University Physics IV · Particle Physics · 14.8

Electroweak Unification & the Higgs

Two forces with almost nothing in common turn out to be one gauge structure read from two directions. This is where the mixing angle comes from, why the W and Z weigh what they do while the photon weighs nothing, and why the model can predict their mass ratio but not a single fermion mass.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Weak and electromagnetic forces look nothing alike — one short-ranged, parity-violating and feeble at low energy, the other infinite-ranged and everywhere — yet the electroweak theory says they are two readings of one gauge structure, SU(2)_L × U(1)_Y. The price of writing that down is that gauge invariance forbids mass terms: an explicit M²W_μW^μ is not invariant, and a Dirac mass pairs a left-handed doublet with a right-handed singlet, which the symmetry forbids too. The Higgs mechanism buys the masses back without surrendering the symmetry.

A scalar doublet whose potential has its minimum away from zero picks a vacuum; the Lagrangian keeps the full symmetry while the vacuum breaks it down to the electromagnetic U(1) that leaves the photon massless, and three of the four scalar degrees of freedom become the longitudinal polarisations the W⁺, W⁻ and Z need in order to be massive. What comes out is genuinely predictive: one angle θ_W ties the photon and the Z to W³ and B, fixes M_W = M_Z cos θ_W at tree level, and fixes every neutral-current coupling with no new freedom. What it costs is that each fermion mass is a separate Yukawa coupling put in by hand — and because the up-type and down-type Yukawa matrices are diagonalised by different rotations, their misalignment survives as the CKM matrix: three angles, one phase, and the model's only quark-sector source of CP violation.

Simple definition: Electroweak unification says the photon and the Z are orthogonal mixtures of the same two gauge fields, W³ of SU(2)_L and B of U(1)_Y, rotated by one weak mixing angle θ_W, with the Higgs field's vacuum value giving mass to three of the four bosons and leaving the photon massless.
Example: The measured masses fix that angle with no freedom left: cos θ_W = M_W/M_Z = 80.377/91.1876 = 0.8815, so sin²θ_W = 0.2230 and θ_W = 28.2° — the same angle that then fixes the Z's coupling to every fermion.

Neutral mixing of W³ and BA = B cos θ_W + W³ sin θ_WZ = −B sin θ_W + W³ cos θ_W

One rotation in field space. Whatever combination is not the photon has to be the Z, so a neutral current with no charge exchange is unavoidable.

A is the photon field, Z the neutral weak boson; θ_W is the weak mixing angle, tan θ_W = g′/g.

Electric charge built from two couplingse = g sin θ_W = g′ cos θ_W

Electromagnetism stops being fundamental — its coupling is the altitude of the right triangle whose legs are g and g′.

g is the SU(2)_L coupling, g′ the U(1)_Y coupling, e the positron charge; all three are dimensionless.

Boson masses from the vacuum valueM_W = ½ g vM_Z = ½ v √(g² + g′²) = M_W / cos θ_Wm_γ = 0

Two couplings and one scale deliver 80.4 and 91.2 GeV. The photon stays massless because the chosen vacuum leaves Q unbroken.

v = (√2 G_F)^(−1/2) = 246.22 GeV is the Higgs vacuum expectation value; masses in GeV.

The Higgs potential and its scalarV = μ²|φ|² + λ|φ|⁴v = √(−μ²/λ)m_H = √(2λ) v

The three flat directions become longitudinal W and Z polarisations; the one radial direction that costs energy is the 2012 scalar.

λ > 0 and μ² < 0 for a broken vacuum; μ² in GeV², λ dimensionless. m_H = 125.25 GeV gives λ = 0.129.

Fermion masses as free Yukawa couplingsm_f = y_f v / √2

Nine charged-fermion masses are nine inputs. What is predicted is that the Higgs couples to each in proportion to m_f — and that is testable.

y_f is one free dimensionless coupling per fermion; m_f in GeV. yₜ = 0.992 while yₑ = 2.94 × 10⁻⁶.

CKM matrix and its single phaseV = UᵤL† U_dLΣᵢ |Vᵤᵢ|² = 1

Cross-generation weak decay lives entirely in the mismatch between the two Yukawa diagonalisations, and so does all quark CP violation.

A unitary 3×3 matrix of dimensionless entries: three real angles and one CP-violating phase survive quark rephasing.

Two forces, one gauge structure

Fermi's four-fermion contact term describes beta decay well at low energy and becomes nonsense at high energy: its cross-section grows without limit as s and breaks unitarity above a few hundred GeV. Replacing the contact term by an exchanged massive vector boson repairs that, but a mass term M²W_μW^μ is not gauge invariant, so the mass cannot simply be written in. The electroweak answer is to begin from a larger symmetry, SU(2)_L × U(1)_Y, whose four gauge fields are W¹, W², W³ and B. The first two combine into the charged W± that raise and lower weak isospin. The other two are both electrically neutral, and neither one on its own can be the photon, because the photon is defined by what it couples to — electric charge, Q = T₃ + Y/2, which mixes weak isospin with hypercharge. So the physical neutral states must be mixtures, and the theory therefore predicted a second neutral boson driving a current with no charge exchange. Gargamelle found exactly that in 1973, a decade before either the W or the Z was produced.

The mixing angle is a rotation, and its value is measured

Write the physical neutral fields as one orthogonal rotation of W³ and B: A = B cos θ_W + W³ sin θ_W and Z = −B sin θ_W + W³ cos θ_W. Demanding that A couple to Q with strength e, and not couple to the neutrino at all, fixes the angle in terms of the two gauge couplings — tan θ_W = g′/g — and forces e = g sin θ_W = g′ cos θ_W. Two things follow. Every Z coupling is then determined: for each fermion it is proportional to T₃ − Q sin²θ_W, with nothing left to adjust. And θ_W itself is not predicted; it is an input, measured in several independent ways — from the W-to-Z mass ratio, from forward-backward asymmetries at the Z pole, from neutrino-nucleon scattering, from parity violation in caesium atoms and in polarised electron scattering. That those determinations agree once each is quoted in a stated scheme, across orders of magnitude in momentum transfer, is the real content of the claim that the two forces are one.

Breaking the symmetry with the vacuum, not the Lagrangian

Add one complex scalar doublet φ with the potential V = μ²|φ|² + λ|φ|⁴. With λ > 0 and μ² < 0 the minimum is not at φ = 0 but on a sphere of radius v/√2, where v = √(−μ²/λ) = 246.22 GeV — a number fixed by the muon lifetime through v = (√2 G_F)^(−1/2). The Lagrangian keeps the full symmetry; the vacuum picks one point on the sphere and does not. Counting is the whole argument. The doublet carries four real degrees of freedom. Three of them are flat directions of the potential, so their quanta would be massless Goldstone modes — but a gauge transformation rotates them away, and they reappear as the longitudinal polarisation that each of W⁺, W⁻ and Z must have to be massive: a massless vector has two polarisations, a massive one has three. The fourth, radial direction costs energy, and its quantum is the Higgs boson, m_H = √(2λ)v = 125.25 GeV, so λ = 0.129. One generator combination leaves the chosen vacuum untouched, and that unbroken U(1) is electromagnetism — which is why the photon alone stays massless.

M_W = M_Z cos θ_W is a prediction, and it misses by 0.7%

A doublet has a consequence other scalar choices do not share: at tree level ρ = M_W²/(M_Z² cos²θ_W) is exactly 1, so M_W = M_Z cos θ_W. Test it. Feed α = 1/137.036, G_F = 1.1664 × 10⁻⁵ GeV⁻² and M_Z = 91.1876 GeV into the tree relation M_W²(1 − M_W²/M_Z²) = πα/(√2 G_F) and it returns M_W = 80.94 GeV against a measured 80.377 GeV. The prediction is out by 0.56 GeV, or 0.70% — hundreds of times the experimental uncertainty. That is not a failure; it is the loop expansion becoming visible. Collect the corrections into one factor, writing πα/[√2 G_F (1 − Δr)] on the right, and the data demand Δr = 0.036. Δr carries a term growing as mₜ² and one growing only as ln m_H, so the LEP precision fits pinned the top mass near 170 GeV before Fermilab produced a top quark in 1995, and later bounded the Higgs mass from above before the LHC found it at 125 GeV. A relation accurate to 0.7% and wrong in a calculable way is worth far more than one accurate to 10%.

Fermion masses are permitted, not predicted

A Dirac mass term joins a left-handed field to a right-handed one. Here those live in different representations — left-handed fields in SU(2) doublets, right-handed fields as singlets — so m ψ̄_L ψ_R is not gauge invariant and cannot be written down at all. The scalar doublet rescues it: the combination ψ̄_L φ ψ_R is invariant, and once φ takes its vacuum value the term becomes a mass, m_f = y_f v/√2, with y_f a free dimensionless number. Nothing in the theory sets y_f. The top quark needs yₜ = √2 × 172.7/246.22 = 0.992, an order-one coupling; the electron needs yₑ = 2.94 × 10⁻⁶, smaller by a factor of 338 000, which is exactly the ratio of their masses and therefore explains nothing. What the mechanism does predict is that the Higgs couples to each fermion in proportion to that fermion's own mass, so the decay pattern of a 125 GeV scalar is fixed with no freedom left. Measured couplings to the τ, b, t, W and Z now sit on that proportionality line over three decades in mass, and that is the part of the mechanism genuinely under test.

CKM: what survives when two Yukawa matrices disagree

With three generations the Yukawa couplings are matrices, one 3×3 for up-type quarks and one for down-type. Diagonalising each — which is what finding mass eigenstates means — takes its own unitary rotation of the left-handed fields, UᵤL and U_dL. The charged weak current couples an up-type field to a down-type field in the same doublet, so it survives that pair of rotations only as the product V = UᵤL† U_dL. Had the two Yukawa matrices been diagonalised by the same rotation, V would be the identity and every quark would decay inside its own generation. They are not, and V is the CKM matrix. Unitarity gives it testable sum rules: the first row returns 0.97435² + 0.22500² + 0.00369² = 0.999997. Its magnitudes are steeply hierarchical — near 1 on the diagonal, about 0.22 for one generation step, 10⁻² to 10⁻³ for two. Once every absorbable phase is rotated into the quark fields, four parameters remain: three angles and one phase, and that lone phase is the whole quark-sector source of CP violation.

Change one variable at a time

Make the relationship visible.

Interactive model

Higgs vacuum value v246 GeV

SU(2)_L coupling g0.655

U(1)_Y coupling g′0.250

Open at g′ = 0.250: θ_W is only 20.9° and 1/α reads 230. Raise g′ to 0.350 and the vertical leg lands on its dashed target, θ_W reaches 28.1°, and 1/α falls to 132 — still 4% from 137.04, which is the one-loop correction this tree-level triangle omits.

MIXING ANGLE θ_W20.89 °

sin²θ_W0.1272

M_W = v g / 280.56 GeV

M_Z = M_W / cos θ_W86.23 GeV

Live interpretationMIXING ANGLE θ_W: 20.89 °. sin²θ_W: 0.1272. M_W = v g / 2: 80.56 GeV. M_Z = M_W / cos θ_W: 86.23 GeV

Catch the common trap

Explain before calculating.

Misconception check

“The Higgs gives things their mass” flattens two different mechanisms into one sentence and then explains neither. The W and Z masses are genuine outputs: once g, g′ and v are fixed, M_W = gv/2 and M_Z = M_W/cos θ_W follow with nothing left to adjust, which is precisely why the tree-level ratio can be tested and found 0.7% off. Fermion masses are not outputs at all. Each is a separate Yukawa coupling chosen to reproduce the measured value through m_f = y_f v/√2, running from yₜ = 0.992 down to yₑ = 2.94 × 10⁻⁶; the mechanism explains only why such a term is allowed to exist in a theory whose gauge symmetry would otherwise forbid it. And the mass you can actually weigh is mostly not this at all: roughly 99% of a proton's mass is QCD binding energy, so the Higgs field accounts for about 1% of your own.

Tree level says M_W = M_Z cos θ_W. The effective sin²θ_W = 0.2315 gives 79.94 GeV against a measured 80.377 GeV. What is that 0.44 GeV gap?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe measured boson masses are M_W = 80.377 GeV and M_Z = 91.1876 GeV. Find cos θ_W, sin²θ_W and θ_W from the tree-level relation, then compare your sin²θ_W with the effective leptonic value 0.23155 extracted from Z-pole asymmetries.

The tree relation M_W = M_Z cos θ_W rearranges directly: cos θ_W = 80.377/91.1876 = 0.88145.
Square it and subtract from one: sin²θ_W = 1 − 0.88145² = 1 − 0.77695 = 0.22305.
Take the root and the inverse sine: sin θ_W = 0.47228, so θ_W = 28.18°.
Compare: 0.23155 − 0.22305 = 0.00850, a 3.8% difference. Both are quoted to five figures with far smaller errors, so this is not two measurements disagreeing — it is two definitions of the same angle, one from the mass ratio and one from the Z couplings, separated by radiative corrections.

Answercos θ_W = 0.8815, sin²θ_W = 0.2230, θ_W = 28.2°. The effective value 0.23155 sits 3.8% higher; quoting sin²θ_W without naming the scheme is meaningless at this precision.

MediumThe first row of the CKM matrix measures |V_ud| = 0.97435, |Vᵤₛ| = 0.22500, |V_ub| = 0.00369. (a) Test row unitarity. (b) Extract the Cabibbo angle and check it against |V_ud|. (c) A charmed meson can decay by c → s or c → d; with |V_cs| = 0.97349 and |V_cd| = 0.22486, estimate the ratio of those two rates.

(a) Unitarity requires the squared magnitudes of a row to sum to 1: 0.97435² + 0.22500² + 0.00369² = 0.949358 + 0.050625 + 0.000014 = 0.999997.
The deficit is 3 × 10⁻⁶, consistent with 1. This is a real test: a fourth generation, or mixing into anything outside the three doublets, would steal probability from this row.
(b) In the two-generation corner |Vᵤₛ| = sin θ_C, so θ_C = arcsin 0.22500 = 13.00°. Unitarity then wants |V_ud| = cos θ_C = 0.97436, against the measured 0.97435 — agreement to one part in 100 000.
(c) Rates go as the squared matrix element, so Γ(c → d)/Γ(c → s) = (|V_cd|/|V_cs|)² = (0.22486/0.97349)² = 0.23098² = 0.0534.
That is Cabibbo suppression: about 5.3% before any phase-space difference between a strange and a down final state is folded in.

AnswerRow sum 0.999997, unitary within 3 × 10⁻⁶; θ_C = 13.00°, consistent with |V_ud| = cos θ_C; the Cabibbo-suppressed channel runs at about 5.3% of the favoured one.

HardPredict M_W at tree level from α = 1/137.036, G_F = 1.16638 × 10⁻⁵ GeV⁻² and M_Z = 91.1876 GeV, using M_W²(1 − M_W²/M_Z²) = πα/(√2 G_F). Compare with the measured 80.377 GeV, then find the correction Δr the data demand in M_W²(1 − M_W²/M_Z²) = πα/[√2 G_F (1 − Δr)].

Evaluate the right-hand side once. πα = 3.14159/137.036 = 0.0229253, and √2 G_F = 1.41421 × 1.16638 × 10⁻⁵ = 1.64951 × 10⁻⁵ GeV⁻², so A² = 0.0229253/(1.64951 × 10⁻⁵) = 1389.8 GeV², i.e. A = 37.28 GeV.
Write x = M_W². Then x(1 − x/M_Z²) = A² becomes the quadratic x² − M_Z²x + A²M_Z² = 0, whose roots are x = ½M_Z²[1 ± √(1 − 4A²/M_Z²)].
With M_Z² = 8315.2 GeV²: 4A²/M_Z² = 5559.2/8315.2 = 0.66857, so √(1 − 0.66857) = √0.33143 = 0.57570. Take the upper root, the one that is a heavy W rather than a light one: x = 4157.6 × 1.57570 = 6551.1 GeV², so M_W = 80.94 GeV.
Measured is 80.377 GeV. The tree prediction overshoots by 0.56 GeV, which is 0.70% — enormous against a 10 MeV experimental uncertainty, so the discrepancy is real physics, not measurement error.
Now solve for Δr. At the measured mass, M_W²(1 − M_W²/M_Z²) = 6460.5 × (1 − 6460.5/8315.2) = 6460.5 × 0.22305 = 1441.0 GeV². So 1 − Δr = 1389.8/1441.0 = 0.96447, giving Δr = 0.0355.
Δr is dominated by the running of α from zero momentum up to M_Z, partly cancelled by a term rising as mₜ². That mₜ² sensitivity is why fitting Δr located the top mass to within about 20 GeV years before a top quark was ever produced.

AnswerTree level gives M_W = 80.94 GeV, 0.56 GeV (0.70%) above the measured 80.377 GeV; absorbing the gap requires Δr = 0.036.