Decays of the Z boson

The decay of the Z boson is the weak neutral current caught in the act. A Z at rest — all $91.2$ GeV of it — lives barely $3\times10^{-25}$ seconds and then splits into a fermion and its antiparticle, $Z\to f\bar f$ , choosing among every flavour of quark and lepton except the top, which is too heavy. The calculation mirrors that of the W boson with one decisive difference: the W couples only to left-handed chiral states, but the Z couples to both left- and right-handed states, with strengths encoded in the vector and axial-vector couplings $c_V$ and $c_A$ . Those couplings are not free parameters. Each one is fixed by the fermion's weak isospin, its electric charge, and a single universal number — the weak mixing angle $\theta_W$ . That is what makes Z decay a flagship result rather than a bookkeeping exercise. Once $\sin^2\theta_W$ is measured in one place, every partial width follows from $\Gamma\propto(c_V^2+c_A^2)$ , so the entire decay pattern — how often the Z makes jets, charged leptons, or invisible neutrinos — is a parameter-free prediction of electroweak unification. Summing the partial widths over all accessible fermions, with a colour factor of three and a QCD correction for the quark channels, predicts a total width $\Gamma_Z\approx2.5$ GeV split roughly $70\%$ hadrons, $20\%$ invisible, $10\%$ charged leptons. This article assembles that prediction from the neutral-current vertex, and shows how the invisible slice alone told LEP that nature has exactly three light neutrinos.

Key takeaways

The Z decays to every fermion–antifermion pair except $t\bar t$ (since $m_t>m_Z$ ), through the weak neutral current, with partial width $\Gamma(Z\to f\bar f)=\dfrac{g_Z^2 m_Z}{48\pi}(c_V^2+c_A^2)$ .
Unlike the W, the Z couples to both chiralities; the vector/axial couplings are $c_V=T_3-2Q\sin^2\theta_W$ and $c_A=T_3$ , fixed entirely by weak isospin, charge, and the weak mixing angle.
Quark channels carry a colour factor of $N_c=3$ and a QCD correction $[1+\alpha_s(m_Z)/\pi]$ for gluon emission; the top quark is kinematically forbidden.
Summing all channels gives $\Gamma_Z\approx2.495$ GeV, split into $\approx69.9\%$ hadrons, $\approx20.0\%$ invisible (neutrinos), $\approx10.1\%$ charged leptons.
The invisible width fixes the number of light neutrino generations: LEP found $N_\nu=2.984\pm0.008$ — exactly three.
Key idea: once the weak mixing angle is fixed, every Z partial width follows from the fermion's vector and axial-vector couplings via $\Gamma\propto(c_V^2+c_A^2)$ , so the Z's entire decay pattern is a parameter-free prediction of the unified electroweak theory.

Where this lives today

The Z's decays are the working substance of precision electroweak physics. Between 1989 and 1995 the LEP collider at CERN produced some seventeen million Z bosons at the resonance peak, and reconstructing their decays measured the mass to $m_Z=91.1876\pm0.0021$ GeV and the total width to $\Gamma_Z=2.4955\pm0.0023$ GeV — keV-level knowledge of a particle that lives a fraction of a yoctosecond. The invisible width gave $N_\nu=2.984\pm0.008$ , and the leptonic asymmetries pinned $\sin^2\theta_{\text{eff}}=0.23153$ — the input that made the Standard Model predictive enough to forecast the top-quark and Higgs masses before either was seen. A long-standing few-sigma tension between $\sin^2\theta_W$ from the leptonic asymmetry $A_\ell$ and from the b-quark asymmetry $A_{FB}^b$ remains one of the sharpest loose threads in the precision data. The Z decay is also an everyday tool. At the LHC, $Z\to\mu^+\mu^-$ and $Z\to e^+e^-$ are the cleanest "standard candles" in the detector — used to calibrate momentum and energy scales, measure luminosity, and validate reconstruction — because their rate and lineshape are known so exactly. The invisible channel $Z\to\nu\bar\nu$ , tagged as missing energy recoiling against a jet or photon, is a primary dark-matter search mode. And the proposed FCC-ee and CEPC "Tera-Z" factories aim to make $10^{12}$ Z bosons — a hundred thousand times the LEP sample — to push the partial widths and asymmetries to where a deviation would be an unmistakable crack in the Standard Model.

The physical picture

A Z boson at rest is a lump of pure electroweak field energy with nowhere to stay. It carries no conserved charge to protect it, so it decays the instant it can, coupling to a fermion current and splitting into a particle–antiparticle pair. Which pair? Any it can afford: every charged lepton, every neutrino, and every quark lighter than half the Z mass — only the top quark, at $173$ GeV, is barred, because two of them weigh nearly four Z masses. At each vertex the Z speaks through the neutral current with a vector piece $c_V$ and an axial piece $c_A$ , and because the coupling touches both chiralities, two helicity configurations survive in the massless limit: a left-handed fermion with a right-handed antifermion (weight $c_L^2$ ), and its mirror (weight $c_R^2$ ). The same-chirality combinations vanish, killed by $P_LP_R=0$ . The first figure is that vertex.

Figure

The Z decay vertex Z → f f̄ through the weak neutral current. The vertex factor is −ig_Z/2 γ^μ(c_V − c_A γ⁵) with c_V = T₃ − 2Q sin²θ_W and c_A = T₃. Two helicity pairings survive: f_L f̄_R (weight c_L²) and f_R f̄_L (weight c_R²). For a charged lepton c_V = −0.038 while c_A = −0.5, so the vector coupling nearly cancels

The observable that turns this cartoon into physics is not any single decay but the pattern of them. Because each channel's rate is $\propto c_V^2+c_A^2$ with couplings fixed by known quantum numbers, the Z hands out its decays in fractions the theory nails in advance: roughly seven in ten Z bosons make a quark–antiquark pair that hadronises into jets, two in ten vanish into neutrinos we never see, and one in ten makes a clean charged-lepton pair. The second figure is that breakdown — the partial widths and branching ratios a real experiment reports, adding up to the measured $\Gamma_Z=2.4955$ GeV.

Figure

The Z partial widths and branching ratios. Hadrons dominate at 1744 MeV (69.9 percent), the invisible neutrino channels take 499 MeV (20.0 percent, three flavours), and charged leptons take 252 MeV (10.1 percent), summing to the total width Γ_Z = 2495 MeV. The invisible slice divided by the per-neutrino width 166 MeV gives exactly three light neutrino generations

The core equations

The starting point is the neutral-current vertex. Where the charged current is purely left-handed, the neutral current mixes a vector and an axial-vector piece, so one Lorentz structure carries two independent couplings — the essential W–Z distinction.

-i\frac{g_Z}{2}\,\gamma^\mu\big(c_V-c_A\gamma^5\big),\qquad g_Z=\frac{g_W}{\cos\theta_W} \qquad (1)

where: $g_Z$ — neutral-current coupling, related to $g_W$ by the mixing angle; $c_V,c_A$ — the fermion's vector and axial-vector couplings; $\gamma^5$ — the chirality matrix, so $c_V$ multiplies the parity-even current and $c_A$ the parity-odd one. Reading it: set $c_V=c_A$ and only $\tfrac12(1-\gamma^5)$ survives, recovering the W-like limit; in general $c_V\neq c_A$ and both chiralities couple. The couplings are not inputs — they drop out of the electroweak structure once the Z is written as the mixture of the $W^3$ and $B$ gauge fields orthogonal to the photon, fixed by three known quantum numbers.

c_V=T_3-2Q\sin^2\theta_W,\qquad c_A=T_3,\qquad c_L=T_3-Q\sin^2\theta_W,\quad c_R=-Q\sin^2\theta_W \qquad (2)

where: $T_3$ — weak isospin ( $+\tfrac12$ for up-type quarks and neutrinos, $-\tfrac12$ for down-type quarks and charged leptons); $Q$ — electric charge in units of $e$ ; $\sin^2\theta_W\approx0.231$ ; $c_L,c_R$ — the chiral couplings, with $c_V=c_L+c_R$ , $c_A=c_L-c_R$ . Reading it: a right-handed neutrino has $Q=T_3=0$ , so all its couplings vanish — the Z ignores it entirely. Squaring the amplitude, summing over final spins, and averaging over the massive Z's three polarisations collapses to a single compact number (masses neglected). The two chiral pairings each contribute, summing to exactly $c_V^2+c_A^2$ .

\langle|\mathcal M|^2\rangle=\frac{1}{3}\,g_Z^2\,m_Z^2\,(c_V^2+c_A^2) \qquad (3)

where: the $\tfrac13$ averages over the Z's three polarisations; $m_Z=91.19$ GeV. Reading it: the couplings enter symmetrically as $c_V^2+c_A^2$ because in the massless limit the interference term that would distinguish them is antisymmetric and integrates away — the rate cannot tell a vector coupling from an axial one of the same size. Folding this matrix element into two-body phase space ( $|\vec p|=m_Z/2$ ) gives the partial width for one channel — the master formula of the subject. For quarks a colour factor and a QCD correction multiply it.

\Gamma(Z\to f\bar f)=N_c\,\frac{g_Z^2 m_Z}{48\pi}\,(c_V^2+c_A^2)\Big[1+\frac{\alpha_s(m_Z)}{\pi}\Big]_{\text{quarks}} \qquad (4)

where: $N_c=3$ for quarks, $1$ for leptons; $g_Z^2\approx0.55$ ; the bracket $[1+\alpha_s/\pi]\approx1.038$ with $\alpha_s(m_Z)\approx0.118$ applies only to hadronic channels. Reading it: the prefactor $g_Z^2 m_Z/48\pi\approx0.332$ GeV is universal; all channel variation lives in $N_c$ and $c_V^2+c_A^2$ . The total width is the sum over every open channel, and each channel's share is its branching ratio — the numbers the second figure plots.

\Gamma_Z=\sum_f \Gamma(Z\to f\bar f),\qquad B(Z\to f\bar f)=\frac{\Gamma(Z\to f\bar f)}{\Gamma_Z} \qquad (5)

where: the sum runs over $\nu_e,\nu_\mu,\nu_\tau$ ; $e,\mu,\tau$ ; and the quarks $u,d,s,c,b$ (no top). Reading it: with the quark flavours weighted by colour, the hadronic channels dominate the sum, which is why the Z is overwhelmingly a source of jets. Finally, the neutrino channels are invisible, so their combined width is inferred by subtraction, and dividing by the per-neutrino width counts the light generations.

\Gamma_{\text{inv}}=\Gamma_Z-\Gamma_{\text{had}}-3\,\Gamma_{\ell\ell},\qquad N_\nu=\frac{\Gamma_{\text{inv}}}{\Gamma(Z\to\nu\bar\nu)} \qquad (6)

where: $\Gamma_{\text{had}}$ — total hadronic width; $\Gamma_{\ell\ell}$ — one charged-lepton width; $\Gamma(Z\to\nu\bar\nu)\approx166$ MeV per neutrino. Reading it: the invisible width is a direct, model-independent headcount of the light neutrinos — a number no other experiment could pin so cleanly.

Derivation AThe neutral-current couplings from the mixing of the gauge fields

The goal is to derive equation (2), $c_V=T_3-2Q\sin^2\theta_W$ and $c_A=T_3$ , from the structure of electroweak unification, showing the Z couplings are not free inputs but are dictated by weak isospin, electric charge, and one mixing angle. We start from the $SU(2)_L\times U(1)_Y$ currents, identify the photon as one combination of the neutral gauge fields, take the Z as the orthogonal one, and read off its couplings to left- and right-handed fermions.

Step 1. The gauge theory has an isospin triplet $W^{1,2,3}_\mu$ coupling with strength $g_W$ to weak isospin $T^a$ , and a hypercharge field $B_\mu$ coupling with $g'$ to $Y=2(Q-T_3)$ . Only left-handed fermions carry $T_3$ ; right-handed ones are singlets ( $T_3=0$ ) feeling only hypercharge. Why: this left-right asymmetry is the origin of parity violation, and why the Z ends up with different couplings to the two chiralities.

Step 2. The neutral fields $W^3_\mu$ and $B_\mu$ mix by the weak angle into the photon and Z: $A_\mu=\sin\theta_W W^3_\mu+\cos\theta_W B_\mu$ , $Z_\mu=\cos\theta_W W^3_\mu-\sin\theta_W B_\mu$ , with $\tan\theta_W=g'/g_W$ . The photon is chosen to couple to the charge $Q=T_3+Y/2$ with strength $e=g_W\sin\theta_W$ . Why: defining the photon to couple to $Q$ is the whole content of unification — it forces the orthogonal Z to couple to whatever is left over.

Step 3. What is left over is $T_3-Q\sin^2\theta_W$ . The Z current gives coupling $c_L=T_3-Q\sin^2\theta_W$ to left-handed fields and $c_R=-Q\sin^2\theta_W$ to right-handed ones (which have $T_3=0$ , so only the hypercharge remnant survives), with overall strength $g_Z=g_W/\cos\theta_W$ . Why: the right-handed coupling is leakage through the mixing angle — nonzero only because the photon and Z share the $B_\mu$ field, which is why the Z, unlike the W, touches right-handed fermions at all.

Step 4. Repackage into vector and axial form: $c_V=c_L+c_R=T_3-2Q\sin^2\theta_W$ and $c_A=c_L-c_R=T_3$ . Check a lepton: $T_3=-\tfrac12$ , $Q=-1$ , so $c_A=-\tfrac12$ , $c_V=-\tfrac12+2\sin^2\theta_W=-0.038$ . Check a neutrino: $c_V=c_A=+\tfrac12$ . Why: the near-cancellation in the charged-lepton $c_V$ — an accident of $\sin^2\theta_W\approx\tfrac14$ — makes forward-backward asymmetries exquisitely sensitive to the mixing angle.

Go deeper · Derivation B · 1 minThe partial width from the neutral-current vertex

Derivation BThe partial width from the neutral-current vertex

The objective is to derive the master formula (4), $\Gamma(Z\to f\bar f)=\tfrac{g_Z^2 m_Z}{48\pi}(c_V^2+c_A^2)$ per colour, by squaring the amplitude from vertex (1), averaging over the Z's spin, summing over final spins, and integrating over two-body phase space with massless fermions. The key steps are the massive-vector polarisation sum and the Dirac trace, which conspire to leave only $c_V^2+c_A^2$ .

Step 1. Write the amplitude $\mathcal M=-\tfrac{g_Z}{2}\,\epsilon_\mu(p)\,\bar u(p_1)\gamma^\mu(c_V-c_A\gamma^5)v(p_2)$ , with $\epsilon_\mu$ the Z polarisation. Squaring and summing over final spins turns the spinor bilinears into a trace via $\sum u\bar u=\slashed p_1$ , $\sum v\bar v=\slashed p_2$ (masses neglected). Why: spinor completeness removes all reference to spin states, leaving only momenta and couplings — the only things a rate can depend on.

Step 2. Average over the Z's three polarisations using $\sum_{\text{pol}}\epsilon_\mu\epsilon_\nu^*=-g_{\mu\nu}+p_\mu p_\nu/m_Z^2$ . The trace $\mathrm{Tr}[\slashed p_1\gamma^\mu(c_V-c_A\gamma^5)\slashed p_2\gamma^\nu(c_V-c_A\gamma^5)]$ splits into a symmetric part $\propto(c_V^2+c_A^2)$ and an antisymmetric $\epsilon$ -tensor part $\propto c_Vc_A$ . Why: the antisymmetric piece is the parity-violating interference, and it is about to disappear.

Step 3. Contract the trace with the polarisation sum. The antisymmetric $c_Vc_A$ term vanishes against the symmetric tensor $-g_{\mu\nu}+p_\mu p_\nu/m_Z^2$ , and the $p_\mu p_\nu/m_Z^2$ piece contracts to zero for massless final states. What survives is $\langle|\mathcal M|^2\rangle=\tfrac13 g_Z^2 m_Z^2(c_V^2+c_A^2)$ , using $2p_1\!\cdot p_2=m_Z^2$ . Why: this is equation (3) — the width depends only on $c_V^2+c_A^2$ , blind to the relative sign of the couplings even though they violate parity.

Step 4. Insert into the two-body decay formula $\Gamma=\tfrac{|\vec p|}{8\pi m_Z^2}\langle|\mathcal M|^2\rangle$ with $|\vec p|=m_Z/2$ : $\Gamma=\tfrac{1}{16\pi m_Z}\cdot\tfrac13 g_Z^2 m_Z^2(c_V^2+c_A^2)=\tfrac{g_Z^2 m_Z}{48\pi}(c_V^2+c_A^2)$ . Multiply by $N_c=3$ for quarks. Why: this compact result is the entire subject — every Z partial width is this one number times the channel's $c_V^2+c_A^2$ and colour, so the whole pattern is fixed by equation (2).

Go deeper · Derivation C · 1 minThe total width and the branching-ratio pattern

Derivation CThe total width and the branching-ratio pattern

The aim is to assemble the total width $\Gamma_Z\approx2.5$ GeV and the branching ratios ( $\approx70\%$ hadronic, $\approx20\%$ invisible, $\approx10\%$ leptonic) by summing the master formula over all open channels, showing this pattern is a pure prediction of $\sin^2\theta_W$ . We evaluate the universal prefactor once, tabulate $c_V^2+c_A^2$ for each fermion type, apply colour and QCD corrections to quarks, and add up.

Step 1. Fix the universal prefactor $\Gamma_0=\tfrac{g_Z^2 m_Z}{48\pi}$ . With $g_Z^2=4\sqrt2\,G_F m_Z^2=0.549$ and $m_Z=91.19$ GeV, $\Gamma_0=0.332$ GeV. Every partial width is $\Gamma_0$ times $N_c(c_V^2+c_A^2)$ . Why: pulling out $\Gamma_0$ separates the one dimensionful scale from the dimensionless couplings, so only $\sin^2\theta_W$ and quantum numbers remain.

Step 2. Tabulate the coupling factors with $\sin^2\theta_W=0.231$ : neutrinos $c_V^2+c_A^2=0.50$ ; charged leptons $0.25$ ; up-type $0.29$ ; down-type $0.37$ . Multiply by $\Gamma_0$ (and $N_c=3$ for quarks): each $\nu\bar\nu$ gives $166$ MeV, each $\ell^+\ell^-$ $83$ MeV, each $u\bar u/c\bar c$ $285$ MeV, each $d\bar d/s\bar s/b\bar b$ $368$ MeV. Why: the down-type quarks are the largest channels because their $c_V=-0.35$ adds constructively to $c_A$ , whereas the charged leptons are smallest because their $c_V\approx0$ nearly cancels.

Step 3. Sum the visible channels. Three neutrinos give $\Gamma_{\text{inv}}=499$ MeV; three charged leptons give $250$ MeV; the five accessible quark flavours (top forbidden) give a tree-level hadronic total of $1675$ MeV. Apply the QCD correction $[1+\alpha_s/\pi]=1.038$ to the hadronic part, raising it to $\approx1740$ MeV. Why: the QCD factor is not optional bookkeeping — it shifts the hadronic width by nearly $4\%$ , comparable to the entire charged-lepton width, and must be included to match the data.

Step 4. Total: $\Gamma_Z=499+250+1740\approx2490$ MeV, against the measured $2495.5$ MeV — per-cent agreement with no free parameter beyond $\sin^2\theta_W$ . The branching ratios follow: hadrons $69.9\%$ , invisible $20.0\%$ , charged leptons $10.1\%$ . Why: a single measured angle, fed through equations (2) and (4), reproduces the entire decay budget of the Z, and the residual per-cent discrepancy is the size of the electroweak radiative corrections computed in the next article.

Limiting cases

Case	Result	Meaning
Neutrino channel ( $Q=0$ )	$c_V=c_A=T_3=\tfrac12$ ; $c_V^2+c_A^2=\tfrac12$	maximal, purely left-handed coupling; $166$ MeV per species, the invisible width
Charged lepton	$c_V=-0.038\approx0$ , $c_A=-\tfrac12$	vector coupling nearly cancels; width is almost pure axial, $83$ MeV
Down-type quark	$c_V^2+c_A^2=0.37$ , $N_c=3$	largest single channel, $368$ MeV — colour times a large coupling
Top quark	$m_t>m_Z$ : channel closed	the only fermion the Z cannot decay to; $\Gamma(Z\to t\bar t)=0$
$\sin^2\theta_W\to\tfrac14$	charged-lepton $c_V\to0$ exactly	leptonic forward-backward asymmetry would vanish; measures the angle
W-like limit $c_V=c_A$	only left-handed pairing, weight $c_L^2$ ; $c_R=0$	recovers the pure $V-A$ charged current — the Z reduces to a W-type coupling

Common trapThe Z couples only to left-handed fermions, like the W

It does not — this is the single most common error in Z physics. The W couples purely to left-handed states through the $V-A$ current $\gamma^\mu(1-\gamma^5)$ , but the Z couples to both chiralities, with generally unequal strengths $c_L=T_3-Q\sin^2\theta_W$ and $c_R=-Q\sin^2\theta_W$ . The right-handed coupling is nonzero for every charged fermion because the Z inherits a piece of the hypercharge field $B_\mu$ it shares with the photon. What is forbidden — the grain of truth the misconception distorts — are the same-chirality pairings $f_L\bar f_L$ and $f_R\bar f_R$ , which vanish because $P_LP_R=0$ . So in massless kinematics only $f_L\bar f_R$ and $f_R\bar f_L$ survive, but both survive, weighted by $c_L^2$ and $c_R^2$ . The neutrino is the sole exception: with $Q=0$ its $c_R$ vanishes, so the Z couples to it left-handed only — the one place it genuinely behaves like the W.

Worked examples

1 · The total Z width from first principles (the flagship number)

Using $\sin^2\theta_W=0.231$ , $g_Z^2=0.549$ , and $m_Z=91.19$ GeV, predict $\Gamma_Z$ and compare with the measured $2.4955$ GeV.

Step 1. Fix the universal prefactor: $\Gamma_0=\dfrac{g_Z^2 m_Z}{48\pi}=\dfrac{0.549\times91.19}{150.80}=\mathbf{0.332\ GeV}$ . Every partial width is $\Gamma_0\,N_c\,(c_V^2+c_A^2)$ .

Step 2. Compute the coupling factors. Neutrinos: $c_V=c_A=\tfrac12$ , so $c_V^2+c_A^2=0.50$ . Charged leptons: $c_V=-0.038$ , $c_A=-0.5$ , so $0.251$ . Up-type: $c_V=0.192$ , $c_A=0.5$ , so $0.287$ . Down-type: $c_V=-0.346$ , $c_A=-0.5$ , so $0.370$ .

Step 3. Assemble the channels. Invisible (3 neutrinos): $3\times0.332\times0.50=\mathbf{0.498\ GeV}$ . Charged leptons (3 flavours): $3\times0.332\times0.251=\mathbf{0.250\ GeV}$ . Hadrons, tree level: up-type $2\times3\times0.332\times0.287=0.571$ plus down-type $3\times3\times0.332\times0.370=1.104$ , i.e. $\mathbf{1.675\ GeV}$ .

Step 4. Apply the QCD correction to the hadronic part: $1.675\times1.038=1.738$ GeV. Total: $\Gamma_Z=0.498+0.250+1.738=\mathbf{2.486\ GeV}$ . The measured value is $2.4955$ GeV, so the tree-plus-QCD prediction is low by only $0.4\%$ — the residual is the electroweak radiative correction. A single mixing angle predicts $\Gamma_Z\approx2.49$ GeV against $2.4955$ measured, with no other free parameter.

2 · The invisible width and counting neutrino generations

From the measured $\Gamma_Z=2.4955$ GeV, hadronic width $\Gamma_{\text{had}}=1.7444$ GeV, and single charged-lepton width $\Gamma_{\ell\ell}=0.0840$ GeV, extract the invisible width and the number of light neutrino species.

Step 1. The invisible width is what remains after subtracting everything visible: $\Gamma_{\text{inv}}=\Gamma_Z-\Gamma_{\text{had}}-3\Gamma_{\ell\ell}=2.4955-1.7444-3(0.0840)=2.4955-1.7444-0.2520=\mathbf{0.4991\ GeV}$ .

Step 2. The Standard-Model width per neutrino species is $\Gamma(Z\to\nu\bar\nu)=\Gamma_0\times0.50=\mathbf{0.166\ GeV}$ . Each invisible flavour contributes exactly this much.

Step 3. Divide: $N_\nu=\dfrac{\Gamma_{\text{inv}}}{\Gamma(Z\to\nu\bar\nu)}=\dfrac{0.4991}{0.1662}=\mathbf{3.00}$ . The invisible width holds room for precisely three light neutrinos and no more.

Step 4. The full LEP analysis, using the ratio method $N_\nu=(\Gamma_{\text{inv}}/\Gamma_{\ell\ell})(\Gamma_{\ell\ell}/\Gamma_\nu)_{\text{SM}}$ to cancel common uncertainties, gives $N_\nu=2.984\pm0.008$ . The Z's invisible decays count exactly three light neutrino generations from a single decay width. The tiny $2\sigma$ shortfall below $3$ is a known effect of photon-radiation modelling in the fit, not evidence for fewer than three neutrinos.

3 · Why the leptonic vector coupling is a magnifying glass on $\sin^2\theta_W$

The forward-backward asymmetry in $Z\to\ell^+\ell^-$ is $A_\ell=\dfrac{2c_Vc_A}{c_V^2+c_A^2}$ . Show why it is extraordinarily sensitive to the weak mixing angle, and evaluate it.

Step 1. For a charged lepton, $c_V/c_A=(T_3-2Q\sin^2\theta_W)/T_3=1-4\sin^2\theta_W$ . With $\sin^2\theta_W=0.231$ this is $1-0.924=\mathbf{0.076}$ — a small number sitting near zero, which is the whole point.

Step 2. Write $A_\ell=\dfrac{2(c_V/c_A)}{1+(c_V/c_A)^2}=\dfrac{2(0.076)}{1+0.0058}=\mathbf{0.150}$ . Because $c_V/c_A$ is near zero, $A_\ell\approx2(1-4\sin^2\theta_W)$ , so a $1\%$ change in $\sin^2\theta_W$ shifts $A_\ell$ by about $8\%$ .

Step 3. That amplification is why leptonic asymmetries, not the total width, give the sharpest measurement of the mixing angle. The measured $A_\ell=0.1465$ corresponds to an effective $\sin^2\theta_{\text{eff}}^{\text{lept}}=0.23153$ , known to five decimals. The near-cancellation of the leptonic $c_V$ turns a modest asymmetry into a precision probe: measuring $A_\ell$ pins $\sin^2\theta_W$ tighter than any width could.

Physics of the portrait

The Z boson's decay pattern is the most complete vindication of one three-page paper. The couplings that fix every branching ratio were written down, in exactly the form used above, by a young theorist who was trying to do something else entirely.

Origins · 1 minSteven Weinberg and how this was discovered

Steven Weinberg

1933–2021 · the 1967 model of leptons that fixed the neutral-current couplings

Steven Weinberg was an American theoretical physicist whose 1967 paper "A Model of Leptons" — barely three pages long — became the most-cited article in the history of particle physics. He had set out to build a theory of the strong interactions using spontaneously broken symmetry, and realised the machinery fit the weak interaction instead. By gauging $SU(2)_L\times U(1)_Y$ and breaking it with a scalar field, he unified the electromagnetic and weak forces for leptons, predicting a heavy neutral boson — the Z — alongside the charged W. Crucially, he did not leave its couplings free: they emerged fixed by weak isospin, electric charge, and a single mixing angle, in precisely the form $c_V=T_3-2Q\sin^2\theta_W$ , $c_A=T_3$ that governs every Z decay. At the time there was no evidence for a weak neutral current, and the paper attracted almost no attention for years. 't Hooft's 1971 proof of renormalisability, the 1973 discovery of the neutral current at Gargamelle, and the 1983 discovery of the W and Z themselves turned the neglected model into the electroweak Standard Model. Weinberg shared the 1979 Nobel Prize with Glashow and Salam. When LEP later measured the Z's partial widths and asymmetries to parts per thousand and found them exactly where his couplings placed them, it was the most stringent confirmation any theory of matter has received.

History

1961

— Glashow builds the SU(2)×U(1) structure with a neutral heavy partner to the W

1967

— Weinberg fixes the neutral-current couplings through the mixing angle and predicts the Z; Salam gives the model independently the next year

1973

— the Gargamelle bubble chamber at CERN observes the weak neutral current

1983

— the UA1 and UA2 experiments discover the Z through Z → e⁺e⁻ and μ⁺μ⁻

1989

— LEP measures the Z lineshape and invisible width, fixing the number of light neutrinos at three

Going further

Texts. Thomson, Modern Particle Physics, §15.4 (decays of the Z) and §16 (electroweak precision at LEP) · Halzen & Martin, Quarks and Leptons, §15.4–15.5 (the neutral current and Z couplings) · Griffiths, Introduction to Elementary Particles, 2nd ed., §10.4–10.5 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.2–20.4 (the electroweak Lagrangian and neutral current) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29 (electroweak symmetry breaking).

Provenance. The neutral-current couplings $c_V=T_3-2Q\sin^2\theta_W$ , $c_A=T_3$ are owned by aq-pp-neutral-current and aq-pp-weinberg-angle; the gauge-field mixing that produces them by aq-pp-electroweak-unification; the chiral projectors and helicity structure by aq-pp-weyl-chirality; and the two-body phase-space measure by aq-pp-particle-decay-rates. This article synthesises them into the full width and branching ratios. A verifier should confirm: (1) $c_V,c_A$ for each fermion from equation (2) with $\sin^2\theta_W=0.231$ ; (2) $\langle|\mathcal M|^2\rangle=\tfrac13 g_Z^2 m_Z^2(c_V^2+c_A^2)$ from the trace and polarisation sum; (3) $\Gamma_0=g_Z^2 m_Z/48\pi=0.332$ GeV and per-channel widths $\nu:166$ , $\ell:83$ , up: $285$ , down: $368$ MeV; (4) $\Gamma_Z\approx2.49$ GeV after colour and QCD corrections vs $2.4955$ measured; (5) branching ratios $69.9\%/20.0\%/10.1\%$ ; (6) $N_\nu=\Gamma_{\text{inv}}/\Gamma_\nu=3$ .

Prerequisites. Electroweak Unification · The Weak Neutral Current · The Weak Mixing Angle · Particle Decay Rates and Phase Space · Weyl Chirality and Projectors. Leads to. The Z Resonance at LEP · The Number of Generations · Electroweak Radiative Corrections.

Ask the tutor: Want the polarisation sum and Dirac trace done line by line, or how the LEP lineshape turns into $N_\nu=3$ ?

Why does the Z decay to so many more hadrons than leptons?

Two effects stack up. First, quarks come in three colours, so every quark channel carries a factor $N_c=3$ . Second, there are five accessible quark flavours ( $u,d,s,c,b$ ) against three charged leptons, and the down-type quarks in particular have a large coupling factor $c_V^2+c_A^2=0.37$ because their vector and axial couplings add constructively, whereas the charged leptons have the smallest factor, $0.25$ , because their vector coupling nearly cancels. Multiply colour by flavour count by coupling strength and the hadronic channels sweep up about $70\%$ of all Z decays — which is why the Z is a copious source of jets and a clean leptonic decay is comparatively rare and precious.

What exactly is the "invisible width" and how is it measured if nothing is seen?

The invisible width is the part of the Z's decay rate that goes into neutrinos, $Z\to\nu\bar\nu$ , which leave no trace in any detector. Since you cannot see these decays, you measure everything else — the hadronic and charged-lepton widths — precisely, and the total width from the shape of the Z resonance as you scan the beam energy across the peak. The invisible part is then what is left over: $\Gamma_{\text{inv}}=\Gamma_Z-\Gamma_{\text{had}}-3\Gamma_{\ell\ell}$ . That result, about $499$ MeV, divided by the calculated per-neutrino width of $166$ MeV, gives three. It is a beautiful piece of physics: an unmeasurable channel counted by the hole it leaves in a measurable total.

How does counting the Z's invisible decays tell us there are three neutrinos?

Each light neutrino species adds a fixed $\approx166$ MeV to the total width, so the Z's width acts like a scale that weighs the number of neutrino types: more species means a wider resonance. LEP measured it precisely enough to read off $N_\nu=2.984\pm0.008$ , cleanly excluding a fourth light neutrino that couples normally to the Z. The word "light" matters — this counts only neutrinos below $m_Z/2\approx45$ GeV, since a heavier one could not be produced. It does not forbid heavy sterile neutrinos that do not couple to the Z, but for ordinary light neutrinos the answer is definitively three.

Why does the partial width depend on c_V² + c_A² and not on c_V and c_A separately?

Because in the massless limit the decay rate cannot distinguish a vector coupling from an axial one of the same magnitude. When you square the amplitude and sum over spins, the vector–axial cross term $c_Vc_A$ comes attached to an antisymmetric tensor (the Levi-Civita $\epsilon$ -symbol) encoding parity violation. That term integrates to zero once you average over the Z spin and integrate over the symmetric two-body phase space, leaving only the parity-even $c_V^2+c_A^2$ . The parity-violating information is not lost — it reappears in angular distributions like the forward-backward asymmetry, which is odd under the exchange the total rate is even under. So the rate sees $c_V^2+c_A^2$ , but the angular shape sees the interference $c_Vc_A$ , which is what makes asymmetries such sharp probes of the couplings.

Is the Z its own antiparticle, and does that affect its decays?

Yes. The Z is electrically neutral and carries no conserved quantum number distinguishing particle from antiparticle, so it is its own antiparticle — like the photon. This fits its decays: it goes to $f\bar f$ for any fermion, treating particle and antiparticle symmetrically, because there is no Z charge to conserve. It is the essential contrast with the W, which is charged ( $W^+$ , $W^-$ are distinct) and always decays to pairs with net charge such as $W^-\to e^-\bar\nu_e$ . The Z's self-conjugacy is also why it mediates purely "neutral current" processes that change no flavour and no charge — the smoking-gun prediction of electroweak unification.

Why is the Z so short-lived compared to the muon or the pion?

Because its decay is not suppressed by a heavy propagator. The muon and charged pion decay through a virtual W boson, and the enormous $M_W^2\approx(80\text{ GeV})^2$ in the propagator denominator throttles the rate by a factor of order $(m/M_W)^4$ — which is why they live microseconds to nanoseconds. The Z, by contrast, decays as a real particle straight into a fermion pair at a single vertex, with no propagator suppression. Its width is therefore of order $g_Z^2 m_Z\sim$ a few GeV, and its lifetime $\tau=\hbar/\Gamma_Z\approx2.6\times10^{-25}$ s — so short that the Z never travels a measurable distance and is only ever seen as a resonant bump in the production cross section, never as a track.

Decays of the Z boson

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · The total Z width from first principles (the flagship number)

2 · The invisible width and counting neutrino generations

3 · Why the leptonic vector coupling is a magnifying glass on $\sin^2\theta_W$

Physics of the portrait

Going further

Before this

Where this leads

Decays of the Z boson

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · The total Z width from first principles (the flagship number)

2 · The invisible width and counting neutrino generations

3 · Why the leptonic vector coupling is a magnifying glass on sin⁡2θW\sin^2\theta_Wsin2θW​

Physics of the portrait

Going further

Before this

Where this leads

3 · Why the leptonic vector coupling is a magnifying glass on $\sin^2\theta_W$