University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.6

Trace technology: Casimir's trick, contractions and γ⁵

Every rate with fermions in it sums |M|² over spins that nobody measured. This topic is the toolkit that makes that sum cheap: Casimir's trick closes each spinor chain into a trace, and a handful of identities, γ⁵ and the ε tensor among them, reduce any trace to dot products of momenta with every sign fixed once.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A fermion line in a Feynman amplitude is a spinor chain ū(p′)Γu(p), where Γ is a vertex or a string of vertices and propagator numerators. It depends on the spins chosen, but an unpolarised measurement sums |M|² over all of them, and that sum needs no spinor. Complex conjugation reverses the chain into ū(p)Γ̄u(p′), with Γ̄ = γ⁰Γ†γ⁰, and the spin sums Σ u ū = p̸ + m and Σ v v̄ = p̸ − m (UAQM-04.08, quoted) close it into Tr[Γ(p̸ + m)Γ̄(p̸′ + m′)]. That is Casimir's trick; the rest is algebra on traces with only the Clifford algebra and cyclicity.

Odd strings of gammas trace to zero, a contracted pair γ^μ ⋯ γ_μ shortens a string by two, every even trace unfolds into pairings of metrics, and γ⁵ anticommutes with every γ^μ and so kills each trace with fewer than four other gammas, turning four into −4i ε^μνρσ. The answer is a polynomial in invariants such as p⋅p′ and m², plus, when γ⁵ appears an odd number of times, an ε term that in an unpolarised two-body rate survives only by meeting a second ε, the pair becoming dot products again. The method stops in three places. It needs summed spins: a polarised measurement needs a spin projector or helicity amplitudes.

The contraction identities count four values of μ, so Unit 12 redoes them in d dimensions. And γ⁵ and ε are four-dimensional objects whose properties cannot all survive that continuation, a subtlety left to later courses.

Simple definition: Trace technology reduces a spin-summed |M|² to traces of gamma matrices by Casimir's trick, Σ|ū(p′)Γu(p)|² = Tr[Γ(p̸ + m)Γ̄(p̸′ + m′)], and then reduces those traces to dot products of momenta using the Clifford algebra alone.
Example: With Γ = 1 the trace is 4(p⋅p′ + m²) = 2(4m² − t); with Γ = iγ⁵ it is 4(p⋅p′ − m²) = −2t. For an electron, m = 0.511 MeV, at −t = 1.00 MeV² they give 4.09 MeV² and 2.00 MeV², and the 8m² = 2.09 MeV² between them is everything γ⁵ changed.

Casimir's trickΣ over s, s′ of |ū(p′, s′) Γ u(p, s)|² = Tr[Γ (p̸ + m) Γ̄ (p̸′ + m′)], with Γ̄ = γ⁰Γ†γ⁰

Derived for any 4 × 4 Γ, propagator numerators inside it included. An external v(p) enters through Σ v v̄ = p̸ − m, so each antifermion flips the sign of its mass.

p, p′: four-momenta (GeV); m, m′: masses (GeV); ū u = 2m, so for a dimensionless Γ the sum is in GeV²

Contraction around three slashesγ^μ a̸ b̸ c̸ γ_μ = −2 c̸ b̸ a̸

Derived from UAQM-04.03's γ^μ a̸ γ_μ = −2a̸, quoted, by one anticommutation per added slash, passing γ^μ a̸ b̸ γ_μ = 4a⋅b. Note the reversed order.

a, b, c: any four-vectors, a̸ = γ^ν a_ν; γ_μ = g_μν γ^ν; true in four dimensions only

Even traces by recursionTr(a̸₁a̸₂ ⋯ a̸₂ₙ) = Σ over k = 2 to 2n of (−1)^k (a₁⋅aₖ) Tr(the string without a̸₁ and a̸ₖ)

Derived by carrying a̸₁ round the string and using cyclicity. Six slashes give 5 × 3 = 15 products of dot products, eight give 105; odd strings trace to zero.

a₁ to a₂ₙ: four-vectors; Tr 1 = 4; its four-slash case 4[(a⋅b)(c⋅d) − (a⋅c)(b⋅d) + (a⋅d)(b⋅c)] is UAQM-04.03's

A trace with γ⁵ and four gammasTr(γ⁵ γ^μ γ^ν γ^ρ γ^σ) = −4i ε^μνρσ

Derived. With fewer gammas the trace vanishes: Tr γ⁵ = Tr(γ⁵ γ^μ γ^ν) = 0. With ε⁰¹²³ = −1 and the same γ⁵ it reads +4i: check conventions before borrowing a result.

ε^μνρσ: totally antisymmetric, ε⁰¹²³ = +1, so ε₀₁₂₃ = −1; γ⁵ = iγ⁰γ¹γ²γ³ (UAQM-05.05); all dimensionless

Contracting two ε tensorsε^μνρσ ε_μναβ = −2(δ^ρ_α δ^σ_β − δ^ρ_β δ^σ_α)

Derived. The minus signs are Minkowski's, from ε₀₁₂₃ = −1. Two ε's from two γ⁵ traces always end here, and the pair becomes ordinary dot products.

δ^ρ_α: the unit matrix on indices; one more contraction gives −6δ^σ_β, a full one −24; dimensionless

The vector spin-sum tensorTr[γ^μ (p̸ + m) γ^ν (p̸′ + m)] = 4[p^μ p′^ν + p^ν p′^μ − g^μν (p⋅p′ − m²)]

Derived. Its 00 component with E′ = E is UAQM-05.09's Mott trace, 8E²(1 − β² sin²(θ/2)); Unit 9 contracts two such tensors for e⁺e⁻ → μ⁺μ⁻.

Casimir's trick with γ^μ at one vertex and γ^ν at its barred partner; p, p′ and m in GeV, the tensor in GeV²

Casimir's trick, for any chain

Units are natural, ħ = c = 1, with the metric (+, −, −, −), and spinors keep Advanced QM's ū u = 2m, so ū(p′)Γu(p) is in GeV for a dimensionless Γ. Reading a fermion line against its arrow gives a chain ū(p′)Γu(p), where Γ may be one vertex or a string such as γ^ν(k̸ + m)γ^μ with a propagator numerator inside. Its complex conjugate is a reversed chain: [ū′Γu]* = u†Γ†γ⁰u′ = ū(γ⁰Γ†γ⁰)u′ = ūΓ̄u′, using γ⁰γ⁰ = 1. Barring reverses a string and conjugates its numbers, and since γ⁰γ^μ†γ⁰ = γ^μ (UAQM-04.03, quoted), the bar of γ^ν(k̸ + m)γ^μ is γ^μ(k̸ + m)γ^ν. With indices written out, ū′ₐ Γ_ab u_b ū_c Γ̄_cd u′_d contains (u ū)_bc and (u′ū′)_da, and UAQM-04.08's Σ u ū = p̸ + m, quoted, closes them into Tr[Γ(p̸ + m)Γ̄(p̸′ + m′)], derived. An external antifermion uses Σ v v̄ = p̸ − m instead. UAQM-04.08 met this for one vertex; here it is the general rule: two fermion lines give a product of two traces, and a closed fermion loop is a trace already, carrying topic 7.04's −1.

Contract first, then trace

UAQM-04.03 derived γ^μ γ_μ = 4 and γ^μ a̸ γ_μ = −2a̸, quoted; with a̸ replaced by γ^ν it reads γ^μ γ^ν γ_μ = −2γ^ν. Each longer contraction costs one more anticommutation, c̸γ_μ = 2c_μ − γ_μ c̸. Two slashes: γ^μ a̸ b̸ γ_μ = 2b̸a̸ − γ^μ a̸ γ_μ b̸ = 2b̸a̸ + 2a̸b̸ = 4a⋅b. Three: γ^μ a̸ b̸ c̸ γ_μ = 2c̸a̸b̸ − 4(a⋅b)c̸, and writing a̸b̸ = 2a⋅b − b̸a̸ turns this into −2c̸b̸a̸, with the order reversed. All derived, and all four-dimensional: the 4 in γ^μ γ_μ counts the values of μ, and in Unit 12's d dimensions γ^μ a̸ γ_μ becomes −(d − 2)a̸. The payoff is that a contracted pair removes two gammas before any trace is expanded. Inside a fermion line γ^μ(p̸ + m)γ_μ = −2p̸ + 4m, so g_μν Tr[γ^μ(p̸ + m)γ^ν(p̸′ + m)] = Tr[(−2p̸ + 4m)(p̸′ + m)] = −8p⋅p′ + 16m² in one line. Dimension check: GeV², as the uncontracted trace was.

Every even trace is a sum over pairings

UAQM-04.03 derived Tr 1 = 4, Tr(a̸b̸) = 4a⋅b, the vanishing of every odd trace, and the four-slash trace 4[(a⋅b)(c⋅d) − (a⋅c)(b⋅d) + (a⋅d)(b⋅c)], all quoted. Longer traces follow from one move. Carry a̸₁ to the right through the other 2n − 1 slashes: each step a̸₁a̸ₖ = 2a₁⋅aₖ − a̸ₖa̸₁ leaves 2(a₁⋅aₖ) times the string without a̸₁ and a̸ₖ, with sign (−1)^k; at the end a̸₁ sits last with an overall minus. Cyclicity makes that last string the original trace, so twice the trace is the sum, derived. For six slashes, Tr(a̸b̸c̸d̸e̸f̸) = (a⋅b)Tr(c̸d̸e̸f̸) − (a⋅c)Tr(b̸d̸e̸f̸) + (a⋅d)Tr(b̸c̸e̸f̸) − (a⋅e)Tr(b̸c̸d̸f̸) + (a⋅f)Tr(b̸c̸d̸e̸): five four-slash traces of three terms, 15 in all, and 105 for eight. A free check: with every slash equal to a̸ the signs sum to 1 − 1 + 1 − 1 + 1 = 1, giving 4(a⋅a)³, which is Tr[(a̸a̸)³] because a̸a̸ = a⋅a. The count grows as (2n − 1)!!, which is why contracting first, and shortening strings with p̸u = μ where a spinor sits, is worth the effort.

γ⁵ in a trace: zero until four gammas, then ε

Quoted from UAQM-05.05: γ⁵ = iγ⁰γ¹γ²γ³, (γ⁵)² = 1, γ⁵† = γ⁵ and γ⁵γ^μ = −γ^μ γ⁵. Three traces follow, all derived. Tr γ⁵ = Tr(γ⁵γ⁰γ⁰) = Tr(γ⁰γ⁵γ⁰) = −Tr γ⁵, by cyclicity and one anticommutation, so it is zero. Tr(γ⁵ γ^μ γ^ν) is zero too: for μ = ν it is a multiple of Tr γ⁵, and for μ ≠ ν insert γ^λ γ^λ = ±1 with λ different from both and carry one γ^λ round the trace past three anticommuting factors, a sign that forces zero. Tr(γ⁵ γ^μ γ^ν γ^ρ γ^σ) survives only when all four indices differ, since a repeat reduces it to the previous case, and it changes sign when any two are swapped, so it is C ε^μνρσ. One evaluation fixes C: γ⁰γ¹γ²γ³ = −iγ⁵, so Tr(γ⁵γ⁰γ¹γ²γ³) = −i Tr(γ⁵γ⁵) = −4i, and with ε⁰¹²³ = +1, C = −4i. In a fermion line the working rule is γ⁵(p̸ + m) = (−p̸ + m)γ⁵. That one sign is why Σ|ū′iγ⁵u|² = Tr[(p̸ − m)(p̸′ + m)] = 4(p⋅p′ − m²) while Σ|ū′u|² = 4(p⋅p′ + m²): they differ by exactly 8m².

Two ε tensors make dot products again

A trace with one γ⁵ is imaginary and antisymmetric. Moving γ⁵ to the front by cyclicity, Tr[γ^μ p̸ γ^ν γ⁵ p̸′] = −4i ε^αμβν p′_α p_β, derived; with masses kept the result is the same, because every mass term leaves fewer than four gammas beside γ⁵. Contracted with anything symmetric in μ and ν, such as the vector tensor without γ⁵, it gives zero, so in an unpolarised two-body rate an ε term survives only by meeting a second ε. Their product is a determinant of Kronecker deltas, and in Minkowski space each non-zero term of ε^μνρσ ε_μνρσ equals −1, because lowering four different indices costs three minus signs. So ε^μνρσ ε_μνρσ = −4! = −24; with one index free only the 3! orderings of the other three survive, ε^μνρσ ε_μνρτ = −6δ^σ_τ; and with two free, ε^μνρσ ε_μναβ = −2(δ^ρ_α δ^σ_β − δ^ρ_β δ^σ_α), all derived. In Euclidean space all three would be positive. A forgotten minus here reverses the sign of every interference term built from two γ⁵ traces.

Checking a trace, and what it is for

Trace results are identities, checked by dimension, limits and brute force rather than by a measurement. Take a fermion scattered elastically off a static source with |p′| = |p| = βE. The three spin sums, derived from the traces above, are 8E²(1 − β² cos²(θ/2)) for Γ = 1, 8E²β² sin²(θ/2) for Γ = iγ⁵, and 8E²(1 − β² sin²(θ/2)) for Γ = γ⁰, the last being UAQM-05.09's Mott trace. Each is in GeV². As β → 0 the first and third tend to 8m² = 2 × (2m)², two spin states each with ū′u → 2m, while the pseudoscalar sum falls as |q|², because γ⁵ joins the large and small components of a slow spinor. As β → 1 the mass term drops out: Γ = 1 and Γ = iγ⁵ give the same 8E² sin²(θ/2), and the γ⁰ sum becomes 8E² cos²(θ/2), zero for backscattering because a vector vertex cannot flip the helicity of a massless fermion. The scalar and pseudoscalar sums depend only on t, and a sum over explicit spinors at random momenta reproduces every trace here to rounding. Measurement meets them through the rates they feed: topic 7.07's widths, and Unit 9's σ(e⁺e⁻ → μ⁺μ⁻) = 86.8 nb/(s in GeV²), quoted ahead.

Change one variable at a time

Make the relationship visible.

Interactive model

Fermion speed β0.80

Scattering angle θ120 °

Set β to 0: solid and faint sit at 1 and dashed at 0, the slow limit where ū′u and ū′γ⁰u both tend to 2m. Raise β to 0.98: dashed climbs to meet solid, since γ⁵ only reversed the m² term, and faint falls near 0 at 180°, where a vector vertex cannot flip a fast fermion's helicity.

Γ = 1 SUM ÷ 8E²0.840

Γ = iγ⁵ SUM ÷ 8E²0.480

Γ = γ⁰ SUM ÷ 8E²0.520

−t ÷ m²5.33

Live interpretationΓ = 1 SUM ÷ 8E²: 0.840. Γ = iγ⁵ SUM ÷ 8E²: 0.480. Γ = γ⁰ SUM ÷ 8E²: 0.520. −t ÷ m²: 5.33

Catch the common trap

Explain before calculating.

Misconception check

The commonest slip is to write the complex conjugate of a spinor chain as the same chain with Γ replaced by Γ† or Γ*, in the same order. Conjugation does more: ū(p′)Γu(p) becomes ū(p)Γ̄u(p′), so the spinors swap ends, a string of matrices reverses, and the γ⁰ hidden in each ū turns Γ† into Γ̄ = γ⁰Γ†γ⁰, which differs from Γ† by a factor −1 for each γ⁵ and each spatial γⁱ it contains. The sign is not cosmetic. For the pseudoscalar vertex iγ⁵, using Γ† = −iγ⁵ in place of Γ̄ = +iγ⁵ gives Tr[iγ⁵(p̸ + m)(−iγ⁵)(p̸′ + m)] = 4(m² − p⋅p′) = 2t, negative at every non-zero momentum transfer: an impossible value for a sum of squared moduli, and one that would reappear in topic 7.07 as a negative decay rate. The check that catches it is free. A spin sum of |…|² must be positive, and with Γ = 1 at p′ = p it must equal 8m².

A pseudoscalar vertex iγ⁵ connects a fermion of mass m and four-momentum p to the same fermion with four-momentum p′, and t = (p′ − p)². With spinors normalised to ū u = 2m, what does Casimir's trick give for the sum over both spins of |ū(p′) iγ⁵ u(p)|²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

Bars: 1 is its own bar, and the bar of iγ⁵ is γ⁰(−iγ⁵)γ⁰ = +iγ⁵. The sums are Tr[(p̸ + m)(p̸′ + m)] and Tr[iγ⁵(p̸ + m)iγ⁵(p̸′ + m)] = Tr[(p̸ − m)(p̸′ + m)], the second after moving one γ⁵ through p̸ + m.
Odd traces vanish and Tr(a̸b̸) = 4a⋅b, so the sums are 4(p⋅p′ + m²) and 4(p⋅p′ − m²).
Kinematics: m² = 11164.0 MeV², E² = |p|² + m² = 51164.0 MeV², so E = 226.19 MeV, and p⋅p′ = E² − |p|² cos θ = 51164.0 − 20000.0 = 31164.0 MeV².
Scalar: 4(31164.0 + 11164.0) = 1.6931 × 10⁵ MeV². Pseudoscalar: 4(31164.0 − 11164.0) = 8.000 × 10⁴ MeV². Difference: 8.931 × 10⁴ MeV² = 8m².
Checks: t = −4|p|² sin²(θ/2) = −4.000 × 10⁴ MeV², and 2(4m² − t) = 2(44656 + 40000) = 1.6931 × 10⁵ MeV² while −2t = 8.000 × 10⁴ MeV², as invariants should. Both are positive, and at θ = 0, where t = 0, the scalar sum would fall to 8m² and the pseudoscalar sum to zero.

AnswerΣ|ū′u|² = 1.693 × 10⁵ MeV² and Σ|ū′iγ⁵u|² = 8.000 × 10⁴ MeV²; they differ by 8m² = 8.931 × 10⁴ MeV².

MediumAn electron, m = 0.511 MeV, with E = 1.500 MeV scatters elastically off a heavy nucleus through θ = 90.0°, E′ = E. With L^μν = Σ ū′γ^μ u (ū′γ^ν u)* over both spins, find the charge-density sum L⁰⁰, the contraction g_μν L^μν by the contraction identity, and from them L¹¹ + L²² + L³³.

The bar of γ^ν is γ⁰γ^ν†γ⁰ = γ^ν (UAQM-04.03), so L^μν = Tr[γ^μ(p̸ + m)γ^ν(p̸′ + m)]. Odd traces vanish, the quoted four-γ trace handles p̸ and p̸′, and m² Tr(γ^μ γ^ν) = 4m² g^μν; hence L^μν = 4[p^μ p′^ν + p^ν p′^μ − g^μν(p⋅p′ − m²)].
Kinematics: |p| = √(1.500² − 0.511²) = 1.4103 MeV, β² = |p|²/E² = 0.88395, and at 90° p⋅p′ = E² − |p|² cos θ = 2.2500 MeV²; m² = 0.26112 MeV².
L⁰⁰ = 4(2E² − p⋅p′ + m²) = 4(4.5000 − 2.2500 + 0.2611) = 10.044 MeV². Mott check: 8E²(1 − β² sin²45°) = 18.000 × 0.558027 = 10.044 MeV².
Contract first: γ^μ(p̸ + m)γ_μ = −2p̸ + 4m, so g_μν L^μν = Tr[(−2p̸ + 4m)(p̸′ + m)] = −8p⋅p′ + 16m² = −18.000 + 4.178 = −13.822 MeV². The tensor agrees: 4[2p⋅p′ − 4(p⋅p′ − m²)].
With g = diag(1, −1, −1, −1), g_μν L^μν = L⁰⁰ − (L¹¹ + L²² + L³³), so the spatial sum is 10.0445 + 13.8221 = 23.867 MeV², matching 4[2|p|² cos θ + 3(p⋅p′ − m²)] = 12 × 1.98888 MeV² = 23.867 MeV² at cos θ = 0.
Limit: as |p| → 0, p⋅p′ → m², so L⁰⁰ → 8m² while the spatial sum vanishes as |p|²: a slow electron's vector current is all charge density.

AnswerL⁰⁰ = 10.04 MeV², g_μν L^μν = −13.82 MeV² and L¹¹ + L²² + L³³ = 23.87 MeV²; at E = 2.94m the spin-summed spatial current is 2.38 times the charge density.

HardTwo massless fermion lines each carry one γ⁵: A^μν = Tr[γ^μ p̸ γ^ν γ⁵ p̸′] and B_μν = Tr[γ_μ k̸ γ_ν γ⁵ k̸′]. In the centre-of-momentum frame p and k collide head-on with E = 3.00 GeV each, p′ makes θ = 60.0° with p, and k′ is opposite p′. Evaluate A^μν B_μν with the ε identities, compare it with the same contraction without γ⁵, and show that their sum does not depend on θ.

Cyclicity puts γ⁵ in front: A^μν = Tr(γ⁵ p̸′ γ^μ p̸ γ^ν) = −4i ε^αμβν p′_α p_β, and likewise B_μν = −4i ε_γμδν k′^γ k^δ. Each is imaginary and antisymmetric in μ and ν.
Reordering (α, μ, β, ν) to (μ, ν, α, β) is an odd permutation, so ε^αμβν = −ε^μναβ; the lower ε gives a second minus that cancels the first. With (−4i)² = −16 and ε^μναβ ε_μνγδ = −2(δ^α_γ δ^β_δ − δ^α_δ δ^β_γ), A^μν B_μν = 32[(p⋅k)(p′⋅k′) − (p⋅k′)(p′⋅k)].
Without γ⁵ the quoted four-γ trace gives 4[p^μ p′^ν + p^ν p′^μ − g^μν p⋅p′] for each line, and contracting the two tensors with g^μν g_μν = 4 leaves 32[(p⋅k)(p′⋅k′) + (p⋅k′)(p′⋅k)].
Kinematics: p⋅k = p′⋅k′ = 2E² = 18.0 GeV² and p⋅k′ = p′⋅k = E²(1 + cos θ) = 13.5 GeV². So A^μν B_μν = 32(324 − 182.25) = 4536 GeV⁴, and without γ⁵ the result is 32(324 + 182.25) = 1.620 × 10⁴ GeV⁴.
Sum: 64(p⋅k)(p′⋅k′) = 2.074 × 10⁴ GeV⁴ = 16s² with s = 4E² = 36.0 GeV², with no θ in it; the difference, 64(p⋅k′)(p′⋅k) = 1.166 × 10⁴ GeV⁴ = 16u² with u = (p − k′)² = −27.0 GeV², carries all the angle.
Checks: two traces of GeV² each give GeV⁴; at θ = 0, p′ = p and ε^αμβν p_α p_β = 0, and indeed 32(324 − 324) = 0; a brute-force sum with explicit 4 × 4 Dirac matrices returns 4536 GeV⁴ to rounding.

AnswerA^μν B_μν = 4.54 × 10³ GeV⁴ and 1.62 × 10⁴ GeV⁴ without γ⁵; their sum, 2.07 × 10⁴ GeV⁴ = 16s², is independent of θ.