The electroweak gauge group SU(2)xU(1)

The gauge principle is almost too tidy — demand a symmetry locally and a force falls out. A single $U(1)$ phase forces the photon; the eight rotations of colour $SU(3)$ force the gluons. The weak interaction refuses to cooperate. Run the same machine on the group $SU(2)$ and it hands you three massless gauge bosons — but experiment shows only two are charged, the third is not the photon, and all three are heavy. Worse, the weak force is maximally chiral: it touches left-handed particles and ignores right-handed ones outright. No single simple group reproduces that. The resolution, found by Glashow in 1961, is that the correct symmetry is a product: $SU(2)_L \times U(1)_Y$ . The $SU(2)_L$ factor acts only on left-handed doublets of weak isospin — its generators are the Pauli matrices $T = \tfrac12\boldsymbol\sigma$ , so it demands three gauge fields $W^1, W^2, W^3$ . A second, independent $U(1)_Y$ generated by weak hypercharge $Y$ demands a fourth field $B$ . The two charged combinations $W^\pm$ reproduce the observed $V{-}A$ charged currents to the letter. The neutral fields $W^3$ and $B$ then mix, and only that mixture — rotated by the weak angle $\theta_W$ — sorts into the massless photon coupling to $Q = T^3 + Y/2$ and a heavy neutral $Z$ that couples to everything. This article builds that group from the ground up and shows why nature needed two factors, not one.

Key takeaways

The electroweak symmetry is a product group $SU(2)_L \times U(1)_Y$ : three $W$ fields from $SU(2)_L$ plus one $B$ field from hypercharge $U(1)_Y$ — four gauge fields in all.
$SU(2)_L$ acts only on left-handed doublets ( $T = \tfrac12\boldsymbol\sigma$ ); right-handed fermions are weak-isospin singlets ( $T = 0$ ) blind to the $W$ fields — the source of parity violation.
The charged combinations $W^\pm_\mu = \tfrac{1}{\sqrt2}(W^1_\mu \mp iW^2_\mu)$ reproduce the $V{-}A$ current $\tfrac{g}{\sqrt2}\,\bar\nu\gamma^\mu\tfrac12(1-\gamma^5)e$ exactly.
Electric charge is not put in by hand: $Q = T^3 + Y/2$ (Gell-Mann–Nishijima) ties the two factors together and fixes every fermion's hypercharge.
The neutral $W^3$ and $B$ mix by the weak angle into the photon and $Z$ , forcing $e = g\sin\theta_W = g'\cos\theta_W$ with $\sin^2\theta_W \approx 0.231$ .
The theory predicts a weak neutral current (found 1973) and the boson masses $M_W = M_Z\cos\theta_W$ (found 1983) — from group structure alone.
Key idea: one simple group cannot be the weak force; the product $SU(2)_L \times U(1)_Y$ acting only on left-handed doublets forces four gauge fields whose charged pair gives the $V{-}A$ currents and whose neutral pair mixes into the photon and the $Z$ .

The core insight

A single gauge group cannot reproduce the weak interaction. Demanding local invariance under the product $SU(2)_L \times U(1)_Y$ — with $SU(2)_L$ acting only on left-handed doublets — forces four gauge fields: the charged pair $W^\pm$ reproduces the observed $V{-}A$ currents, and the neutral pair $W^3, B$ mixes through the weak angle into the massless photon (coupling to $Q = T^3 + Y/2$ ) and the massive $Z$ .

Where this lives today

The electroweak group is the most precisely tested structure in physics. At LEP the $Z$ was produced by the millions, and the shape of its resonance pinned $M_Z=91.1876\pm0.0021\ \mathrm{GeV}$ , the number of light neutrino species at exactly three, and the effective mixing angle $\sin^2\theta_W^{\rm eff}=0.23155$ to five digits — every one a check on $SU(2)_L\times U(1)_Y$ . The couplings, the mixing angle, and the boson masses are over-constrained, so the global electroweak fit tests whether the whole product-group structure is self-consistent — and it is, to the per-mille level. That fit predicted the top-quark mass from radiative corrections before the top was found in 1995, and boxed in the Higgs mass before 2012. The frontier now is stress-testing it: the 2022 CDF measurement of $M_W$ sits several standard deviations high and, if confirmed, would signal physics beyond the minimal group. Neutral currents — Glashow's prediction — are today the everyday tool of neutrino experiments and of atomic and electron-scattering parity tests, each reading $\sin^2\theta_W$ at a different energy and tracing its predicted running. The search for an extra $Z'$ is, at heart, a search for a factor beyond $SU(2)_L\times U(1)_Y$ .

The physical picture

A weak-isospin doublet is a two-component column that the $SU(2)_L$ rotations stir together — the upper and lower members differ by one unit of electric charge, like $(\nu_e, e^-)_L$ . The decisive experimental fact is chirality: only the left-handed projection of a fermion sits in such a doublet. The right-handed electron, up quark, and down quark are singlets — one-component objects the $SU(2)_L$ generators cannot touch, so they never emit or absorb a $W$ . This lopsidedness is not bolted on afterwards; it is written into the representation content of the group, and it is exactly why the weak interaction violates parity. The first figure lays out that content and recovers electric charge from the two quantum numbers $T^3$ and $Y$ .

The three $W$ fields are not yet the particles a detector sees. Two of them, $W^1$ and $W^2$ , pair into the charged $W^+$ and $W^-$ that step between doublet members — the carriers of beta decay, reproducing the $V{-}A$ current established by aq-pp-v-minus-a. The neutral $W^3$ is the subtle one: it is not the photon, and cannot be, because it couples only to left-handed particles while electromagnetism is parity-symmetric. The escape is the second factor. The hypercharge field $B$ is also neutral, and the two mix: a rotation by the weak angle $\theta_W$ turns $(W^3, B)$ into the physical photon and $Z$ . The second figure is that mixing, with the real angle and masses that make it quantitative.

Figure

The four gauge fields sort into the observed bosons. W¹ and W² combine into the charged W± that carry the V−A currents; the neutral W³ and B rotate by the weak angle θ_W into the massless photon and the heavy Z. With sin²θ_W ≈ 0.231 (θ_W ≈ 28.7°) the photon couples to electric charge through e = g sinθ_W = g′ cosθ_W, and the masses obey M_W = M_Z cosθ_W, i.e. 80.4 = 91.2 × 0.881 GeV

The core equations

The whole structure is packed into one object: the covariant derivative for the product group. It carries two independent couplings — $g$ for $SU(2)_L$ and $g'$ for $U(1)_Y$ — and one gauge field per generator, giving three $W$ 's and one $B$ .

D_\mu = \partial_\mu + ig\,T^i W^i_\mu + ig'\,\frac{Y}{2}\,B_\mu,\qquad T^i = \tfrac12\sigma^i,\quad [T^i,T^j] = i\epsilon^{ijk}T^k \qquad (1)

where: $g,\,g'$ — the $SU(2)_L$ and $U(1)_Y$ gauge couplings; $T^i=\tfrac12\sigma^i$ — the three weak-isospin generators (Pauli matrices), acting only on left-handed doublets; $W^i_\mu$ ( $i=1,2,3$ ) — the three $SU(2)_L$ gauge fields; $B_\mu$ — the single hypercharge field; $Y$ — weak hypercharge. Reading it: on a right-handed singlet $T^i=0$ , so only the $B$ term survives — right-handed fermions feel hypercharge but never the $W$ 's. The two factors are welded together by a single relation that recovers ordinary electric charge from the two internal quantum numbers. It is the Gell-Mann–Nishijima formula, and it is what makes the theory electroweak rather than two unrelated forces.

Q = T^3 + \frac{Y}{2} \qquad (2)

where: $Q$ — electric charge in units of $e$ ; $T^3$ — the third component of weak isospin ( $\pm\tfrac12$ for doublet members, $0$ for singlets); $Y$ — weak hypercharge. Reading it: (2) defines the hypercharge of every multiplet once its charges are known — lepton doublet $Y=-1$ , $e_R$ has $Y=-2$ , quark doublet $Y=+\tfrac13$ — and it is the bridge that forces the neutral fields to mix in just the way that leaves electromagnetism parity-symmetric. The first two $W$ fields are not mass eigenstates; the electrically charged physical fields are their complex combinations, and inserting them into the $SU(2)_L$ current produces exactly the $V{-}A$ charged current.

W^\pm_\mu = \frac{1}{\sqrt2}\big(W^1_\mu \mp iW^2_\mu\big),\qquad \mathcal L_{CC} = -\frac{g}{\sqrt2}\Big(W^+_\mu\,\bar\nu_L\gamma^\mu e_L + W^-_\mu\,\bar e_L\gamma^\mu \nu_L\Big) \qquad (3)

where: $\bar\nu_L\gamma^\mu e_L = \bar\nu\gamma^\mu\tfrac12(1-\gamma^5)e$ — the left-handed (vector-minus-axial) current; $g/\sqrt2$ — the charged-current coupling. Reading it: $W^+$ raises the doublet ( $e_L\to\nu_L$ ) and $W^-$ lowers it, so these are the carriers of beta decay; the $\tfrac12(1-\gamma^5)$ projector is automatic because only $\psi_L$ was placed in the doublet. The two neutral fields mix by an orthogonal rotation through the weak angle $\theta_W$ , and demanding that one combination couple to electric charge $Q$ fixes that angle and welds the two couplings to $e$ .

\begin{pmatrix} A_\mu \\ Z_\mu \end{pmatrix} = \begin{pmatrix} \cos\theta_W & \sin\theta_W \\ -\sin\theta_W & \cos\theta_W \end{pmatrix}\begin{pmatrix} B_\mu \\ W^3_\mu \end{pmatrix},\qquad e = g\sin\theta_W = g'\cos\theta_W,\quad \tan\theta_W = \frac{g'}{g} \qquad (4)

where: $A_\mu$ — the photon; $Z_\mu$ — the neutral weak boson; $\theta_W$ — the weak (Weinberg) mixing angle. Reading it: the photon is the particular mixture that couples to $Q=T^3+Y/2$ with universal strength $e$ ; the orthogonal mixture is the $Z$ . The detailed derivation of $\theta_W$ and its running is owned by aq-pp-weinberg-angle. Working the neutral couplings through (2) and (4) gives the $Z$ current — chiral, and different for every fermion — alongside the parity-even photon current.

\mathcal L_{NC} = e\,A_\mu\sum_f Q_f\,\bar f\gamma^\mu f \;+\; \frac{g}{\cos\theta_W}\,Z_\mu\sum_f \bar f\gamma^\mu\big(T^3_f - Q_f\sin^2\theta_W\big)f \qquad (5)

where: the sum runs over all fermions $f$ ; $T^3_f=\pm\tfrac12$ for left-handed, $0$ for right-handed. Reading it: the photon coupling is pure $Q$ (no chirality), while the $Z$ coupling carries $T^3$ (so it distinguishes left from right) minus a charge-dependent piece $Q\sin^2\theta_W$ — this weak neutral current, developed in aq-pp-neutral-current, was the theory's boldest prediction. Finally, matching the low-energy charged current to Fermi's four-fermion theory, and the mass relation forced by the mixing, connect the group to measured numbers.

\frac{G_F}{\sqrt2} = \frac{g^2}{8M_W^2},\qquad M_W = M_Z\cos\theta_W \qquad (6)

where: $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ — the Fermi constant; $M_W,\,M_Z$ — the boson masses. Reading it: the first relation is why the weak force is weak at low energy (the heavy $W$ propagator suppresses it); the second is a parameter-free prediction of the group structure once $\theta_W$ is measured — the masses that were confirmed in 1983. The masses themselves come from the Higgs mechanism (aq-pp-higgs-mechanism), recalled here by name.

Derivation ACharged currents from local SU(2) on left-handed doublets

The goal is to start from the raw $SU(2)_L$ part of the covariant derivative acting on a left-handed doublet and show that the physical charged bosons $W^\pm$ , together with the $V{-}A$ current structure, emerge with no extra input — the chirality projector $\tfrac12(1-\gamma^5)$ appearing automatically because only the left-handed field was placed in the doublet. We expand the interaction, combine the off-diagonal fields into raising and lowering bosons, and read off the current.

Step 1. Write the doublet and the interaction. For the first lepton generation $\chi_L=(\nu_L,\,e_L)^\top$ with $T^i=\tfrac12\sigma^i$ , the $SU(2)_L$ interaction inside $\bar\chi_L\,i\gamma^\mu D_\mu\,\chi_L$ is $-g\,\bar\chi_L\gamma^\mu T^i\chi_L\,W^i_\mu$ . Only $W^1$ and $W^2$ change the electric charge of the state, since $\sigma^1,\sigma^2$ are off-diagonal. Why: the off-diagonal Pauli matrices are precisely the operators that turn an $e_L$ into a $\nu_L$ and back, so they must assemble into the charge-raising and -lowering bosons.

Step 2. Combine the off-diagonal fields. Using $W^\pm_\mu=\tfrac{1}{\sqrt2}(W^1_\mu\mp iW^2_\mu)$ and $\sigma^\pm=\tfrac12(\sigma^1\pm i\sigma^2)$ , the sum $W^1_\mu\sigma^1+W^2_\mu\sigma^2 = \sqrt2\,(W^+_\mu\sigma^+ + W^-_\mu\sigma^-)$ . Since $T^1W^1+T^2W^2=\tfrac12(W^1\sigma^1+W^2\sigma^2)$ , the charged part of the interaction is $-\tfrac{g}{\sqrt2}(W^+_\mu\,\bar\chi_L\gamma^\mu\sigma^+\chi_L + W^-_\mu\,\bar\chi_L\gamma^\mu\sigma^-\chi_L)$ . Why: the raising operator $\sigma^+=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ picks the lower member out and deposits it in the upper slot, so $\bar\chi_L\gamma^\mu\sigma^+\chi_L=\bar\nu_L\gamma^\mu e_L$ — a single term stepping $e\to\nu$ .

Step 3. Expose the $V{-}A$ structure. Because $\nu_L=\tfrac12(1-\gamma^5)\nu$ and $e_L=\tfrac12(1-\gamma^5)e$ , and using $[\tfrac12(1-\gamma^5)]^2=\tfrac12(1-\gamma^5)$ together with $\gamma^\mu\tfrac12(1+\gamma^5)=\tfrac12(1-\gamma^5)\gamma^\mu$ , the current collapses to $\bar\nu_L\gamma^\mu e_L=\bar\nu\gamma^\mu\tfrac12(1-\gamma^5)e$ . Why: the left-handed projector, inserted once by the doublet assignment, propagates through the gamma-matrix algebra to give exactly the vector-minus-axial form measured in muon and beta decay — the $V{-}A$ law is a theorem here, not an input.

Step 4. Match to Fermi theory. At energies far below $M_W$ the $W$ propagator contracts to a point, $1/(q^2-M_W^2)\to -1/M_W^2$ , so two charged currents joined by a $W$ reduce to a four-fermion contact term with strength $G_F/\sqrt2 = g^2/8M_W^2$ . Why: this is the quantitative payoff — the historic weakness of the weak force is not a small coupling ( $g\approx0.65$ is comparable to $e$ ) but the large mass $M_W$ in the denominator, and this relation is how $g$ was first extracted from the measured $G_F$ .

Go deeper · Derivation B · 2 minWhy the neutral W is not the photon and the mixing that fixes it

Derivation BWhy the neutral W is not the photon and the mixing that fixes it

The purpose is to prove that the neutral member of the $SU(2)_L$ triplet, $W^3$ , cannot be the photon, and to show that mixing it with the hypercharge field $B$ through a single rotation is the unique way to get a field coupling to electric charge $Q=T^3+Y/2$ . The angle is not free — demanding a parity-symmetric, charge-proportional photon coupling fixes it and welds $g,g'$ to $e$ . We collect the neutral part of the covariant derivative, impose the photon condition, and read off both $e=g\sin\theta_W$ and the $Z$ coupling.

Step 1. Collect the neutral interaction. The neutral part of $D_\mu$ is $ig\,T^3W^3_\mu + ig'\tfrac{Y}{2}B_\mu$ . If $W^3$ alone were the photon it would couple to $T^3$ , which is $\pm\tfrac12$ for left-handed fermions and $0$ for right-handed ones — a maximally parity-violating coupling. Why: electromagnetism is exactly parity-symmetric and couples right-handed electrons as strongly as left-handed ones, so a field coupling to $T^3$ is disqualified as the photon on the spot — the $U(1)_Y$ factor is not optional.

Step 2. Rotate the neutral fields. Write $W^3_\mu = A_\mu\sin\theta_W + Z_\mu\cos\theta_W$ and $B_\mu = A_\mu\cos\theta_W - Z_\mu\sin\theta_W$ , an orthogonal rotation by an unknown angle $\theta_W$ . Substitute to find the coefficient of $A_\mu$ : $g\,T^3\sin\theta_W + g'\tfrac{Y}{2}\cos\theta_W$ . Why: any physical relabelling of two neutral fields is an orthogonal rotation; the one free angle $\theta_W$ is the single knob we will turn to make $A_\mu$ behave like the photon.

Step 3. Impose the photon condition. For $A_\mu$ to couple as $e\,Q=e(T^3+\tfrac{Y}{2})$ we need the $T^3$ and $\tfrac{Y}{2}$ pieces to carry the same strength $e$ : $g\sin\theta_W = e$ and $g'\cos\theta_W = e$ . Both hold simultaneously only if $\tan\theta_W=g'/g$ , and then the $A_\mu$ coupling is exactly $e(T^3+\tfrac{Y}{2})=eQ$ . Why: this is the crux — requiring the photon to see electric charge and nothing else fixes the mixing angle in terms of the two couplings and makes $e=g\sin\theta_W=g'\cos\theta_W$ ; the electroweak unification is this single equation.

Step 4. Read off the $Z$ . The orthogonal coefficient of $Z_\mu$ is $g\,T^3\cos\theta_W - g'\tfrac{Y}{2}\sin\theta_W$ ; using $g'=g\tan\theta_W$ and $\tfrac{Y}{2}=Q-T^3$ this becomes $\tfrac{g}{\cos\theta_W}\big(T^3 - Q\sin^2\theta_W\big)$ . Why: the leftover neutral field is forced to exist and to couple in this chiral, charge-shifted way — a brand-new weak neutral current, distinct from both the photon and the charged $W$ , whose 1973 discovery in neutrino scattering was the first confirmation that the group really is $SU(2)_L\times U(1)_Y$ .

Go deeper · Derivation C · 2 minFixing the hypercharges and the charge-cancellation miracle

Derivation CFixing the hypercharges and the charge-cancellation miracle

This derivation shows that the Gell-Mann–Nishijima relation $Q=T^3+Y/2$ is not merely bookkeeping but a constraint that assigns a definite hypercharge to every multiplet, and that the resulting assignments make the electric charges of one fermion generation sum to zero — the condition that quietly cancels the gauge anomalies which would otherwise render the theory inconsistent. We assign $Y$ multiplet by multiplet, then add the charges over a generation and watch quarks and leptons conspire.

Step 1. Assign lepton hypercharges. For the left doublet $(\nu,e)_L$ both members must share one $Y$ (they are one $SU(2)_L$ object); with $Q=0,-1$ and $T^3=+\tfrac12,-\tfrac12$ , equation (2) gives $Y=-1$ for both. The right-handed electron is a singlet ( $T^3=0$ ) with $Q=-1$ , so $Y=-2$ . Why: the doublet's two members are forced to the same hypercharge because $U(1)_Y$ cannot tell them apart, and the split into their different charges comes entirely from $T^3$ — the two factors sharing the work.

Step 2. Assign quark hypercharges. The left doublet $(u,d)_L$ has $Q=+\tfrac23,-\tfrac13$ and $T^3=+\tfrac12,-\tfrac12$ , giving $Y=+\tfrac13$ . The singlets follow: $u_R$ with $Q=+\tfrac23$ has $Y=+\tfrac43$ , and $d_R$ with $Q=-\tfrac13$ has $Y=-\tfrac23$ . Why: the fractional quark hypercharges are not chosen — they are read straight off (2) once the fractional charges are known, and their oddness relative to the leptons is exactly what the next step needs.

Step 3. Sum the electric charges over a generation. Counting the three colours of each quark: $\sum Q = \underbrace{3(+\tfrac23)}_{u} + \underbrace{3(-\tfrac13)}_{d} + \underbrace{0}_{\nu} + \underbrace{(-1)}_{e} = 2 - 1 + 0 - 1 = 0$ . Why: this vanishing is not a coincidence of the charges — it is required for the cancellation of the triangle (gauge) anomalies that would otherwise spoil renormalisability, and it holds only because there are exactly three colours and the quark charges are $\tfrac23,-\tfrac13$ .

Step 4. Draw the consequence. The anomaly-freedom condition ties the lepton and quark sectors together: remove the quarks and the leptons alone are anomalous; the factor of three from colour and the fractional charges are precisely what is needed to cancel. Why: the electroweak group is only consistent as a quantum theory because each generation pairs a colour-triplet of fractionally charged quarks with a colourless lepton doublet — the charge quantisation we observe is a fingerprint of that cancellation, and it is why the proton charge is exactly opposite the electron's.

Limiting cases

Case	Result	Meaning
Right-handed fermion	$T^i=0$ , only $B$ couples	singlets are blind to the $W$ 's — the origin of parity violation
Drop $U(1)_Y$ ( $g'\to0$ )	$\theta_W\to0$ , $A_\mu\to B_\mu$ , $Z\to W^3$	photon becomes pure hypercharge; the chiral $W^3$ cannot be electromagnetism — the group fails
Unmixed limit ( $\theta_W\to0$ )	$e\to0$ , no photon coupling to $T^3$ sector	electromagnetism switches off; mixing is what makes the neutral fields physical
Low energy ( $q^2\ll M_W^2$ )	$G_F/\sqrt2 = g^2/8M_W^2$	four-fermion Fermi theory; weakness is the heavy propagator, not a small $g$
Massless-gauge (pre-Higgs)	four massless fields, $SU(2)_L\times U(1)_Y$ exact	the unbroken theory; masses require spontaneous breaking (Higgs)
Photon combination	couples to $Q=T^3+Y/2$ , parity-even	the particular mixture the Gell-Mann–Nishijima relation singles out

Common trapThe neutral W³ is the photon

It is not. The single most common error is to identify the neutral member of the $SU(2)_L$ triplet, $W^3$ , with the photon and stop there. Two facts forbid it. First, $W^3$ couples to weak isospin $T^3$ , which is nonzero only for left-handed fermions — so a " $W^3$ photon" would ignore right-handed electrons entirely and violate parity maximally, whereas the real photon couples to left and right identically. Second, the photon must couple to electric charge $Q=T^3+Y/2$ , an object that mixes both quantum numbers, so it cannot be built from $SU(2)_L$ alone. The physical photon is the specific rotated combination $A_\mu=B_\mu\cos\theta_W+W^3_\mu\sin\theta_W$ , and the orthogonal combination is a genuinely new particle, the $Z$ , with its own chiral neutral current. Equivalently: $SU(2)_L$ by itself is not the electroweak group — the hypercharge $U(1)_Y$ is indispensable, and it is the mixing of the two neutral fields, not either one alone, that yields electromagnetism.

Worked examples

1 · Predicting the W and Z masses from the group (fully worked)

Given $\alpha=1/137.036$ , $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ , and the measured $\sin^2\theta_W=0.231$ , predict $M_W$ and $M_Z$ at tree level and compare with the values found at CERN.

Step 1. Assemble the mass formula. Combine $G_F/\sqrt2=g^2/8M_W^2$ with $g^2=e^2/\sin^2\theta_W=4\pi\alpha/\sin^2\theta_W$ to eliminate $g$ : $M_W^2 = \dfrac{g^2}{4\sqrt2\,G_F} = \dfrac{\pi\alpha}{\sqrt2\,G_F\,\sin^2\theta_W}$ . It is convenient to define the electroweak scale $A_0^2=\dfrac{\pi\alpha}{\sqrt2\,G_F}$ , so that $M_W=A_0/\sin\theta_W$ .

Step 2. Evaluate $A_0$ . Numerator $\pi\alpha=\pi/137.036=0.022925$ . Denominator $\sqrt2\,G_F=1.41421\times1.1664\times10^{-5}=1.6497\times10^{-5}\ \mathrm{GeV}^{-2}$ . Thus $A_0^2=0.022925/(1.6497\times10^{-5})=1389.7\ \mathrm{GeV}^2$ , giving $A_0=37.28\ \mathrm{GeV}$ — a pure number built from $\alpha$ and $G_F$ alone.

Step 3. Divide by the mixing. With $\sin\theta_W=\sqrt{0.231}=0.4806$ and $\cos\theta_W=\sqrt{0.769}=0.8769$ : $M_W=A_0/\sin\theta_W=37.28/0.4806=\mathbf{77.6\ GeV}$ , and $M_Z=M_W/\cos\theta_W=77.6/0.8769=\mathbf{88.5\ GeV}$ .

Step 4. Compare and interpret. The measured masses are $M_W=80.4\ \mathrm{GeV}$ and $M_Z=91.2\ \mathrm{GeV}$ — the tree-level prediction is low by about $3.7\%$ . That gap is not a failure but a triumph: it is filled precisely by the one-loop radiative corrections (the parameter $\Delta r$ , dominated by the heavy top quark and the Higgs), and demanding consistency of the full electroweak fit is how the top mass was anticipated before its discovery. The bare group structure predicts $M_W\approx77.6$ and $M_Z\approx88.5\ \mathrm{GeV}$ ; loop corrections raise them to the observed $80.4$ and $91.2\ \mathrm{GeV}$ .

2 · Hypercharge bookkeeping and the charge sum

Use $Q=T^3+Y/2$ to find the hypercharge of every fermion field in the first generation, then verify the electric charges sum to zero.

Step 1. Leptons. Left doublet $(\nu_e,e^-)_L$ : $Q=0$ with $T^3=+\tfrac12\Rightarrow Y=2(Q-T^3)=2(0-\tfrac12)=\mathbf{-1}$ ; the $e_L$ member gives $2(-1+\tfrac12)=-1$ , the same $Y$ , as it must. Singlet $e^-_R$ : $T^3=0$ , $Q=-1\Rightarrow Y=\mathbf{-2}$ .

Step 2. Quarks. Left doublet $(u,d)_L$ : $u$ gives $Y=2(\tfrac23-\tfrac12)=\tfrac13$ , $d$ gives $2(-\tfrac13+\tfrac12)=\tfrac13$ , so $Y=\mathbf{+\tfrac13}$ . Singlets: $u_R$ gives $Y=2(\tfrac23)=\mathbf{+\tfrac43}$ ; $d_R$ gives $Y=2(-\tfrac13)=\mathbf{-\tfrac23}$ .

Step 3. Sum the charges, counting colour. Quarks come in three colours, leptons in one: $\sum Q = 3(+\tfrac23)+3(-\tfrac13)+0+(-1) = 2-1+0-1=\mathbf{0}$ .

Step 4. Interpret. The vanishing sum is the gauge-anomaly cancellation condition. It is why the theory is renormalisable, and it forces the quark and lepton charges into exact proportion — the reason the proton and electron charges are equal and opposite to better than $10^{-21}$ . Every hypercharge is fixed by $Q=T^3+Y/2$ , and $\sum_{\text{gen}}Q=0$ ties quarks to leptons through anomaly freedom.

Physics of the portrait

The electroweak gauge group has a clear author. Before anyone knew how to give the $W$ and $Z$ their masses, the group itself — $SU(2)\times U(1)$ with its mixing angle — was written down, and the man who wrote it built it precisely to solve the neutral-current puzzle this article turns on.

Origins · 1 minSheldon Glashow and how this was discovered

Sheldon Glashow

b. 1932 · the SU(2)×U(1) structure of the weak and electromagnetic interactions

Sheldon Lee Glashow, working in 1961, saw what a single gauge group could not do. A pure $SU(2)$ theory gives three gauge bosons, but its neutral member is chiral and cannot be the photon, while a pure $U(1)$ has no charged carriers at all. In Partial-symmetries of weak interactions he enlarged the symmetry to the product $SU(2)\times U(1)$ , adding a second neutral field and the mixing angle — now the weak or Weinberg angle — that rotates the two neutral fields into the photon and a new heavy neutral boson. The construction predicted a weak neutral current, then unknown. Glashow's group was a symmetry without a mechanism: it could not make the $W$ and $Z$ massive without spoiling gauge invariance. That gap closed in 1967–68 when Weinberg and Salam grafted the Higgs mechanism onto exactly his group, turning a classification into a predictive theory. The neutral current appeared in Gargamelle in 1973, and the $W$ and $Z$ at CERN in 1983 — at the masses the mixing angle demanded. Glashow, Weinberg, and Salam shared the 1979 Nobel Prize, four years before the bosons were seen.

History

1961

— Glashow proposes the SU(2)×U(1) gauge group and the weak mixing angle for a unified weak and electromagnetic force

1964

— Higgs, Brout, and Englert devise spontaneous breaking of a gauge symmetry

1967

— Weinberg applies the Higgs mechanism to Glashow's group, giving masses to the W and Z

1973

— the Gargamelle bubble chamber at CERN discovers the predicted weak neutral current

1983

— UA1 and UA2 discover the W and Z bosons at the masses the mixing angle requires

Going further

Texts. Thomson, Modern Particle Physics, §15.2–15.3 (the electroweak gauge group and its currents) · Halzen & Martin, Quarks and Leptons, Ch. 13–15 (the standard electroweak model, where $A_0=37.3\ \mathrm{GeV}$ is defined) · Griffiths, Introduction to Elementary Particles, 2nd ed., §10.7–10.9 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.1–20.2 (the Glashow–Weinberg–Salam theory) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29. The key results assembled here are: $W^\pm=\tfrac{1}{\sqrt2}(W^1\mp iW^2)$ gives the $V{-}A$ charged current; $Q=T^3+Y/2$ yields lepton-doublet $Y=-1$ , $e_R$ $Y=-2$ , quark-doublet $Y=+\tfrac13$ ; $e=g\sin\theta_W=g'\cos\theta_W$ from the photon condition; $A_0=37.28\ \mathrm{GeV}$ , tree-level $M_W=77.6$ , $M_Z=88.5\ \mathrm{GeV}$ ; and $\sum_{\rm gen}Q=0$ .

Prerequisites. The Local Gauge Principle · Weyl Chirality · Weak Isospin and Hypercharge · V−A Structure of the Weak Interaction. Leads to. Electroweak Unification · The Weinberg Angle · The Higgs Mechanism · The Weak Neutral Current.

Ask the tutor: Ask me to walk the neutral sector line by line — "show how W³ and B mix into the photon and Z" — and we will rotate the fields, impose the photon condition, and read off $e=g\sin\theta_W$ and the full $Z$ coupling together.

Why does the weak interaction only act on left-handed particles?

Because that is how the fields are placed into the group. In building the theory one chooses to put only the left-handed projection of each fermion into an $SU(2)_L$ doublet, while the right-handed projection is a singlet with $T=0$ . The $SU(2)_L$ generators then physically cannot act on right-handed fields — a singlet has nothing for them to rotate. This is an input, motivated entirely by experiment: the 1957 discovery of parity violation showed the weak charged current is purely $V{-}A$ , i.e. it couples only to left-handed particles and right-handed antiparticles. The gauge group encodes that fact in its representation content, which is why the " $L$ " subscript is not decoration but the defining feature of the weak force.

What is weak hypercharge, and how is it different from electric charge?

Weak hypercharge $Y$ is the conserved quantum number of the $U(1)_Y$ factor — the second, independent Abelian symmetry of the electroweak group. It is not the same as electric charge; the two are related by $Q=T^3+Y/2$ , which blends hypercharge with the third component of weak isospin. Crucially, $Y$ is the same for both members of a left-handed doublet (they are one $U(1)_Y$ object), whereas their electric charges differ by one unit — the difference coming entirely from $T^3=\pm\tfrac12$ . Electric charge only emerges after the neutral fields mix: the photon couples to $Q$ , but the underlying gauge charges are $T^3$ (for $SU(2)_L$ ) and $Y$ (for $U(1)_Y$ ).

Why are there exactly four gauge fields?

The counting rule is one gauge field per generator. $SU(2)$ has three generators — the three Pauli matrices $T^i=\tfrac12\sigma^i$ — so it demands three fields $W^1,W^2,W^3$ . $U(1)_Y$ has a single generator, hypercharge, so it demands one field $B$ . Three plus one is four. After the symmetry is spontaneously broken these four reorganise into the four observed neutral-and-charged bosons: the two charged $W^\pm$ , the massive neutral $Z$ , and the massless photon $\gamma$ . The count is conserved through the mixing — four fields in, four physical bosons out — even though the physical bosons are combinations, not the original gauge fields.

If SU(2)ₗ already has a neutral field W³, why add U(1)ᵧ at all?

Because $W^3$ cannot be the photon and $SU(2)_L$ alone cannot describe electromagnetism. The $W^3$ couples to weak isospin $T^3$ , which is zero for right-handed fermions, so a theory with $W^3$ as the only neutral field would have a maximally parity-violating "electromagnetism" — flatly contradicting experiment, where the photon couples to left- and right-handed electrons identically. Adding the hypercharge $U(1)_Y$ supplies a second neutral field $B$ , and mixing $W^3$ with $B$ produces one combination (the photon) that couples to the parity-even charge $Q=T^3+Y/2$ and an orthogonal one (the $Z$ ) that carries the leftover chiral coupling. Without the second factor there is simply no massless, parity-symmetric photon.

Is the weak force actually weak, given that g is not small?

No — the coupling is not small. The $SU(2)_L$ coupling $g\approx0.65$ is larger than the electromagnetic $e\approx0.30$ ; indeed $e=g\sin\theta_W$ , so electromagnetism is a fraction of the weak coupling. The apparent weakness at everyday energies comes entirely from the large $W$ mass: a low-energy process picks up the $W$ propagator factor $1/M_W^2$ , and with $M_W=80\ \mathrm{GeV}$ that suppression gives the tiny Fermi constant $G_F/\sqrt2=g^2/8M_W^2$ . At energies at or above $M_W$ the suppression disappears and the two interactions become comparable — the sense in which they are "unified".

Where do the W and Z masses come from in this picture?

Not from the gauge group itself. Local gauge invariance forbids a hand-written mass term $\tfrac12 M^2 W_\mu W^\mu$ — it is not gauge invariant, exactly as for the photon in aq-pp-local-gauge-principle — so the unbroken $SU(2)_L\times U(1)_Y$ theory has four massless fields. The masses come from the Higgs mechanism: a scalar field acquires a vacuum expectation value that spontaneously breaks the symmetry down to the $U(1)$ of electromagnetism, giving mass to three of the four combinations ( $W^\pm$ and $Z$ ) while leaving the photon massless. The relation $M_W=M_Z\cos\theta_W$ is a direct prediction of that mechanism acting on this group, developed in aq-pp-higgs-mechanism.