The weak mixing angle

The electroweak sector opens with an awkward surplus: two neutral gauge fields, but the wrong two. The unbroken group $SU(2)_L \times U(1)_Y$ supplies $W^{(3)}_\mu$ from the weak isospin triplet and $B_\mu$ from hypercharge, each massless and each coupling in a way that matches nothing observed — $W^{(3)}$ talks only to left-handed states, $B$ talks to a hypercharge no experiment measures directly. The photon we know couples to electric charge and ignores handedness; the $Z$ we know is massive and couples to both chiralities, but unequally. Neither $W^{(3)}$ nor $B$ is either of these. The single idea that repairs the whole picture is that the physical fields are a rotation of the gauge fields by one angle, the weak mixing angle $\theta_W$ . Rotate $(B, W^{(3)})$ by $\theta_W$ and you get $(A, Z)$ : a massless combination that couples exactly to electric charge $Q$ , and an orthogonal massive combination that is the $Z$ . The moment you demand that the massless field reproduce ordinary QED — neutrino neutral, electron charge $-e$ — the angle stops being free bookkeeping and starts making predictions. It ties the electromagnetic coupling to the two gauge couplings through $e = g_W\sin\theta_W = g'\cos\theta_W$ , fixes every fermion's hypercharge through $Y = 2(Q - I_W^{(3)})$ , and hands you the complete list of $Z$ couplings with no further input. One measured number, $\sin^2\theta_W \approx 0.231$ , then predicts the entire neutral-current spectrum and the ratio $M_W/M_Z$ . This article takes the rotation as the organizing idea and works out everything that flows from it.

Key takeaways

The photon $A_\mu$ and the $Z$ boson $Z_\mu$ are orthogonal mixtures of the neutral gauge fields $B_\mu$ (hypercharge) and $W^{(3)}_\mu$ (weak isospin), set by a single angle $\theta_W$ .
Requiring the massless combination to be ordinary electromagnetism forces $e = g_W\sin\theta_W = g'\cos\theta_W$ and the charge rule $Y = 2(Q - I_W^{(3)})$ .
Every $Z$ coupling is then fixed: $c_L = I_W^{(3)} - Q\sin^2\theta_W$ , $c_R = -Q\sin^2\theta_W$ , equivalently $c_V = I_W^{(3)} - 2Q\sin^2\theta_W$ , $c_A = I_W^{(3)}$ .
The same angle sets the mass ratio $M_W = M_Z\cos\theta_W$ , so $\rho = M_W^2/(M_Z^2\cos^2\theta_W) = 1$ at tree level.
Measured, $\sin^2\theta_W \approx 0.231$ (scheme-dependent at the fourth decimal), giving $\theta_W \approx 28.7°$ ; this is why the $Z$ couples to both chiralities and the neutral current violates parity but not maximally.
Key idea: because the electroweak group has two neutral gauge fields, the physical photon and Z are a single rotation of them; forcing that rotation to recover electromagnetism ties all the couplings together and predicts every Z coupling.

The core insight

Because there are two neutral gauge fields in the electroweak group, the physical neutral bosons (photon and Z) are a rotation of them by a single angle; requiring this rotation to recover ordinary electromagnetism ties all the electroweak couplings together and predicts every Z coupling.

Where this lives today

The weak mixing angle is now one of the most precisely determined parameters in physics, and its value is a battleground. At the $Z$ pole, LEP and SLC turned the angle into a fourth-decimal quantity by measuring asymmetries built from the very couplings of Worked Example 1: the effective leptonic value is $\sin^2\theta_{\rm eff}^{\,\ell} = 0.23155 \pm 0.00004$ . Because $\theta_W$ is a ratio of running couplings, it is scale-dependent, and testing that running is a live program: atomic parity violation in cesium probes it near zero momentum transfer, Qweak at Jefferson Lab measured the proton's weak charge, and the MOLLER (JLab) and P2 (Mainz) experiments aim to pin the low-energy value to per-mille precision — each a test of whether new heavy particles distort the predicted curve. Two anomalies keep the subject unsettled. The NuTeV neutrino-scattering determination came out about three standard deviations from the Standard Model, still debated in terms of nuclear and QCD effects. And in 2022 the CDF collaboration reported $M_W = 80.4335$ GeV, in sharp tension with $M_W = M_Z\cos\theta_W$ folded through the global electroweak fit — a result other experiments do not confirm. The through-line is that a single angle over-constrains masses, couplings, and asymmetries; every new measurement is a consistency test of the rotation this article is about, and any crack would be a signpost to physics beyond the Standard Model.

The physical picture

Start with the two neutral fields the gauge group gives you, $B_\mu$ and $W^{(3)}_\mu$ , and treat them as perpendicular axes. They are not the fields you can build a detector for, because neither is a mass eigenstate once the Higgs field takes its vacuum value: the mass term the Higgs generates is off-diagonal in the $(B, W^{(3)})$ basis, mixing the two. Diagonalizing that mass matrix is a rotation by $\theta_W$ : one eigenvector comes out massless (the photon), the other massive (the $Z$ ). The first figure is that rotation drawn literally — two frames sharing an origin, tilted by $\theta_W$ , with the photon axis carrying the coupling relation that fixes the angle.

Figure

The weak mixing as a rotation by θ_W ≈ 28.7° of the gauge fields B and W³ into the physical photon A and Z boson, with the coupling relation e = g_W sinθ_W = g′ cosθ_W that pins the angle down

The payoff of the rotation is that the angle is not just a label on a diagram — it dictates how strongly the $Z$ couples to every species of matter. Because the $Z$ inherits a piece of $W^{(3)}$ (which sees only left-handed weak isospin $I_W^{(3)}$ ) and a piece of $B$ (which sees hypercharge, hence electric charge $Q$ ), each fermion's $Z$ coupling is a specific combination $I_W^{(3)} - Q\sin^2\theta_W$ for its left-handed part and $-Q\sin^2\theta_W$ for its right-handed part. The second figure plots the vector and axial couplings that result, and the single most striking feature it shows is an accident of the number $0.231$ : the charged lepton's vector coupling nearly vanishes.

Figure

Predicted Z couplings in the vector–axial plane for sin²θ_W = 0.2312: g_A = I_W³ is quantized to ±1/2, while g_V = I_W³ − 2Q sin²θ_W spreads horizontally, and the charged lepton sits almost on the g_V = 0 line at g_V = −0.038

The core equations

The whole structure hangs on the rotation, so it comes first. The physical neutral bosons are defined as the orthogonal combinations of the gauge fields that diagonalize the neutral mass matrix, and the mixing angle is the rotation angle that does it.

A_\mu = B_\mu\cos\theta_W + W^{(3)}_\mu\sin\theta_W, \qquad Z_\mu = -B_\mu\sin\theta_W + W^{(3)}_\mu\cos\theta_W \qquad (1)

where: $A_\mu$ — the photon (massless eigenvector); $Z_\mu$ — the $Z$ boson (massive eigenvector); $B_\mu$ — the $U(1)_Y$ hypercharge gauge field; $W^{(3)}_\mu$ — the third component of the $SU(2)_L$ gauge triplet; $\theta_W$ — the weak mixing angle. The transformation is an orthogonal rotation, so it preserves the kinetic terms; the whole content is the choice of $\theta_W$ that removes the mass-matrix cross term. Fixing $\theta_W$ is not a free choice once you insist the massless combination be the physical photon. Demanding that $A_\mu$ couple to the electromagnetic current — charge $Q$ , blind to chirality — forces a relation between the electromagnetic coupling $e$ and the two gauge couplings.

e = g_W\sin\theta_W = g'\cos\theta_W, \qquad \tan\theta_W = \frac{g'}{g_W}, \qquad \frac{1}{e^2} = \frac{1}{g_W^2} + \frac{1}{g'^2} \qquad (2)

where: $e$ — the electromagnetic (proton-charge) coupling; $g_W$ — the $SU(2)_L$ coupling; $g'$ — the $U(1)_Y$ coupling. The angle is literally the ratio of the two gauge couplings; the last form shows $e$ is smaller than either $g_W$ or $g'$ , so electromagnetism is the weakest-looking face of a unified interaction. The same demand that $A_\mu$ reproduce QED fixes how electric charge is built from the two conserved quantum numbers of the gauge group — the Gell-Mann–Nishijima relation, now derived rather than assumed.

Q = I_W^{(3)} + \frac{Y}{2} \quad\Longleftrightarrow\quad Y = 2\left(Q - I_W^{(3)}\right) \qquad (3)

where: $Q$ — electric charge in units of $e$ ; $I_W^{(3)}$ — third component of weak isospin ( $\pm\tfrac12$ for left-handed doublets, $0$ for right-handed singlets); $Y$ — weak hypercharge. Read right-to-left it is a prediction: measure a fermion's charge and its isospin and its hypercharge is fixed, which is how the entire hypercharge assignment of the Standard Model is pinned down. With the photon accounted for, the orthogonal combination — the $Z$ — has its couplings completely determined. Substituting the rotation into the neutral currents gives the coupling of the $Z$ to a fermion of definite chirality, and hence the vector/axial split.

c_L = I_W^{(3)} - Q\sin^2\theta_W, \quad c_R = -Q\sin^2\theta_W, \qquad c_V = I_W^{(3)} - 2Q\sin^2\theta_W, \quad c_A = I_W^{(3)} \qquad (4)

where: $c_L, c_R$ — the couplings of the $Z$ to the left- and right-handed components of a fermion; $c_V = c_L + c_R$ and $c_A = c_L - c_R$ — the vector and axial couplings. The axial coupling $c_A = I_W^{(3)}$ is purely isospin and does not see $\theta_W$ at all; the entire angle-dependence lives in the vector coupling, which is why $\sin^2\theta_W$ is measured through vector-sensitive observables like asymmetries. The overall strength of these couplings is set by a single neutral-current coupling constant, which is why the $Z$ interaction is often written with $g_Z$ out front.

g_Z = \frac{g_W}{\cos\theta_W} = \frac{e}{\sin\theta_W\cos\theta_W}, \qquad \mathcal{L}_{\rm NC} = g_Z\, Z_\mu \sum_f \bar f\,\gamma^\mu\!\left(c_L P_L + c_R P_R\right) f \qquad (5)

where: $g_Z$ — the neutral-current coupling; $P_L = \tfrac12(1-\gamma^5)$ , $P_R = \tfrac12(1+\gamma^5)$ — chirality projectors; the sum runs over all fermions $f$ . Because $g_Z = g_W/\cos\theta_W > g_W$ , the $Z$ actually couples more strongly per unit coupling than the $W$ , a fact partly hidden by the $Z$ 's larger mass. Finally, the same rotation that produced a massless photon fixes the ratio of the massive gauge-boson masses, because the Higgs field feeds the $(B, W^{(3)})$ mass matrix whose eigenvalues are $0$ and $M_Z$ .

M_W = M_Z\cos\theta_W, \qquad \rho \equiv \frac{M_W^2}{M_Z^2\cos^2\theta_W} = 1 \ \ (\text{tree level}) \qquad (6)

where: $M_W, M_Z$ — the $W$ and $Z$ masses; $\rho$ — the ratio that equals one for any theory whose electroweak symmetry breaks with Higgs doublets. This turns the mixing angle into something directly measurable from masses alone: $\cos^2\theta_W = M_W^2/M_Z^2$ , the "on-shell" definition of the angle.

Derivation AThe rotation that makes a massless photon and fixes the couplings

The goal is to show that the weak mixing angle is not put in by hand but is forced: starting from the $SU(2)_L\times U(1)_Y$ covariant derivative acting on a fermion, we rotate the two neutral gauge fields by an unknown angle, demand that one combination be massless and couple exactly to electric charge $Q$ (ordinary QED with the neutrino neutral), and read off both the coupling relation $e = g_W\sin\theta_W = g'\cos\theta_W$ and the charge rule $Y = 2(Q - I_W^{(3)})$ , along with the $Z$ couplings of the orthogonal combination, all from that single requirement.

Step 1. Write the neutral part of the gauge covariant derivative, $D_\mu \supset -i g_W W^{(3)}_\mu T^3 - i g' \tfrac{Y}{2} B_\mu$ , where $T^3$ acts as $I_W^{(3)}$ on the fermion and $Y$ is its hypercharge. Why: these are the only two neutral generators in the group, so any neutral coupling a fermion can have is a linear combination of $I_W^{(3)}$ and $Y$ — the raw material the rotation will reorganize.

Step 2. Insert the inverse rotation $B_\mu = A_\mu\cos\theta_W - Z_\mu\sin\theta_W$ and $W^{(3)}_\mu = A_\mu\sin\theta_W + Z_\mu\cos\theta_W$ . Collect the terms multiplying $A_\mu$ : the fermion couples to the photon as $-i\big(g_W\sin\theta_W\, I_W^{(3)} + g'\cos\theta_W\,\tfrac{Y}{2}\big)A_\mu$ . Why: the photon is whichever combination the fermions couple to through electric charge, so this bracket must collapse to $e\,Q$ .

Step 3. For that bracket to equal $e\,Q = e\big(I_W^{(3)} + \tfrac{Y}{2}\big)$ for every fermion — regardless of its individual $I_W^{(3)}$ and $Y$ — the coefficient of $I_W^{(3)}$ and the coefficient of $\tfrac{Y}{2}$ must separately match, giving $g_W\sin\theta_W = e$ and $g'\cos\theta_W = e$ . Equating them yields $\tan\theta_W = g'/g_W$ and the charge identity $Q = I_W^{(3)} + Y/2$ , i.e. $Y = 2(Q - I_W^{(3)})$ . Why: this is the crux — one physical demand (the photon is electromagnetism) simultaneously fixes the angle, unifies the couplings, and determines every hypercharge.

Step 4. Now collect the $Z_\mu$ terms: $-i\big(g_W\cos\theta_W\,I_W^{(3)} - g'\sin\theta_W\,\tfrac{Y}{2}\big)Z_\mu$ . Substitute $g_W = e/\sin\theta_W$ , $g' = e/\cos\theta_W$ , and $\tfrac{Y}{2} = Q - I_W^{(3)}$ , and simplify using $\sin^2\theta_W + \cos^2\theta_W = 1$ . The bracket becomes $\dfrac{e}{\sin\theta_W\cos\theta_W}\big(I_W^{(3)} - Q\sin^2\theta_W\big) = g_Z\big(I_W^{(3)} - Q\sin^2\theta_W\big)$ . Why: the orthogonal combination's coupling falls out for free once the photon is fixed — no new parameter enters, which is the source of the model's predictive power.

Step 5. Read off the chiral couplings. For a left-handed fermion $T^3 = I_W^{(3)}$ , giving $c_L = I_W^{(3)} - Q\sin^2\theta_W$ ; for a right-handed fermion $T^3 = 0$ (it is an $SU(2)$ singlet), leaving $c_R = -Q\sin^2\theta_W$ . Why: the right-handed part still couples to the $Z$ — through its hypercharge, hence its charge — which is exactly why the $Z$ , unlike the $W$ , is not purely left-handed. *As a check: setting $Q=0$ , $I_W^{(3)}=+\tfrac12$ gives the neutrino $c_L=\tfrac12$ , $c_R=0$ (left-handed only); setting $I_W^{(3)}=-\tfrac12$ , $Q=-1$ gives the electron $c_L = -\tfrac12+\sin^2\theta_W$ , $c_R = \sin^2\theta_W$ , reproducing $c_V = -\tfrac12+2\sin^2\theta_W$ , $c_A = -\tfrac12$ .*

Go deeper · Derivation B · 1 minThe mass ratio and why rho equals one

Derivation BThe mass ratio and why rho equals one

The aim is to derive $M_W = M_Z\cos\theta_W$ directly from the Higgs mechanism, showing that the very same rotation angle that diagonalizes the couplings also diagonalizes the gauge-boson mass matrix, and that the doublet structure of the Higgs field forces the tree-level relation $\rho = M_W^2/(M_Z^2\cos^2\theta_W) = 1$ — a relation so robust it became one of the sharpest tests of the Standard Model and a window onto physics beyond it.

Step 1. Take the Higgs kinetic term $|D_\mu\phi|^2$ with $\phi$ a complex $SU(2)$ doublet of hypercharge $Y=1$ , and let $\phi$ acquire the vacuum value $\langle\phi\rangle = \tfrac{1}{\sqrt2}(0, v)^{\mathsf T}$ . The gauge-boson masses come from replacing $\phi$ by its vev inside the covariant derivative. Why: masses for the gauge bosons are exactly the terms quadratic in the gauge fields that survive when the Higgs sits still at its minimum.

Step 2. The charged combinations $W^\pm_\mu = (W^1_\mu \mp i W^2_\mu)/\sqrt2$ pick up $M_W = \tfrac12 g_W v$ . Why: the off-diagonal isospin generators rotate the vev, producing a clean diagonal mass for the charged $W$ with no mixing.

Step 3. The neutral fields get a $2\times2$ mass matrix in the $(W^{(3)}_\mu, B_\mu)$ basis, $\dfrac{v^2}{8}\begin{pmatrix} g_W^2 & -g_W g' \\ -g_W g' & g'^2 \end{pmatrix}$ . Its determinant is zero, so one eigenvalue vanishes: that massless eigenvector is the photon, and the nonzero eigenvalue is $\tfrac{v^2}{4}(g_W^2 + g'^2)$ , giving $M_Z = \tfrac12 v\sqrt{g_W^2 + g'^2}$ . Why: a zero eigenvalue is the mathematical statement that a combination stays massless — electromagnetism is unbroken — and the eigenvector angle is precisely $\theta_W$ .

Step 4. Form the ratio: $\dfrac{M_W}{M_Z} = \dfrac{\tfrac12 g_W v}{\tfrac12 v\sqrt{g_W^2+g'^2}} = \dfrac{g_W}{\sqrt{g_W^2+g'^2}} = \cos\theta_W$ , using $\tan\theta_W = g'/g_W$ from (2). Hence $M_W = M_Z\cos\theta_W$ and $\rho = M_W^2/(M_Z^2\cos^2\theta_W) = 1$ . Why: the mass ratio and the coupling ratio are governed by the same angle, which is the deep reason the electroweak sector is over-constrained and therefore testable. *As a numerical check: with the on-shell values $M_Z = 91.1876$ GeV and $\cos^2\theta_W = 1 - \sin^2\theta_W = 0.7769$ , the relation predicts $M_W = 91.1876\times\sqrt{0.7769} = 80.37$ GeV, matching the measured $M_W = 80.38$ GeV.*

Limiting cases

Case	Result	Meaning
$\sin^2\theta_W \to 0$ ( $g' \to 0$ )	$A_\mu \to W^{(3)}_\mu$ , $Z_\mu \to -B_\mu$	no mixing; electromagnetism would be pure $SU(2)$ and the neutrino would carry charge — excluded
$\sin^2\theta_W = \tfrac14$	$c_V^{\,\ell} = I_W^{(3)} - 2Q\sin^2\theta_W = 0$ for the charged lepton	the electron's vector coupling would vanish exactly; nature sits close ( $\sin^2\theta_W \approx 0.231$ ), so $c_V^{\,\ell}$ is small but nonzero
$\sin^2\theta_W \to 1$ ( $g_W \to \infty$ )	$A_\mu \to B_\mu$ , $Z_\mu \to W^{(3)}_\mu$	the photon would be pure hypercharge; also excluded by measured couplings
Right-handed fermion	$c_R = -Q\sin^2\theta_W$	the $Z$ still couples to right-handed matter through charge — the origin of parity violation that is not maximal
Neutrino ( $Q=0$ )	$c_L = I_W^{(3)}$ , $c_R = 0$	the $Z$ couples to neutrinos purely left-handed, exactly like the $W$ ; the invisible width counts $N_\nu$
$\theta_W$ running	$\sin^2\theta_W(\mu)$ varies with scale $\mu$	it is a coupling ratio, so it runs; low-energy and $Z$ -pole values differ by percent-level, all predicted

Common trapThe Z is just the W³ of SU(2)_L

It is not, and believing it costs you the entire neutral-current phenomenology. If the $Z$ were $W^{(3)}$ , it would couple only to left-handed weak isospin, so its coupling to right-handed electrons would be exactly zero and the neutral current would violate parity maximally, like the charged current. Experiment says otherwise: the $Z$ couples to right-handed fermions with strength $c_R = -Q\sin^2\theta_W$ , nonzero for anything charged, and its parity violation is partial. The reason is that the $Z$ is the mixture $-B\sin\theta_W + W^{(3)}\cos\theta_W$ ; the $B$ (hypercharge) piece it carries is what reaches right-handed states. The photon is the orthogonal mixture, and it is massless precisely because it is the combination the Higgs vacuum leaves untouched. Collapsing " $Z$ " to " $W^{(3)}$ " throws away the $B$ admixture and with it every quantitative prediction the angle makes.

Worked examples

1 · The charged lepton's vanishing vector coupling

Compute the vector and axial couplings of the $Z$ to the electron using $\sin^2\theta_W = 0.2312$ , and explain why the smallness of the result makes the leptonic asymmetries such a sensitive probe of the angle.

Step 1. The electron's left-handed component lives in a doublet with $I_W^{(3)} = -\tfrac12$ and it has charge $Q = -1$ . The axial coupling is pure isospin: $c_A = I_W^{(3)} = \mathbf{-\tfrac12}$ , independent of $\theta_W$ .

Step 2. The vector coupling is $c_V = I_W^{(3)} - 2Q\sin^2\theta_W = -\tfrac12 - 2(-1)(0.2312) = -0.5 + 0.4624 = \mathbf{-0.0376}$ . The two terms nearly cancel because $2\sin^2\theta_W = 0.4624$ is close to $\tfrac12$ .

Step 3. The near-cancellation is the whole point. The asymmetries sensitive to $c_V$ scale as $A_\ell \propto 2c_V c_A/(c_V^2 + c_A^2)$ , so with $c_V^{\,\ell}$ tiny, a $0.001$ shift in $\sin^2\theta_W$ moves $c_V$ by $0.002$ — a $5\%$ change in a $0.0376$ quantity. A small quantity that responds by a large fraction is the ideal ruler, which is why the LEP/SLC asymmetries measure $\sin^2\theta_W$ to the fourth decimal even though $c_V^{\,\ell} \approx -0.038$ is barely above zero.

2 · Predicting the W mass from the angle

Given $M_Z = 91.1876$ GeV and the on-shell value $\sin^2\theta_W = 1 - M_W^2/M_Z^2$ , use the mass relation to check internal consistency, and contrast with the $Z$ -pole effective value.

Step 1. The on-shell definition is $\cos^2\theta_W = M_W^2/M_Z^2$ , so $M_W = M_Z\cos\theta_W = M_Z\sqrt{1-\sin^2\theta_W}$ . Taking the on-shell $\sin^2\theta_W = 0.22305$ gives $\cos\theta_W = \sqrt{0.77695} = 0.88145$ .

Step 2. Then $M_W = 91.1876 \times 0.88145 = \mathbf{80.37\ GeV}$ , matching the measured $M_W = 80.38$ GeV to the quoted precision — the relation is self-consistent by construction in this scheme.

Step 3. The subtlety is scheme dependence. The effective leptonic angle measured at the $Z$ pole is $\sin^2\theta_{\rm eff}^{\,\ell} = 0.23155$ , larger than the on-shell $0.22305$ by about $0.008$ . They are the same physical parameter in different renormalization schemes, related by calculable radiative corrections (dominated by the top quark and Higgs loops). Quoting a single " $\sin^2\theta_W$ " without a scheme is meaningless at the fourth decimal; the on-shell value feeds mass relations, the effective value feeds $Z$ couplings, and the $\overline{\rm MS}$ value $\hat s_Z^2 = 0.23122$ is the theorist's default.

3 · The invisible width and counting neutrino generations

Use the partial-width formula $\Gamma(Z\to f\bar f) = \dfrac{G_F M_Z^3}{6\pi\sqrt2}\,N_c\,(c_V^2 + c_A^2)$ to compute the $Z$ decay rate into one neutrino species and into an electron pair, then show how the measured invisible width counts the light neutrino generations.

Step 1. Evaluate the prefactor with $G_F = 1.16637\times10^{-5}\ \mathrm{GeV^{-2}}$ and $M_Z = 91.1876$ GeV: $\dfrac{G_F M_Z^3}{6\pi\sqrt2} = \dfrac{(1.16637\times10^{-5})(7.582\times10^{5})}{26.657} = 0.3318\ \mathrm{GeV} = \mathbf{331.8\ MeV}$ .

Step 2. For a single neutrino ( $N_c = 1$ , $Q = 0$ ): $c_V = c_A = \tfrac12$ , so $c_V^2 + c_A^2 = \tfrac12$ and $\Gamma_\nu = 331.8\times0.5 = \mathbf{165.9\ MeV}$ at tree level. For the electron: $c_V^2 + c_A^2 = (-0.0376)^2 + (-0.5)^2 = 0.2514$ , giving $\Gamma_{ee} = 331.8\times0.2514 = \mathbf{83.4\ MeV}$ , close to the measured $83.9$ MeV.

Step 3. Three neutrino species contribute an invisible width $\Gamma_{\rm inv} = 3\times165.9 = 497.7$ MeV at tree level (the precise Standard Model value, with radiative corrections, is $\Gamma_\nu^{\rm SM} = 167.2$ MeV per species). The measured invisible width is $\Gamma_{\rm inv}^{\rm meas} = 499.0\pm1.5$ MeV. Dividing, $N_\nu = \Gamma_{\rm inv}^{\rm meas}/\Gamma_\nu^{\rm SM} = 499.0/167.2 = \mathbf{2.984\pm0.008}$ . The $Z$ lineshape, through couplings fixed entirely by $\sin^2\theta_W$ , declares that there are exactly three light neutrino generations — one of the most quoted results in all of particle physics.

Physics of the portrait

The weak mixing angle is a monument to a particular style of theoretical physics: guess the symmetry, let it dictate the particle content, and accept the constraints it forces on you. The mixing was written down years before the boson it predicted was seen, and the name it carries belongs to the paper that made it a complete, predictive model of leptons.

Origins · 1 minSteven Weinberg and how this was discovered

Steven Weinberg

1933–2021 · architect of electroweak unification and the weak mixing angle

Steven Weinberg's 1967 paper "A Model of Leptons" is three pages long and among the most cited in the history of physics. Its move was to take Sheldon Glashow's 1961 gauge structure $SU(2)\times U(1)$ — which already contained a neutral partner to the charged weak bosons and a mixing between two neutral fields — and combine it with spontaneous symmetry breaking à la the newly understood Higgs mechanism, so that the $W$ and $Z$ became massive while the photon stayed massless. That single step turned a schematic idea into a machine that predicted the existence of the $Z$ , its mass in terms of a mixing angle, and the entire pattern of neutral-current couplings. Abdus Salam reached an equivalent formulation independently and lectured on it in 1968, which is why the framework is often called Glashow–Weinberg–Salam and the angle the "Weinberg" or "Weinberg–Salam" angle. For a while the model was ignored, until in 1971 Gerard 't Hooft and Martinus Veltman proved that spontaneously broken gauge theories are renormalizable — abruptly the model had finite predictions and became worth taking seriously. Neutral currents appeared in Gargamelle's bubble-chamber film in 1973, and the $W$ and $Z$ themselves at CERN's SppS in 1983, with masses obeying $M_W = M_Z\cos\theta_W$ . Glashow, Salam, and Weinberg shared the 1979 Nobel Prize; 't Hooft and Veltman the 1999 Prize. Weinberg argued forcefully in his later decades that the mixing angle's measured value is a clue — still unexplained — to whatever unifies the forces above it.

History

1961

— Glashow proposes the SU(2)×U(1) structure with a neutral boson mixing two neutral fields

1967

— Weinberg's "A Model of Leptons" introduces the mixing angle and predicts a massive Z

1971

— 't Hooft and Veltman prove spontaneously broken gauge theories renormalizable

1973

— Gargamelle at CERN observes the weak neutral current

1983

— UA1 and UA2 discover the W and Z with masses obeying M_W = M_Z cosθ_W

Going further

Texts. Thomson, Modern Particle Physics, §15.3 (electroweak unification) and §15.3.1 (couplings of the Z), Eqs. (15.17)–(15.38) and Table 15.1 · Halzen & Martin, Quarks and Leptons, Ch. 13 (the Glashow–Weinberg–Salam model) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.2 (the Glashow–Weinberg–Salam theory) · Griffiths, Introduction to Elementary Particles, 2nd ed., §9.2 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29. The key numerical anchors for this article are: $e = g_W\sin\theta_W = g'\cos\theta_W$ and $Y = 2(Q-I_W^{(3)})$ from the photon-coupling requirement; $c_V^{\,e} = -0.0376$ , $c_A^{\,e} = -0.5$ at $\sin^2\theta_W = 0.2312$ ; $M_W = 80.37$ GeV from $M_Z = 91.1876$ GeV and on-shell $\sin^2\theta_W = 0.22305$ ; and $\Gamma_\nu = 165.9$ MeV tree level with $N_\nu = 2.984$ from $\Gamma_{\rm inv} = 499.0$ MeV.

Prerequisites. Weak Isospin and Hypercharge · The Electroweak Gauge Group · Chiral Structure of the Weak Interaction · Chirality and Projection Operators. Leads to. The Weak Neutral Current · Z Boson Decays · The Higgs Mechanism · Electroweak Radiative Corrections.

Ask the tutor: Ask me to derive any single Z coupling from scratch, or to trace how a per-mille shift in $\sin^2\theta_W$ ripples through the W mass, the asymmetries, and the invisible width.

Is the weak mixing angle a fundamental constant or a derived quantity?

It is derived from the two gauge couplings, since $\tan\theta_W = g'/g_W$ , so it is not an independent input beyond $g_W$ and $g'$ . But because $g_W$ and $g'$ are themselves free parameters of the Standard Model — not predicted by anything — the angle is effectively an experimental input too. What makes it special is leverage: once $\sin^2\theta_W$ is fixed by any one measurement, it predicts a huge range of others (all $Z$ couplings, the $W/Z$ mass ratio, neutral-current cross sections). So it behaves like a fundamental constant in practice, while remaining a ratio of couplings in principle.

Why do different sources quote sin²θ_W as 0.231, 0.2312, 0.2315, or 0.223?

Because $\theta_W$ is a renormalized coupling ratio, and its numerical value depends on the renormalization scheme and the energy scale. The on-shell scheme defines $\sin^2\theta_W = 1 - M_W^2/M_Z^2 \approx 0.2231$ purely from measured masses. The effective leptonic angle at the $Z$ pole, which enters the $Z$ couplings and asymmetries, is $\sin^2\theta_{\rm eff}^{\,\ell} \approx 0.23155$ . The $\overline{\rm MS}$ value $\hat s_Z^2 \approx 0.23122$ is common in theory papers. These differ at the third and fourth decimals by calculable radiative corrections, dominated by top-quark and Higgs loops. They are all the same physical parameter; you just have to state the scheme before comparing numbers.

If the electron's vector coupling is nearly zero, why isn't the Z–electron interaction nearly zero?

Because the axial coupling is not small. The full $Z$ –electron coupling involves both $c_V \approx -0.038$ and $c_A = -0.5$ , and the decay rate goes as $c_V^2 + c_A^2 \approx 0.25$ , completely dominated by the axial part. What the small $c_V$ suppresses is specifically the parity-conserving vector piece and the interference observables built from it — the forward–backward and left–right asymmetries, which scale as $c_V c_A$ . So the $Z$ decays to electrons at a healthy rate, but the tiny asymmetries it produces are exquisitely sensitive to the exact value of $\sin^2\theta_W$ , which is why they became the premier way to measure the angle.

What is the ρ parameter and why does it equal one?

The $\rho$ parameter is $M_W^2/(M_Z^2\cos^2\theta_W)$ , the ratio that tests the mass relation of Equation (6). It equals one at tree level for any theory in which electroweak symmetry is broken by Higgs fields in $SU(2)$ doublets (and singlets), because the doublet structure ties the charged and neutral masses together. If the Higgs sector had, say, triplets, $\rho$ would generically differ from one. Radiative corrections shift $\rho$ slightly above one, and that shift is dominated by the large top–bottom mass splitting, so measuring $\rho$ precisely was historically one of the ways the top-quark mass was predicted before it was discovered. It remains a sharp null test for exotic Higgs sectors.

How does the Z boson violate parity if it couples to both chiralities?

Parity violation does not require coupling to only one chirality — it requires coupling to the two chiralities unequally. The $Z$ couples to left-handed fermions with strength $c_L = I_W^{(3)} - Q\sin^2\theta_W$ and to right-handed ones with $c_R = -Q\sin^2\theta_W$ , and these are different (unless $I_W^{(3)} = 0$ ). A parity transformation swaps left and right, so an interaction that treats them differently is not parity-symmetric. The measure of the violation is the axial coupling $c_A = c_L - c_R = I_W^{(3)}$ : nonzero means parity is violated, and its size relative to $c_V$ sets how strongly. The $W$ , with $c_R = 0$ , violates parity maximally; the $Z$ violates it partially.

Could the mixing angle have any value, or is 0.231 special?

Within the Standard Model, nothing predicts the value — it is whatever $g'/g_W$ happens to be, an input from experiment. But grand unified theories, which embed $SU(2)\times U(1)$ into a larger simple group like $SU(5)$ or $SO(10)$ , do predict a value: at the unification scale they typically give $\sin^2\theta_W = 3/8 = 0.375$ . Running that prediction down to the electroweak scale with the renormalization group brings it into the neighborhood of the measured $0.231$ , historically taken as encouraging evidence for unification. So $0.231$ is not special within the Standard Model, but it may be a low-energy shadow of a much simpler structure at high energy.