Spin-averaged matrix elements

Every cross section a collider actually measures is spin-blind. The electron and positron beams at a machine like LEP or a B-factory are unpolarised — a given electron is equally likely to be spin-up or spin-down — and an ordinary detector counts the outgoing muons without asking which way their spins point. Yet the matrix element $\mathcal{M}$ is defined for one definite spin configuration at a time. To connect the calculable amplitude to the countable rate you must collapse the spin information the experiment does not carry: average over the spin states of the two initial particles (each prepared at random) and sum over those of the two final particles (the detector accepts all of them). The result is the spin-averaged squared matrix element $\langle|\mathcal{M}|^2\rangle$ , the single number that feeds every unpolarised cross-section and decay-rate formula. This article is the workhorse of the QED-calculation chapter. It takes the amplitude for $e^+e^-\to\mu^+\mu^-$ from aq-pp-ee-to-mumu, the Dirac spinor solutions from aq-pp-dirac-solutions, and the helicity language of aq-pp-spin-helicity-states, and welds them into a procedure that never requires writing a single spinor component. The engine is Casimir's trick: the completeness relations $\sum_s u\bar u = \slashed{p}+m$ turn the spin sum into a trace of gamma matrices, which the trace technology of aq-pp-trace-techniques evaluates in a few lines. Run the machine once and the famous result falls out — $\langle|\mathcal{M}|^2\rangle = e^4(1+\cos^2\theta)$ — the angular distribution that is one of the cleanest confirmations of QED and of the pointlike, spin-half nature of leptons.

Key takeaways

Real beams and detectors are spin-blind, so an observable rate needs $\langle|\mathcal{M}|^2\rangle$ : average over the initial spins, sum over the final spins. For two spin-half initial particles the averaging factor is $\tfrac14$ .
Casimir's trick converts the otherwise messy sum over spinor components into a single trace using the completeness relations $\sum_s u^s\bar u^s = \slashed{p}+m$ and $\sum_s v^s\bar v^s = \slashed{p}-m$ .
The trace theorems for gamma matrices then evaluate everything algebraically — no explicit spinors, no reference frame chosen until the very end.
For $e^+e^-\to\mu^+\mu^-$ at $E\gg m$ the machine gives $\langle|\mathcal{M}|^2\rangle = e^4(1+\cos^2\theta)$ , and hence $\dfrac{d\sigma}{d\Omega}=\dfrac{\alpha^2}{4s}(1+\cos^2\theta)$ and $\sigma=\dfrac{4\pi\alpha^2}{3s}=\dfrac{86.8\ \mathrm{nb}}{s\,[\mathrm{GeV}^2]}$ .
The same answer emerges from helicity amplitudes: chirality conservation kills 12 of the 16 combinations, leaving four spin-1 amplitudes that scale as $1\pm\cos\theta$ .
Key idea: when beams and detectors are blind to spin you average over initial spins and sum over final spins, and Casimir's trick turns that double spin sum into gamma-matrix traces — but the physics behind the $1+\cos^2\theta$ distribution is clearest once each surviving amplitude is recognised as a definite spin-1 helicity state.

The core insight

When beams and detectors are blind to spin, you average over initial spins and sum over final spins; Casimir's trick turns that double sum into a trace of gamma matrices, and the physics behind the resulting angular distribution is best understood by first isolating each helicity amplitude and recognising it as a definite total-spin state.

Where this lives today

The $1+\cos^2\theta$ distribution and the $86.8\ \mathrm{nb}/s$ normalisation are not museum pieces — they remain the reference against which real data are calibrated. The famous $R$ ratio, $R=\sigma(e^+e^-\to\text{hadrons})/\sigma(e^+e^-\to\mu^+\mu^-)$ , uses exactly this spin-averaged muon cross section as its denominator, and its steps at flavour thresholds are the cleanest evidence for the number of quark colours and the fractional quark charges. At the $Z$ pole, LEP turned the tiny departures from the symmetric QED curve into precision physics: $Z^0$ exchange adds a parity-violating axial current, so the forward-backward asymmetry $A_{FB}$ becomes non-zero and pins down the weak mixing angle $\sin^2\theta_W$ — at bottom, a controlled distortion of the matrix element computed here. Modern calculations push the same completeness-to-trace logic to millions of terms inside automated tools like FORM, FeynCalc, and MadGraph. And polarised machines run the logic in reverse: dropping the $\tfrac14$ average and selecting individual helicity amplitudes turns the spin sum back into a spin-resolved probe of new physics hiding in the chiral structure of the couplings.

The physical picture

Picture the collision in its centre-of-momentum frame: an electron and a positron rush together along a common axis, annihilate into a single virtual photon, and that photon materialises as a muon pair flying apart at some angle. Because the photon carries spin 1, it can only be made from — and only decay into — a fermion pair whose spins add up to one unit along a direction. This single fact, spin-1 in the middle, controls everything. It tells you which of the sixteen possible in/out helicity combinations can contribute at all (only four), and it tells you the shape of the angular distribution (the overlap of an initial spin-1 state with a rotated final spin-1 state). The first figure is that geometry.

Figure

Electron and positron annihilate through a virtual photon of spin 1; of the 16 helicity combinations only 4 survive, all with the two spins aligned so the pair carries S_z = ±1, and each surviving amplitude scales as 1 ± cosθ

The angular distribution the detector actually records is what you get after averaging those four amplitudes with equal weight. Two of them peak forward (they scale as $1+\cos\theta$ , so their square is $(1+\cos\theta)^2$ , largest at $\theta=0$ ) and two peak backward (as $(1-\cos\theta)^2$ ). Adding the four squared amplitudes and dividing by four for the initial-spin average gives the smooth, forward-backward-symmetric shape $1+\cos^2\theta$ — twice as much rate straight ahead or straight back as sideways. That $2:1$ ratio, and the exact curve, are a direct fingerprint of a spin-1 exchange between spin-half pointlike particles. The second figure is the punchline: the observable curve and its decomposition into the two chiral pieces.

Figure

Tree-level QED angular distribution dσ/dΩ ∝ 1 + cos²θ for e⁺e⁻ → μ⁺μ⁻ at E ≫ m: forward and backward are twice the transverse rate, and the curve is the sum of a forward-peaking ½(1+cosθ)² and a backward-peaking ½(1−cosθ)² helicity term; the total cross section is σ = 4πα²/3s = 86.8 nb divided by s in GeV²

The core equations

The starting point is the tree-level amplitude for $e^+e^-\to\mu^+\mu^-$ , a single $s$ -channel photon exchange from aq-pp-ee-to-mumu: a product of an electron current and a muon current, tied together by the photon propagator $-ig_{\mu\nu}/q^2$ with $q^2=s$ .

i\mathcal{M} = \frac{i e^2}{s}\,\big[\bar v(p_2)\gamma^\mu u(p_1)\big]\,\big[\bar u(p_3)\gamma_\mu v(p_4)\big] \qquad (1)

where: $u(p_1),\,v(p_2)$ are the incoming electron and positron spinors; $u(p_3),\,v(p_4)$ the outgoing muon and antimuon; $e$ the charge with $e^2=4\pi\alpha$ ; $s=(p_1+p_2)^2$ the squared centre-of-mass energy. Reading it: two Dirac currents joined by a photon, with every spin label hidden inside the spinors — exactly what the averaging must remove. An unpolarised observable does not know the individual spins, so we average the squared amplitude over the $(2s_1+1)(2s_2+1)=4$ initial spin states and sum over the final ones. The averaging factor $\tfrac14$ is the whole content of the word "unpolarised".

\langle|\mathcal{M}|^2\rangle \equiv \frac{1}{2}\frac{1}{2}\sum_{s_1,s_2}\sum_{s_3,s_4}|\mathcal{M}|^2 = \frac14\sum_{\text{spins}}|\mathcal{M}|^2 \qquad (2)

where: the two $\tfrac12$ factors average over the electron and positron spins; the final spins are summed with unit weight. Reading it: divide by the initial spin states you did not control, add up the final states you cannot resolve — never divide by the final-state count. What makes this tractable is the completeness of the Dirac solutions: summing the outer product of a spinor with its conjugate over both spins gives a simple $4\times4$ matrix, not a mess of components. These two relations are the only spinor facts the rest of the calculation needs.

\sum_{s}u^{s}(p)\,\bar u^{s}(p) = \slashed{p}+m, \qquad \sum_{s}v^{s}(p)\,\bar v^{s}(p) = \slashed{p}-m \qquad (3)

where: $\slashed{p}=\gamma^\mu p_\mu$ ; $m$ the particle mass; the sum runs over the two spin states. Reading it: the spin sum is a projector onto positive- (or negative-) energy solutions carrying no free spin index — which is why it can be pushed inside a trace. Applying (3) to a squared current is Casimir's trick: a modulus-squared bilinear $|\bar u(p_3)\Gamma v(p_4)|^2$ , summed over spins, becomes a single trace, because $\sum u\bar u$ and $\sum v\bar v$ close the spinor indices into a loop.

\sum_{\text{spins}}\big[\bar u(p_3)\Gamma v(p_4)\big]\big[\bar u(p_3)\Gamma' v(p_4)\big]^{*} = \mathrm{Tr}\!\big[(\slashed{p}_3+m)\,\Gamma\,(\slashed{p}_4-m)\,\bar\Gamma'\big] \qquad (4)

where: $\Gamma,\Gamma'$ are strings of gamma matrices (here $\gamma^\mu,\gamma^\nu$ ); $\bar\Gamma'\equiv\gamma^0\Gamma'^{\dagger}\gamma^0$ is the Dirac adjoint, with $\bar\gamma^\nu=\gamma^\nu$ . Reading it: the conjugate current becomes a "reversed" string sandwiched by projectors, and the whole spin sum is now pure Dirac algebra. For our process the double spin sum factorises into two independent traces — one for the electron line, one for the muon line — contracted through the photon indices. These are the leptonic tensors.

L_e^{\mu\nu}=\mathrm{Tr}\!\big[(\slashed{p}_2-m_e)\gamma^\mu(\slashed{p}_1+m_e)\gamma^\nu\big],\quad L^{(\mu)}_{\mu\nu}=\mathrm{Tr}\!\big[(\slashed{p}_3+m_\mu)\gamma_\mu(\slashed{p}_4-m_\mu)\gamma_\nu\big] \qquad (5)

where: $L_e^{\mu\nu}$ carries the electron and positron momenta, $L^{(\mu)}_{\mu\nu}$ the muon momenta; the spin average is $\tfrac14(e^4/s^2)L_e^{\mu\nu}L^{(\mu)}_{\mu\nu}$ . Reading it: the problem splits into "what the electrons did" times "what the muons did", coupled only through the two Lorentz indices. Evaluating the traces and contracting the two tensors — done in full below — gives the compact, frame-independent result and its centre-of-mass form. This is the equation every unpolarised QED calculation of this process rests on.

\langle|\mathcal{M}|^2\rangle = \frac{8e^4}{s^2}\big[(p_1\!\cdot\!p_3)(p_2\!\cdot\!p_4)+(p_1\!\cdot\!p_4)(p_2\!\cdot\!p_3)\big] \;\xrightarrow{\ E\gg m\ }\; e^4\big(1+\cos^2\theta\big) \qquad (6)

where: $p_i\!\cdot\!p_j$ are Lorentz-invariant momentum products; $\theta$ is the electron-to-muon angle in the centre-of-mass frame. Reading it: manifestly Lorentz invariant on the left, it collapses to the memorable $1+\cos^2\theta$ once masses are dropped and a frame is chosen.

Derivation AFrom completeness relations to a single trace

The goal of this derivation is to justify Casimir's trick end to end: to show precisely how the sum over the four Dirac spin states of a squared current bilinear becomes a trace of gamma matrices, with no approximation and no frame chosen. We start from the Dirac adjoint of a bilinear, insert the completeness relations of equation (3), and close the spinor indices into a loop. The payoff is equation (4) — the identity that converts an intractable component-by-component spin sum into a few lines the trace theorems finish off.

Step 1. Take one muon-line current $j=\bar u(p_3)\gamma^\mu v(p_4)$ and its conjugate $j'^{*}=[\bar u(p_3)\gamma^\nu v(p_4)]^{*}$ . A number equals its own transpose, and the conjugate of a bilinear is $[\bar u\,\Gamma\, v]^{*}=\bar v\,\bar\Gamma\, u$ with $\bar\Gamma=\gamma^0\Gamma^\dagger\gamma^0$ ; for $\Gamma=\gamma^\nu$ this is again $\gamma^\nu$ . So the product is $j\,j'^{*}=[\bar u(p_3)\gamma^\mu v(p_4)][\bar v(p_4)\gamma^\nu u(p_3)]$ . Why: writing the complex conjugate as a genuine "return trip" bilinear is what lets the two factors be chained head-to-tail, which is the geometric content of a trace.

Step 2. Restore the Dirac indices explicitly: $j\,j'^{*}=\bar u_a(p_3)(\gamma^\mu)_{ab} v_b(p_4)\,\bar v_c(p_4)(\gamma^\nu)_{cd} u_d(p_3)$ , where repeated indices are summed. Every factor is now an ordinary number and can be reordered freely. Group the spinors that carry the same momentum together: $u_d(p_3)\bar u_a(p_3)$ and $v_b(p_4)\bar v_c(p_4)$ . Why: the indices reveal that the spinors already form the outer products that the completeness relations collapse — the trick is only visible once the bilinears are opened up into components.

Step 3. Now sum over the muon and antimuon spins and apply equation (3): $\sum_{s_3}u_d(p_3)\bar u_a(p_3)=(\slashed{p}_3+m_\mu)_{da}$ and $\sum_{s_4}v_b(p_4)\bar v_c(p_4)=(\slashed{p}_4-m_\mu)_{bc}$ . Substituting, the summed product becomes $(\slashed{p}_3+m_\mu)_{da}(\gamma^\mu)_{ab}(\slashed{p}_4-m_\mu)_{bc}(\gamma^\nu)_{cd}$ . Why: the completeness relations are precisely the statement that a spin sum of an outer product is a projector — this is where the physics of "sum over unobserved spins" enters the algebra.

Step 4. Read the index structure: $d\to a\to b\to c\to d$ is a closed loop, so the summed expression is exactly $\mathrm{Tr}[(\slashed{p}_3+m_\mu)\gamma^\mu(\slashed{p}_4-m_\mu)\gamma^\nu]$ — equation (5) and the muon half of (4). The identical argument on the electron line gives $L_e^{\mu\nu}=\mathrm{Tr}[(\slashed{p}_2-m_e)\gamma^\mu(\slashed{p}_1+m_e)\gamma^\nu]$ . Why: the closed index loop is the definition of a trace, so the four-spin sum is now two traces — purely mechanical from here.

Go deeper · Derivation B · 1 minEvaluating the traces and contracting the tensors

Derivation BEvaluating the traces and contracting the tensors

The goal here is to carry the two leptonic-tensor traces of equation (5) to a number and contract them, arriving at the frame-independent result of equation (6) and its centre-of-mass form. We work in the high-energy limit $E\gg m$ so mass terms drop, use the standard trace theorems (odd numbers of gamma matrices vanish; the four-gamma trace expands into three metric products), then insert the centre-of-mass momenta. This is the calculation that produces the celebrated $1+\cos^2\theta$ .

Step 1. Drop the masses (justified because $m_e^2/s\sim10^{-11}$ at collider energies, and $m_\mu$ is kept only in the worked example on threshold). The electron tensor is $L_e^{\mu\nu}=\mathrm{Tr}[\slashed{p}_2\gamma^\mu\slashed{p}_1\gamma^\nu]$ . The four-gamma trace theorem, $\mathrm{Tr}[\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\nu]=4(g^{\alpha\mu}g^{\beta\nu}-g^{\alpha\beta}g^{\mu\nu}+g^{\alpha\nu}g^{\mu\beta})$ , gives $L_e^{\mu\nu}=4\big(p_2^\mu p_1^\nu+p_2^\nu p_1^\mu-g^{\mu\nu}(p_1\!\cdot\!p_2)\big)$ . Why: the trace theorems are the entire computational content of Casimir's trick — they turn Dirac matrices into ordinary tensors built from the external momenta.

Step 2. By identical algebra the muon tensor is $L^{(\mu)}_{\mu\nu}=4\big(p_{3\mu}p_{4\nu}+p_{3\nu}p_{4\mu}-g_{\mu\nu}(p_3\!\cdot\!p_4)\big)$ . Contract the two: the metric-times-metric piece gives $g^{\mu\nu}g_{\mu\nu}=4$ , and after expanding the nine products the terms proportional to $(p_1\!\cdot\!p_2)(p_3\!\cdot\!p_4)$ cancel identically ( $-1-1-1-1+4=0$ ). Why: that cancellation is the mathematical shadow of current conservation, $q_\mu L^{\mu\nu}=0$ — the longitudinal photon does not contribute, and it is reassuring to see it drop out on its own.

Step 3. What survives is $L_e^{\mu\nu}L^{(\mu)}_{\mu\nu}=32\big[(p_1\!\cdot\!p_3)(p_2\!\cdot\!p_4)+(p_1\!\cdot\!p_4)(p_2\!\cdot\!p_3)\big]$ . Insert the average and couplings: $\langle|\mathcal{M}|^2\rangle=\tfrac14\,(e^4/s^2)\,(32)[\cdots]=\dfrac{8e^4}{s^2}\big[(p_1\!\cdot\!p_3)(p_2\!\cdot\!p_4)+(p_1\!\cdot\!p_4)(p_2\!\cdot\!p_3)\big]$ , which is the invariant form of equation (6). Why: this is the form you keep for any process or frame — the angular dependence is not yet committed to, so it is the right object to record and to check against other calculations.

Step 4. Choose the centre-of-mass frame with beam energy $E$ : $p_1=(E,E\hat z)$ , $p_2=(E,-E\hat z)$ , $p_3=(E,E\hat n)$ , $p_4=(E,-E\hat n)$ with $\hat z\!\cdot\!\hat n=\cos\theta$ and $s=4E^2$ . Then $p_1\!\cdot\!p_3=p_2\!\cdot\!p_4=E^2(1-\cos\theta)$ and $p_1\!\cdot\!p_4=p_2\!\cdot\!p_3=E^2(1+\cos\theta)$ , so $\langle|\mathcal{M}|^2\rangle=\dfrac{8e^4}{16E^4}E^4\big[(1-\cos\theta)^2+(1+\cos\theta)^2\big]=e^4(1+\cos^2\theta)$ . Why: the two squared factors are exactly the forward- and backward-peaking helicity pieces of the second figure, so this last line is where the trace calculation and the helicity picture visibly meet.

Go deeper · Derivation C · 2 minThe helicity route to the same answer

Derivation CThe helicity route to the same answer

The goal of this derivation is to reproduce $\langle|\mathcal{M}|^2\rangle=e^4(1+\cos^2\theta)$ without traces at all, by resolving the amplitude into definite helicity channels so the abstract result acquires a transparent physical meaning. In the massless limit chirality is conserved at every QED vertex and coincides with helicity, which annihilates most of the sixteen combinations; the survivors are rotations of a spin-1 state, and their squares are the Wigner $d$ -functions. This is the calculation that explains why the distribution looks the way it does.

Step 1. At $E\gg m$ the vector current $\bar u\gamma^\mu u$ connects only spinors of the same chirality, and for the physical spinors chirality equals helicity. The incoming electron and positron must therefore have opposite helicities (an $R$ electron pairs with an $L$ positron, or vice versa), and likewise for the outgoing pair. Of the $2^4=16$ helicity assignments this leaves only $2\times2=4$ non-zero amplitudes. Why: chirality conservation is the selection rule that does for helicities what the trace theorems do for tensors — it discards, for free, everything that cannot contribute.

Step 2. Read the surviving states as angular momentum. With the electron beam as the quantisation axis, an $R$ electron plus $L$ positron carries $S_z=+1$ ; the $L$ / $R$ arrangement carries $S_z=-1$ . Both are the $m=\pm1$ substates of a total spin-1 system — precisely the spin content of the intermediate photon. The final muon pair, measured along its own flight direction at angle $\theta$ , is likewise a spin-1 state with $S_{z'}=\pm1$ . Why: identifying initial and final states as spin-1 is what turns a scattering amplitude into a rotation matrix element, importing all the machinery of angular momentum.

Step 3. The amplitude is then the overlap of the initial $|1,m\rangle$ with the rotated final $|1,m'\rangle$ , i.e. a Wigner $d$ -function $d^{1}_{m'm}(\theta)$ . Explicitly $d^{1}_{1,1}=d^{1}_{-1,-1}=\tfrac12(1+\cos\theta)$ and $d^{1}_{1,-1}=d^{1}_{-1,1}=\tfrac12(1-\cos\theta)$ , so the four amplitudes scale as $1+\cos\theta,\,1-\cos\theta,\,1-\cos\theta,\,1+\cos\theta$ — exactly the table in the first figure. Why: the $d$ -functions are the model-independent statement of how a spin-1 object looks when viewed from a rotated axis, so the angular shape is fixed by spin alone, not by the details of QED.

Step 4. Square and average with the factor $\tfrac14$ : $\langle|\mathcal{M}|^2\rangle=\tfrac14 e^4[(1+\cos\theta)^2+(1-\cos\theta)^2+(1-\cos\theta)^2+(1+\cos\theta)^2]=\tfrac14 e^4\cdot 4(1+\cos^2\theta)=e^4(1+\cos^2\theta)$ . Why: getting the identical answer by two completely different routes — blind trace algebra and explicit angular momentum — is the strongest possible check that both the method and the bookkeeping are right.

Limiting cases

Case	Result	Meaning
Ultra-relativistic, $E\gg m$	$\langle\|\mathcal{M}\|^2\rangle=e^4(1+\cos^2\theta)$	chirality = helicity; 4 of 16 combinations survive; the clean $1+\cos^2\theta$ shape
Muon mass kept	$\propto\beta\big[(1+\tfrac{m^2}{E^2})+(1-\tfrac{m^2}{E^2})\cos^2\theta\big]$	angular distribution flattens as energy drops toward threshold
Threshold, $E\to m_\mu$	isotropic, rate $\to0$ as $\beta\to0$	pair produced at rest in an $S$ -wave; spin decouples from angle
Electron mass restored	correction of order $m_e^2/s\sim10^{-11}$	utterly negligible at any collider; the $\tfrac14$ average is unaffected
Include $Z^0$ exchange	forward-backward asymmetry $A_{FB}\neq0$	the parity-violating weak current breaks the QED symmetry of the curve
Polarised beams	drop the $\tfrac14$ ; keep selected helicities	isolates a single $d$ -function; how helicity is measured experimentally

Common trapYou do not average over the final-state spins

The single most common bookkeeping error is to treat initial and final spins symmetrically — to write $\tfrac{1}{16}\sum_{\text{spins}}|\mathcal{M}|^2$ , dividing by all sixteen combinations. That is wrong. You average over the initial spins because the beam delivered one unknown configuration at random, weighting each by probability $\tfrac12$ per particle. You sum over the final spins because the detector accepts every one — they are distinct outcomes that all count toward the same rate, not alternatives to be averaged. The correct prefactor is therefore $\tfrac12\cdot\tfrac12=\tfrac14$ , from the two initial spins only. A related slip is to imagine the $\tfrac14$ counts photon polarisations: it does not — the photon is virtual and internal, already accounted for by the propagator. Get the prefactor wrong and every cross section is silently off by a fixed factor.

Worked examples

1 · The muon-pair cross section at 10 GeV (fully worked)

An $e^+e^-$ collider runs at $\sqrt{s}=10\ \mathrm{GeV}$ , in the continuum below the $\Upsilon$ resonances and far from the $Z$ . Using the spin-averaged result, find the total cross section for $e^+e^-\to\mu^+\mu^-$ and the differential rate straight ahead and sideways.

Step 1. Convert the spin-averaged matrix element into a differential cross section. For a massless $2\to2$ process in the centre of mass, $\dfrac{d\sigma}{d\Omega}=\dfrac{1}{64\pi^2 s}\langle|\mathcal{M}|^2\rangle$ . Inserting $\langle|\mathcal{M}|^2\rangle=e^4(1+\cos^2\theta)$ with $e^2=4\pi\alpha$ gives $e^4=16\pi^2\alpha^2$ , so $\dfrac{d\sigma}{d\Omega}=\dfrac{16\pi^2\alpha^2}{64\pi^2 s}(1+\cos^2\theta)=\dfrac{\alpha^2}{4s}(1+\cos^2\theta)$ .

Step 2. Integrate over solid angle. With $\int(1+\cos^2\theta)\,d\Omega=2\pi\int_{-1}^{1}(1+x^2)\,dx=2\pi\cdot\tfrac{8}{3}=\dfrac{16\pi}{3}$ , the total is $\sigma=\dfrac{\alpha^2}{4s}\cdot\dfrac{16\pi}{3}=\dfrac{4\pi\alpha^2}{3s}$ . Restoring $(\hbar c)^2=0.3894\ \mathrm{GeV^2\,mb}$ and $\alpha=1/137.036$ gives the standard pocket formula $\sigma=\dfrac{86.8\ \mathrm{nb}}{s\,[\mathrm{GeV}^2]}$ .

Step 3. Evaluate at $s=100\ \mathrm{GeV}^2$ : $\boldsymbol{\sigma=86.8/100=0.868\ \mathrm{nb}=868\ \mathrm{pb}}$ . The differential rate at $90^\circ$ is $\dfrac{d\sigma}{d\Omega}=\dfrac{\alpha^2}{4s}(1)=\dfrac{(1/137.036)^2}{400\ \mathrm{GeV}^2}=1.33\times10^{-7}\ \mathrm{GeV}^{-2}=\boldsymbol{51.8\ \mathrm{pb/sr}}$ , while straight ahead or straight back $(1+\cos^2\theta=2)$ it is twice that, $\boldsymbol{103.7\ \mathrm{pb/sr}}$ .

Step 4. Sanity check: $51.8\ \mathrm{pb/sr}$ times the shape integral $16\pi/3=16.76\ \mathrm{sr}$ gives $868\ \mathrm{pb}$ , matching the closed form. The clean pointlike-lepton prediction is $\sigma=868\ \mathrm{pb}$ with a $2:1$ forward-to-transverse ratio — the number experiments normalise their $R$ -ratio measurements against.

2 · Keeping the muon mass near threshold

Repeat close to threshold, at $\sqrt{s}=0.5\ \mathrm{GeV}$ , where the muon mass $m_\mu=0.1057\ \mathrm{GeV}$ is no longer negligible. What is the angular distribution, and how does the total cross section change?

Step 1. With the muon mass retained (electron still massless), the trace calculation gives $\dfrac{d\sigma}{d\Omega}=\dfrac{\alpha^2}{4s}\,\beta\Big[\big(1+\tfrac{m_\mu^2}{E^2}\big)+\big(1-\tfrac{m_\mu^2}{E^2}\big)\cos^2\theta\Big]$ , where $E=\sqrt{s}/2$ is the beam energy and $\beta=\sqrt{1-m_\mu^2/E^2}$ is the muon speed. As $m_\mu\to0$ this returns the massless $1+\cos^2\theta$ .

Step 2. Put in the numbers: $E=0.25\ \mathrm{GeV}$ , so $m_\mu^2/E^2=(0.1057/0.25)^2=0.1786$ and $\beta=\sqrt{1-0.1786}=0.906$ . The bracket is $[1.179+0.821\cos^2\theta]$ : still peaked forward and back, but the forward-to-transverse ratio has softened from $2.00$ to $2.00/1.179=\boldsymbol{1.70}$ .

Step 3. The total cross section picks up the phase-space and spin factor $\sigma=\dfrac{4\pi\alpha^2}{3s}\,\beta\,\dfrac{3-\beta^2}{2}$ . Here $\beta(3-\beta^2)/2=0.906\times(3-0.821)/2=0.906\times1.089=0.987$ , so $\sigma=\dfrac{86.8\ \mathrm{nb}}{0.25}\times0.987=347.2\times0.987=\boldsymbol{343\ \mathrm{nb}}$ .

Step 4. Read the limit: exactly at threshold $E=m_\mu$ , so $\beta\to0$ (the rate vanishes) and $m_\mu^2/E^2\to1$ (the bracket becomes $2+0\cdot\cos^2\theta$ , perfectly isotropic). The pair is produced at rest in a pure $S$ -wave, where there is no direction for the angular distribution to remember — the $\cos^2\theta$ term is the relativistic imprint of the muons' motion, and it fades exactly as they slow down.

Physics of the portrait

Spin averaging is one of those techniques that carries a person's name for a single clever observation that made an entire industry of calculations routine. The observation is that a sum over unobserved spins is a completeness relation, and a completeness relation folded into a squared amplitude is a trace.

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

Casimir's trick

the identity turning a spin sum ∑ u ū = p̸ + m into a gamma-matrix trace · relativistic quantum mechanics of spin-half scattering

Hendrik Brugt Gerhard Casimir (1909–2000) was a Dutch theoretical physicist of extraordinary range — a student of Paul Ehrenfest at Leiden, an assistant to Niels Bohr and Wolfgang Pauli, later research director of Philips. He is best remembered for the Casimir effect, the attraction of two uncharged plates from vacuum fluctuations, and for the Casimir operators that label the representations of a Lie algebra. But buried in his early work on relativistic electron scattering is the manipulation every particle physicist now calls "Casimir's trick": recognising that the sum over the two spin states of the outer product $u^s(p)\bar u^s(p)$ is not a bookkeeping headache but the projector $\slashed{p}+m$ , so a spin-summed squared amplitude collapses into a trace of gamma matrices. The move looks almost trivial once seen, yet it is what makes unpolarised cross-section calculation tractable by hand: without it, evaluating $\langle|\mathcal{M}|^2\rangle$ means writing out four-component spinors and summing sixteen terms by brute force; with it, the answer is a few trace theorems. When Feynman's diagrams arrived at the end of the 1940s the trick slotted straight in as the standard way to handle external fermion spins, and Bjorken and Drell's 1964 textbook fixed the trace technology into the form students still learn.

History

1928

— Dirac's equation yields four spinor solutions, two spins each for particle and antiparticle

1933

— Casimir recasts the sum over spin states as a trace of gamma matrices

1948

— Feynman's diagrammatic rules make the squared amplitude a routine trace computation

1949

— Dyson systematises the perturbation expansion and the handling of external spins

1964

— Bjorken and Drell codify the trace theorems in a standard textbook

Going further

Texts. Peskin & Schroeder, An Introduction to Quantum Field Theory, §5.1 (the $e^+e^-\to\mu^+\mu^-$ calculation, eqs 5.10–5.13) and §5.1 trace technology · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.7 ("Casimir's trick") and §7.8 (trace theorems) · Halzen & Martin, Quarks and Leptons, §6.2–6.4 · Thomson, Modern Particle Physics, §6.3 ("Spin in electron–positron annihilation"), Fig. 6.9 and Problem 6.6 · Schwartz, Quantum Field Theory and the Standard Model, §13.

Provenance. The amplitude of equation (1) is recalled by name from aq-pp-ee-to-mumu; the four Dirac solutions and their completeness relations (equation 3) are owned by aq-pp-dirac-solutions; the helicity/chirality language of the third derivation is developed in aq-pp-spin-helicity-states. The gamma-matrix trace theorems used in the second derivation are proved in aq-pp-trace-techniques (forward reference, quoted not re-derived here). A verifier should confirm: (1) the completeness relations $\sum u\bar u=\slashed{p}+m$ , $\sum v\bar v=\slashed{p}-m$ ; (2) the contraction $L_e^{\mu\nu}L^{(\mu)}_{\mu\nu}=32[(p_1\!\cdot\!p_3)(p_2\!\cdot\!p_4)+(p_1\!\cdot\!p_4)(p_2\!\cdot\!p_3)]$ ; (3) the centre-of-mass reduction to $e^4(1+\cos^2\theta)$ ; (4) $\sigma=4\pi\alpha^2/3s=86.8\ \mathrm{nb}/s[\mathrm{GeV}^2]$ , giving $0.868$ nb at $\sqrt{s}=10$ GeV; (5) the mass-corrected $\sigma=(4\pi\alpha^2/3s)\beta(3-\beta^2)/2=343$ nb at $\sqrt{s}=0.5$ GeV.

Prerequisites. Spin and Helicity States · Dirac Spinor Solutions · The Process $e^+e^-\to\mu^+\mu^-$ . Leads to. Trace Techniques · Differential Cross Sections · Chirality in QED.

Ask the tutor: Want me to run Casimir's trick line by line on a process you choose — Bhabha, Compton, or $e^+e^-\to Z\to\mu^+\mu^-$ with the forward-backward asymmetry — and show where the traces differ?

Why average over the initial spins but sum over the final ones?

Because they describe two different physical situations. The beams deliver particles whose spins you did not prepare: each electron is up or down with equal probability, so the observable rate is the probability-weighted mean over those inputs — an average, weight $\tfrac12$ per particle. The final muons are different: the detector counts a muon whichever way its spin points, so up and down are two separate events that both add to the same number. You add their rates — a sum, weight $1$ each. Mixing these up, by dividing by the final-state count, is the classic error; the prefactor for two spin-half initial particles is $\tfrac14$ , full stop.

What is Casimir's trick really doing, in one sentence?

It exploits the fact that a sum over an unobserved spin is a completeness relation. Summing $u^s(p)\bar u^s(p)$ over the two spins does not give a messy spin-dependent object — it gives the clean $4\times4$ projector $\slashed{p}+m$ , which carries no free spin index. So it can be slid inside a matrix product, and the chain of spinors and gamma matrices closes into a loop — the definition of a trace. A sum that looked like sixteen explicit spinor evaluations becomes two gamma-matrix traces the theorems finish in a few lines. The trick is the bridge from "sum over spins" to "take a trace".

Why do only four of the sixteen helicity combinations contribute?

Because the QED vertex is a vector current $\bar\psi\gamma^\mu\psi$ , and at high energy such a current connects only fermions of the same chirality — which for a relativistic particle equals helicity. This forces the electron and positron to opposite helicities (two choices) and, independently, the muon and antimuon to opposite helicities (two more): two times two is four. The other twelve need the current to flip a chirality, impossible without a mass insertion, so their amplitudes are suppressed by $m/E$ and vanish in the massless limit. This chiral structure is what makes QED and the whole Standard Model a theory of definite-helicity currents.

Where does the 1 + cos²θ come from physically?

From angular momentum, not from the details of QED. The intermediate photon has spin 1, so both the annihilating pair and the produced pair must form spin-1 states. The amplitude is the overlap between an initial spin-1 state quantised along the beam and a final one quantised along the muon direction at angle $\theta$ — the Wigner rotation functions $d^1_{m'm}(\theta)=\tfrac12(1\pm\cos\theta)$ . Squaring and averaging the four surviving channels gives $1+\cos^2\theta$ . Any spin-1 exchange between pointlike spin-half particles gives this same shape — which is why measuring it confirms that muons are pointlike and spin-half.

How does the spin-averaged matrix element become a cross section?

Through the standard $2\to2$ phase-space formula. In the centre of mass for massless final states, $d\sigma/d\Omega=\langle|\mathcal{M}|^2\rangle/(64\pi^2 s)$ ; keeping a final-state mass multiplies by the speed ratio $\beta=|\vec p_f|/|\vec p_i|$ . Inserting $\langle|\mathcal{M}|^2\rangle=e^4(1+\cos^2\theta)$ with $e^2=4\pi\alpha$ gives $(\alpha^2/4s)(1+\cos^2\theta)$ ; integrating gives $\sigma=4\pi\alpha^2/3s$ . Converting with $(\hbar c)^2=0.3894\ \mathrm{GeV^2\,mb}$ yields the pocket formula $86.8\ \mathrm{nb}/s[\mathrm{GeV}^2]$ . The matrix element is the only field-theoretic input; the rest is kinematics.

What changes when the weak interaction is included?

Near the $Z$ mass the process also proceeds through $Z^0$ exchange, and the two amplitudes interfere. The $Z$ couples through a mix of vector and axial currents, $\bar\psi\gamma^\mu(g_V-g_A\gamma^5)\psi$ , and the axial piece violates parity. In the spin-averaged square this adds a term linear in $\cos\theta$ on top of the QED $1+\cos^2\theta$ , so the distribution is no longer forward-backward symmetric; the resulting asymmetry $A_{FB}$ measures the weak mixing angle. Mechanically nothing new is needed — Casimir's trick still applies with $\gamma^\mu(g_V-g_A\gamma^5)$ in place of $e\gamma^\mu$ , and the extra trace containing $\gamma^5$ produces the parity-odd $\epsilon^{\mu\nu\rho\sigma}$ term that becomes the asymmetry.