Properties of the W bosons

The W boson is the heavy, charged mediator of the weak force — a spin-1 particle of mass about 80 GeV whose wavefunction is a plane wave carrying a polarisation four-vector, $W^\mu=\epsilon^\mu_\lambda\,e^{-ip\cdot x}$ . Everything distinctive about it flows from one fact: it is massive. A massless gauge boson like the photon has only two transverse polarisations; a massive vector has three — the transverse circular modes $\epsilon_\pm$ with $S_z=\pm1$ plus a longitudinal mode $\epsilon_L$ with $S_z=0$ . That third degree of freedom is the fingerprint of mass, and it governs both how the W decays and how weak amplitudes behave at high energy. This article synthesises the weak-interaction head — the V−A vertex from aq-pp-v-minus-a, the propagator's polarisation sum from aq-pp-w-boson-propagator, the spinors from aq-pp-dirac-solutions, the two-body rate from aq-pp-particle-decay-rates — and asks what the W actually does. Two consequences organise the rest. First, the decay $W^-\to e^-\bar\nu_e$ shows exactly how a polarisation vector enters a matrix element: the amplitude factorises into $\epsilon_\mu$ dotted with the leptonic V−A current, and averaging over the three polarisations makes the decay isotropic with $\langle|\mathcal M|^2\rangle=\tfrac13 g_W^2 m_W^2$ . Second, summing every open channel fixes the total width — lepton universality makes the three leptonic widths equal, quark channels add a colour factor of three and CKM elements that sum to unity by unitarity, and a small QCD enhancement completes the count, giving $\Gamma_W\approx2.1$ GeV and a lifetime near $3\times10^{-25}$ s. The same longitudinal mode also forces electroweak unification: without Z exchange, $e^+e^-\to W^+W^-$ would violate unitarity.

Key takeaways

The W is a massive spin-1 particle described by a plane wave times a polarisation four-vector $\epsilon^\mu_\lambda$ with three states — two transverse ( $\pm$ ) and one longitudinal (L) — the longitudinal mode being the signature of its mass.
Its decay matrix element is $\mathcal M=\frac{g_W}{\sqrt2}\,\epsilon_\mu(p)\,\bar u(p_e)\,\tfrac12\gamma^\mu(1-\gamma^5)\,v(p_{\bar\nu})$ — the polarisation dotted into the leptonic V−A current.
Each W polarisation gives a distinct spin-1 angular distribution $(1\pm\cos\theta)^2$ and $\sin^2\theta$ ; averaging over the three yields isotropic decay and $\langle|\mathcal M|^2\rangle=\tfrac13 g_W^2 m_W^2$ .
The leptonic partial width is $\Gamma(W^-\to e^-\bar\nu_e)=\dfrac{g_W^2 m_W}{48\pi}=\dfrac{G_F m_W^3}{6\sqrt2\,\pi}\approx0.23$ GeV; summing all channels gives $\Gamma_W\approx2.1$ GeV, BR $(W\to$ hadrons $)\approx67.5\%$ , and $\tau_W\approx3\times10^{-25}$ s.
W-pair production $e^+e^-\to W^+W^-$ would violate unitarity without the Z-exchange diagram, foreshadowing electroweak unification.
Key idea: a massive spin-1 W has three polarisations, and its decay amplitude is just the dot product of its polarisation vector with the V−A lepton current; averaging over polarisations gives isotropic, calculable decay rates whose sum fixes the W width.

The core insight

A massive spin-1 W boson has three polarisations, and its decay amplitude is simply the dot product of its polarisation vector with the V−A lepton current; averaging over polarisations gives isotropic, calculable decay rates whose sum fixes the W width.

Where this lives today

The properties derived here are the daily currency of collider physics. Because the W decays before it can be seen, every measurement reconstructs its decay products: the leptonic channel $W\to\ell\nu$ shows up as one high-energy charged lepton plus large missing transverse momentum — the fingerprint of the escaping neutrino — the signature by which the W was discovered and by which its mass is still measured. The two-thirds hadronic branching ratio computed above is why $W\to q\bar q'$ jets fill top-quark and Higgs events. The W mass itself is a precision observable: entered into the global electroweak fit alongside the Z, top, and Higgs masses, it tests the Standard Model at the loop level — which is why the 2022 CDF II result, several standard deviations above the fit, set off intense scrutiny and remains unresolved against the ATLAS and CMS measurements. At the high-energy frontier, the longitudinal polarisation $\epsilon_L\to p^\mu/m_W$ ties the W to the Higgs: the growth of $W_LW_L$ scattering is tamed by Higgs exchange, so measuring it at the LHC directly probes whether the Higgs mechanism works as the Standard Model demands. The humble polarisation four-vector of this article is, in the end, a test of electroweak symmetry breaking.

The physical picture

A W boson is created off-shell inside a weak vertex, propagates, and — if real, as at a collider — decays within about $10^{-25}$ s into a fermion–antifermion pair. To describe its spin we attach a polarisation four-vector to the plane wave, and the first thing to get right is how many such vectors there are. A four-vector has four components, but the physical field must satisfy $p_\mu\epsilon^\mu=0$ , which removes one, leaving three physical states: the two transverse circular modes carried over from the photon plus one extra — the longitudinal mode — that only a massive particle can have. The first figure fixes this counting and shows the polarisation sum that encodes it.

Figure

The massive W has three polarisation states — two transverse with S_z = ±1 and one longitudinal with S_z = 0 — while the massless photon keeps only the two transverse modes. The polarisation sum Σ ε^μ ε*^ν = −g^μν + p^μ p^ν/m_W² carries the extra p^μp^ν term precisely because of that third state

Once the states are counted, the decay is almost a formality: the polarisation vector threads through the V−A vertex and meets the outgoing spinors. In the ultra-relativistic limit the weak current couples only to left-handed particles and right-handed antiparticles, so the electron and antineutrino from $W^-\to e^-\bar\nu_e$ emerge in a definite spin-1 configuration and the angular distribution is a pure Wigner rotation function. Averaged over the three polarisations the decay is isotropic, and the total rate over all channels is the single number the second figure makes concrete.

Figure

The W total width Γ_W ≈ 2.10 GeV divides by channel: each leptons pair eν, μν, τν takes 10.8% and the six quark combinations take 67.5% together, with the leptonic partial width Γ(W→eν) = g_W² m_W/48π ≈ 0.227 GeV. The short lifetime τ = ħ/Γ_W ≈ 3.1×10⁻²⁵ s follows directly

The core equations

A massive vector field carries its spin in a polarisation four-vector. The plane-wave solution and the transversality condition from the field equation reduce four components to three physical states — the starting point for everything else.

W^\mu(x)=\epsilon^\mu_\lambda\,e^{-ip\cdot x},\qquad p_\mu\,\epsilon^\mu_\lambda=0,\qquad \lambda\in\{+,-,L\} \qquad (1)

where: $\epsilon^\mu_\lambda$ — polarisation four-vector of state $\lambda$ ; $p^\mu$ — W four-momentum with $p^2=m_W^2$ ; the constraint $p_\mu\epsilon^\mu=0$ removes the timelike component, leaving three states. Reading: a spin-1 particle needs a direction attached, and mass permits one more than masslessness does. In the rest frame the three polarisations are the familiar spin-1 states. Boosting to momentum $p^\mu=(E,0,0,|\vec p|)$ leaves the transverse vectors alone but stretches the longitudinal one — whose high-energy limit is the key to weak unitarity.

\epsilon^\mu_\pm=\mp\tfrac{1}{\sqrt2}\,(0,1,\pm i,0),\qquad \epsilon^\mu_L=\tfrac{1}{m_W}\,(|\vec p|,0,0,E)\;\xrightarrow{\,E\gg m_W\,}\;\frac{p^\mu}{m_W} \qquad (2)

where: $\epsilon^\mu_\pm$ — transverse circular modes ( $S_z=\pm1$ ); $\epsilon^\mu_L$ — longitudinal mode ( $S_z=0$ ). Reading: at high energy $\epsilon_L\approx p^\mu/m_W$ grows without bound, so amplitudes with longitudinal W's threaten to blow up — the seed of the argument that forces the Z to exist. Squaring the field means summing over its polarisations — the completeness relation of a massive vector, the object that appears in every W propagator and every spin-summed rate.

\sum_{\lambda}\epsilon^\mu_\lambda\,\epsilon^{*\nu}_\lambda=-g^{\mu\nu}+\frac{p^\mu p^\nu}{m_W^2} \qquad (3)

where: the $-g^{\mu\nu}$ is the photon-like transverse part; the $p^\mu p^\nu/m_W^2$ is the longitudinal addition. Reading: send $m_W\to\infty$ or contract with a conserved massless current and the extra term drops — why light fermions barely feel the longitudinal mode. The decay amplitude is where the polarisation meets matter: the V−A vertex from aq-pp-v-minus-a supplies the $\tfrac12\gamma^\mu(1-\gamma^5)$ structure, and the Dirac spinors from aq-pp-dirac-solutions the external legs.

\mathcal M(W^-\to e^-\bar\nu_e)=\frac{g_W}{\sqrt2}\,\epsilon_\mu(p)\,\bar u(p_e)\,\tfrac12\gamma^\mu(1-\gamma^5)\,v(p_{\bar\nu}) \qquad (4)

where: $g_W$ — weak coupling ( $g_W\approx0.65$ ); $\epsilon_\mu(p)$ — W polarisation; $\bar u,v$ — electron and antineutrino spinors. Reading: the amplitude is the W polarisation dotted into the leptonic charged current $j^\mu=\bar u\,\tfrac12\gamma^\mu(1-\gamma^5)\,v$ . Squaring, summing over final spins, and averaging over the three initial polarisations collapses all angular structure into one constant — the hallmark of a spin-averaged spin-1 decay.

\langle|\mathcal M|^2\rangle=\frac{1}{3}\sum_{\lambda}\sum_{\text{spins}}|\mathcal M|^2=\frac{1}{3}\,g_W^2\,m_W^2 \qquad (5)

where: the factor $\tfrac13$ averages over the W's three polarisations; $2\,p_e\cdot p_{\bar\nu}=m_W^2$ for massless products. Reading: no angular dependence survives the polarisation average, so the decay is isotropic and the rate is a pure number. Feeding the constant amplitude into the two-body rate from aq-pp-particle-decay-rates gives the leptonic partial width, equivalently written through the Fermi constant.

\Gamma(W^-\to e^-\bar\nu_e)=\frac{|\vec p|}{8\pi m_W^2}\langle|\mathcal M|^2\rangle=\frac{g_W^2\,m_W}{48\pi}=\frac{G_F\,m_W^3}{6\sqrt2\,\pi} \qquad (6)

where: $|\vec p|=m_W/2$ (massless pair); $G_F/\sqrt2=g_W^2/8m_W^2$ recalled from aq-pp-w-boson-propagator. Reading: the two forms agree and give $\approx0.227$ GeV; the total width is this times the number of open channels.

Derivation AWhy the W has exactly three polarisation states

The goal is to show, without hand-waving, that a massive spin-1 field carries precisely three physical polarisations — one more than the photon — and to derive the completeness relation (3) that packages them. We start from the four components of a Lorentz four-vector, impose the transversality condition the massive field equation forces, count what survives in the rest frame, and verify the polarisation sum $-g^{\mu\nu}+p^\mu p^\nu/m_W^2$ reduces to the photon case when the mass is removed. The point is that the "extra" longitudinal state is not optional bookkeeping — mass demands it, and it reshapes both the propagator and the decay rate.

Step 1. Begin with the free massive-vector (Proca) field equation $(\partial^2+m_W^2)W^\mu-\partial^\mu(\partial_\nu W^\nu)=0$ . Taking $\partial_\mu$ of the whole equation and using $\partial_\mu\partial^\mu(\partial_\nu W^\nu)=\partial^2(\partial_\nu W^\nu)$ , the two terms cancel except for $m_W^2\,\partial_\mu W^\mu=0$ , so a massive field automatically satisfies $\partial_\mu W^\mu=0$ . In momentum space this is $p_\mu\epsilon^\mu=0$ . Why: mass supplies the constraint; the photon's $m=0$ makes it empty, which is why gauge freedom must remove the photon's extra state by hand instead.

Step 2. Count the survivors in the rest frame $p^\mu=(m_W,0,0,0)$ . The constraint $p_\mu\epsilon^\mu=0$ reads $m_W\,\epsilon^0=0$ , so $\epsilon^0=0$ and the polarisation is purely spatial — a three-vector with three independent directions. Choose $\epsilon_\pm=\mp\tfrac1{\sqrt2}(0,1,\pm i,0)$ and $\epsilon_L=(0,0,0,1)$ ; these are the $S_z=\pm1,0$ eigenstates. Why: four components minus one constraint leaves three, and in the rest frame they are just the ordinary spin-1 multiplet — the longitudinal state on equal footing with the transverse pair.

Step 3. Build the polarisation sum. Using orthonormality $\epsilon_\lambda\cdot\epsilon^*_{\lambda'}=-\delta_{\lambda\lambda'}$ and the fact that the three $\epsilon_\lambda$ together with $p^\mu/m_W$ form a complete basis, the identity $\sum_\lambda\epsilon^\mu_\lambda\epsilon^{*\nu}_\lambda-\frac{p^\mu p^\nu}{m_W^2}=-g^{\mu\nu}$ must hold, giving $\sum_\lambda\epsilon^\mu_\lambda\epsilon^{*\nu}_\lambda=-g^{\mu\nu}+\frac{p^\mu p^\nu}{m_W^2}$ . Why: the four basis vectors resolve the metric, so removing $p^\mu/m_W$ from $-g^{\mu\nu}$ leaves exactly the sum over the three physical states — the extra term is the longitudinal mode made explicit.

Step 4. Check the limits. As $m_W\to\infty$ (or against a conserved current $p_\mu j^\mu=0$ ) the $p^\mu p^\nu/m_W^2$ term drops and the sum becomes the transverse $-g^{\mu\nu}$ ; the longitudinal contribution to any light-fermion process is suppressed by $m_f^2/m_W^2$ . Why: this is the precise sense in which "the W is almost like a photon" for everyday decays — the third state contributes negligibly unless something (a heavy top, high energy) couples to it strongly.

Go deeper · Derivation B · 1 minThe leptonic decay rate of the W

Derivation BThe leptonic decay rate of the W

The goal is to compute $\Gamma(W^-\to e^-\bar\nu_e)$ from first principles and land on the compact result $g_W^2 m_W/48\pi$ , verifying that the answer is a pure constant with no leftover angular dependence. We square the amplitude (4), sum over final spins with the massless completeness relations, contract with the polarisation sum (3), and insert the constant spin-averaged amplitude into the two-body phase-space formula — a clean, parameter-light prediction for a single W channel, checkable against experiment to the percent level.

Step 1. Square and sum over final spins. With massless leptons $\sum u\bar u=\slashed p_e$ and $\sum v\bar v=\slashed p_{\bar\nu}$ , so $\sum_{\text{spins}}|\mathcal M|^2=\frac{g_W^2}{2}\cdot\frac14\,\epsilon_\mu\epsilon^*_\nu\,\mathrm{Tr}\!\left[\slashed p_e\gamma^\mu(1-\gamma^5)\,\slashed p_{\bar\nu}\gamma^\nu(1-\gamma^5)\right]$ . The trace evaluates to $8\left[p_e^\mu p_{\bar\nu}^\nu+p_e^\nu p_{\bar\nu}^\mu-g^{\mu\nu}(p_e\cdot p_{\bar\nu})-i\epsilon^{\alpha\mu\beta\nu}p_{e\alpha}p_{\bar\nu\beta}\right]$ . Why: the projector $(1-\gamma^5)$ selects the left-handed current, and the antisymmetric $\epsilon$ -tensor piece — the parity-violating part — drops out once we sum symmetrically over W polarisations.

Step 2. Contract with the polarisation sum (3). Averaging over the three W polarisations replaces $\epsilon_\mu\epsilon^*_\nu\to\frac13\left(-g_{\mu\nu}+p_\mu p_\nu/m_W^2\right)$ . The $-g_{\mu\nu}$ part gives $\frac13\cdot g_W^2\cdot 2\,(p_e\cdot p_{\bar\nu})$ , while the $p_\mu p_\nu/m_W^2$ part vanishes: using $p=p_e+p_{\bar\nu}$ , $p\cdot p_e=p\cdot p_{\bar\nu}=p_e\cdot p_{\bar\nu}=m_W^2/2$ , the numerator $2(p_e\cdot p_{\bar\nu})^2-m_W^2(p_e\cdot p_{\bar\nu})=0$ . Why: momentum conservation makes the longitudinal piece cancel exactly — the light leptons do not feel the third polarisation, the $m_f^2/m_W^2$ suppression of Derivation A.

Step 3. Assemble the spin-averaged amplitude. With $2\,p_e\cdot p_{\bar\nu}=(p_e+p_{\bar\nu})^2=m_W^2$ , the surviving term is $\langle|\mathcal M|^2\rangle=\frac13\,g_W^2\,m_W^2$ — a constant, independent of decay angle. Why: the isotropy is the content of the polarisation average — a single polarisation decays anisotropically, but an unpolarised sample cannot pick out a direction.

Step 4. Insert into the two-body rate. For a decay into two massless particles $\Gamma=\dfrac{|\vec p|}{8\pi m_W^2}\langle|\mathcal M|^2\rangle$ with $|\vec p|=m_W/2$ , giving $\Gamma=\dfrac{m_W/2}{8\pi m_W^2}\cdot\dfrac13 g_W^2 m_W^2=\dfrac{g_W^2 m_W}{48\pi}$ . Using $g_W^2=4\sqrt2\,G_F m_W^2$ this equals $\dfrac{G_F m_W^3}{6\sqrt2\,\pi}$ . Why: both forms are the same physics through two couplings — $g_W$ from the gauge theory, $G_F$ from low-energy phenomenology — and their agreement cross-checks the electroweak framework.

Go deeper · Derivation C · 1 minSumming the channels to the total width

Derivation CSumming the channels to the total width

The goal is to turn the single leptonic width into the full $\Gamma_W\approx2.1$ GeV by counting every fermion pair the W can decay into, weighting quarks by colour and CKM factors and adding the leading QCD correction. We equate the three leptonic channels by universality, collapse the quark sum to an integer count by CKM unitarity, apply the colour factor, and finish with the QCD enhancement — predicting not just the width but the branching ratios, in particular the two-thirds dominance of hadronic decays that every collider W-tag relies on.

Step 1. Count the leptonic channels. The W decays to $e\bar\nu_e$ , $\mu\bar\nu_\mu$ , $\tau\bar\nu_\tau$ , and because the weak coupling $g_W$ is the same for all generations (lepton universality), each contributes an equal width $\Gamma_{e\nu}$ . That is three identical units. Why: universality is tested to sub-percent precision, so one channel stands in for all three.

Step 2. Count the quark channels. Kinematically open are $u\bar d,\,u\bar s,\,u\bar b,\,c\bar d,\,c\bar s,\,c\bar b$ (the top is far too heavy). Each quark pair carries a colour factor of 3 and a CKM weight $|V_{ij}|^2$ . Summing over the down-type quarks for a fixed up-type quark uses unitarity: $\sum_j|V_{uj}|^2=\sum_j|V_{cj}|^2=1$ . So the two up-type quarks contribute $3\times(1+1)=6$ leptonic units. Why: CKM unitarity erases all the mixing-angle detail — the quark sector behaves as two extra "colourful lepton generations," which is why the count is a clean integer.

Step 3. Add the QCD correction. Real gluon emission and virtual gluon loops multiply each quark channel by $\kappa_{QCD}=1+\alpha_s/\pi+\cdots\approx1.038$ at the W scale ( $\alpha_s(m_W)\approx0.12$ ). The total is $\Gamma_W=(3+6\kappa_{QCD})\,\Gamma_{e\nu}$ . Why: leptonic channels get no such factor — only coloured final states radiate gluons — so QCD tips the balance further toward hadrons.

Step 4. Evaluate and read off branching ratios. With $\Gamma_{e\nu}\approx0.227$ GeV, $\Gamma_W\approx(3+6\times1.038)\times0.227=9.23\times0.227\approx2.10$ GeV. Each leptonic branching ratio is $1/9.23\approx10.8\%$ (32.5% for all leptons) and the hadronic fraction is $6\kappa_{QCD}/9.23\approx67.5\%$ . Why: these are exactly what the second figure shows and experiment confirms — the width and the two-thirds hadronic ratio drop out of nothing but channel counting.

Limiting cases

Case	Result	Meaning
Massless limit $m_W\to0$	third polarisation decouples; $\sum\epsilon\epsilon\to-g^{\mu\nu}$	the W would behave like a photon with only two transverse states — mass is what adds the longitudinal mode
Light-fermion decay	longitudinal contribution $\sim m_f^2/m_W^2\to0$	electrons and light quarks barely feel $\epsilon_L$ ; the rate is set by the transverse modes
Polarisation-averaged decay	$\langle\|\mathcal M\|^2\rangle=\tfrac13 g_W^2 m_W^2$ , isotropic	no preferred direction survives the average over three W spin states
Single-polarisation decay	$(1\pm\cos\theta)^2$ or $\sin^2\theta$	a polarised W decays anisotropically — pure Wigner $d^1_{\lambda\lambda'}(\theta)$ rotation functions
High-energy $W_LW_L$ scattering	amplitude would grow as $s/m_W^2$	unitarity is saved only by Z (and Higgs) exchange — the longitudinal mode forces electroweak unification
Low-energy limit $\|q^2\|\ll m_W^2$	propagator $\to ig_{\mu\nu}/m_W^2$	recovers Fermi's contact theory; $G_F/\sqrt2=g_W^2/8m_W^2$

Common trapThe W has only two polarisation states like the photon

It has three. The photon is massless, and gauge invariance strips it to two transverse polarisations — there is no rest frame in which to define a third. The W is massive, so it has a rest frame, and there it is an ordinary spin-1 object with the full multiplet $S_z=+1,0,-1$ : two transverse and one longitudinal. Dropping the longitudinal mode is not harmless. It supplies the $p^\mu p^\nu/m_W^2$ term in the propagator and polarisation sum, it grows as $p^\mu/m_W$ at high energy, and it is why $e^+e^-\to W^+W^-$ would violate unitarity without a Z boson. Get the count wrong and you get the decay rate, the propagator, and the whole high-energy structure of the weak interaction wrong. The one place it genuinely drops out is the decay to light fermions, where its contribution is suppressed by $m_f^2/m_W^2$ — but that is a result of the calculation, not a licence to forget the state exists.

Worked examples

1 · The leptonic partial width from the coupling (fully worked)

Given the weak coupling $g_W^2=4\sqrt2\,G_F m_W^2$ with $G_F=1.166\times10^{-5}\ \mathrm{GeV}^{-2}$ and $m_W=80.4$ GeV, compute $\Gamma(W^-\to e^-\bar\nu_e)$ two ways and compare with the measured $\approx0.226$ GeV.

Step 1. Fix the coupling. $g_W^2=4\sqrt2\times(1.166\times10^{-5})\times(80.4)^2=5.657\times1.166\times10^{-5}\times6464=\mathbf{0.426}$ , so $g_W=0.653$ and $\alpha_W=g_W^2/4\pi\approx1/29$ — larger than the electromagnetic $\alpha\approx1/137$ .

Step 2. Use the coupling form of (6): $\Gamma_{e\nu}=\dfrac{g_W^2 m_W}{48\pi}=\dfrac{0.426\times80.4}{48\pi}=\dfrac{34.25}{150.8}=\mathbf{0.227\ GeV}=227\ \mathrm{MeV}$ .

Step 3. Cross-check with the Fermi-constant form: $\Gamma_{e\nu}=\dfrac{G_F m_W^3}{6\sqrt2\,\pi}=\dfrac{1.166\times10^{-5}\times(80.4)^3}{6\sqrt2\,\pi}=\dfrac{6.06}{26.66}=\mathbf{0.227\ GeV}$ , identical. The measured leptonic width is $\approx0.226$ GeV, so the tree-level prediction is good to under 1%. Γ(W→eν) = 0.227 GeV, in agreement with experiment.

2 · Total width, branching ratios, and lifetime

Using $\Gamma_{e\nu}=0.227$ GeV and $\kappa_{QCD}=1.038$ , find $\Gamma_W$ , the leptonic and hadronic branching ratios, and the W lifetime.

Step 1. Count channels: three leptonic units plus $3\times2=6$ colour-weighted quark units, each quark unit boosted by $\kappa_{QCD}$ . Effective count $N=3+6\kappa_{QCD}=3+6.228=\mathbf{9.23}$ .

Step 2. Total width: $\Gamma_W=N\,\Gamma_{e\nu}=9.23\times0.227=\mathbf{2.10\ GeV}$ (measured $2.085$ GeV).

Step 3. Branching ratios: each lepton BR $=1/9.23=\mathbf{10.8\%}$ (leptons together $32.5\%$ ); hadrons $=6\kappa_{QCD}/9.23=6.228/9.23=\mathbf{67.5\%}$ .

Step 4. Lifetime: $\tau_W=\hbar/\Gamma_W=(6.582\times10^{-25}\ \mathrm{GeV\,s})/(2.10\ \mathrm{GeV})=\mathbf{3.1\times10^{-25}\ s}$ . Γ_W ≈ 2.10 GeV, BR(hadrons) ≈ 67.5%, τ_W ≈ 3×10⁻²⁵ s.

3 · The decay angular distribution of a polarised W

A sample of $W^-$ is fully polarised with $S_z=+1$ along the beam axis. In the ultra-relativistic V−A limit, what angular distribution does the emitted electron follow, and where is it most likely to go?

Step 1. In the V−A limit the electron is left-handed and the antineutrino right-handed, so the pair combines into a spin-1 state with projection $\lambda'=-1$ along the electron direction. The decay amplitude is then the Wigner rotation matrix $d^1_{+1,-1}(\theta)=\tfrac12(1-\cos\theta)$ .

Step 2. Square it: $\dfrac{d\Gamma}{d\Omega}\propto|d^1_{+1,-1}(\theta)|^2=\tfrac14(1-\cos\theta)^2$ . The rate vanishes at $\theta=0$ (forward) and peaks at $\theta=\pi$ (backward). The electron is emitted preferentially opposite the W spin, following $(1-\cos\theta)^2$ — a purely quantum-mechanical spin-1 signature. Averaging over the three polarisations $(1-\cos\theta)^2+(1+\cos\theta)^2+2\sin^2\theta$ gives a flat distribution, consistent with Example 1.

Physics of the portrait

The W was predicted two decades before anyone could make one. Turning the prediction into a measured 80-GeV particle took a purpose-built proton–antiproton collider and a physicist who bet his career on it.

Origins · 1 minCarlo Rubbia and how this was discovered

Carlo Rubbia

1934– · the experimental discovery of the W and Z bosons at CERN in 1983

By the late 1970s the Glashow–Weinberg–Salam electroweak theory had predicted the W and Z with definite masses near 80 and 91 GeV, but no accelerator could reach them. Carlo Rubbia's decisive idea was not a new machine but a new use of an old one: collide protons against antiprotons in the same ring, delivering the energy needed to make an 80-GeV particle head-on. With Simon van der Meer's stochastic cooling to pack the fragile antiproton beam into usable density, CERN converted its Super Proton Synchrotron into the SppS collider, and Rubbia's UA1 experiment — with the independent UA2 — hunted the rare events in which a quark and antiquark annihilate into a real W. In January 1983 UA1 announced a handful of clean $W\to e\nu$ events — an isolated high-energy electron recoiling against missing transverse momentum from the unseen neutrino, at exactly the predicted mass — with the Z following months later. It was direct confirmation that the weak force is carried by heavy vector bosons with precisely the mass, spin, and couplings the theory required, and Rubbia and van der Meer shared the 1984 Nobel Prize barely a year later. Everything in this article — the mass, the three polarisations, the width, the branching ratios — became, with the SppS, a set of numbers to be measured rather than merely computed.

History

1957

— parity violation is observed and the V−A structure of the weak charged current is established

1968

— the Weinberg–Salam electroweak model predicts massive charged W and neutral Z bosons

1983

— UA1 and UA2 at CERN's SppS collider discover the W boson at about 80 GeV

1984

— Rubbia and van der Meer receive the Nobel Prize for the discovery

2022

— the CDF II experiment reports a high-precision, and controversially anomalous, W-mass measurement

Going further

Texts. Thomson, Modern Particle Physics, §15.1 (properties of the W bosons), §15.1.1 (W decay), §15.1.2 (W-pair production), Appendix D (polarisation four-vectors) · Halzen & Martin, Quarks and Leptons, Ch. 12–13 (weak interactions and the W) · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 9–10 · Peskin & Schroeder, An Introduction to QFT, §20.2 (spontaneously broken gauge symmetry) · Schwartz, QFT and the Standard Model, Ch. 29. The key ingredients assembled in this article are: the V−A vertex $\tfrac12\gamma^\mu(1-\gamma^5)$ ; the massive polarisation sum $-g^{\mu\nu}+p^\mu p^\nu/m_W^2$ and the coupling relation $G_F/\sqrt2=g_W^2/8m_W^2$ ; the external spinors and their completeness $\sum u\bar u=\slashed p$ ; the two-body rate formula $\Gamma=|\vec p|\langle|\mathcal M|^2\rangle/8\pi m^2$ ; and CKM unitarity $\sum_j|V_{ij}|^2=1$ . The central results are: (1) three polarisations from $p_\mu\epsilon^\mu=0$ ; (2) $\langle|\mathcal M|^2\rangle=\tfrac13 g_W^2 m_W^2$ from the trace plus polarisation sum; (3) $\Gamma_{e\nu}=g_W^2 m_W/48\pi=G_F m_W^3/6\sqrt2\pi\approx0.227$ GeV; (4) $\Gamma_W=(3+6\kappa_{QCD})\Gamma_{e\nu}\approx2.10$ GeV with BR(hadrons) $\approx67.5\%$ and $\tau_W\approx3\times10^{-25}$ s.

Prerequisites. The W-Boson Propagator · The V−A Charged Current · Dirac Equation Solutions · Particle Decay Rates · The CKM Matrix. Leads to. Electroweak Unification · Z-Boson Decays · Measuring the W Properties · The Weinberg Angle.

Ask the tutor: Want to see exactly how the $p^\mu p^\nu/m_W^2$ term cancels in the decay but blows up in $W_LW_L$ scattering — the same longitudinal mode with two opposite fates?

Why does the W have three polarisations but the photon only two?

Because the W is massive and the photon is not. A massive spin-1 particle has a rest frame, and there it is an ordinary $S=1$ object with three states $S_z=+1,0,-1$ ; the constraint $p_\mu\epsilon^\mu=0$ removes only the timelike component, leaving three physical polarisations. The photon has no rest frame; gauge invariance removes an extra state, so only the two transverse helicities survive. The "missing" third state of the photon is exactly the one the W keeps — its longitudinal mode — and that mode is what the mass supplies.

What is the longitudinal polarisation and why do people fuss over it?

The longitudinal mode $\epsilon_L$ is the $S_z=0$ state, polarised along the motion. At rest it is unremarkable, but under a boost its polarisation vector becomes $\epsilon_L^\mu=(|\vec p|,0,0,E)/m_W$ , which at high energy approaches $p^\mu/m_W$ and grows without bound. Processes involving longitudinal W's — like $W_LW_L$ scattering — then have amplitudes that rise with energy and would eventually violate unitarity, cured only by Z and Higgs exchange. So the longitudinal mode is the thread that ties the W to the Higgs mechanism.

Why is the weak force "weak" if the coupling g_W is larger than the electric charge?

It is a mass effect, not a coupling effect. Numerically $g_W\approx0.65$ exceeds the electric charge $e\approx0.30$ , so the weak vertex is intrinsically stronger than the electromagnetic one. What makes weak processes rare at low energy is the W's mass in the propagator: every amplitude is suppressed by $1/(q^2-m_W^2)\approx-1/m_W^2$ when the momentum transfer is small. Beta decay is slow because the W is heavy, not because the coupling is feeble — and at energies comparable to $m_W$ , as at a collider, the weak interaction is anything but weak.

Why do hadronic decays dominate the W width?

Pure counting. The W decays to three lepton pairs and to quark pairs from two up-type quarks ( $u,c$ ; top is too heavy). Each quark pair carries a colour factor of three, and summing the CKM elements for the allowed down-type partners gives one by unitarity — so quarks contribute $3\times2=6$ units against the leptons' 3. A small QCD enhancement ( $\kappa_{QCD}\approx1.038$ ) nudges the quark side higher, giving a hadronic fraction $6\kappa_{QCD}/(3+6\kappa_{QCD})\approx67.5\%$ — just over two-thirds, entirely from colour and generation counting.

Where does the QCD correction κ_QCD come from?

When the W decays to a quark pair the quarks can radiate gluons — real emission ( $W\to q\bar q'g$ ) and virtual loops both correct the rate. Summed to leading order this multiplies each quark channel by $\kappa_{QCD}=1+\alpha_s/\pi+\cdots$ , which at the W scale ( $\alpha_s\approx0.12$ ) is about 1.038 — a 4% enhancement. Leptonic channels get no such factor because leptons carry no colour. Small as it is, it shifts the hadronic branching ratio and total width by enough to matter at present precision.

How is the W actually detected if it lives only 10⁻²⁵ seconds?

You never see the W itself — you reconstruct its decay products. In the golden channel $W\to\ell\nu$ the detector records one isolated high-energy charged lepton and a large imbalance in transverse momentum, because the neutrino escapes unseen. That "missing transverse momentum" opposite a lone lepton is the classic W signature, exactly how UA1 found the first W events in 1983. Hadronic decays $W\to q\bar q'$ appear as jet pairs whose invariant mass reconstructs to $m_W$ . The lifetime never enters directly; the decay is effectively instantaneous, so all the physics is in the final-state kinematics.