Trace techniques

Ask a collider what it measures and the honest answer is a rate — how often a given final state comes out. To predict that rate you square the quantum amplitude and, because real detectors do not read out the spin of every particle, you sum over the spins you cannot see and average over the ones you did not prepare. For a process like $e^+e^-\to\mu^+\mu^-$ the amplitude is a string of fermion currents $\bar u\,\Gamma\,u$ built from Dirac spinors and $\gamma$ -matrices. Each current is just a complex number, but its spin-summed square is a sum over many spinor components paired against many more — a bookkeeping nightmare if you enumerate spin states by hand, especially when masses cannot be dropped. Trace techniques are the escape. The completeness relations of Dirac spinors, $\sum_s u^s\bar u^s=\slashed{p}+m$ and $\sum_s v^s\bar v^s=\slashed{p}-m$ , let you replace every explicit spin sum with a covariant $4\times4$ matrix. Slide those in and the fermion line literally closes on itself: the head-to-tail chain of indices becomes a trace of a product of $\gamma$ -matrices. From there a compact set of purely algebraic theorems — the trace of an odd number of $\gamma$ 's vanishes, $\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu}$ , and a four-matrix rule — grinds the whole thing down to dot products of four-momenta with no spinor in sight. The method scales, handles massive particles gracefully, and is what every symbolic engine runs internally. This article assembles the machinery and turns the crank once, fully, on the flagship QED annihilation.

Key takeaways

A spin-summed squared current $\sum_{\text{spins}}|\bar u(p')\Gamma u(p)|^2$ becomes a single trace $\mathrm{Tr}[(\slashed{p}'+m')\Gamma(\slashed{p}+m)\bar\Gamma]$ — this is Casimir's trick.
The engine is the pair of completeness relations $\sum_s u^s\bar u^s=\slashed{p}+m$ (particles) and $\sum_s v^s\bar v^s=\slashed{p}-m$ (antiparticles — note the sign).
A short list of trace theorems does all the work: odd number of $\gamma$ 's gives $0$ ; $\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu}$ ; and a four- $\gamma$ rule.
The chiral partner $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\varepsilon^{\mu\nu\rho\sigma}$ carries every parity-violating effect in the weak sector.
For $e^+e^-\to\mu^+\mu^-$ the average factorizes into an electron trace times a muon trace, giving $\overline{|\mathcal{M}|^2}\propto(1+\cos^2\theta)$ and $\sigma=4\pi\alpha^2/3s\approx0.87$ nb at $\sqrt{s}=10$ GeV.
Key idea: because each fermion current is just a number, the spin-summed squared amplitude can be rearranged so the spinor sums collapse — via $\sum u\bar u=\slashed{p}+m$ — into traces of products of $\gamma$ -matrices, evaluated by a handful of purely algebraic theorems.

The core insight

Because each fermion current is just a number, the spin-summed squared amplitude can be rearranged so that the spinor sums collapse — via $\sum u\bar u=\slashed{p}+m$ — into traces of products of gamma matrices, which are evaluated by a handful of purely algebraic theorems.

Where this lives today

Trace techniques are the computational bedrock of collider phenomenology. Every cross section quoted for the LHC begins life as a spin sum turned into traces, and because modern processes carry traces with hundreds of terms, the work is delegated to symbolic engines — FORM, FeynCalc, and their descendants — whose innermost routines are nothing but the theorems above, extended to $d$ dimensions for dimensional regularization. The $e^+e^-\to\mu^+\mu^-$ result worked here is the literal normalization of the ratio $R=\sigma(e^+e^-\to\text{hadrons})/\sigma(e^+e^-\to\mu^+\mu^-)$ , whose measured steps at flavour thresholds were among the cleanest early evidence for three quark colours. The chiral trace $\mathrm{Tr}(\gamma^5\cdots)$ drives the forward–backward asymmetry measured at LEP to pin down the weak mixing angle, and the same massive-fermion traces feed the vacuum-polarization pieces of the muon $g\!-\!2$ , where the standing tension between measurement and theory rides on getting exactly these calculations right to many loops.

The physical picture

A Feynman amplitude reads a fermion line backwards, against the arrows: start at the outgoing adjoint spinor $\bar u$ , walk through the vertices $\Gamma$ , end at the incoming spinor $u$ . That string of matrices sandwiched between two spinors is one number. Square it and sum over the spins you do not measure, and the completeness relation stitches the free end of one copy to its conjugate — the open line closes into a loop, and a closed loop of matrix indices is precisely a trace. The first figure makes that stitching visual: a spin sum on the left, a single closed trace on the right.

Figure

Summing over unobserved spins closes the fermion line into a loop. The completeness relation ∑ₛ u ū = γ·p + m converts the spinor sum into a trace Tr((γ·p₂ − m) γ^μ (γ·p₁ + m) γ^ν) of 4×4 matrices — no explicit spinor survives

Nothing in that picture is approximated — it is an exact rewriting. The payoff comes when the trace is evaluated and every spinor disappears, leaving a Lorentz scalar built from four-momenta. For $e^+e^-\to\mu^+\mu^-$ the result is a clean $(1+\cos^2\theta)$ angular law — muon pairs prefer the beam axis and shun right angles to it, a shape confirmed at every $e^+e^-$ collider from SPEAR to LEP. The second figure is that prediction, the quantity an experimenter actually plots.

Figure

The trace machinery delivers the e⁺e⁻ → μ⁺μ⁻ angular distribution dσ/dΩ ∝ 1 + cos²θ, twice as large along the beam (θ = 0, 180°) as across it (θ = 90°), integrating to the total σ = 4πα²/3s ≈ 0.87 nb at √s = 10 GeV

The core equations

Everything begins with the completeness (or "spin-sum") relations for on-shell Dirac spinors. Summing the outer product $u\bar u$ over the two spin states of a particle gives a covariant projector; the antiparticle sum flips the mass sign. These two lines are the only place spinors enter the whole method.

\sum_{s=1,2} u^s(p)\,\bar u^s(p)=\slashed{p}+m,\qquad \sum_{s=1,2} v^s(p)\,\bar v^s(p)=\slashed{p}-m \qquad (1)

where: $u^s,v^s$ — particle/antiparticle spinors, $\bar u^s u^{s'}=2m\,\delta^{ss'}$ ; $\slashed{p}\equiv\gamma^\mu p_\mu$ ; $m$ — fermion mass. Read it as: a spin sum of $u\bar u$ is not a number but the on-shell projector matrix $\slashed{p}\pm m$ . To square a current you need the complex conjugate of $\bar u_a\Gamma u_b$ , and for spinor bilinears it equals a "barred" bilinear read in reverse, with Dirac adjoint $\bar\Gamma=\gamma^0\Gamma^\dagger\gamma^0$ — which for the QED vertex is the vertex back again.

\big[\bar u(p_a)\,\Gamma\,u(p_b)\big]^{*}=\bar u(p_b)\,\bar\Gamma\,u(p_a),\qquad \bar\Gamma=\gamma^0\Gamma^\dagger\gamma^0,\qquad \overline{\gamma^\mu}=\gamma^\mu,\quad \overline{\gamma^\mu\gamma^5}=\gamma^\mu\gamma^5 \qquad (2)

where: $\Gamma$ — any string of $\gamma$ -matrices; $\dagger$ — Hermitian conjugate. Read it as: conjugating a current reverses the spinor order and replaces $\Gamma$ by $\bar\Gamma$ , which for the vector and axial-vector vertices is unchanged. Combine (1) and (2) and one exact identity emerges — Casimir's trick. The spin sum of the squared current, a quadruple sum over spinor components, equals a single trace with no spinors at all — the theorem the whole method rests on.

\sum_{\text{spins}}\big|\bar u(p')\,\Gamma\,u(p)\big|^{2}=\mathrm{Tr}\!\big[(\slashed{p}'+m')\,\Gamma\,(\slashed{p}+m)\,\bar\Gamma\big] \qquad (3)

where: $p,m$ and $p',m'$ — incoming and outgoing fermion momenta and masses; the trace runs over the suppressed Dirac indices. Read it as: attach $\slashed{p}+m$ to the incoming leg, $\slashed{p}'+m'$ to the outgoing leg, close the loop, take the trace. The trace is evaluated by theorems that never require writing a matrix: an odd number of $\gamma$ 's vanishes, the identity has trace $4$ , and two $\gamma$ 's give the metric. These three lines dispatch most of every calculation.

\mathrm{Tr}(\mathbb{1})=4,\qquad \mathrm{Tr}(\gamma^{\mu_1}\cdots\gamma^{\mu_n})=0\ \ (n\ \text{odd}),\qquad \mathrm{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu} \qquad (4)

where: $g^{\mu\nu}$ — Minkowski metric $\mathrm{diag}(+,-,-,-)$ . Read it as: the special case $\mathrm{Tr}(\slashed{a}\slashed{b})=4\,a\cdot b$ is the atom into which every longer trace decomposes. The four-matrix theorem is the workhorse for any tree-level cross section — a pairwise-metric expansion with an alternating sign, and what turns the electron and muon tensors below into dot products.

\mathrm{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4\big(g^{\mu\nu}g^{\rho\sigma}-g^{\mu\rho}g^{\nu\sigma}+g^{\mu\sigma}g^{\nu\rho}\big) \qquad (5)

where: indices are contracted with whatever momenta the slashes carry. Read it as: three ways to pair four indices, minus sign on the "crossed" pairing $\mu\rho,\nu\sigma$ . The companion chiral theorem $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\,\varepsilon^{\mu\nu\rho\sigma}$ (with $\varepsilon^{0123}=+1$ ) supplies every parity-odd term in the weak interaction. Running the machine on $e^+e^-\to\mu^+\mu^-$ gives the master result — a single Lorentz-invariant function of momentum dot products from which every observable follows.

\overline{|\mathcal{M}|^{2}}=\frac{8e^{4}}{q^{4}}\Big[(p_1\!\cdot\! p_3)(p_2\!\cdot\! p_4)+(p_1\!\cdot\! p_4)(p_2\!\cdot\! p_3)+m_\mu^{2}\,(p_1\!\cdot\! p_2)\Big] \qquad (6)

where: $p_1,p_2$ — incoming $e^-,e^+$ ; $p_3,p_4$ — outgoing $\mu^-,\mu^+$ ; $q=p_1+p_2$ ; $e$ — electron charge; the bar is the $1/4$ spin average. Read it as: the answer is manifestly frame-independent — no spinor, no spin index, just invariants — exactly what trace techniques buy you.

Derivation AFrom a spin sum to a single trace — Casimir's trick

The goal is to prove equation (3): that the spin-summed modulus-squared of one fermion current, a fourfold sum over spinor components paired against their conjugates, is identically equal to one trace of a product of $4\times4$ matrices with every explicit spinor eliminated. This is the single most important manipulation in perturbative QED, QCD and the electroweak theory, because it converts the intractable enumeration of spin states into mechanical matrix algebra — exactly, with no approximation of any kind.

Step 1. Write the modulus squared as a product of the current and its conjugate, $|\bar u(p')\Gamma u(p)|^2=[\bar u(p')\Gamma u(p)]\,[\bar u(p')\Gamma u(p)]^{*}$ , and apply the adjoint identity (2) to the second factor: $[\bar u(p')\Gamma u(p)]^{*}=\bar u(p)\,\bar\Gamma\,u(p')$ with $\bar\Gamma=\gamma^0\Gamma^\dagger\gamma^0$ . Why: the complex conjugate of a $c$ -number equals its Hermitian conjugate, and slipping $\gamma^0\gamma^0=\mathbb{1}$ between factors re-expresses the conjugate as a genuine Dirac bilinear rather than a raw dagger.

Step 2. Restore the suppressed Dirac indices so every object is an ordinary number: $\bar u(p')_a\,\Gamma_{ab}\,u(p)_b\;\bar u(p)_c\,\bar\Gamma_{cd}\,u(p')_d$ . Because these all commute, reorder to place the two $u(p')$ adjacent and the two $u(p)$ adjacent: $u(p')_d\,\bar u(p')_a\;\Gamma_{ab}\;u(p)_b\,\bar u(p)_c\;\bar\Gamma_{cd}$ . Why: spinor components commute, so we may freely cycle them into the pattern $u\bar u\ldots u\bar u$ the completeness relation recognizes.

Step 3. Sum over both spins and apply the completeness relations (1): $\sum_{s'}u(p')_d\,\bar u(p')_a=(\slashed{p}'+m')_{da}$ and $\sum_{s}u(p)_b\,\bar u(p)_c=(\slashed{p}+m)_{bc}$ . The chain becomes $(\slashed{p}'+m')_{da}\,\Gamma_{ab}\,(\slashed{p}+m)_{bc}\,\bar\Gamma_{cd}$ . Why: the completeness relations trade an explicit spin sum of outer products $u\bar u$ for the covariant projector $\slashed{p}\pm m$ , once per fermion leg.

Step 4. Read the index pattern: $d\to a$ through the projector, $a\to b$ through $\Gamma$ , $b\to c$ through the second projector, $c\to d$ through $\bar\Gamma$ , with $d$ summed at both ends. A chain whose first and last indices are identified and summed is by definition a trace, so $\sum_{\text{spins}}|\bar u(p')\Gamma u(p)|^{2}=\mathrm{Tr}[(\slashed{p}'+m')\Gamma(\slashed{p}+m)\bar\Gamma]$ . Why: the closed index loop is the fermion line closing on itself — that ring of contracted indices is exactly "the trace".

Go deeper · Derivation B · 2 minThe trace theorems from the Clifford algebra

Derivation BThe trace theorems from the Clifford algebra

The goal is to derive the theorems that make trace evaluation purely algebraic: that the trace of an odd number of gamma matrices vanishes, that two gamma matrices trace to four times the metric, and the four-matrix theorem — using nothing but the Clifford anticommutator, the cyclic property of the trace, and the anticommuting fifth matrix. None of these need a specific representation; they follow from the algebra alone, which is why the same theorems hold in every convention.

Step 1. Odd numbers vanish. Insert $\mathbb{1}=(\gamma^5)^2$ at the front of $\mathrm{Tr}[\gamma^{\mu_1}\cdots\gamma^{\mu_n}]$ and anticommute one $\gamma^5$ rightward through all $n$ matrices; each pass costs a sign because $\gamma^5$ anticommutes with every $\gamma^\mu$ , so we accumulate $(-1)^n$ . Cyclicity then brings the trailing $\gamma^5$ back to meet the leading one, giving $\mathrm{Tr}=(-1)^n\mathrm{Tr}$ , which forces the trace to zero for odd $n$ . Why: $\gamma^5$ acts as a parity counter on the length of the string, and an object equal to minus itself is zero.

Step 2. Two matrices. Trace the defining relation $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}\mathbb{1}$ : the right side is $2g^{\mu\nu}\,\mathrm{Tr}(\mathbb{1})=8g^{\mu\nu}$ , and cyclicity makes the two left traces equal, so $2\,\mathrm{Tr}(\gamma^\mu\gamma^\nu)=8g^{\mu\nu}$ , hence $4g^{\mu\nu}$ . Why: the anticommutator converts an unknown two-matrix trace into the known $\mathrm{Tr}(\mathbb{1})=4$ in one line.

Step 3. Four matrices. Anticommute the leftmost $\gamma^\mu$ rightward through the other three using $\gamma^\mu\gamma^\nu=2g^{\mu\nu}-\gamma^\nu\gamma^\mu$ at each hop. Each hop peels off a metric factor times a two-matrix trace ( $=4g$ ) with an alternating sign, and the term with $\gamma^\mu$ at the far right cancels by cyclicity. The three survivors give $\mathrm{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4(g^{\mu\nu}g^{\rho\sigma}-g^{\mu\rho}g^{\nu\sigma}+g^{\mu\sigma}g^{\nu\rho})$ . Why: each anticommutation trades a four-matrix trace for a metric times a two-matrix trace, closing the recursion after three moves.

Step 4. The chiral partners. Repeating with a $\gamma^5$ inside gives $\mathrm{Tr}(\gamma^5)=0$ and $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu)=0$ , while four gammas plus a $\gamma^5$ cannot reduce to metrics and instead give the totally antisymmetric $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\,\varepsilon^{\mu\nu\rho\sigma}$ . Why: the Levi-Civita symbol is the unique rank-4 invariant antisymmetric in all four indices, so any parity-odd trace is proportional to it. *Note: the $\gamma^5$ normalization convention used throughout is $\varepsilon^{0123}=+1$ .*

Go deeper · Derivation C · 1 minThe electron and muon tensors for e+e- annihilation

Derivation CThe electron and muon tensors for e+e- annihilation

The goal is to assemble Casimir's trick and the trace theorems into the tree-level cross section for $e^+e^-\to\mu^+\mu^-$ , the process where every idea in this article is exercised at once. A single virtual photon connects an electron current to a muon current; squaring and spin-averaging factorizes into an electron tensor times a muon tensor, each an independent four-gamma trace, and contracting them over the shared photon indices produces the Lorentz-invariant master formula (6) with no spinor surviving.

Step 1. The lowest-order amplitude is $\mathcal{M}=\dfrac{e^2}{q^2}\,[\bar v(p_2)\gamma^\mu u(p_1)]\,[\bar u(p_3)\gamma_\mu v(p_4)]$ , with $q=p_1+p_2$ . Averaging initial and summing final spins gives $\overline{|\mathcal{M}|^2}=\tfrac14\sum_{\text{spins}}|\mathcal{M}|^2=\dfrac{e^4}{4q^4}\,L_e^{\mu\nu}\,L^{(\mu)}_{\mu\nu}$ , the two tensors being separate spin sums. Why: the $1/4$ averages over the two electron and two positron initial spins, and the amplitude factorizes because the two lines meet only through the photon propagator.

Step 2. The electron tensor is a spin sum of the current squared, so Casimir's trick (3) with $\Gamma=\gamma^\mu$ , $\bar\Gamma=\gamma^\nu$ gives $L_e^{\mu\nu}=\mathrm{Tr}[(\slashed{p}_2-m_e)\gamma^\mu(\slashed{p}_1+m_e)\gamma^\nu]$ . The three-gamma cross terms vanish by (4), and the two- and four-gamma theorems give $L_e^{\mu\nu}=4[\,p_1^\mu p_2^\nu+p_1^\nu p_2^\mu-(p_1\!\cdot\!p_2+m_e^2)g^{\mu\nu}\,]$ , dropping the mass term as $m_e\to0$ . Why: the incoming positron carries $\slashed{p}_2-m_e$ , the electron $\slashed{p}_1+m_e$ , and the theorems evaluate the string with no explicit spinor.

Step 3. The muon tensor has identical Dirac structure with outgoing labels and the muon mass kept: $L^{(\mu)}_{\mu\nu}=\mathrm{Tr}[(\slashed{p}_3+m_\mu)\gamma_\mu(\slashed{p}_4-m_\mu)\gamma_\nu]=4[\,p_{3\mu}p_{4\nu}+p_{3\nu}p_{4\mu}-(p_3\!\cdot\!p_4+m_\mu^2)g_{\mu\nu}\,]$ . Why: the muon line is the same calculation with $1\to3$ , $2\to4$ ; only the retained mass differs, since $m_\mu=207\,m_e$ is rarely negligible at low energy.

Step 4. Contract the two symmetric tensors. Every index pairing becomes a metric contraction or a momentum dot product, and the sum collapses (dropping $m_e$ ) to $L_e^{\mu\nu}L^{(\mu)}_{\mu\nu}=32[(p_1\!\cdot\!p_3)(p_2\!\cdot\!p_4)+(p_1\!\cdot\!p_4)(p_2\!\cdot\!p_3)+m_\mu^2(p_1\!\cdot\!p_2)]$ , giving $\overline{|\mathcal{M}|^2}=\dfrac{8e^4}{q^4}[\ldots]$ , equation (6). Why: contracting two tensors made only of momenta and metrics yields nothing but dot products — a manifest Lorentz scalar, the whole point of the method. *The CM-frame reduction yields $1+\cos^2\theta$ and $\sigma=4\pi\alpha^2/3s\approx0.87$ nb at $\sqrt{s}=10$ GeV.*

Limiting cases

Case	Result	Meaning
Odd number of $\gamma$ 's	$\mathrm{Tr}=0$	half the candidate terms die before any algebra
Two slashes	$\mathrm{Tr}(\slashed{a}\slashed{b})=4\,a\!\cdot\!b$	the atom into which every longer trace decomposes
Massless fermions ( $m\to0$ )	keep only $\slashed{p}$ projectors	ultra-relativistic limit; chirality conserved at vertices
One $\gamma^5$ present	$-4i\,\varepsilon_{\mu\nu\rho\sigma}$ appears	parity-odd term — forward–backward asymmetry, weak interference
Unpolarized average	factor $\tfrac{1}{(2s_1+1)(2s_2+1)}=\tfrac14$	spin average over two spin- $\tfrac12$ initial particles

Common trap"The trace is cyclic, so I can reorder the matrices however I like."

Cyclic is not the same as arbitrary. $\mathrm{Tr}(ABC)=\mathrm{Tr}(BCA)=\mathrm{Tr}(CAB)$ — you may rotate the whole string — but you may not swap non-adjacent factors, because gamma matrices do not commute: $\mathrm{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)\neq\mathrm{Tr}(\gamma^\mu\gamma^\rho\gamma^\nu\gamma^\sigma)$ in general. The order is fixed by the fermion line read backwards against the arrows: the projector for the adjoint (outgoing) spinor first, then the vertices in the order the line passes them, then the projector for the incoming spinor. Two further traps: antiparticle legs carry $\slashed{p}-m$ , not $\slashed{p}+m$ , so a dropped sign silently corrupts every mass term; and with a $\gamma^5$ present the reversal identity picks up a sign that must be tracked.

Worked examples

1 · The full e⁺e⁻ → μ⁺μ⁻ cross section (fully worked)

Starting from the master formula (6), find the angular distribution and total cross section for $e^+e^-\to\mu^+\mu^-$ at $\sqrt{s}=10$ GeV, neglecting both lepton masses.

Step 1. Centre-of-mass kinematics. With beams along $z$ and muons at angle $\theta$ , take $p_1=(E,0,0,E)$ , $p_2=(E,0,0,-E)$ , $p_3=(E,E\sin\theta,0,E\cos\theta)$ , $p_4=(E,-E\sin\theta,0,-E\cos\theta)$ , so $q^2=s=4E^2$ . The needed invariants are $p_1\!\cdot\!p_3=p_2\!\cdot\!p_4=E^2(1-\cos\theta)$ and $p_1\!\cdot\!p_4=p_2\!\cdot\!p_3=E^2(1+\cos\theta)$ .

Step 2. Insert into (6) with $m_\mu\to0$ : the bracket is $E^4(1-\cos\theta)^2+E^4(1+\cos\theta)^2=2E^4(1+\cos^2\theta)$ , so $\overline{|\mathcal{M}|^2}=\dfrac{8e^4}{(4E^2)^2}\cdot 2E^4(1+\cos^2\theta)=\mathbf{e^4(1+\cos^2\theta)}$ . Muons emerge twice as often along the beam ( $\theta=0,\pi$ ) as across it ( $\theta=\pi/2$ ).

Step 3. Convert to a differential cross section. In the CM frame with a massless final state, $\dfrac{d\sigma}{d\Omega}=\dfrac{\overline{|\mathcal{M}|^2}}{64\pi^2 s}$ . Using $e^4=(4\pi\alpha)^2=16\pi^2\alpha^2$ gives $\dfrac{d\sigma}{d\Omega}=\dfrac{16\pi^2\alpha^2}{64\pi^2 s}(1+\cos^2\theta)=\mathbf{\dfrac{\alpha^2}{4s}(1+\cos^2\theta)}$ .

Step 4. Integrate. Since $\int(1+\cos^2\theta)\,d\Omega=2\pi\int_0^\pi(1+\cos^2\theta)\sin\theta\,d\theta=\dfrac{16\pi}{3}$ , the total is $\sigma=\dfrac{\alpha^2}{4s}\cdot\dfrac{16\pi}{3}=\mathbf{\dfrac{4\pi\alpha^2}{3s}}$ .

Step 5. Put in numbers. At $\sqrt{s}=10$ GeV, $s=100$ GeV $^2$ . With $\alpha=1/137$ and $(\hbar c)^2=3.894\times10^{5}\ \mathrm{nb\,GeV^2}$ , $\sigma=\dfrac{4\pi(1/137)^2}{3\times100}\times3.894\times10^5\ \mathrm{nb}=\mathbf{0.87\ nb}$ — the "point cross section" $\sigma_{\text{pt}}=86.8\ \mathrm{nb}/(s/\mathrm{GeV}^2)$ against which the $e^+e^-\to$ hadrons rate is normalized to give the ratio $R$ .

2 · A contraction-identity trace (warm-up)

Evaluate $\mathrm{Tr}[(\slashed{p}+m)\gamma^\mu(\slashed{p}'+m)\gamma_\mu]$ , a trace with a summed (contracted) Lorentz index that appears in self-energy and total-rate calculations.

Step 1. Expand the product into four traces. The two terms linear in $m$ contain three gamma matrices and vanish by theorem (4), leaving $\mathrm{Tr}[\slashed{p}\,\gamma^\mu\slashed{p}'\gamma_\mu]+m^2\,\mathrm{Tr}[\gamma^\mu\gamma_\mu]$ .

Step 2. Use the contraction identity $\gamma^\mu\slashed{p}'\gamma_\mu=-2\slashed{p}'$ (itself a consequence of the Clifford algebra) in the first term: $\mathrm{Tr}[\slashed{p}(-2\slashed{p}')]=-2\,\mathrm{Tr}(\slashed{p}\slashed{p}')=-2\cdot 4\,(p\!\cdot\!p')=\mathbf{-8\,(p\!\cdot\!p')}$ .

Step 3. The second term uses $\gamma^\mu\gamma_\mu=4\,\mathbb{1}$ , so $m^2\,\mathrm{Tr}(4\,\mathbb{1})=16m^2$ . Adding the pieces gives $\mathrm{Tr}[(\slashed{p}+m)\gamma^\mu(\slashed{p}'+m)\gamma_\mu]=\mathbf{16m^2-8\,(p\!\cdot\!p')}$ — done without writing a single $4\times4$ matrix, the whole content being $\mathrm{Tr}(\slashed{a}\slashed{b})=4a\!\cdot\!b$ plus two contraction identities.

Physics of the portrait

Trace techniques are one of those rare tools whose difficulty is entirely front-loaded: prove the completeness relations and the trace theorems once, and every cross section thereafter is turning a crank. The device that collapses the spin sum into a trace is old enough to carry a name.

Origins · 1 minCasimir's trick and how this was discovered

Casimir's trick

the spin-sum-to-trace identity that eliminates every explicit spinor · spin sums as gamma-matrix traces in QED

The manoeuvre of replacing a spin sum by a trace over gamma matrices is traditionally attributed to Hendrik Brugt Gerhard Casimir (1909–2000), the Dutch physicist better known for the Casimir effect and for directing Philips Research. A student of Paul Ehrenfest at Leiden who worked with Pauli and Bohr, Casimir left his name on the "trick" of using completeness relations to turn spinor sums into traces — Griffiths' textbook labels it exactly so. The raw materials came from Paul Dirac, whose 1928 equation introduced the four gamma matrices and the algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ that every trace theorem flows from, and the method became indispensable with the diagrammatic revolution of Feynman and Dyson around 1949, which recast every QED amplitude as a product of the spinor bilinears Casimir's trick devours. When parity violation forced the axial vertex $\gamma^\mu\gamma^5$ into the weak currents in the late 1950s, the same machinery absorbed it through one extra theorem, $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\varepsilon^{\mu\nu\rho\sigma}$ . What began as a way to avoid summing spinors by hand is today the innermost loop of nearly every perturbative Standard Model calculation.

History

1928

— Dirac's relativistic equation introduces the gamma matrices and the algebra behind every trace theorem

1949

— Feynman and Dyson cast QED amplitudes as products of spinor bilinears, making spin sums routine

1958

— the V−A theory of the weak interaction puts the axial vertex and gamma-five traces at centre stage

1963

— Veltman's Schoonschip becomes the first computer program to evaluate gamma-matrix traces symbolically

1995

— Peskin and Schroeder codify the trace-theorem toolkit for a generation of students

Going further

Texts. Thomson, Modern Particle Physics, §6.5 (Trace techniques) and §6.5.1–6.5.4 · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.7 (Casimir's trick) and Ch. 7 · Halzen & Martin, Quarks and Leptons, §6.2–6.4 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §5.1 and Appendix A.3 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 13. The key ingredients assembled here are: the completeness relations $\sum u\bar u=\slashed{p}+m$ and $\sum v\bar v=\slashed{p}-m$ ; the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ and the $\gamma^5$ convention $\varepsilon^{0123}=+1$ ; the $1/4$ spin-averaging prefactor for two spin- $\tfrac{1}{2}$ initial particles; and the $e^+e^-\to\mu^+\mu^-$ master formula (6), whose CM reduction gives $1+\cos^2\theta$ and $\sigma=4\pi\alpha^2/3s\approx0.87$ nb at $\sqrt{s}=10$ GeV.

Prerequisites. Gamma matrices and the Clifford algebra · Solutions of the Dirac equation · Spinor normalisation · Spin averaging of matrix elements · $e^+e^-\to\mu^+\mu^-$ . Leads to. Compton scattering · The Rosenbluth formula · Deep inelastic scattering · Chirality in QED.

Ask the tutor: Want to see Casimir's trick turn the Compton or Bhabha amplitude into traces step by step, or compare it head-to-head with the helicity-amplitude method on the same process?

Why does a spin sum turn into a trace, exactly?

Because a spinor bilinear $\bar u\Gamma u$ is a single complex number, and a number equals its own transpose. Writing the modulus squared with explicit Dirac indices exposes a closed chain: an index runs out of one spinor, through the vertex, into the projector $\slashed p+m$ , through the second vertex, and back to where it started. A chain of matrices whose first and last indices are contracted together is the definition of a trace. So the trace is not an analogy or an approximation — it is literally the closed loop of indices that the fermion line becomes once you sum over the spins at its ends.

Why does the trace of an odd number of gamma matrices vanish?

Insert $\mathbb{1}=(\gamma^5)^2$ at the front and drag one $\gamma^5$ through the string to the back. Because $\gamma^5$ anticommutes with every $\gamma^\mu$ , moving it past $n$ matrices costs $(-1)^n$ ; cyclicity then slides that trailing $\gamma^5$ back to annihilate the leading one, leaving $\mathrm{Tr}=(-1)^n\mathrm{Tr}$ . For odd $n$ the trace equals minus itself, which only zero can do. This is a huge labour saver: in any expansion, every term with an odd number of gammas — usually about half — vanishes before you compute anything.

Casimir's trick or helicity amplitudes — which should I use?

They compute the same physics by opposite routes. Casimir's trick sums over spins first and works with traces of $4\times4$ matrices; it is unbeatable for unpolarized rates and keeps masses effortlessly, since $m$ just rides along in $\slashed p+m$ . The helicity-amplitude method instead computes each fixed-helicity amplitude as a small complex number and squares at the end; it shines for many-particle massless processes, where few helicity configurations survive and the traces would otherwise balloon. Rule of thumb: few legs or heavy particles favour traces; many massless legs favour helicity amplitudes.

Where does the gamma-five trace actually show up?

Anywhere parity is violated, which means the entire weak sector. The charged-current vertex is $\gamma^\mu(1-\gamma^5)$ , so squaring a weak amplitude produces traces containing a $\gamma^5$ , and the only non-vanishing new structure is $\mathrm{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\varepsilon^{\mu\nu\rho\sigma}$ . That antisymmetric tensor is what generates forward–backward asymmetries — the tendency for a produced fermion to prefer the hemisphere along, rather than against, the incoming particle — measured with high precision at LEP to extract the weak mixing angle. In pure QED and QCD, where vertices are vector-like, the $\gamma^5$ terms cancel in spin-summed rates and you rarely meet them until you compute an interference or an anomaly.

Where does the factor of 1/4 out front come from?

It is the spin average over the initial state. The incoming electron has either of two spin projections, and so does the positron, giving four equally likely combinations an unpolarized beam presents. Since you cannot predict which occurs, the meaningful quantity is the average, which divides the summed-over-all-spins result by $(2s_e+1)^2=4$ . Final-state spins are different: the detector sums over every final spin it can reach, so you sum with no dividing factor. The prescription is: average initial ( $\times1/4$ here), sum final.

When can I drop the electron mass but not the muon mass?

Whenever the energy scale $\sqrt{s}$ dwarfs the mass. At a $10$ GeV collider, $m_e=0.511$ MeV gives $m_e^2/s\sim10^{-9}$ , so the projector $\slashed p_1+m_e$ truncates to $\slashed p_1$ with no measurable error. The muon, at $m_\mu=105.7$ MeV, has $m_\mu^2/s\sim10^{-4}$ — small at $10$ GeV but not always ignorable, and near threshold $\sqrt{s}\approx2m_\mu$ it dominates the phase space. The trace method makes this painless: keep every mass symbolically in $\slashed p+m$ and drop the small ones only at the end, so no derivation is redone when the approximation changes.