University Physics IV · Particle Physics · 14.6

The Quark Model & Hadron Spectroscopy

By 1962 the particle table had outgrown the word elementary. This is the lesson where you stop memorising hadrons and start generating them: put u, d and s on a charge-and-hypercharge lattice, multiply the triplets, and watch the octets, decuplets and nonets fall out — one missing site and all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

By 1961 the hadron list had passed a hundred entries with no principle sorting them. Gell-Mann and Ne'eman noticed that plotting each hadron by its third isospin component I₃ and its hypercharge Y = B + S does not scatter the states: they land on hexagons with a doubled centre, on triangles, on lattices of exactly eight or ten sites. Those are the weight diagrams of SU(3), the group you get by treating u, d and s as three interchangeable states of one object.

Multiplying the fundamental triplet reproduces the observed patterns and nothing else — 3 ⊗ 3̄ = 8 ⊕ 1 for mesons, 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 for baryons — and charge is not an independent label but the fixed combination Q = I₃ + Y/2. The pattern also left one site empty and set its mass by the equal spacing of the rows; the Ω⁻ turned up in 1964 within a percent of it. Two costs come with the win.

Antisymmetry fails for the Δ⁺⁺ = uuu unless quarks carry a further three-valued label, so the flavour model cannot stand without colour. And turning patterns into masses needs an effective constituent mass of a few hundred MeV per quark, fitted to the very spectrum it then describes. The quark model is an exact classifier and a parametrisation of masses — not a theory of them.

Simple definition: The quark model builds every hadron as a colour-singlet bound state — three quarks for a baryon, a quark and an antiquark for a meson — and sorts them into SU(3) flavour multiplets by their isospin and hypercharge.
Example: The Ξ⁻ is dss: B = 1 and S = −2 give Y = −1, so Q = I₃ + Y/2 = −1/2 − 1/2 = −1, which is exactly the charge measured, on a state of mass 1321.7 MeV/c².

Gell-Mann–Nishijima relationQ = I₃ + Y/2, Y = B + S

Charge is not an independent label: fix I₃ and Y, and Q is forced.

Q in units of e; I₃, B and S are pure numbers. In full, Y = B + S + C + B̃ + T.

SU(3) flavour products3 ⊗ 3̄ = 8 ⊕ 1 · 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1

Predicts nonets, octets and decuplets — and forbids every other multiplet size.

u, d, s as one triplet; the numbers count states, and 10 + 8 + 8 + 1 = 27.

Baryon antisymmetry and the colour singletψ = ψ_space ψₛₚᵢₙ ψ_flavour ψ_colour, ψ_colour = ε_abc q^a q^b q^c

Δ⁺⁺ = uuu with L = 0 and S = 3/2 is symmetric in the first three factors, so colour must supply the minus sign.

a, b, c run over the three colours; ε_abc flips sign under any exchange.

Constituent mass with hyperfine termM = Σᵢ mᵢ + A Σ_(i<j) (Sᵢ · Sⱼ)/(mᵢ mⱼ)

The 1/(mᵢmⱼ) is a product of magnetic moments: heavier pairs split less.

mᵤ ≈ m_d ≈ 310, mₛ ≈ 490 MeV/c²; A fitted. ⟨Sᵢ⋅Sⱼ⟩ = +ħ²/4 or −3ħ²/4.

Decuplet equal spacingM(nₛ) ≈ 1232 + 150 nₛ MeV/c²

Put the Ω⁻ at 1684 MeV before it existed; it was found at 1672.5 MeV.

nₛ = 0 to 3 strange quarks; each row averaged over its isospin multiplet.

Gell-Mann–Okubo, the 1/2⁺ octet(M_N + M_Ξ)/2 = (3M_Λ + M_Σ)/4

1128.6 against 1135.1 MeV: the octet's spacing is fixed to 0.6% too.

Multiplet averages / MeV c⁻²: N 938.9, Ξ 1318.3, Λ 1115.7, Σ 1193.1.

Two axes that sort two hundred hadrons

Isospin came first: the proton and neutron differ by 1.3 MeV out of 939, so treat them as one particle carrying a two-valued label, I₃ = ±1/2. Strangeness came next, invented to explain why kaons and lambdas are always produced in pairs by the strong force and then decay slowly and singly — S is conserved by the strong interaction and violated by the weak. Combine the two into the hypercharge Y = B + S and plot every hadron at its point (I₃, Y). The spin-1/2 baryons land on eight sites: a hexagon with p and n along the top at Y = +1, the three Σ and the Λ across the middle at Y = 0, and the two Ξ below at Y = −1, the centre doubly occupied by Λ and Σ⁰. The spin-3/2 baryons land on a triangle of ten. Charge never needs an axis of its own, because Q = I₃ + Y/2 reads it off the two you already have.

Three quarks, and the multiplets that follow

Put u at (I₃, Y) = (+1/2, +1/3), d at (−1/2, +1/3) and s at (0, −2/3) — a small triangle — and Q = I₃ + Y/2 returns +2/3, −1/3 and −1/3 as required. Antiquarks sit at the reflected points. The patterns are then vector addition. A meson is qq̄, so its site is the sum of a quark site and an antiquark site, and the nine sums fall on a hexagon with a triply occupied centre: 3 ⊗ 3̄ = 8 ⊕ 1, the nonet. A baryon is qqq, giving 27 combinations that sort by exchange symmetry into 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 — a fully symmetric decuplet, two mixed octets and an antisymmetric singlet, and 10 + 8 + 8 + 1 = 27. That arithmetic is the content of the model. Nothing in it permits a light-quark baryon multiplet of six or twelve states, and none has ever been found.

The empty corner and the Ω⁻

In 1962 the 3/2⁺ decuplet had nine of its ten sites filled: four Δ at 1232 MeV with S = 0, three Σ* at 1385 with S = −1, two Ξ* at 1533 with S = −2. The apex, S = −3, was empty. Read the rows: 1385 − 1232 = 153 MeV and 1533 − 1385 = 148 MeV, a constant step near 150 MeV each time a light quark is swapped for a strange one. Extrapolating gives 1533 + 150 = 1683 MeV, and Q = I₃ + Y/2 with I₃ = 0 and Y = 1 − 3 = −2 gives Q = −1. Better still, at that mass every strangeness-conserving strong channel is shut — the lightest, Ξ⁰K⁻, needs 1809 MeV — so the state has to decay weakly and will leave a visible track. Brookhaven found it in 1964: mass 1672.5 MeV/c², charge −1, lifetime 8.2 × 10⁻¹¹ s. The octet obeys the same discipline through Gell-Mann–Okubo, (M_N + M_Ξ)/2 = (3M_Λ + M_Σ)/4, which gives 1128.6 against 1135.1 MeV.

The Δ⁺⁺ forces a fourth label

Write the Δ⁺⁺ out. Its content is uuu, its spin is 3/2, and in the ground state there is no orbital angular momentum, so L = 0. Take the factors in turn. The spatial part with L = 0 is symmetric under swapping any two quarks. The flavour part, three identical u, is symmetric. The J_z = +3/2 state has all three spins up, so the spin part is symmetric too. Their product is symmetric — and three identical spin-1/2 fermions may not occupy a symmetric state. Either quarks are not fermions, or a factor is missing. The missing factor is colour: a label taking three values, combined as ε_abc q^a q^b q^c, which changes sign under any exchange and restores the antisymmetry. Δ⁻ = ddd and Ω⁻ = sss make the identical demand. Colour would be a cheap patch if this were its only evidence, but two independent measurements confirm the count of three: the ratio R of hadron to muon-pair production in e⁺e⁻ annihilation is three times the naive quark sum, reaching 10/3 above the charm threshold, and the π⁰ → γγ rate goes as N_c² and comes out a factor of nine above the colourless prediction.

Spin splits what flavour cannot

Classification says nothing about mass, so the model adds one term: M = Σᵢmᵢ + A Σ_(i<j)(Sᵢ⋅Sⱼ)/(mᵢmⱼ), the colour analogue of a magnetic dipole–dipole interaction, which is where the 1/(mᵢmⱼ) comes from. Its power shows wherever flavour is degenerate. The ρ and the π have the same quark content and differ only in whether the pair's spins are parallel or antiparallel — yet they differ by 637 MeV, more than four pion masses. The K* and K, with one quark replaced by the heavier s, differ by only 396 MeV, and the ratio 637/396 = 1.61 is mₛ/mᵤ read straight off the spectrum. The Λ and the Σ⁰ are both uds and sit on the same lattice site, and the model still separates them: the Λ carries its ud pair in a spin singlet, the Σ⁰ in a triplet, so the Λ lies lower, 1115.7 against 1192.6 MeV. A 77 MeV gap between two states of identical quark content is spin doing the work.

Fitted, not derived

Be clear about which half of this lesson is theorem and which is fit. The multiplet structure is group theory: given three flavours, 8 ⊕ 1 and 10 ⊕ 8 ⊕ 8 ⊕ 1 are forced, and Q = I₃ + Y/2 then holds site by site once Y is defined as B + S. The masses are another matter. The constituent values mᵤ ≈ 310 and mₛ ≈ 490 MeV/c² are fitted to the spectrum they are then used to describe, and they are not the masses in the QCD Lagrangian, which are roughly 2.2, 4.7 and 93 MeV/c² for u, d and s. The hyperfine constant A is fitted as well, and the η–η′ mixing angle is a third free number. The framework is also non-relativistic, which sits awkwardly with a 310 MeV/c² quark confined to about 1 fm, where the uncertainty principle alone gives it p ≈ ħ/r ≈ 200 MeV/c. The quark model organises the spectrum and predicts gaps inside it; computing a hadron mass from first principles waits for lattice QCD.

Change one variable at a time

Make the relationship visible.

Interactive model

Third isospin component I₃0.5

Strangeness S0

Hold S = 0 and sweep I₃ across the top row: Q climbs from −1 to +2 in unit steps, with Δ⁺⁺ at the right-hand end. Then drive S down to −3 — the rows narrow to a single site, the Ω⁻ at Q = −1, and the mass readout climbs 150 MeV a step.

HYPERCHARGE Y = B + S1.0

CHARGE Q = I₃ + Y/21.0 e

STRANGE QUARKS nₛ0

EQUAL-SPACING MASS1232 MeV/c²

Live interpretationHYPERCHARGE Y = B + S: 1.0. CHARGE Q = I₃ + Y/2: 1.0 e. STRANGE QUARKS nₛ: 0. EQUAL-SPACING MASS: 1232 MeV/c²

Catch the common trap

Explain before calculating.

The Δ⁺⁺ is a uuu state with J^P = 3/2⁺ and no orbital angular momentum between the quarks. Its existence was the argument that forced a new quantum number into the quark model. What exactly was the problem it exposed?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA baryon is produced with charge Q = −1 and strangeness S = −2. Use the Gell-Mann–Nishijima relation Q = I₃ + (B + S)/2 to find its third isospin component, deduce its quark content, and name the isospin multiplet it belongs to.

A baryon has B = +1, so the hypercharge is Y = B + S = 1 + (−2) = −1.
Gell-Mann–Nishijima: Q = I₃ + Y/2, so −1 = I₃ + (−1)/2 = I₃ − 1/2, giving I₃ = −1/2.
S = −2 needs two s quarks, since each carries S = −1. Their charges total −2/3, so the third quark must carry −1 − (−2/3) = −1/3: a d quark. The content is dss.
I₃ = −1/2 with a partner at +1/2 means an isospin doublet, and the partner swaps d for u. Check it: uss has I₃ = +1/2 and Y = −1, so Q = 1/2 − 1/2 = 0.

AnswerI₃ = −1/2 and the content is dss: the Ξ⁻, on the bottom row of the spin-1/2 baryon octet, paired with the Ξ⁰ (uss). Measured masses 1321.7 and 1314.9 MeV/c², a 0.5% split typical of an isospin doublet.

MediumThe J^P = 3/2⁺ decuplet rows have multiplet-average masses 1232 MeV/c² (Δ, S = 0), 1385 (Σ*, S = −1) and 1533 (Ξ*, S = −2). In 1962 the fourth row was empty. Predict the mass, the charge and the dominant decay mode of the state that belongs there.

Each step down the triangle swaps one light quark for a strange one, so the rows run S = 0, −1, −2, −3. The missing row has three strange quarks: the state is sss.
Row gaps: 1385 − 1232 = 153 MeV and 1533 − 1385 = 148 MeV. The spacing is constant to 3%, which is what a single additive constituent-mass step predicts. Take the mean, 150.5 MeV.
Extrapolate one more row: M = 1533 + 150.5 = 1683.5 MeV/c², or equivalently M = 1232 + 3 × 150.5 = 1683.5 MeV/c².
Charge: B = 1 and S = −3 give Y = −2, and an sss state has I = I₃ = 0, so Q = 0 + (−2)/2 = −1.
Decay: the lightest strangeness-conserving strong channel is Ξ⁰K⁻, needing 1314.9 + 493.7 = 1808.6 MeV — far above 1684. Every strong channel is shut, so it must decay weakly, one unit of strangeness at a time, with a lifetime near 10⁻¹⁰ s and a track centimetres long.
Measured: Brookhaven found the Ω⁻ in 1964 at 1672.5 MeV/c², charge −1, lifetime 8.2 × 10⁻¹¹ s.

AnswerM ≈ 1684 MeV/c², Q = −1, and no strong channel open, so it decays weakly. The Ω⁻ came in at 1672.5 MeV/c² with τ = 8.2 × 10⁻¹¹ s — the prediction was 0.7% high.

HardVector and pseudoscalar meson masses: M_ρ = 775, M_π = 138, M_K* = 892, M_K = 496 MeV/c². The hyperfine term is A(S₁⋅S₂)/(m₁m₂). Extract mₛ/mᵤ from these four numbers, then use it to predict the Σ* − Σ mass gap from the Δ − N gap of 293 MeV.

For a qq̄ pair ⟨S₁⋅S₂⟩ is +ħ²/4 in the spin triplet and −3ħ²/4 in the singlet, a difference of exactly ħ². So the vector-minus-pseudoscalar gap isolates the hyperfine term: M_V − M_P = A/(m₁m₂), with the constituent masses themselves cancelling.
ρ and π are both light-quark pairs: M_ρ − M_π = 775 − 138 = 637 MeV = A/mᵤ². K* and K carry one s: 892 − 496 = 396 MeV = A/(mᵤ mₛ).
Divide the two: mₛ/mᵤ = 637/396 = 1.61. With mᵤ = 310 MeV/c² that gives mₛ ≈ 499 MeV/c², the value quark-model fits use.
The same operator acts in baryons. Summing over pairs, Σ⟨Sᵢ⋅Sⱼ⟩ = ½[J(J+1) − 9/4]ħ², which is +3ħ²/4 for J = 3/2 and −3ħ²/4 for J = 1/2. With three light quarks that gives M_Δ − M_N = (3/2)A_B/mᵤ².
For Σ* and Σ, both uus with the two u quarks locked in a spin triplet, the same sum leaves the uu term unchanged and shifts only the two u–s terms: M_Σ* − M_Σ = (3/2)A_B/(mᵤ mₛ). The unknown A_B cancels in the ratio, so (M_Δ − M_N)/(M_Σ* − M_Σ) = mₛ/mᵤ.
Predict: M_Σ* − M_Σ = 293/1.61 = 182 MeV, against a measured 1385 − 1193 = 192 MeV.

Answermₛ/mᵤ = 1.61, so mₛ ≈ 500 MeV/c² if mᵤ = 310. The predicted Σ* − Σ gap is 182 MeV against 192 MeV measured — 5% low, from one ratio fitted to mesons and carried into baryons.