The Standard Model overview

The Standard Model is the most precisely tested theory in the history of science, and its entire content fits on a napkin: a gauge group, a list of matter fields, and one scalar. Everything you have ever touched is built from the first of its three generations — the electron, the electron-neutrino, the up-quark and the down-quark — bound into atoms by just three forces. The genius of the framework is not the list but the organising principle behind it: the forces are not action-at-a-distance but the exchange of spin-1 gauge bosons, dictated by demanding that the theory be invariant under local gauge transformations of the group $SU(3)_C\times SU(2)_L\times U(1)_Y$ . Colour $SU(3)_C$ gives the eight gluons of the strong force; the electroweak $SU(2)_L\times U(1)_Y$ gives the photon of electromagnetism and the massive $W^\pm$ and $Z$ of the weak interaction. Two facts turn this skeleton into the world. First, the gauge symmetry forbids explicit masses, yet the $W$ , $Z$ and every fermion plainly have mass — resolved by a single spin-0 field, the Higgs, whose non-zero vacuum value hides the electroweak symmetry and hands out masses in proportion to how strongly each particle couples to it. Second, the same short-distance theory that looks like point-like Fermi contact at low energy resolves, at the electroweak scale $v\approx246\ \mathrm{GeV}$ , into the exchange of heavy bosons about eighty times the proton's mass. This article is the map of the whole territory — the group, the particles, the mass mechanism and the numbers — with the individual forces, the Dirac equation and the Higgs mechanism developed in the articles it points to.

Key takeaways

Matter is twelve spin-½ fermions — six quarks and six leptons in three generations that differ only in mass — each with an antiparticle.
The forces are gauge interactions of $SU(3)_C\times SU(2)_L\times U(1)_Y$ : eight gluons (strong), the photon (electromagnetic), and the $W^\pm,Z$ (weak).
Electric charge is fixed by Gell-Mann–Nishijima, $Q=T_3+\tfrac{1}{2}Y$ , from weak isospin and hypercharge — not put in by hand.
The Higgs field has a vacuum value $v\approx246\ \mathrm{GeV}$ ; boson masses follow as $m_W=\tfrac12 gv$ , $m_Z=m_W/\cos\theta_W$ , and fermion masses as $m_f=y_f v/\sqrt2$ .
Only ~1 % of the proton's mass is Higgs-given quark mass; the rest is QCD binding energy — mass and the Higgs are not the same story.
Gravity ( $\sim10^{-37}$ relative strength), dark matter, neutrino mass and the matter–antimatter asymmetry all lie outside the Standard Model.
Key idea: the Standard Model is a gauge theory of $SU(3)_C\times SU(2)_L\times U(1)_Y$ acting on three generations of fermions, in which forces are the exchange of gauge bosons and mass is the imprint of a single scalar field's vacuum value.

The core insight

The Standard Model is a gauge theory of $SU(3)_C\times SU(2)_L\times U(1)_Y$ acting on three generations of spin-½ fermions: the forces are the exchange of the gauge bosons that the symmetry demands, and every particle's mass is set by how strongly it couples to one scalar field whose vacuum value $v\approx246\ \mathrm{GeV}$ hides the electroweak symmetry.

Where this lives today

The Standard Model is simultaneously the most successful and the most frustrating theory in physics: it fits essentially every laboratory measurement, yet explains none of the biggest questions. On the success side, the Large Hadron Collider completed the roster in 2012 by finding the Higgs at $125\ \mathrm{GeV}$ , and now measures its couplings to check that they scale with mass as equation (4) demands — so far they do. Precision frontiers push on the cracks: the muon's anomalous magnetic moment $(g-2)$ , measured at Fermilab to parts per billion, tests whether virtual Standard Model particles fully account for the muon's wobble, and the $W$ mass is measured to better than one part in ten thousand to test the $\rho=1$ relation of worked example 2. On the failure side, four facts sit plainly outside the framework: neutrinos have mass (oscillations prove it, the minimal model forbids it), the Universe is 5/6 dark matter, the matter–antimatter asymmetry dwarfs the model's CP violation, and gravity is absent. It is a theory everyone believes and no one thinks is final — so good it has become the backdrop against which the search for what lies beyond is defined.

The physical picture

The Standard Model is best read as a table before it is read as an equation. Twelve fermions fall into three columns — the generations — each a heavier photocopy of the last, and each carrying one up-type quark, one down-type quark, one charged lepton and one neutrino. Alongside them sit the force carriers: the gluon and photon (massless), the $W^\pm$ and $Z$ (heavy), and the lone scalar Higgs. The first figure is that table, showing the one feature no cartoon captures — masses ranging over eleven orders of magnitude, from sub-eV neutrinos to the 173 GeV top.

Figure

The Standard Model particle content: three generations of quarks (blue) and leptons (green), the spin-1 gauge bosons (amber) and the spin-0 Higgs (red, m_H = 125 GeV); masses span from sub-eV neutrinos to the top quark at 173 GeV, and electric charge Q is fixed by weak isospin and hypercharge

The table hides the mechanism that makes it work. Gauge symmetry, imposed exactly, would leave every particle massless — yet the $W$ and $Z$ weigh in near 100 GeV. The resolution is that the electroweak symmetry is hidden, not absent: a scalar field settles into a vacuum with non-zero value $v$ , and the bosons that couple to it acquire mass by "eating" its extra modes. Exactly one neutral combination survives massless — the photon — while the orthogonal one becomes the heavy $Z$ . The mixing that sorts massless from massive is a single rotation by the weak mixing angle $\theta_W$ , and it locks the $W$ and $Z$ masses into a fixed ratio. The second figure is that relation drawn as geometry.

Figure

Electroweak mixing as a rotation by the weak angle θ_W ≈ 28.2°: the neutral W³ and B fields rotate into the massless photon and the massive Z, so the boson masses form a right triangle with m_W = m_Z cos θ_W = 91.19 × 0.881 = 80.4 GeV and sin²θ_W = 1 − m_W²/m_Z² = 0.223

The core equations

The Standard Model is defined by a single Lagrangian, and although writing every term fills a page, its architecture is four pieces: gauge-boson kinetic terms, fermion terms built from the covariant derivative, the Higgs sector, and the Yukawa couplings joining fermions to the Higgs. The schematic form below is the whole theory in outline.

\mathcal{L}_{\mathrm{SM}}=-\tfrac14 F^{a}_{\mu\nu}F^{a\,\mu\nu}+\bar\psi\,i\gamma^\mu D_\mu\psi+|D_\mu\phi|^2-V(\phi)-\big(\bar\psi_L\,y\,\phi\,\psi_R+\text{h.c.}\big) \qquad (1)

where: $F^a_{\mu\nu}$ — field-strength tensors of the twelve gauge bosons ( $8$ gluons $+\,W^{1,2,3}+B$ ); $D_\mu=\partial_\mu-ig_sG_\mu-igW_\mu-ig'\tfrac{Y}{2}B_\mu$ — the covariant derivative carrying the three couplings $g_s,g,g'$ ; $\psi$ — the fermion fields, left-handed as $SU(2)$ doublets and right-handed as singlets; $\phi$ — the Higgs doublet; $V(\phi)$ — its "Mexican-hat" potential; $y$ — the Yukawa coupling matrices. Reading it: forces enter only through $D_\mu$ , so gauge invariance dictates every interaction vertex. The gauge group forbids the theory from putting charges in by hand — instead electric charge is a bookkeeping identity relating the third component of weak isospin $T_3$ and the hypercharge $Y$ . This is the electroweak form of the Gell-Mann–Nishijima relation.

Q=T_3+\tfrac12\,Y \qquad (2)

where: $Q$ — electric charge in units of $|e|$ ; $T_3$ — third component of weak isospin ( $\pm\tfrac12$ inside an $SU(2)_L$ doublet, $0$ for a singlet); $Y$ — weak hypercharge, the $U(1)_Y$ label. Reading it: fix each multiplet's $Y$ once, and the charges of every member — and both chiralities — follow. It is a constraint, not a free choice. When the Higgs takes its vacuum value $v$ , the kinetic term $|D_\mu\phi|^2$ produces mass terms for exactly the bosons that couple to it. The charged $W^\pm$ and the neutral $Z$ become heavy while one neutral combination, the photon, stays massless.

m_W=\tfrac12\,g\,v,\qquad m_Z=\tfrac12 v\sqrt{g^2+g'^2}=\frac{m_W}{\cos\theta_W},\qquad m_\gamma=0 \qquad (3)

where: $g,g'$ — the $SU(2)_L$ and $U(1)_Y$ couplings; $v\approx246\ \mathrm{GeV}$ — the Higgs vacuum value; $\cos\theta_W=g/\sqrt{g^2+g'^2}$ — the weak mixing angle. Reading it: the mass ratio $m_W/m_Z=\cos\theta_W$ is a prediction, not an input, and it holds to a fraction of a percent. Fermions get their mass from the same field through the Yukawa term. There is no mass without a coupling to the Higgs, and the enormous spread of fermion masses is entirely a spread of dimensionless Yukawa couplings.

m_f=\frac{y_f\,v}{\sqrt2},\qquad v=\big(\sqrt2\,G_F\big)^{-1/2}\approx246\ \mathrm{GeV} \qquad (4)

where: $y_f$ — the fermion's Yukawa coupling; $v$ — the same vacuum value, here written in terms of the Fermi constant $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ measured in muon decay. Reading it: the top quark has $y_t\approx1$ (it couples to the Higgs at full strength) while the electron has $y_e\approx3\times10^{-6}$ — the "flavour puzzle" of why is unexplained.

Derivation AElectroweak symmetry breaking and the gauge-boson masses

The goal is to show, from the single Higgs kinetic term $|D_\mu\phi|^2$ evaluated at the vacuum, how three of the four electroweak gauge bosons become massive while one stays exactly massless, and why the survivors obey the rigid relation $m_W=m_Z\cos\theta_W$ . We take the Higgs doublet to its vacuum value, expand the covariant derivative, and read the coefficients of the quadratic boson terms as masses — the whole Higgs mechanism at boson level.

Step 1. Write the Higgs as an $SU(2)_L$ doublet with hypercharge $Y=+1$ and let it settle to its vacuum value $\phi=\tfrac{1}{\sqrt2}\binom{0}{v}$ , the minimum of the Mexican-hat potential $V(\phi)=-\mu^2\phi^\dagger\phi+\lambda(\phi^\dagger\phi)^2$ with $v=\mu/\sqrt\lambda$ . Why: choosing a single point in the degenerate circle of minima is what breaks the symmetry — the Lagrangian keeps it, the vacuum does not.

Step 2. Insert this $\phi$ into $|D_\mu\phi|^2$ with $D_\mu\phi=\big(\partial_\mu-ig\,\tfrac{\sigma^a}{2}W^a_\mu-ig'\tfrac12 B_\mu\big)\phi$ . The derivative annihilates the constant $v$ , leaving only the gauge pieces, so the term becomes $\tfrac{v^2}{8}\big|\,gW^a_\mu\sigma^a+g'B_\mu\,\big|^2$ acting on $\binom{0}{1}$ . Why: a constant field has no kinetic energy of its own, so every surviving term is a boson mass term generated purely by the coupling to the vacuum.

Step 3. Collect the charged combination $W^\pm_\mu=(W^1_\mu\mp iW^2_\mu)/\sqrt2$ . Its coefficient is $\tfrac14 g^2v^2\,W^+_\mu W^{-\,\mu}$ , i.e. a mass $m_W=\tfrac12 gv$ . Why: the two charged bosons are degenerate because $SU(2)$ treats $W^1$ and $W^2$ identically, so a single number sets both.

Step 4. In the neutral sector the term is $\tfrac{v^2}{8}(gW^3_\mu-g'B_\mu)^2$ — a mass only for the combination $Z_\mu\propto gW^3_\mu-g'B_\mu$ , while the orthogonal combination $A_\mu\propto g'W^3_\mu+gB_\mu$ has no mass term at all. Normalising gives $m_Z=\tfrac12 v\sqrt{g^2+g'^2}$ , $m_A=0$ , and with $\cos\theta_W=g/\sqrt{g^2+g'^2}$ the ratio $m_W/m_Z=\cos\theta_W$ . Why: the massless direction is forced to exist because $U(1)_{\mathrm{em}}$ is left unbroken — the photon is the guaranteed survivor, and the mass ratio is its fingerprint. *As a check: $m_W=\tfrac12 gv$ , $m_Z=\tfrac12 v\sqrt{g^2+g'^2}$ , $m_\gamma=0$ ; on-shell $m_W/m_Z=\cos\theta_W$ gives $80.4/91.19=0.881$ , so $\sin^2\theta_W=0.223$ .*

Go deeper · Derivation B · 1 minThe electroweak scale from muon decay

Derivation BThe electroweak scale from muon decay

The aim is to pin the number $v\approx246\ \mathrm{GeV}$ — the scale that sets every boson and fermion mass — to a laboratory measurement, the muon lifetime, without ever knowing $g$ separately. We match the low-energy limit of $W$ exchange to the point-like four-fermion interaction Fermi wrote down in 1934, and show that the coupling $g$ cancels, leaving $v$ expressed purely through the Fermi constant $G_F$ .

Step 1. At momentum transfer $q^2\ll m_W^2$ the $W$ propagator $1/(q^2-m_W^2)$ collapses to the constant $-1/m_W^2$ , so the charged-current amplitude looks like a contact interaction with strength $G_F/\sqrt2=g^2/(8m_W^2)$ . Why: when the exchanged boson is far heavier than the energy available, its exchange is indistinguishable from a zero-range contact term — this is why beta decay looked point-like for decades.

Step 2. Substitute the mass from the previous derivation, $m_W^2=\tfrac14 g^2v^2$ . Then $\dfrac{G_F}{\sqrt2}=\dfrac{g^2}{8\cdot\tfrac14 g^2v^2}=\dfrac{1}{2v^2}$ . Why: the coupling $g$ appears in both the vertex and the mass, so it cancels — the low-energy strength depends only on the vacuum value $v$ .

Step 3. Solve for $v$ : $v=\big(\sqrt2\,G_F\big)^{-1/2}$ . Inserting $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ gives $v=(\,1.414\times1.1664\times10^{-5}\,)^{-1/2}=246.2\ \mathrm{GeV}$ . Why: a single beta-decay measurement, made long before the $W$ was seen, already fixed the scale at which the electroweak symmetry breaks.

Step 4. Read the significance: $v$ is not adjustable — it is the same $v$ that sets $m_W$ , $m_Z$ , $m_H$ and every Yukawa mass, so the 100 GeV clustering of the heavy bosons and the top quark is one scale showing up everywhere. Why: the coincidence " $W$ , $Z$ , $H$ , $t$ all near 100 GeV" is not a coincidence — they all read off the same vacuum value. *As a check: $G_F/\sqrt2=1/(2v^2)\Rightarrow v=(\sqrt2 G_F)^{-1/2}=246.2\ \mathrm{GeV}$ ; combining with $m_W=\tfrac12 gv$ and $g=e/\sin\theta_W$ recovers $m_W\approx80\ \mathrm{GeV}$ .*

Go deeper · Derivation C · 1 minElectric charge from weak isospin and hypercharge

Derivation CElectric charge from weak isospin and hypercharge

The purpose is to demonstrate that the fractional charges of quarks and the integer charges of leptons are not inputs but consequences: once each multiplet's hypercharge is fixed, $Q=T_3+\tfrac12 Y$ assigns every charge, and — crucially — must give the same charge to a fermion's left- and right-handed pieces even though they live in different $SU(2)$ representations. We run one full generation through the formula.

Step 1. Place the left-handed lepton pair in an $SU(2)_L$ doublet $\binom{\nu_e}{e}_L$ with hypercharge $Y=-1$ , so $T_3=+\tfrac12$ for the neutrino and $-\tfrac12$ for the electron. Then $Q(\nu)=\tfrac12+\tfrac12(-1)=0$ and $Q(e)=-\tfrac12+\tfrac12(-1)=-1$ . Why: the doublet's two members must differ in charge by exactly one unit because $T_3$ differs by one — the $W^\pm$ that turns one into the other carries that unit.

Step 2. The right-handed electron is an $SU(2)$ singlet, $T_3=0$ , and must reproduce $Q=-1$ , forcing its hypercharge to $Y=-2$ : $Q=0+\tfrac12(-2)=-1$ . Why: charge is chirality-blind even though the weak force is not — the hypercharge assignments are tuned precisely so both handednesses of the electron carry the same $-1$ .

Step 3. Place the left-handed quarks in a doublet $\binom{u}{d}_L$ with $Y=+\tfrac13$ : $Q(u)=\tfrac12+\tfrac12\cdot\tfrac13=+\tfrac23$ and $Q(d)=-\tfrac12+\tfrac16=-\tfrac13$ . Why: the same machinery that gave integer lepton charges gives thirds for quarks — the fraction is the hypercharge $\tfrac13$ , ultimately tied to the three colours.

Step 4. Check anomaly-free consistency: summing $Q$ over one generation of colour-counted fermions gives $3(\tfrac23-\tfrac13)+(0-1)=0$ , the condition that lets the quantum theory stay gauge-invariant. Why: the charges are not merely assigned but constrained — the sum-to-zero is what makes the Standard Model a consistent quantum theory, and it ties lepton and quark charges together. *As a check: $Q=T_3+\tfrac12 Y$ gives $(\nu,e,u,d)=(0,-1,+\tfrac23,-\tfrac13)$ for both chiralities, and the per-generation charge sum $=0$ (anomaly cancellation).*

Limiting cases

Case	Result	Meaning
Unbroken symmetry, $v\to0$	all gauge bosons massless, all fermions massless	exact $SU(2)_L\times U(1)_Y$ ; the world before electroweak symmetry breaking
Low energy, $E\ll m_W$	weak interaction $\to$ Fermi contact, strength $G_F$	$W,Z$ exchange looks point-like; beta decay and muon decay
No hypercharge, $g'\to0$	$\theta_W\to0$ , $Z=W^3$ , photon $=B$ decouples	electromagnetism would not mix with the weak neutral current
High energy, $E\gg m_W$	electroweak symmetry effectively restored	$W,Z$ as "light" as the photon; the regime probed at the LHC
Low $Q^2$ QCD	$\alpha_s\to$ large, colour confinement	quarks and gluons trapped in hadrons; no free colour
High $Q^2$ QCD	$\alpha_s\to0$ , asymptotic freedom	quarks act nearly free inside protons at short distance
Gravity	$G_N$ effect $\sim10^{-37}$ of electromagnetism	negligible per particle; simply absent from the Standard Model

Common trap“The Higgs field gives everything its mass.”

Only partly, and for your mass hardly at all. The Higgs mechanism supplies the intrinsic masses of the fundamental particles — the $W$ , $Z$ , the charged fermions and the quarks — through their Yukawa couplings. But you, and almost all visible matter, are made of protons and neutrons, and a proton's mass of $938\ \mathrm{MeV}$ is not mostly the mass of its quarks. The three valence quarks contribute only about $2\times2.2+4.7\approx9\ \mathrm{MeV}$ , roughly 1 %; the other 99 % is the energy of the confined gluon field and the relativistic motion of quarks, converted to mass by $E=mc^2$ . So the Higgs sets the quark masses, but the strong interaction — not the Higgs — is what makes the proton (and you) heavy. Switching off the Higgs would leave the proton almost as massive as it is, while making the electron massless and atoms impossible; the everyday statement "the Higgs gives mass" conflates two very different accounts of where mass comes from.

Worked examples

1 · The electroweak scale and the Yukawa hierarchy (fully worked)

From the measured Fermi constant $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ , find the Higgs vacuum value $v$ , then the Yukawa couplings of the top quark ( $m_t=173\ \mathrm{GeV}$ ) and the electron ( $m_e=0.511\ \mathrm{MeV}$ ). What do they reveal?

Step 1. The vacuum value follows from equation (4): $v=(\sqrt2\,G_F)^{-1/2}$ . Compute the product $\sqrt2\,G_F=1.4142\times1.1664\times10^{-5}=1.6495\times10^{-5}\ \mathrm{GeV}^{-2}$ , so $v^2=1/(1.6495\times10^{-5})=6.062\times10^{4}\ \mathrm{GeV}^2$ and $v=\mathbf{246.2\ GeV}$ .

Step 2. Invert $m_f=y_f v/\sqrt2$ to get $y_f=\sqrt2\,m_f/v$ . For the top: $y_t=\sqrt2\times173/246.2=244.6/246.2=\mathbf{0.99}$ . The top quark couples to the Higgs at essentially full strength — a coupling of order one, hinting it plays a special role in electroweak dynamics.

Step 3. For the electron: $y_e=\sqrt2\times0.511\times10^{-3}/246.2=\mathbf{2.9\times10^{-6}}$ . The electron's coupling is a third of a millionth of the top's.

Step 4. Read the result: the fermion masses span eleven orders of magnitude entirely because the dimensionless Yukawa couplings do — from $y_t\approx1$ down to $y_e\approx3\times10^{-6}$ (and smaller still for neutrinos). The Standard Model measures these couplings but does not explain their pattern; that unexplained spread is the flavour puzzle. v = 246 GeV, y_t ≈ 1, y_e ≈ 3×10⁻⁶.

2 · The W mass from the weak mixing angle

Given $m_Z=91.19\ \mathrm{GeV}$ and the on-shell $\sin^2\theta_W=0.223$ , predict $m_W$ and state what a deviation would mean.

Step 1. From equation (3), $m_W=m_Z\cos\theta_W$ . Compute $\cos\theta_W=\sqrt{1-0.223}=\sqrt{0.777}=0.8815$ , so $\theta_W=28.2^\circ$ .

Step 2. Then $m_W=91.19\times0.8815=\mathbf{80.4\ GeV}$ , matching the measured $80.38\ \mathrm{GeV}$ to a fraction of a percent.

Step 3. This agreement is a genuine test: the parameter $\rho\equiv m_W^2/(m_Z^2\cos^2\theta_W)$ equals 1 at tree level precisely because the Higgs sits in a doublet. A measured $\rho\ne1$ would signal new physics (extra Higgs multiplets, new heavy particles in loops), which is why the $W$ mass is measured ever more precisely. m_W = 80.4 GeV, ρ = 1.

3 · How much of the proton's mass is "Higgs mass"?

The proton $(uud)$ has mass $938.3\ \mathrm{MeV}$ . Using $m_u=2.2\ \mathrm{MeV}$ and $m_d=4.7\ \mathrm{MeV}$ , estimate the fraction that is Higgs-given quark mass.

Step 1. Sum the valence-quark (Yukawa) masses: $2m_u+m_d=2\times2.2+4.7=9.1\ \mathrm{MeV}$ .

Step 2. Fraction of the proton mass: $9.1/938.3=\mathbf{0.97\%}$ .

Step 3. The remaining ~99 % is QCD binding energy — gluon field energy and relativistic quark kinetic energy — with no Higgs involvement. The lesson: the Higgs explains why the electron and quarks have mass at all, but the strong force explains why matter is heavy. ≈ 1 % Higgs, ≈ 99 % QCD.

Physics of the portrait

The Standard Model has many parents, but its decisive line — the unification of the electromagnetic and weak forces into a single gauge theory with a Higgs — runs through one 1967 paper of barely three pages.

Origins · 1 minSteven Weinberg and how this was discovered

Steven Weinberg

1933–2021 · electroweak unification and the Standard Model

Steven Weinberg's 1967 paper "A Model of Leptons" — three pages, later the most-cited in particle physics — fused the scattered ingredients of the era into a theory. Sheldon Glashow had proposed the $SU(2)\times U(1)$ structure in 1961 but could not give the $W$ and $Z$ mass without spoiling the symmetry; the Brout–Englert–Higgs mechanism of 1964 supplied the missing idea of a scalar vacuum value; and Weinberg (with Abdus Salam independently) put them together, applying the Higgs mechanism to the electroweak group to make the $W$ and $Z$ heavy while keeping the photon massless and the theory renormalisable. He predicted the weak mixing angle, neutral currents mediated by a $Z$ , and the mass scale of the heavy bosons — all later confirmed at Gargamelle (1973) and the CERN $Sp\bar{p}S$ (1983). Weinberg, Glashow and Salam shared the 1979 Nobel Prize, before the $W$ and $Z$ were even seen. Weinberg insisted always that the goal was not a model but a final theory — the Standard Model, he knew, was a spectacular way-station, not the destination.

History

1961

— Glashow proposes the SU(2) × U(1) structure of the electroweak force

1964

— Brout, Englert and Higgs propose the mass-generating mechanism

1967

— Weinberg's "A Model of Leptons" unifies the weak and electromagnetic forces

1983

— the W and Z bosons are discovered at CERN's proton–antiproton collider

2012

— the Higgs boson is discovered at the LHC by ATLAS and CMS

Going further

Texts. Thomson, Modern Particle Physics, Ch. 1 and Ch. 15–17 (the Standard Model and electroweak unification) · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 1–2 (historical picture and dynamics) · Halzen & Martin, Quarks and Leptons, Ch. 13–15 · Peskin & Schroeder, An Introduction to Quantum Field Theory, Ch. 20 (spontaneously broken gauge theories) · Schwartz, Quantum Field Theory and the Standard Model, Part IV. All numerical values are 2022 PDG values: $m_Z=91.19\ \mathrm{GeV}$ , $m_W=80.38\ \mathrm{GeV}$ , $m_H=125.25\ \mathrm{GeV}$ , $m_t=172.8\ \mathrm{GeV}$ , $G_F=1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ , $v=246.2\ \mathrm{GeV}$ , $\sin^2\theta_W=0.223$ (on-shell), $\alpha_s(m_Z)=0.118$ .

Prerequisites. Natural units · Special-relativistic kinematics · Non-relativistic quantum mechanics review. Leads to. The particle zoo · Fundamental forces · The Dirac equation · Quantum electrodynamics · The Higgs mechanism.

Ask the tutor: Want to see any single piece — the Higgs mechanism, the running couplings, or how $Q=T_3+\tfrac12 Y$ falls out of the group theory — worked in full detail?

Why are there exactly three generations?

No one knows — it is one of the deepest open questions. The theory accommodates three generations but does not require them; a single-generation world would be consistent (though with no CP violation, hence no matter–antimatter asymmetry). Experiment does settle the count at three light neutrinos: the $Z$ decay width measured at LEP matches exactly three species, $N_\nu=2.984\pm0.008$ . A fourth generation with a light neutrino is thus ruled out, but why nature stopped at three, and why each is a heavier copy, remains a genuine mystery.

Is gravity part of the Standard Model?

No. The Standard Model describes three of the four forces — electromagnetic, weak and strong — but not gravity. This is a hard problem, not an oversight: gravity's natural quantum description (a spin-2 graviton) is non-renormalisable, so the perturbative methods that make the other forces calculable break down at high energy. Fortunately gravity is so feeble at particle scales — about $10^{-37}$ of electromagnetism between two protons — that ignoring it costs nothing in any accelerator experiment. Uniting the two is the goal of quantum-gravity programmes such as string theory, and remains unsolved.

What does it really mean to say “the Higgs gives mass”?

It means the Higgs field — not the particle — has a non-zero value everywhere in the vacuum, and particles that interact with it behave as though they have inertia. A large coupling (the top quark) means strongly "slowed" and heavy; zero coupling (the photon) means sailing through, massless. The Higgs boson is just the quantum ripple of that field, the evidence it is really there. Crucially this only sets the intrinsic masses of fundamental particles; the mass of composite objects like protons comes overwhelmingly from the strong force, so the Higgs is responsible for far less of your body weight than the slogan suggests.

Why are the W and Z heavy while the photon is massless?

Because of how the Higgs vacuum value couples to the four electroweak bosons. Three combinations of the $SU(2)_L\times U(1)_Y$ bosons couple to the vacuum and "eat" the Higgs field's extra components, gaining mass — these are the $W^+$ , $W^-$ and $Z$ . But one specific neutral combination, $A_\mu=B\cos\theta_W+W^3\sin\theta_W$ , does not couple to the vacuum at all, so it remains exactly massless: that is the photon. The masslessness is guaranteed because the electromagnetic $U(1)$ symmetry is left unbroken, and an unbroken gauge symmetry always protects its boson's zero mass. The mixing angle $\theta_W$ that decides which combination survives is the same one that fixes $m_W=m_Z\cos\theta_W$ .

How many free parameters does the Standard Model have?

About nineteen, ignoring neutrino masses: three gauge couplings, the two Higgs-potential parameters (equivalently $v$ and $m_H$ ), nine charged-fermion masses, four CKM quark-mixing parameters, and the strong-CP angle $\theta_{\mathrm{QCD}}$ . Adding the neutrino sector pushes the count to about twenty-six. That so much of nature follows from two dozen numbers is remarkable — but they are inputs, measured not predicted, and their patterns are precisely what a deeper theory is expected to explain.

What can the Standard Model not explain?

Four things stand out. It cannot accommodate neutrino masses in its minimal form, yet oscillations prove neutrinos are massive. It offers no dark-matter candidate, though dark matter outweighs ordinary matter five to one. Its built-in CP violation is far too small to explain why the Universe is matter rather than equal matter and antimatter. And it excludes gravity. It is also silent on why its parameters take the values they do. These gaps are not signs the theory is wrong within its domain (it is spectacularly right there) but that it is incomplete — which is why the search for physics beyond it is the field's central occupation.