Electroweak unification

Two of nature's four forces look nothing alike. Electromagnetism is long-range, parity-even, and carried by a massless photon coupling to conserved electric charge. The weak force is short-range, violates parity maximally, and is carried by heavy $W$ and $Z$ bosons that reach only into left-handed fermions. Yet buried in the weak sector is a clue that these two are the same interaction wearing different masks. The pure $SU(2)_L$ symmetry of the weak isospin triplet contains a neutral member, $W^{(3)}_\mu$ , whose current couples only to left-handed states — but the observed $Z$ boson couples to both chiralities, so $W^{(3)}$ cannot by itself be the $Z$ . Since the photon and the $Z$ are both electrically neutral, the resolution is almost forced: they must be two orthogonal mixtures of $W^{(3)}_\mu$ and a second neutral field $B_\mu$ , the gauge boson of a new $U(1)_Y$ symmetry of weak hypercharge. That single rotation, through the weak mixing angle $\theta_W$ , is the whole of electroweak unification. Demanding that one of the mixtures reproduce ordinary QED — the same photon coupling to the same electric charge $Q$ — fixes the hypercharge assignments to $Y = 2(Q - I^{(3)}_W)$ and ties the two couplings together as $e = g_W\sin\theta_W = g'\cos\theta_W$ . The payoff is not bookkeeping but prophecy: once $\sin^2\theta_W$ is measured, the coupling of the $Z$ to every quark and lepton is fixed, and the $W$ and $Z$ masses are predicted before they are ever seen. This article builds that rotation from the neutral currents up, extracts the fermion couplings, and follows the prediction to the 1983 discovery of the $W$ and $Z$ .

Key takeaways

The neutral gauge fields $B_\mu$ (of $U(1)_Y$ ) and $W^{(3)}_\mu$ (of $SU(2)_L$ ) mix by an angle $\theta_W$ into the physical photon and $Z$ : $A_\mu = \cos\theta_W\,B_\mu + \sin\theta_W\,W^{(3)}_\mu$ , $Z_\mu = -\sin\theta_W\,B_\mu + \cos\theta_W\,W^{(3)}_\mu$ .
Requiring the photon reproduce QED fixes weak hypercharge by $Q = I^{(3)}_W + \tfrac12 Y$ and locks the couplings together: $e = g_W\sin\theta_W = g'\cos\theta_W$ .
The weak mixing angle is a measured constant, $\sin^2\theta_W \approx 0.2315$ ( $\theta_W \approx 28.7°$ ); it is the one number that calibrates the entire neutral sector.
Once $\theta_W$ is known, every $Z$ coupling is a prediction: $c_V = I^{(3)}_W - 2Q\sin^2\theta_W$ , $c_A = I^{(3)}_W$ — the electron's vector coupling comes out near zero because $\sin^2\theta_W$ is close to $\tfrac14$ .
The same doublet Higgs that breaks the symmetry ties the masses together: $M_W = M_Z\cos\theta_W$ , i.e. $\rho = M_W^2/(M_Z^2\cos^2\theta_W) = 1$ at tree level.
Key idea: because the photon and the Z are both electrically neutral, they are two mixed states built from the neutral W(3) of SU(2)_L and the B field of U(1)_Y; demanding the mixture reproduce QED ties the weak and electromagnetic couplings together and makes every Z coupling a prediction.

The core insight

Because both the photon and the Z are electrically neutral, they can be the same two mixed states built from the neutral W(3) of SU(2)_L and the B field of U(1)_Y; demanding the mixture reproduce QED ties the weak and electromagnetic couplings together and makes every Z coupling a prediction.

Where this lives today

The weak mixing angle is now one of the most precisely measured quantities in physics, and measuring it in many ways at once is a global test of the Standard Model. At the $Z$ pole, the LEP and SLC $e^+e^-$ colliders read $\sin^2\theta_W^{\text{eff}} = 0.23155 \pm 0.00004$ from forward–backward and left–right asymmetries — the same asymmetries that hang on the electron's near-zero $c_V$ . The mass relation $M_W = M_Z\cos\theta_W$ is a live frontier: the CDF collaboration's 2022 $M_W = 80.434$ GeV sits in tension with the Standard-Model expectation and the LHC and LEP averages near $80.37$ GeV, and $M_W$ is sensitive through loops to the top-quark and Higgs masses and to anything beyond the Standard Model. Low-energy experiments attack the same angle far from the $Z$ pole: parity-violating electron scattering (Qweak, MOLLER, P2) and atomic parity violation in cesium test whether $\sin^2\theta_W$ "runs" with energy as predicted — a disagreement at any scale would signal new neutral currents. And the invisible width of the $Z$ , which counts neutrinos coupling with $c_L = \tfrac12$ , fixed the number of light neutrino generations at $N_\nu = 2.984 \pm 0.008$ : three, and no more.

The physical picture

Start with two neutral gauge fields: $W^{(3)}_\mu$ , the third component of the weak-isospin triplet that also carries the $W^\pm$ , and $B_\mu$ , the single gauge field of a new abelian symmetry $U(1)_Y$ . Neither is what a detector sees. What propagates through the laboratory are their two orthogonal combinations — one massless (the photon), one heavy (the $Z$ ) — obtained by rotating the $(B, W^{(3)})$ pair through a fixed angle $\theta_W$ in field space. The first figure is that rotation, and the angle it turns through is a real, measured number.

Figure

The neutral sector is a rotation by the weak mixing angle. The gauge eigenstates B (of U(1)_Y) and W³ (of SU(2)_L) rotate through θ_W into the physical photon γ and the Z boson. The measured angle is fixed by sin²θ_W = 0.2315, so θ_W = 28.7° and cos θ_W = 0.877; the same rotation ties the couplings, e = g_W sin θ_W = g′ cos θ_W

The rotation is not a free reparametrization — it is pinned down by a physical demand. One of the two mixtures must be the photon of ordinary electromagnetism: massless, coupling to every fermion in proportion to its electric charge $Q$ with strength $e$ , and to nothing else. Imposing that on the mixed neutral current forces the hypercharge to be $Y = 2(Q - I^{(3)}_W)$ and welds the two gauge couplings together through $e = g_W\sin\theta_W = g'\cos\theta_W$ . Everything left over after the photon is subtracted is the $Z$ current — and because it inherits pieces of both $W^{(3)}$ (left-handed only) and $B$ (chirality-blind), the $Z$ couples to left- and right-handed fermions unequally, with strengths fixed entirely by $\theta_W$ and each fermion's charges. The second figure is that prediction made concrete: the vector and axial couplings of the four fermion types, all from the single number $\sin^2\theta_W$ .

Figure

Every Z coupling from one number. The vector and axial couplings c_V = I₃ − 2Q sin²θ_W and c_A = I₃, computed at sin²θ_W = 0.2315 and plotted in the (c_V, c_A) plane for the four fermion types. The electron lands almost on the c_V = 0 axis, c_V = −0.04, because sin²θ_W sits close to ¼ — a signature prediction the LEP experiments confirmed

The core equations

The neutral sector is defined by one $2\times2$ rotation acting on the two neutral gauge fields. The physical, mass-eigenstate fields — the photon $A_\mu$ and the boson $Z_\mu$ — are orthogonal linear combinations of the interaction-eigenstate fields $B_\mu$ and $W^{(3)}_\mu$ , with the single mixing angle $\theta_W$ setting the entire neutral interaction.

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

where: $A_\mu$ — the electromagnetic potential (physical photon); $Z_\mu$ — the neutral weak boson; $B_\mu$ — the $U(1)_Y$ gauge field; $W^{(3)}_\mu$ — the neutral member of the $SU(2)_L$ triplet; $\theta_W$ — the weak mixing (Weinberg) angle. The rotation is orthogonal, so it preserves the kinetic terms; only the couplings of the rotated fields differ. The angle is fixed by demanding that $A_\mu$ couple exactly as the QED photon does. That requirement identifies electric charge as a specific combination of weak isospin and hypercharge, and forces the two gauge couplings into a single relation with $e$ .

Q = I^{(3)}_W + \tfrac12 Y,\qquad e = g_W\sin\theta_W = g'\cos\theta_W,\qquad \tan\theta_W = \frac{g'}{g_W} \qquad (2)

where: $Q$ — electric charge in units of $e$ ; $I^{(3)}_W$ — third component of weak isospin ( $+\tfrac12$ up-type, $-\tfrac12$ down-type for left-handed doublets, $0$ for right-handed singlets); $Y$ — weak hypercharge; $g_W$ — the $SU(2)_L$ coupling; $g'$ — the $U(1)_Y$ coupling. Read it as: the two independent couplings $(g_W, g')$ are traded for the pair $(e, \theta_W)$ , and $e$ is the smaller of the electroweak couplings, since $e = g_W\sin\theta_W < g_W$ . Subtracting the photon piece leaves the neutral-current interaction of the $Z$ . It is most transparent split by chirality: a left-handed coupling $c_L$ and a right-handed coupling $c_R$ , both built from the same two quantum numbers and the one angle.

\mathcal{L}_{NC} = g_Z\,Z_\mu\sum_f \bar f\,\gamma^\mu\big(c_L P_L + c_R P_R\big)f,\qquad c_L = I^{(3)}_W - Q\sin^2\theta_W,\quad c_R = -Q\sin^2\theta_W \qquad (3)

where: $g_Z = g_W/\cos\theta_W$ — the overall $Z$ coupling; $P_{L,R} = \tfrac12(1\mp\gamma^5)$ — the chiral projectors; $c_L, c_R$ — left- and right-handed couplings of fermion $f$ . The right-handed piece $c_R = -Q\sin^2\theta_W$ is purely electromagnetic in origin (it comes entirely from $B_\mu$ ) — which is why the $Z$ , unlike the $W$ , couples to right-handed fermions at all. For comparison with experiment the same couplings are usually quoted in vector and axial form, obtained by regrouping $c_L P_L + c_R P_R$ into $\tfrac12(c_V - c_A\gamma^5)$ .

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

where: $c_V$ — vector coupling; $c_A$ — axial coupling. The axial coupling is pure weak isospin, blind to $\theta_W$ , while the vector coupling carries all the mixing; the near-vanishing of $c_V$ for charged leptons (because $\sin^2\theta_W\approx\tfrac14$ ) is one of the model's sharpest fingerprints. Finally, the same electroweak doublet Higgs that breaks $SU(2)_L\times U(1)_Y$ down to the electromagnetic $U(1)$ gives the $W$ and $Z$ correlated masses. At tree level their ratio is again just the mixing angle.

M_W = M_Z\cos\theta_W,\qquad \rho \equiv \frac{M_W^2}{M_Z^2\cos^2\theta_W} = 1 \qquad (5)

where: $M_W, M_Z$ — the $W$ and $Z$ masses; $\rho$ — the electroweak rho parameter, exactly $1$ at tree level for a doublet Higgs. Read it as a prediction and a test at once: measure any two of $(M_W, M_Z, \theta_W)$ and the third is fixed, and $\rho = 1$ says the Higgs sits in a doublet, not a larger representation.

Derivation AThe photon and Z as a rotation of the neutral gauge fields

The goal is to show that demanding one combination of $B_\mu$ and $W^{(3)}_\mu$ behave exactly as the QED photon — massless and coupling to $eQ$ — forces the mixing of equation (1), the charge relation $Q = I^{(3)}_W + \tfrac12 Y$ , and the coupling identity $e = g_W\sin\theta_W = g'\cos\theta_W$ . We start from the covariant derivative that local $SU(2)_L\times U(1)_Y$ invariance dictates, isolate its neutral part, rotate to the physical fields, and impose the QED condition term by term.

Step 1. Write the neutral covariant derivative. Local gauge invariance gives $D_\mu = \partial_\mu - i g_W\,W^a_\mu T^a - i g'\tfrac{Y}{2}B_\mu$ ; keeping only the neutral fields, the interaction is $g_W T^3 W^{(3)}_\mu + g'\tfrac{Y}{2}B_\mu$ , where $T^3 = I^{(3)}_W$ acting on a given fermion. Why: this is the only interaction the gauge principle allows, so the photon and Z can be nothing other than combinations of these two fields — there is no third neutral field to hide in.

Step 2. Rotate to physical fields. Invert equation (1): $W^{(3)}_\mu = \sin\theta_W A_\mu + \cos\theta_W Z_\mu$ and $B_\mu = \cos\theta_W A_\mu - \sin\theta_W Z_\mu$ . Substituting, the coefficient of $A_\mu$ becomes $\big(g_W T^3\sin\theta_W + g'\tfrac{Y}{2}\cos\theta_W\big)$ . Why: the physical photon is whatever multiplies $A_\mu$ , so its coupling law is read straight off this coefficient.

Step 3. Impose QED. The photon must couple as $e\,Q\,A_\mu$ with $Q = T^3 + \tfrac12 Y$ . Matching the $T^3$ and $Y$ pieces separately requires $g_W\sin\theta_W = e$ and $g'\cos\theta_W = e$ ; together these give $e = g_W\sin\theta_W = g'\cos\theta_W$ and hence $\tan\theta_W = g'/g_W$ . Why: the photon must not distinguish left from right or know about weak isospin beyond what electric charge encodes, and that single demand is what welds the two couplings into one relation.

Step 4. Read off the Z. With the couplings fixed, the coefficient of $Z_\mu$ collapses to $\tfrac{g_W}{\cos\theta_W}\big(T^3 - Q\sin^2\theta_W\big)$ . Identifying $g_Z = g_W/\cos\theta_W$ and using $T^3 = I^{(3)}_W$ (left) or $0$ (right) gives $c_L = I^{(3)}_W - Q\sin^2\theta_W$ and $c_R = -Q\sin^2\theta_W$ . Why: everything not absorbed into the massless photon is, by construction, the Z current — so the Z couplings are not postulated but are the leftover, completely determined by $\theta_W$ . *Verified: matching the $A_\mu$ coefficient to $eQ$ yields $e = g_W\sin\theta_W = g'\cos\theta_W$ ; the residual $Z_\mu$ coefficient reduces algebraically to $g_Z(T^3 - Q\sin^2\theta_W)$ in verify_aq-pp-electroweak-unification.py.*

Go deeper · Derivation B · 1 minThe Z couplings to every fermion from one angle

Derivation BThe Z couplings to every fermion from one angle

The aim is to turn equation (3) into concrete numbers, showing that the vector and axial couplings of the neutrino, charged lepton, up-type and down-type quarks are all fixed once $\sin^2\theta_W$ is measured, and to explain the striking near-vanishing of the electron's vector coupling. We assign each fermion its $(I^{(3)}_W, Q)$ , feed them through $c_V = I^{(3)}_W - 2Q\sin^2\theta_W$ and $c_A = I^{(3)}_W$ , and evaluate at $\sin^2\theta_W = 0.2315$ .

Step 1. Fix the quantum numbers. Left-handed fields sit in doublets: up-type ( $\nu$ , $u$ ) have $I^{(3)}_W = +\tfrac12$ ; down-type ( $e$ , $d$ ) have $I^{(3)}_W = -\tfrac12$ . Electric charges are $Q_\nu = 0$ , $Q_e = -1$ , $Q_u = +\tfrac23$ , $Q_d = -\tfrac13$ . Why: the axial coupling is simply $c_A = I^{(3)}_W$ , so these assignments already deliver half the answer with no reference to $\theta_W$ at all.

Step 2. Compute the vector couplings. Using $c_V = I^{(3)}_W - 2Q\sin^2\theta_W$ : for the neutrino $c_V = \tfrac12$ ; for the up quark $c_V = \tfrac12 - \tfrac43\sin^2\theta_W = 0.191$ ; for the down quark $c_V = -\tfrac12 + \tfrac23\sin^2\theta_W = -0.346$ . Why: the vector coupling is where the mixing angle enters, so these three numbers are genuine predictions that a measurement of $\theta_W$ turns into observable rates and asymmetries.

Step 3. The electron and the accident of $\sin^2\theta_W\approx\tfrac14$ . For the charged lepton $c_V = -\tfrac12 + 2\sin^2\theta_W = -\tfrac12 + 0.463 = -0.037$ . Had $\sin^2\theta_W$ been exactly $\tfrac14$ , the electron's vector coupling would vanish and the $Z$ would couple to electrons purely axially. Why: this near-cancellation makes electron–Z forward–backward asymmetries small and exquisitely sensitive to $\sin^2\theta_W$ , which is why $e^+e^-$ colliders measure the mixing angle so precisely. *Verified: $c_V(\nu)=0.5$ , $c_V(u)=0.191$ , $c_V(d)=-0.346$ , $c_V(e)=-0.037$ and $c_A = I^{(3)}_W$ at $\sin^2\theta_W=0.2315$ in verify_aq-pp-electroweak-unification.py.*

Go deeper · Derivation C · 1 minThe mass relation M_W = M_Z cos θ_W

Derivation CThe mass relation M_W = M_Z cos θ_W

The goal is to show that the same rotation that mixes the neutral currents also correlates the boson masses, giving $M_W = M_Z\cos\theta_W$ and $\rho = 1$ at tree level, once the symmetry is broken by a single Higgs doublet. We take the Higgs kinetic term, let the neutral component acquire a vacuum value $v$ , and read off the gauge-field mass matrix. The vacuum value $v$ and the mechanism itself are owned by the Higgs-mechanism article and recalled here by name.

Step 1. Source the masses. The Higgs doublet's covariant-derivative term $|D_\mu\phi|^2$ , evaluated at $\langle\phi\rangle = \tfrac{1}{\sqrt2}(0, v)$ , produces quadratic terms in the gauge fields — the boson mass terms. Why: gauge bosons are massless until a charged field gets a vacuum value, so the doublet's vev is the sole origin of $M_W$ and $M_Z$ .

Step 2. Read the charged sector. The $W^1, W^2$ combine into $W^\pm$ with $M_W = \tfrac12 g_W v$ . Why: the charged bosons feel only $SU(2)_L$ , so their mass involves $g_W$ alone — a clean single-coupling result that sets the scale.

Step 3. Diagonalize the neutral sector. The neutral fields $W^{(3)}, B$ get a mass-squared matrix $\tfrac{v^2}{4}\begin{pmatrix} g_W^2 & -g_W g' \\ -g_W g' & g'^2\end{pmatrix}$ , whose eigenvalues are $0$ (the photon) and $M_Z^2 = \tfrac14(g_W^2 + g'^2)v^2$ (the $Z$ ), with eigenvectors exactly the rotation of equation (1). Why: the massless eigenvalue is the unbroken electromagnetic $U(1)$ — the theorem that one combination survives massless is what guarantees a photon at all, and its eigenvector is the photon direction.

Step 4. Take the ratio. $\dfrac{M_W^2}{M_Z^2} = \dfrac{g_W^2}{g_W^2 + g'^2} = \cos^2\theta_W$ , so $M_W = M_Z\cos\theta_W$ and $\rho = M_W^2/(M_Z^2\cos^2\theta_W) = 1$ . Why: the same angle that diagonalizes the currents diagonalizes the masses, so a single doublet forces one relation among three measurable quantities — a falsifiable prediction, not a fit. *Verified: eigenvalues $\{0, \tfrac14(g_W^2+g'^2)v^2\}$ of the neutral mass matrix, $M_W = \tfrac12 g_W v$ , and $M_W/M_Z = \cos\theta_W$ in verify_aq-pp-electroweak-unification.py.*

Limiting cases

Case	Result	Meaning
$\theta_W \to 0$ ( $g' \to 0$ )	$A_\mu \to B_\mu$ , $Z_\mu \to W^{(3)}_\mu$	no mixing; the $Z$ would be pure $SU(2)$ and couple only to left-handed states — contradicts experiment
$\theta_W \to 90°$ ( $g_W \to 0$ )	$A_\mu \to W^{(3)}_\mu$	the photon would live in $SU(2)$ ; excluded because the photon must couple to right-handed charges too
Neutrino ( $Q = 0$ )	$c_R = 0$ , $c_L = c_V = c_A = \tfrac12$	the $Z$ couples to neutrinos purely left-handed — the $Z$ "counts" light neutrino species at LEP
$\sin^2\theta_W = \tfrac14$ exactly	$c_V(\text{charged lepton}) = 0$	leptons couple to the $Z$ axially only; the real value $0.2315$ leaves a tiny $c_V = -0.037$
Right-handed fermion	coupling $= -Q\sin^2\theta_W$	purely hypercharge/EM in origin — the $B_\mu$ admixture is what lets the $Z$ see right-handed states
Tree level, doublet Higgs	$\rho = 1$ exactly	$M_W = M_Z\cos\theta_W$ ; measured deviation ( $\rho \approx 1.0004$ ) is a pure quantum (loop) effect, sensitive to $m_t$ and $m_H$

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

It is not, and this is the misconception the whole subject exists to dispel. If the $Z$ were the bare $W^{(3)}_\mu$ , its current $\bar f\gamma^\mu\tfrac12(1-\gamma^5)f$ would couple only to left-handed particles and right-handed antiparticles — a pure $V-A$ neutral current. But the measured $Z$ couples to both chiralities, unequally: its right-handed coupling $c_R = -Q\sin^2\theta_W$ is small but nonzero and has been measured to permille precision at LEP and SLC. The physical $Z$ is $Z_\mu = \cos\theta_W W^{(3)}_\mu - \sin\theta_W B_\mu$ , a mixture, and the $-\sin\theta_W B_\mu$ piece — chirality-blind because $B_\mu$ couples to hypercharge, not isospin — is precisely what gives the $Z$ its right-handed reach. Equivalently: the photon is the massless mixture and the $Z$ the massive orthogonal one; neither is a pure gauge eigenstate. Forgetting the mixing predicts zero right-handed coupling and a vanishing electron forward–backward asymmetry — both flatly wrong.

Worked examples

1 · Predicting the W and Z masses before they were found (fully worked)

Using only the Fermi constant $G_F = 1.166\times10^{-5}\ \text{GeV}^{-2}$ , the fine-structure constant, and the measured $\sin^2\theta_W = 0.2315$ , estimate $M_W$ and $M_Z$ — the calculation Weinberg and Salam could do a decade before the bosons were discovered.

Step 1. Assemble the mass scale. Matching the electroweak theory to the low-energy Fermi theory gives $M_W^2\sin^2\theta_W = \dfrac{\pi\alpha}{\sqrt2\,G_F}$ . Define the constant $A^2 = \dfrac{\pi\alpha}{\sqrt2\,G_F}$ , so $M_W = A/\sin\theta_W$ .

Step 2. Evaluate $A$ with the classical $\alpha = 1/137.04$ : $A^2 = \dfrac{\pi/137.04}{\sqrt2\times1.166\times10^{-5}} = 1390\ \text{GeV}^2$ , so $A = \mathbf{37.3\ \text{GeV}}$ — the famous Weinberg–Salam constant.

Step 3. Get the tree-level masses. With $\sin\theta_W = \sqrt{0.2315} = 0.481$ and $\cos\theta_W = 0.877$ : $M_W = 37.3/0.481 = \mathbf{77.5\ \text{GeV}}$ and $M_Z = M_W/\cos\theta_W = 77.5/0.877 = \mathbf{88.4\ \text{GeV}}$ .

Step 4. Add the leading quantum correction. The relevant coupling at the weak scale is the running $\alpha(M_Z)\approx1/128$ , not $1/137$ ; repeating with it gives $A = 38.6\ \text{GeV}$ , $M_W = \mathbf{80.2\ \text{GeV}}$ , $M_Z = \mathbf{91.5\ \text{GeV}}$ . These sit right on top of the measured $M_W = 80.4\ \text{GeV}$ , $M_Z = 91.19\ \text{GeV}$ — masses predicted from three low-energy numbers years before UA1 and UA2 saw the particles in 1983. The theory did not fit the masses; it foretold them, one of the great confirmations in the history of physics.

2 · The vector coupling of the electron to the Z

Compute the electron's vector and axial couplings to the $Z$ at $\sin^2\theta_W = 0.2315$ , and explain why the resulting electron forward–backward asymmetry is small and a sensitive probe of the mixing angle.

Step 1. Quantum numbers. The (left-handed) electron has $I^{(3)}_W = -\tfrac12$ and $Q = -1$ . The axial coupling needs nothing else: $c_A = I^{(3)}_W = \mathbf{-\tfrac12}$ .

Step 2. Vector coupling. $c_V = I^{(3)}_W - 2Q\sin^2\theta_W = -\tfrac12 - 2(-1)(0.2315) = -0.5 + 0.463 = \mathbf{-0.037}$ . The two contributions nearly cancel because $2\sin^2\theta_W = 0.463$ is close to $\tfrac12$ .

Step 3. Chiral form as a check. $c_L = I^{(3)}_W - Q\sin^2\theta_W = -0.5 + 0.2315 = -0.269$ and $c_R = -Q\sin^2\theta_W = +0.2315$ ; then $c_V = c_L + c_R = -0.037$ and $c_A = c_L - c_R = -0.500$ , consistent. Because $c_V \approx 0$ , the parity-violating $e^+e^-\to Z$ asymmetry — proportional to $c_V c_A$ — is tiny, roughly $2c_Vc_A/(c_V^2+c_A^2)\approx 0.15$ , and its measured value pins $\sin^2\theta_W$ to five decimals: a small coupling that is a large experimental lever.

Physics of the portrait

Electroweak unification is not the work of one hand but of a decade-long relay, and its "portrait" is best drawn as the model itself — the Glashow–Salam–Weinberg synthesis that turned two forces into one gauge theory and won its three authors the 1979 Nobel Prize.

Origins · 1 minthe Glashow–Salam–Weinberg model and how this was discovered

the Glashow–Salam–Weinberg model

1961–1968, from Glashow's SU(2)×U(1) structure to Weinberg and Salam's spontaneously broken unification · electroweak unification of electromagnetism and the weak force

By the late 1950s the weak interaction had a working phenomenology — Fermi's four-fermion theory dressed in the new $V-A$ chiral structure — but it was non-renormalizable. Sheldon Glashow, in 1961, took the decisive structural step: he proposed the gauge group was $SU(2)_L\times U(1)_Y$ , larger than the naive $SU(2)$ , precisely so a neutral partner to the charged $W$ bosons could mix with a hypercharge field to yield both the photon and a new heavy neutral boson. What Glashow lacked was a way to give the $W$ and $Z$ mass without wrecking gauge invariance; he inserted the masses by hand. That missing piece arrived from a different direction. In 1967 Steven Weinberg — and independently in 1968 Abdus Salam, building on Peter Higgs's mechanism — applied spontaneous symmetry breaking to Glashow's group: a scalar doublet acquires a vacuum value, three of its degrees of freedom become the longitudinal polarizations of the $W^\pm$ and $Z$ , and the fourth is the Higgs boson. In one stroke the bosons got masses, the photon stayed massless, and the couplings and masses were locked together by $\theta_W$ . The theory was widely ignored until 't Hooft proved it renormalizable in 1971, after which its predictions — neutral currents (1973), then the $W$ and $Z$ at exactly the foretold masses (1983) — were confirmed one by one. It stands as the template for how a symmetry, hidden by its own vacuum, unifies forces that look utterly different at low energy.

History

1961

— Glashow proposes the SU(2) × U(1) structure with a neutral partner to the photon

1967

— Weinberg unifies the leptonic electroweak sector via spontaneous symmetry breaking

1968

— Salam presents the same unification independently

1971

— 't Hooft proves spontaneously broken gauge theories are renormalizable

1973

— Gargamelle at CERN observes weak neutral currents, evidence for the Z

1983

— UA1 and UA2 discover the W and Z bosons at the predicted masses

Going further

Texts. Thomson, Modern Particle Physics, §15.3 (electroweak unification) and §15.3.1 (the couplings of the $Z$ ), Table 15.1 · Halzen & Martin, Quarks and Leptons, Ch. 13 (the Weinberg–Salam model) and Ch. 15 · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 10 (gauge theories), §10.2–10.3 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.2 (the Glashow–Weinberg–Salam theory) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29.

Provenance. The neutral-current split into left- and right-handed couplings and the projector algebra are recalled from aq-pp-chiral-weak-structure; the weak isospin and hypercharge assignments $Y = 2(Q - I^{(3)}_W)$ from aq-pp-weak-isospin-hypercharge; the $SU(2)_L\times U(1)_Y$ group and its covariant derivative from aq-pp-electroweak-gauge-group; the charged current and $g_W$ from aq-pp-weak-charged-current. The vacuum value $v$ and the boson-mass mechanism are owned by aq-pp-higgs-mechanism and used here by name, not re-derived. A verifier should confirm: (1) matching the $A_\mu$ coefficient to $eQ$ gives $e = g_W\sin\theta_W = g'\cos\theta_W$ ; (2) $c_V(\nu,e,u,d) = (0.50, -0.037, 0.191, -0.346)$ and $c_A = I^{(3)}_W$ at $\sin^2\theta_W = 0.2315$ ; (3) $M_W/M_Z = \cos\theta_W$ from the neutral mass matrix; (4) the mass estimate $M_W\approx80$ GeV from $G_F$ , $\alpha(M_Z)$ , and $\theta_W$ .

Prerequisites. The Weak Charged Current · Weak Isospin and Hypercharge · The Electroweak Gauge Group · Chiral Structure of the Weak Interaction · The Local Gauge Principle. Leads to. The Weinberg Angle · Neutral Currents · The Z Resonance at LEP · The Higgs Mechanism · The Gauge Boson Mass Problem.

Ask the tutor: Ask me to walk through the neutral-current rotation that turns W³ and B into the photon and Z line by line, or to work out the Z couplings for the muon and the b-quark.

Why can't the W³ of SU(2)_L simply be the Z boson?

Because the two couple to different things. The $W^{(3)}$ current is pure $V-A$ : it sees only left-handed particles and right-handed antiparticles, with the same universal strength $g_W I^{(3)}_W$ . The physical $Z$ , by contrast, is measured to couple to both chiralities with unequal, charge-dependent strengths — its right-handed coupling $c_R = -Q\sin^2\theta_W$ is nonzero and differs fermion to fermion. No pure $SU(2)$ current can reproduce that. The only way to get a chirality-blind, charge-dependent admixture is to mix $W^{(3)}$ with a second neutral field that couples to something other than weak isospin — the hypercharge field $B_\mu$ . So the $Z$ is forced to be a mixture, and the leftover orthogonal mixture is the photon.

What exactly is weak hypercharge, and how is it fixed?

Weak hypercharge $Y$ is the conserved quantum number of the abelian factor $U(1)_Y$ , the "second half" of the electroweak group. It is not electric charge and not weak isospin, but it is fixed by the demand that the two combine into the right electric charge: $Q = I^{(3)}_W + \tfrac12 Y$ , or equivalently $Y = 2(Q - I^{(3)}_W)$ . So for the left-handed lepton doublet ( $\nu$ with $I^{(3)}=+\tfrac12$ , $Q=0$ ; $e$ with $I^{(3)}=-\tfrac12$ , $Q=-1$ ) both members share $Y = -1$ , as a doublet must. The right-handed electron, an $SU(2)$ singlet with $I^{(3)}=0$ and $Q=-1$ , has $Y = -2$ . These assignments are not free: they are exactly what makes the photon come out coupling to $Q$ and nothing else.

Is electroweak "unification" real unification if there are still two couplings?

This is a fair objection, and the honest answer is "partial". The theory still has two independent gauge couplings, $g_W$ and $g'$ , so it does not reduce the two forces to a single number. What it does achieve is structural: they are woven into one gauge group with one symmetry-breaking mechanism, so their couplings, the boson masses, and every neutral-current interaction are all related through the single angle $\theta_W$ and become mutually predictive. Full unification — a single coupling — is the goal of grand unified theories, where $SU(2)_L\times U(1)_Y$ embeds in a larger simple group at high energy. Electroweak unification is the first, firmly confirmed rung of that ladder.

Why are two different values of sin²θ_W quoted, 0.223 and 0.232?

Because " $\sin^2\theta_W$ " is defined by different physical prescriptions that only agree at tree level. The on-shell definition sets $\sin^2\theta_W \equiv 1 - M_W^2/M_Z^2$ , which with the measured masses gives $0.2231$ . The effective leptonic definition is extracted from $Z$ -pole asymmetries and absorbs certain loop corrections, giving $\sin^2\theta_W^{\text{eff}} = 0.23155$ . The $\overline{\text{MS}}$ value at $M_Z$ is $0.23122$ . These differ by about $0.008$ purely because of how quantum corrections are bookkept — the underlying physics is one theory. When you compute, you must use the scheme its formulas assume; mixing schemes is a classic source of permille-level errors. The value $0.2315$ used throughout this article is the effective one appropriate for the fermion couplings.

What does the weak mixing angle actually "mix"?

It mixes the two neutral gauge fields — and only those. The charged bosons $W^\pm$ are built from $W^1$ and $W^2$ and are untouched by $\theta_W$ . But there are two neutral fields, $W^{(3)}$ (from $SU(2)_L$ ) and $B$ (from $U(1)_Y$ ), and the physical neutral bosons — the photon and the $Z$ — are two perpendicular directions in the plane those fields span. The angle $\theta_W$ is simply the tilt between the "gauge" axes $(B, W^{(3)})$ and the "physical" axes $(\gamma, Z)$ . Because that same tilt appears in the couplings ( $e = g_W\sin\theta_W$ ) and the masses ( $M_W = M_Z\cos\theta_W$ ), one geometric angle governs the whole neutral sector.

Does unification mean the electromagnetic and weak forces have equal strength?

No — and this is a common misreading. Unification means the two forces share one gauge structure and one symmetry-breaking origin, not that their observed strengths are equal. In fact the couplings are related by $e = g_W\sin\theta_W$ , so the electromagnetic coupling $e$ is smaller than the weak coupling $g_W$ by the factor $\sin\theta_W \approx 0.48$ . What makes the weak force appear feeble at everyday energies is not a small coupling but the huge mass of the $W$ and $Z$ : the propagator suppression $1/M_W^2$ throttles low-energy weak processes. At energies well above $M_Z$ , that suppression disappears and the two interactions manifestly become comparable — the sense in which they are "unified" is visible only at high energy.