The Quark Model; Mesons · Elementary Particle Physics

A hundred hadrons fall onto a plane, and the pattern survives even though the symmetry predicting it is badly broken — members of one multiplet differ by hundreds of MeV.

🎯 Why this matters

The pattern is the evidence here, and only here. Three sections later the same diagrams get read backwards: as statements about what a hadron is made of, rather than about which states sit beside which.

Five sections of this chapter have produced a catalogue. Nine pseudoscalar mesons, nine vector mesons, eight $J^P = 1/2^+$ baryons and nine $3/2^+$ ones, each sorted into isospin multiplets, each with a mass and a width, and no reason for any of it. This section supplies the reason, and it is the shortest sentence in the chapter: hadrons are made of quarks.

4.6 The census, and the proposal

Everything §4.2–4.5 found, sorted by hypercharge and isospin

family	count	the SU(2) multiplets it falls into
pseudoscalar mesons, 0⁻	9	two singlets (η, η′), two doublets (K, K̄), one triplet (π)
vector mesons, 1⁻	9	the same multiplets again: ω and φ, K and K̄, ρ
baryons, 1/2⁺	8	doublets at Y = +1 (N) and Y = −1 (Ξ), a singlet (Λ) and a triplet (Σ) at Y = 0
baryons, 3/2⁺		a quartet at Y = +1 (Δ), a triplet at Y = 0 (Σ), a doublet at Y = −1 (Ξ)

Two nines, an eight and a nine-that-should-be-ten. <strong>Those are not arbitrary numbers</strong>, and in 1964 G. Zweig and M. Gell-Mann independently proposed the reason: baryons are three quarks, mesons a quark and an antiquark, with three kinds of quark available.

📐 Physics you need first — a weight diagram is a ladder with two rungs

§3.8 drew isospin multiplets as one axis: a row of states labelled by $I_z$ , running from $-I$ to $+I$ in integer steps. That worked because SU(2) has exactly one quantity you can measure at the same time as the total isospin.

SU(3) has two. In group language it has rank 2: you can simultaneously diagonalise $I_z$ and the hypercharge $Y$ , but nothing else. So a multiplet is no longer a row — it is a pattern of points in a plane, and the plane’s axes are the two quantities you already know how to compute.

Everything else carries over unchanged:

each point is a state, and the number of points is the dimension of the representation — 3, 8, 10 rather than $2I+1$ ;
the multiplet is closed under the group’s operations, which now move you between points in two directions rather than one;
members share their spin and parity, and would share a mass if the symmetry were exact.

The one genuinely new thing is that the patterns have shape — triangles, hexagons — and the shape is what identifies the representation. You are about to meet a triangle (the quarks), an inverted triangle (the antiquarks), a hexagon with a doubled centre (the octet), and a point (the singlet).

⚠️ Two different SU(3)s, and they are unrelated physics

The book stops to insist on this and the site will too, because the collision is permanent.

SU(3)_f — flavour — is what this section is about: a classification symmetry relating u, d and s. It is approximate, broken even by the strong interaction, and it exists because those three quarks happen to be light.
SU(3) of colour is the exact gauge symmetry of the strong force, the subject of Chapter 6. It has nothing to do with flavour.

Mathematically they are the same group; physically they have nothing in common, and confusing them is the single commonest error in this material. The book adds the subscript f for exactly that reason, and this site keeps it.

Note also the strength of the breaking. Isospin SU(2) is broken by the electromagnetic interaction and by the few-MeV u–d mass difference, so multiplets are degenerate to about 1 %. SU(3)_f is broken by the strong interaction itself, because $m_s \approx 93$ MeV is not small — so its multiplets are spread over hundreds of MeV. Both symmetries are, as §4.8 puts it, accidental.

Table 4.1 — the three lightest quarks

quark	$Q$	$I$	$I_z$	$S$	$\mathcal{B}$	$Y$	mass
d	−1/3	1/2	−1/2	0	1/3	1/3	4.67 MeV
u	+2/3	1/2	+1/2	0	1/3	1/3
s	−1/3	0	0		1/3	−2/3	93.4 MeV

Check the last two columns against Gell-Mann–Nishijima, I_z = Q − Y/2: for the u, +2/3 − 1/6 = +1/2 ✓; for the s, −1/3 + 1/3 = 0 ✓. <strong>Two features are startling and both are real:</strong> the charges are fractional, and the masses of u and d are a few MeV — under 1 % of the nucleon they build. Where the rest of the nucleon's mass comes from is §6.7.

The two fundamental representations

Fig. 4.17(a) — the representation 3: the quarks · J^P = 1/2⁺

Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check I_z = Q − Y/2.

Three states in a triangle: an isospin doublet at Y = +1/3 and a singlet at Y = −2/3. The antiquark representation 3̄ is the same triangle inverted — every charge reversed — and you can see it by pressing the antiparticle button on any of the meson diagrams below.

⚙️ Engineer’s bridge — the plane is a lattice of additive labels

A weight diagram looks exotic and is not. Both axes are quantities you have been adding up since §3.6: $I_z$ and $Y = \mathcal{B} + S$ are additive over the valence quarks .

That single fact does all the work in the next section. Combine a quark at $(I_z, Y)$ with an antiquark at $(I_z', Y')$ and the meson sits at $(I_z + I_z', Y + Y')$ — a vector sum on the lattice. So the nine mesons are just the nine ways of adding one point of the triangle to one point of the inverted triangle, and the hexagon-with-a-doubled-centre that comes out is a consequence of arithmetic, not of group theory.

The engineering reading: a weight diagram is a state space indexed by two independent counters, and combining systems is convolution on that lattice. When two of the nine sums land on the same lattice site — as three of them do at the origin — you get degeneracy, and degeneracy is where the interesting physics lives, because it is precisely where the labels stop distinguishing the states.

Where it breaks: the lattice tells you which sites exist, and how many states sit on each one, but not which linear combinations are physical. That question — what the three states at the origin actually are — is the rest of this page.

Where it breaks: the lattice tells you which labels exist and is silent about which linear combinations are physical — and at the origin, where three states share the same quantum numbers, that silence is the whole problem. The physical π⁰, η and η′ are mixtures determined by dynamics (and, for the η′, by the axial anomaly) that no amount of additive bookkeeping predicts. An address scheme with three occupants at one address needs something outside the scheme to resolve them.

4.7 Mesons: nine states, nine places

\htmlClass{t-a}{\mathbf 3} \otimes \htmlClass{t-b}{\bar{\mathbf 3}} = \htmlClass{t-c}{\mathbf 1} \oplus \htmlClass{t-d}{\mathbf 8}

(4.44)

Bettini p. 154. Nine states in, nine places out — and the split into 1 and 8 is what makes two of them special.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Fig. 4.18 — the octet, and (at the origin, alone) the singlet · J^P = any

Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check I_z = Q − Y/2.

The shape alone, without labels — a hexagon with two states stacked at the centre. Every meson nonet on this page is this figure plus a singlet at the origin, giving three states there in total, and the whole of the rest of the section is about what those three states are.

The pseudoscalar nonet

The ground state of a quark and an antiquark should be an $S$ wave, and §3.3’s table then gives just two possibilities: $^1S_0$ with $J^{PC} = 0^{-+}$ and $^3S_1$ with $1^{--}$ . Those are exactly the two nonets that exist.

Fig. 4.19 — the pseudoscalar mesons, J^P = 0⁻ · J^P = 0⁻

Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check I_z = Q − Y/2.

The six rim states have unambiguous quark contents — read them off as a vector sum on the lattice. The three states stacked at the origin do not: π⁰, η and η′ all have I_z = Y = 0, and only the π⁰ is fixed, by belonging to the isospin triplet. Press the antiparticle button and the diagram maps onto itself, because a meson nonet contains its own antiparticles.

Erratum — the K⁻ in Fig. 4.19

The printed figure labels the K⁻ with the quark content u s̄ — the same content it gives the K⁺ directly above it. Two members of a multiplet cannot have the same quark content, and the K⁻ is the antiparticle of the K⁺, so it must be s ū. The diagram above uses the correct assignment; the K̄⁰ label (s d̄) is right as printed.

You did not need the book to tell you which is which, and that is the point of the lattice. The K⁻ sits at $I_z = -\tfrac12$ , $Y = -1$ , so its strangeness is $S = -1$ : it carries an s quark, not an s̄. Reading the content off the axes is faster than remembering it, and it catches typography like this one.

\begin{aligned} |\htmlClass{t-pi}{\pi^0}\rangle &= \tfrac{1}{\sqrt2}\left|u\bar u - d\bar d\right\rangle && (4.45) \\[3pt] |\htmlClass{t-e1}{\eta_1}\rangle &= \tfrac{1}{\sqrt3}\left|u\bar u + d\bar d + s\bar s\right\rangle && (4.46) \\[3pt] |\htmlClass{t-e8}{\eta_8}\rangle &= \tfrac{1}{\sqrt6}\left|u\bar u + d\bar d - 2s\bar s\right\rangle && (4.47) \end{aligned}

(4.45–4.47)

Bettini pp. 154–155. Three orthogonal states at the same lattice site, and only the first is a physical particle.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — two states, and neither is what you measure

$\pi^0$ is settled: isospin fixes it, and it is a particle. The other two are not.

$\eta_1$ and $\eta_8$ are the states SU(3)_f hands you. The η and η′ are what a bubble chamber hands you. If SU(3)_f were an exact symmetry the two lists would coincide, because states of different representations cannot mix. It is not exact — and, unlike isospin, it is broken by the strong interaction itself. So the physical states are superpositions, with one free angle:

|\eta\rangle = \cos\theta\,|\eta_8\rangle - \sin\theta\,|\eta_1\rangle, \qquad |\eta'\rangle = \sin\theta\,|\eta_8\rangle + \cos\theta\,|\eta_1\rangle.

This is the same structure as every mixing problem in the book — the eigenbasis of the symmetry is not the eigenbasis of the Hamiltonian, so what propagates is a rotation of what the symmetry classifies. Chapter 8 is the same sentence about neutral kaons, and Chapter 10 about neutrinos.

singlet–octet mixing · the two isosinglets of a nonet

mixing angle θ = 35.0°— ideal mixing

state 1 = cos θ |η₈⟩ − sin θ |η₁⟩0.0 % light · 100.0 % ss̄

+0.005 (uū+dd̄)/√2 − 1.000 ss̄

state 2 = sin θ |η₈⟩ + cos θ |η₁⟩100.0 % light · 0.0 % ss̄

+1.000 (uū+dd̄)/√2 + 0.005 ss̄

Written in the flavour basis: n = (uū+dd̄)/√2 and s = ss̄, with η₁ = √(2/3) n + √(1/3) s and η₈ = √(1/3) n − √(2/3) s. The two bars are always orthogonal, so the ss̄ fraction of one is the light-quark fraction of the other — mixing moves content between the two states and creates none. State 1 is the octet-dominated one at θ = 0; which physical particle that turns out to be is not the same in the two nonets — it is the η, the lighter pseudoscalar, and the φ, the heavier vector.

Essentially ideal. Here state 1 is almost pure ss̄ — that is the φ — and state 2 is almost pure (uū+dd̄)/√2, which is the ω, and that is why the ω’s mass is nearly the ρ’s. Ideal mixing is the whole explanation of §4.5’s puzzle: the φ prefers KK̄ because its s and s̄ walk straight into the kaons, while 3π would need them to annihilate first.

The vector nonet, and why the φ behaves as it does

Fig. 4.20 — the vector mesons, J^P = 1⁻ · J^P = 1⁻

Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check I_z = Q − Y/2.

The same nine places, the same quark contents on the rim — and a completely different answer at the centre. Here the mixing is close to ideal, so the two isosinglets separate cleanly into a non-strange ω and an almost purely strange φ.

\htmlClass{t-w}{\omega} \approx \tfrac{1}{\sqrt2}\left|u\bar u + d\bar d\right\rangle, \qquad \htmlClass{t-p}{\phi} \approx \left|s\bar s\right\rangle

(4.48)

Bettini p. 156 — ideal mixing. Two consequences follow immediately, and both were puzzles in §4.5.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

🔢 Worked example — the mixing algebra, and what the masses say

Ideal mixing is not an extra assumption; it is one particular angle, and the angle is fixed by asking for one state to be pure ss̄.

Reproduce it

import numpy as np
R23, R13 = np.sqrt(2/3), np.sqrt(1/3)
print("the two SU(3) isosinglets, in the flavour basis n = (uu+dd)/sqrt2, s = ss")
print(f"   eta_1 = (uu+dd+ss)/sqrt3   ->  n = {R23:+.4f}  s = {R13:+.4f}")
print(f"   eta_8 = (uu+dd-2ss)/sqrt6  ->  n = {R13:+.4f}  s = {-R23:+.4f}")
th = np.degrees(np.arctan(1/np.sqrt(2)))
print("ideal mixing: demand that one physical state have no s at all")
print(f"   tan(theta) = sqrt(1/3)/sqrt(2/3) = 1/sqrt(2)  ->  theta = {th:.3f} deg")
c, s = np.cos(np.radians(th)), np.sin(np.radians(th))
s1 = (c*R13 - s*R23, -c*R23 - s*R13)          # cos t |8> - sin t |1>
s2 = (s*R13 + c*R23, -s*R23 + c*R13)          # sin t |8> + cos t |1>
print("at that angle")
print(f"   state 1:  n = {s1[0]:+.4f}  s = {s1[1]:+.4f}   -> |s|^2 = {s1[1]**2:.3f}, pure s sbar")
print(f"   state 2:  n = {s2[0]:+.4f}  s = {s2[1]:+.4f}   -> |s|^2 = {s2[1]**2:.3f}, no strangeness at all")
print(f"   orthogonality check: {s1[0]*s2[0]+s1[1]*s2[1]:+.1e} ;  norms "
      f"{s1[0]**2+s1[1]**2:.6f} and {s2[0]**2+s2[1]**2:.6f}")
print("what the measured masses say")
print(f"   VECTOR nonet:  m(rho) = 775, m(omega) = 782  ->  differ by {100*(782-775)/775:5.1f} %"
      f"  (ideal mixing)")
print(f"                  m(phi) - m(omega) = {1019-782:.0f} MeV, about 2 x m_s = {2*93.4:.0f} MeV")
print(f"   PSEUDOSCALAR:  m(pi) = 135, m(eta) = 548  ->  differ by {100*(548-135)/135:5.1f} %"
      f"  (NOT ideal)")
print("                  the eta and eta' are both far from pure flavour states")

prints

the two SU(3) isosinglets, in the flavour basis n = (uu+dd)/sqrt2, s = ss
 eta_1 = (uu+dd+ss)/sqrt3   ->  n = +0.8165  s = +0.5774
 eta_8 = (uu+dd-2ss)/sqrt6  ->  n = +0.5774  s = -0.8165
ideal mixing: demand that one physical state have no s at all
 tan(theta) = sqrt(1/3)/sqrt(2/3) = 1/sqrt(2)  ->  theta = 35.264 deg
at that angle
 state 1:  n = -0.0000  s = -1.0000   -> |s|^2 = 1.000, pure s sbar
 state 2:  n = +1.0000  s = -0.0000   -> |s|^2 = 0.000, no strangeness at all
 orthogonality check: +0.0e+00 ;  norms 1.000000 and 1.000000
what the measured masses say
 VECTOR nonet:  m(rho) = 775, m(omega) = 782  ->  differ by   0.9 %  (ideal mixing)
                m(phi) - m(omega) = 237 MeV, about 2 x m_s = 187 MeV
 PSEUDOSCALAR:  m(pi) = 135, m(eta) = 548  ->  differ by 305.9 %  (NOT ideal)
                the eta and eta' are both far from pure flavour states

The mass comparison is the evidence, and it is unusually direct. In the vector nonet the ρ and ω agree to 1 %, which is what two states built from the same light quarks should do — and the φ sits 237 MeV above, close to twice the strange quark mass. Ideal mixing, and SU(3)_f breaking that is just the s quark being heavy.

In the pseudoscalar nonet the same comparison gives 135 against 548 MeV. The η is not a light-quark state at all, and the η′ at 958 MeV is worse. Something beyond quark masses is at work, and the book names it without explaining it: the colour interaction induces continuous transitions among $u\bar u$ , $d\bar d$ and $s\bar s$ , and getting these two states right requires the QCD vacuum of §6.8.

Aside — which J^PC values a qq̄ pair can have

The book closes §4.7 by noting that many more mesons exist beyond these two nonets, in states with non-zero orbital momentum, and that their spin–parities are among those of Table 3.1 — “but not, for example, $J^{PC} = 0^{--}$ , $0^{+-}$ , $1^{-+}$ .”

Those are exactly the three the site’s JPCTable marks as unreachable. A meson found with one of them cannot be a quark and an antiquark, whatever its mass — and hunting for one is a live experimental programme. The lattice in that widget shows two more, $2^{+-}$ and $3^{-+}$ , which the book does not list because only the lowest three are searched for.

🔑 If you remember only three things

Nine places, filled before they were explained. The diagram organised what already existed, and that is what later made a gap in it worth betting on.
What the algebra hands you is not what a detector sees. Two of the three neutral states here are mixtures, and the mixing angle is itself a number to be measured.
A classification can tolerate a broken symmetry. The masses inside a multiplet spread by hundreds of MeV and the multiplet still holds, which is why the scheme outlived every attempt to make it exact.

Where this goes next

§4.8 does the same construction for three quarks — $3 \otimes 3 \otimes 3$ — and finds that it predicts four multiplets where nature shows two. Resolving that discrepancy is how colour was discovered.
§6.5 supplies the dynamical rule that suppresses φ → 3π, and §6.8–6.9 explains the η′ mass and the failure of ideal mixing in the pseudoscalar nonet.
§6.7 answers the question Table 4.1 raises: if u and d weigh a few MeV, where do the other 930 MeV of a proton come from?
Chapter 8 is mixing again, with two states whose superposition changes in flight rather than sitting still.

✅ Check yourself — the quark model and the meson nonets

0/5 answered · 0 correct

1.Why is an SU(3) multiplet drawn in a plane, when an isospin multiplet was drawn on a line?
2.The book insists that SU(3)_f and the SU(3) of colour are unrelated. What is the strongest way to see that they are different physics?
3.The π⁰, η and η′ all sit at $I_z = Y = 0$ . Why is only the π⁰ determined without further input?
4.In the FlavourMixer, set the angle to 35°. What happens, and which puzzle does it solve?
5.In the vector nonet the ρ and ω differ by 1 %; in the pseudoscalar nonet the π and η differ by 300 %. What does the contrast show?