University Physics IV · Particle Physics · 14.2

Quarks, Leptons & Generations

Before you can draw a single Feynman diagram you need to know which fields exist and what each is charged under. This lesson builds that table from gauge quantum numbers rather than from masses, shows why three generations is a measured number and not a derived one, and marks every entry the model simply takes on trust.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The Standard Model does not start from a list of particles but from a list of fields labelled by how they transform under SU(3) × SU(2) × U(1). Carry that through and one generation is fifteen Weyl fields: a left-handed quark doublet in three colours, right-handed u and d singlets in three colours, a left-handed lepton doublet, one right-handed charged lepton, and no right-handed neutrino. Nature then supplies three copies with identical charges and wildly different masses, and those copies are the generations.

That there are exactly three light ones is not a theorem but a lineshape: 499 MeV of the Z's 2495.5 MeV width is explained by no visible channel, and dividing by the predicted width per neutrino species gives 2.984 ± 0.008. The classification buys a great deal — every gauge coupling follows from a field's slot, and anomaly cancellation forces the hypercharges, which is why the proton's charge matches the positron's to a part in 10²¹. What it costs is the mass spectrum.

Each mass is one Yukawa coupling to the Higgs, put in by hand: 0.99 for the top, 2.9×10⁻⁶ for the electron, below 10⁻¹² for the neutrinos. The inventory is complete and predictive about interactions, and silent about why any entry weighs what it does.

Simple definition: The fermion inventory is the Standard Model's list of matter fields, sorted not by mass but by the SU(3), SU(2) and U(1) charges each one carries: six quarks in colour triplets, six colourless leptons, arranged in three identical-looking generations.
Example: The charm quark is the second-generation copy of the up quark: identical charge +2/3, identical colour triplet, identical weak-isospin slot — and a mass of 1.27 GeV against the up quark's 2.2 MeV, a factor of about 580.

Charge from the electroweak slotQ = T₃ + Y/2

Y is not a choice: Y = 2(Q − T₃) gives 1/3 for the quark doublet, 4/3 for u_R, −1 for the lepton doublet, −2 for e_R.

T₃ = ±½ for the two members of a left-handed doublet, 0 for any right-handed singlet; Q and Y in units of e

Chiral projectionψ_L = ½(1 − γ⁵)ψ, ψ_R = ½(1 + γ⁵)ψ

Only ψ_L carries weak isospin, so the W couples to it alone. Chirality equals helicity only in the limit m/E → 0.

Dimensionless projectors: P_L² = P_L, P_R² = P_R, P_L P_R = 0

Weyl fields in one generation3×2 + 3 + 3 + 2 + 1 = 15 (×3 generations = 45)

The ledger any proposed fourth generation or sterile neutrino must be added to; 16 per generation if ν_R exists.

Quark doublet in three colours, u_R, d_R, lepton doublet, e_R; no ν_R

Charge ledger for one generation3(+⅔) + 3(−⅓) + 0 + (−1) = 0

Balances only for three colours and third-integer quark charges — the reason Qₚᵣₒₜₒₙ = −Q_electron to within 10⁻²¹ e.

Sum over u, d, ν and e, each quark counted once per colour

Light neutrino species from the ZN_ν = (Γᵢₙᵥ/Γ_ℓℓ) ÷ (Γ_νν/Γ_ℓℓ)_SM, Γᵢₙᵥ = Γ_Z − Γ_had − 3Γ_ℓℓ

Converts missing width into a species count, N_ν = 2.984 ± 0.008 — valid only for neutrinos below M_Z/2 = 45.6 GeV.

Widths in MeV; the predicted ratio is 1.9912 and Γᵢₙᵥ comes to 499 MeV

Fermion mass as a Yukawa couplingm_f = y_f v/√2, v = 246.22 GeV

Each mass is one fitted number: yₜ = 0.991, yₑ = 2.9×10⁻⁶, y_ν ≈ 3×10⁻¹³ at the 0.05 eV scale.

y_f dimensionless and free; v is the Higgs vacuum expectation value

The inventory is a list of charges, not of masses

Twelve fermions, and the useful way to hold them is by what they are charged under rather than by how heavy they are. Six quarks — u, d, c, s, t, b — each carry colour and sit in a triplet of SU(3); the up-type three have charge +2/3, the down-type three −1/3, and every one carries baryon number 1/3. Six leptons — e, μ, τ and one neutrino each — are colour singlets, so the strong force never touches them: charge −1 for the charged three, 0 for the neutrinos, and a lepton-family number Lₑ, L_μ or L_τ apiece. Antiparticles double the list with every sign flipped. Counting two-component Weyl fields rather than particles, one generation is fifteen: six for the coloured left-handed quark doublet, three each for u_R and d_R, two for the lepton doublet, one for e_R — and no ν_R. Three generations makes forty-five. That total is the ledger any proposed new fermion has to be added to, and the colour label in it is inferred from spectroscopy and jet counting, never read directly off a detector.

Chirality splits every fermion, and only half feels the W

Apply the projectors P_L = ½(1 − γ⁵) and P_R = ½(1 + γ⁵) to a Dirac field and you get two independent pieces, and the Standard Model treats them as different objects. ψ_L goes into a weak-isospin doublet — (u, d)_L and (ν, e)_L — with T₃ = +½ on the upper member and −½ on the lower; ψ_R is an SU(2) singlet with T₃ = 0. The W couples to the doublet index and therefore to left-handed chirality alone, which is the whole origin of parity violation and the reason a right-handed electron cannot emit a W. Do not read chirality as a spin direction: chirality coincides with helicity only in the massless limit, and the mismatch is of order m/E — about 5×10⁻⁴ for a 1 GeV electron, but of order 1 for one at rest. The split carries a price. A Dirac mass term m(ψ̄_L ψ_R + ψ̄_R ψ_L) joins a doublet to a singlet, so it is not gauge invariant, and no fermion in this model is allowed a mass written in by hand.

Hypercharge from Q = T₃ + Y/2, audited by the anomalies

Weak hypercharge is fixed, not chosen: Y = 2(Q − T₃). For the left-handed quark doublet the up member gives 2(2/3 − 1/2) = 1/3 and the down member 2(−1/3 + 1/2) = 1/3 — the same value, as it must be, since Y labels the doublet rather than its members. The right-handed quarks are singlets with T₃ = 0, so Y = 2Q: 4/3 for u_R and −2/3 for d_R. The lepton doublet gets Y = −1 and e_R gets −2. Nothing so far explains why the charges take these particular fractions, and this is where the arithmetic earns its place: quantum consistency demands that certain sums over a generation vanish. Add the electric charges of one generation with colour counted — 3(+2/3) + 3(−1/3) + 0 + (−1) — and the total is exactly zero, which happens only because quarks come in three colours and carry thirds. That is why the proton's charge cancels the electron's; experiment holds the residue below 10⁻²¹ of e.

A generation is a copy, and the copies barely talk

The second and third generations duplicate the first slot for slot: (c, s, μ, ν_μ) and (t, b, τ, ν_τ) carry exactly the charges (u, d, e, νₑ) carry and differ only in mass. Because the gauge charges match, every interaction rate matches once phase space is accounted for — that is lepton universality, and Γ(Z→e⁺e⁻), Γ(Z→μ⁺μ⁻) and Γ(Z→τ⁺τ⁻) agree at 84.0 MeV to a few parts in a thousand. What does not carry across is flavour. The three lepton-family numbers are separately conserved in every charged-lepton process seen so far, so μ⁻ → e⁻γ is absent below a branching ratio of 3.1×10⁻¹³, and the muon must instead decay as μ⁻ → e⁻ + ν̄ₑ + ν_μ, exporting both family labels in neutrinos. Quarks behave differently: the CKM matrix mixes their generations, so s → u proceeds with amplitude |Vᵤₛ| = 0.2243 and strange particles decay weakly in around 10⁻¹⁰ s.

Counting the light generations off the Z lineshape

Nothing in the model says how many copies there should be, so the number is measured. The Z couples to every fermion pair light enough to produce, neutrinos included, and neutrinos leave no signal — so they appear as width that no visible channel accounts for. LEP scanned e⁺e⁻ collisions across √s = 88–94 GeV and fitted the resonance: Γ_Z = 2495.5 ± 2.3 MeV, of which 1744.4 MeV is hadronic and 3 × 83.99 = 252.0 MeV is charged leptons. The remainder, 499 MeV, is invisible. Dividing by the predicted width per species, Γ_νν = 1.9912 Γ_ℓℓ = 167.2 MeV, gives N_ν = 2.984 ± 0.008. Four light species would have put Γ_Z at 2665 MeV, seventy-four standard deviations high, and pulled the peak hadronic cross-section down from 41.5 nb to 36.4 nb. The result is sharp but narrow: it counts only neutrinos below M_Z/2 = 45.6 GeV that couple to the Z with full strength.

Every mass in the table is an input

Because no fermion may carry a bare mass, each one is generated by a Yukawa coupling to the Higgs field, m_f = y_f v/√2 with v = 246.22 GeV. The coupling is a free parameter, one per fermion, fitted to the measured mass and predicted by nothing. The numbers it has to take are absurd. The top quark needs yₜ = 0.991, essentially unity; the electron needs 2.94×10⁻⁶; a neutrino at the 0.05 eV scale that oscillation demands would need about 3×10⁻¹³. The charged fermions alone spread over 5.5 decades, since mₜ/mₑ = 3.4×10⁵, and the full range reaches 12.5 decades. Counting the neutrino sector in, the model carries roughly twenty-six measured parameters, and some twenty of them are the masses and mixings of this one table. So the inventory is predictive about how fermions interact and silent about why any of them weighs what it does. That gap is the flavour problem, and it is open.

Change one variable at a time

Make the relationship visible.

Interactive model

Light neutrino species N3

Width per species Γ(νν)167 MeV

Beam energy √s91.2 GeV

Hold Γ(νν) at 167 MeV and step N from 1 to 6: only N = 3 lays the solid curve on the dashed measurement. Then fix N = 3 and drag Γ(νν) — the fit is just as good elsewhere, which is why counting species needs the predicted width per species, not the missing width alone.

TOTAL WIDTH Γ_Z2.497 GeV

INVISIBLE WIDTH501 MeV

PEAK σ (HADRONS)41.5 nb

σ AT THIS ENERGY41.5 nb

Live interpretationTOTAL WIDTH Γ_Z: 2.497 GeV. INVISIBLE WIDTH: 501 MeV. PEAK σ (HADRONS): 41.5 nb. σ AT THIS ENERGY: 41.5 nb

Catch the common trap

Explain before calculating.

Misconception check

Students read the three generations as energy levels — a muon as an excited electron that ought to relax by emitting a photon. Nothing in the model works that way. Generations are separate fields with identical gauge charges, not states of one system, so there is no μ → e + γ vertex at all; MEG II holds that branching ratio below 3.1×10⁻¹³. The muon does decay, but through the charged weak current as μ⁻ → e⁻ + ν̄ₑ + ν_μ, a three-body process that carries both family labels away in neutrinos and takes 2.2 μs — an eternity by nuclear standards. Excited states also come in ladders converging on a limit; the charged leptons are three isolated masses, 0.511 MeV, 105.7 MeV and 1777 MeV, spaced by factors of 207 and 17, with nothing between them and no fourth rung anywhere below 45.6 GeV.

LEP scanned the Z resonance and found 499 MeV of width unaccounted for by hadronic and charged-lepton decays. Dividing by the Standard Model width per neutrino species gives N_ν = 2.984 ± 0.008. What has that established?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUse Q = T₃ + Y/2 to assign a weak hypercharge to each field of the first generation — the left-handed quark doublet, u_R, d_R, the left-handed lepton doublet and e_R — then add up the electric charges of one whole generation, counting each quark once per colour.

Rearrange to Y = 2(Q − T₃). In a left-handed doublet the upper member has T₃ = +½ and the lower −½; every right-handed field is an SU(2) singlet with T₃ = 0.
Quark doublet: u_L gives Y = 2(2/3 − 1/2) = 1/3, and d_L gives Y = 2(−1/3 + 1/2) = 1/3. The two agree, as they must — hypercharge labels the doublet, not its members.
Right-handed quarks are singlets, so Y = 2Q directly: Y(u_R) = 4/3 and Y(d_R) = −2/3.
Lepton doublet: ν_L gives 2(0 − 1/2) = −1 and e_L gives 2(−1 + 1/2) = −1. The singlet e_R gives Y = 2(−1) = −2.
Charge sum over the generation, three colours per quark: 3(+2/3) + 3(−1/3) + 0 + (−1) = 2 − 1 + 0 − 1 = 0.
Repeat with two colours instead of three and the sum becomes 4/3 − 2/3 − 1 = −1/3, which is not zero. The ledger balances only for exactly three colours.

AnswerY = 1/3 (quark doublet), 4/3 (u_R), −2/3 (d_R), −1 (lepton doublet), −2 (e_R); and Σ Q over the generation is exactly zero, which holds only because quarks come in three colours.

MediumLEP measured the Z total width as Γ_Z = 2495.5 ± 2.3 MeV, the hadronic width as 1744.4 MeV, and each charged-lepton width as 83.99 MeV. The Standard Model predicts Γ_νν = 1.9912 Γ_ℓℓ per neutrino species. Find the number of light neutrino species, then say what Γ_Z would have been for four of them.

Visible width: hadrons plus three charged-lepton channels, 1744.4 + 3(83.99) = 1744.4 + 251.97 = 1996.37 MeV.
Invisible width: Γᵢₙᵥ = 2495.5 − 1996.37 = 499.13 MeV. Nothing was detected here — it is the width the lineshape fit demands and the visible channels cannot supply.
Work in ratios so that common couplings cancel: Γᵢₙᵥ/Γ_ℓℓ = 499.13/83.99 = 5.9427.
Divide by the predicted ratio per species: N_ν = 5.9427/1.9912 = 2.9845. The published LEP value is 2.9840 ± 0.0082.
For four species: Γ_νν = 1.9912 × 83.99 = 167.24 MeV, so Γ_Z would be 1996.37 + 4(167.24) = 2665.3 MeV.
That sits 169.8 MeV above the measured 2495.5 MeV, and the measurement is good to 2.3 MeV — a 74σ excess. Two species would fall 164.7 MeV short, 72σ low.

AnswerN_ν = 2.98, so three light species. Four would push Γ_Z to 2665 MeV, 170 MeV above the measured 2495.5 ± 2.3 MeV — an excess of about 74 standard deviations.

HardTake v = 246.22 GeV, mₜ = 172.57 GeV, mₑ = 0.511 MeV, and the heaviest neutrino at the 0.050 eV floor that atmospheric oscillation sets. Using m_f = y_f v/√2, find the Yukawa coupling of each, and state how many decades of mass the fermion table spans with and without the neutrinos.

Invert the relation: y_f = √2 m_f / v, a pure number with no units to carry.
Top quark: yₜ = 1.41421 × 172.57/246.22 = 244.06/246.22 = 0.991. Of order unity, which makes the top the one fermion whose coupling looks natural.
Electron: yₑ = 1.41421 × 0.000511/246.22 = 7.227×10⁻⁴/246.22 = 2.94×10⁻⁶.
Neutrino, treating the mass as Dirac: 0.050 eV = 5.0×10⁻¹¹ GeV, so y_ν = 1.41421 × 5.0×10⁻¹¹/246.22 = 2.87×10⁻¹³.
Charged-fermion span: mₜ/mₑ = 172.57/5.11×10⁻⁴ = 3.38×10⁵, and log₁₀(3.38×10⁵) = 5.53 — nearly six decades.
Adding the neutrino: mₜ/m_ν = 172.57/5.0×10⁻¹¹ = 3.45×10¹², so log₁₀ = 12.5. Every one of these couplings is fitted, none derived; that is the flavour problem in one line.

Answeryₜ = 0.991, yₑ = 2.94×10⁻⁶, y_ν ≈ 2.9×10⁻¹³. The charged fermions span 5.5 decades and the whole table 12.5 — and the Standard Model predicts none of it.