)⁴ = Z₂² = 0.940² = 0.8836, multiplying the sum of connected, amputated diagrams.
- For ψψ̄ → φφ, two fermion legs and two scalar legs give (√Z₂)²(√Zφ)² = 0.940 × 0.97 = 0.912, which rounds to 0.91 at the precision of Zφ.
- Dimension check: the four spinors of the first process carry mass dimension 4 × ½ = 2, so the rest of a dimensionless four-particle amplitude must carry −2, as a scalar propagator between two dimensionless Yukawa couplings does.
AnswerZ₂ = 0.940. The factors are √Z₂ u(p₁, s₁), √Z₂ v̄(p₂, s₂), √Z₂ ū(p₃, s₃) and √Z₂ v(p₄, s₄), with product Z₂² = 0.884; for ψψ̄ → φφ the product is Z₂Zφ = 0.91.
MediumAn electron (m = 0.511 MeV) moves along +z at 0.600c. In the Dirac representation take u(p, s) = √(E + m)(χ, σ⋅p χ/(E + m)) and v(p, s) = √(E + m)(σ⋅p η/(E + m), η), quoted from UAQM-04. With χ = η = (1, 0), evaluate u†(p, s)u(p, s), u†(p, s)v(−p, s) and ū(p, s)v(−p, s), and use them to explain why the LSZ projection uses ūγ⁰ = u† rather than ū.
- γ = 1/√(1 − 0.600²) = 1.250, so E = 0.6388 MeV, |p| = γβm = 0.3833 MeV, E + m = 1.1498 MeV and |p|/(E + m) = ⅓. Then u(p, s) = √(E + m)(1, 0, ⅓, 0) and, since σ⋅(−p) = −|p|σz, v(−p, s) = √(E + m)(−⅓, 0, 1, 0).
- u†u = (E + m)(1 + 1/9) = 1.1498 × 10/9 = 1.278 MeV = 2E, the normalisation that makes ∫ d³x e(ip⋅x) u†ψ equal √(2E) b.
- u†(p, s)v(−p, s) = (E + m)(−⅓ + ⅓) = 0, and the other η gives zero as well, so the d† term, which the x integral leaves at momentum −p, drops out: the projection is exactly √(2E) b, constant in time for a free field.
- ū = u†γ⁰ reverses the lower pair: ū(p, s) = √(E + m)(1, 0, −⅓, 0), so ū(p, s)v(−p, s) = (E + m)(−⅓ − ⅓) = −2|p| = −0.767 MeV, three quarters of ū u = 2m = 1.022 MeV.
- Projecting with ū alone would therefore mix d†(−p) into the would-be annihilation operator with phase e(2iEt), oscillating with period 2π/(2E) = 2πħ/(1.278 MeV) = 3.24 × 10⁻²¹ s even without interactions. With u† the operator changes only when interactions act, which is what lets its t → ∓∞ limits define in and out operators.
- Limit: at rest |p| = 0, so ū v(−p) = −2|p| vanishes and the two projections agree; the γ⁰ matters only for a moving particle.
Answeru†u = 2E = 1.28 MeV, u†(p, s)v(−p, s) = 0 and ū(p, s)v(−p, s) = −2|p| = −0.767 MeV. Only the u† = ūγ⁰ projection removes the antifermion term, giving √(2E) b(p, s) exactly.
HardIn a model, a fermion of Lagrangian mass m₀ = 1.000 GeV has, below the ψφ threshold, the self-energy Σ(p̸) = A(p²)p̸ + B(p²) with A = −0.020 − 0.012p²/m₀² and B = 0.050m₀ + 0.010p²/m₀. Find the physical mass m and Z₂, show that the numerator of the full propagator at the pole is a multiple of p̸ + m, and say how far wrong Z₂ would be if the p² dependence of Σ were ignored.
- The pole: S′ = i/(p̸(1 − A) − (m₀ + B)), and on the mass shell p̸ has eigenvalue m, so m(1 − A(m²)) = m₀ + B(m²), that is m = (m₀ + B(m²))/(1 − A(m²)).
- Iterate from m = m₀: m = 1.0600/1.0320 = 1.0271 GeV; then m² = 1.0550 GeV² gives 1.06055/1.03266 = 1.0270 GeV, which repeats to five figures, so m = 1.0270 GeV and m² = 1.0547 GeV².
- Residue, derived: 1/Z₂ = 1 − dΣ/dp̸ at p̸ = m, with dΣ/dp̸ = A + 2m²A′ + 2mB′ because Σ depends on p̸ also through p² = p̸². In GeV units A(m²) = −0.03266, 2m²A′ = 2 × 1.0547 × (−0.012) = −0.02531 and 2mB′ = 2 × 1.0270 × 0.010 = 0.02054, so dΣ/dp̸ = −0.03743, a pure number as it must be.
- Z₂ = 1/1.03743 = 0.9639. Since dΣ/dp̸ ≤ 0, Z₂ ≤ 1, as the sum rule ∫ρ₁ dμ² = 1 requires; with every coefficient set to zero, m → m₀ and Z₂ → 1, the tree-level limit.
- Numerator: the inverse of p̸(1 − A) − (m₀ + B) is [p̸(1 − A) + (m₀ + B)]/[p²(1 − A)² − (m₀ + B)²]. At p² = m² the mass condition makes m₀ + B = m(1 − A), so the numerator is (1 − A)(p̸ + m) = 1.0327(p̸ + m), a multiple of Σₛ u ū, which is what lets amputation leave √Z₂ u.
- Ignoring the p² dependence replaces dΣ/dp̸ by A(m²) and gives Z₂ = 1/1.03266 = 0.9684, too large by 0.46%; each √Z₂ would be 0.23% too large and a four-fermion amplitude's Z₂² 0.93% too large.
Answerm = 1.027 GeV and Z₂ = 0.964, with the pole's numerator 1.033(p̸ + m). Dropping the p² dependence of Σ gives Z₂ = 0.968, 0.46% too large.