Nothing forbids a two-quark or a four-quark state. The rule that only mesons and baryons exist is dynamical — those are the combinations the colour force actually binds.
🎯 Why this matters
A dynamical rule has to be computed rather than declared, and a computation can be checked. That is why the section ends on hyperfine splittings: the factors deciding what binds also decide by how much.§4.8 established that hadrons are colour singlets, using exchange symmetry and the Pauli principle. This section asks the harder question: why? Why does nature build and and nothing else — not , not , not a free quark?
The answer turns out to be a sign, and it is computable.
Neutrality, and what it means for colour
An atom is neutral because it holds as many positive charges as negative. The colour version is the colour singlet colour singlet a colour-neutral combination, the only kind that exists as a free particle; qq̄ via 3⊗3̄ ⊃ 1 and qqq via 3⊗3⊗3 ⊃ 1. defined in §6.4 — open in glossary , and there are two ways to build one from quarks.
Mesons are the easy case. A quark is in the 3, an antiquark in the 3̄, and Eq. (6.37) says their product contains a singlet:
Why a meson can exist. The same decomposition that gave eight gluons in §6.3 also guarantees one colourless combination — and that combination is what a meson is.
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.
Baryons need two steps. Eq. (6.40) gives , with the 6 symmetric and the 3̄ antisymmetric. Then
Why a baryon can exist, and why its quarks are forced into an antisymmetric colour pairing. Twenty-seven combinations of three coloured quarks contain exactly one colourless state.
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.
and the crucial structural point: the singlet comes only from the 3̄ ⊗ 3 piece, never from 6 ⊗ 3. So every quark pair inside a baryon sits in the antisymmetric colour 3̄, and couples to the third quark to make the singlet, Eq. (6.42).
exchange symmetry of three quarks · every label below is computed, not looked up
| J | sym. | Jz = +3/2 | Jz = +1/2 | Jz = −1/2 | Jz = −3/2 |
|---|---|---|---|---|---|
| 3/2 | S | ||||
| 1/2 | M,A | — | — | ||
| 1/2 | M,S | — | — |
Jz = +1/2. The badges come from swapping the two quarks in every basis ket and comparing amplitude by amplitude. Notice that M,A and M,S are only ±1 under P₁₂ — they are not eigenstates of the other two exchanges at all, which is exactly what "mixed" means and why neither can carry a baryon by itself.
💡 What this really says — the strangest sentence in the chapter
Two quarks in an antisymmetric colour combination — say — attract a green quark, and attract it exactly as an antigreen charge would.
Read that again. In electrostatics, two like charges in any arrangement whatever still repel a third like charge. Here, two colours in the right antisymmetric combination behave like the anticolour of the third. That is why three quarks can bind at all, and it has no analogue in electromagnetism.
The mechanism is the same non-abelian structure of §6.3: the combination transforms under SU(3) exactly as does, because 3̄ is contained in 3 ⊗ 3. Nothing like it happens in U(1), where the product of two charges is just their sum.
Two constructions, one outcome. Left: a quark and an antiquark of the matching anticolour, summed with equal weight over all three colours — the combination is invariant, so it is colourless in every gauge. Right: three quarks of different colours, where the point is the shaded box. Any pair in the antisymmetric combination (RB − BR) transforms under SU(3) exactly as an antigreen charge does, because 3̄ ⊂ 3 ⊗ 3 — so it attracts the green quark just as an antiquark would. That is what makes three-body binding possible, and it has no counterpart in electromagnetism.
Computing the sign
The book computes the colour factors diagram by diagram from Figs. 6.20 and 6.21, getting for the meson and for the baryon pair. There is a shorter route to the same numbers that also answers a question the book leaves implicit — why do the other channels not bind?
📐 Physics you need first — the Casimir shortcut
For a two-body system, the colour factor of a channel is
Supplied by this site — the book computes each colour factor from its Feynman diagrams instead. The Casimir shortcut gets the same numbers in one line and, unlike the diagrams, tells you why the channels that do not bind do not bind.
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.
where is the quadratic Casimir of each representation — the SU(3) analogue of for angular momentum. The values are , , , .
The derivation is the same trick §6.4’s spin algebra uses two paragraphs from now: write and solve for the cross term. Here , and the colour dot product is what multiplies .
Negative means attractive.
colour factors · which channels attract, and what that predicts
The colour factor of a two-body channel is (C_total − C_a − C_b)/2, with the quadratic Casimirs C₁ = 0, C₃ = C₃̄ = 4/3, C₆ = 10/3, C₈ = 3. Negative means attractive.
| pair | product | channel | colour factor | ||
|---|---|---|---|---|---|
| q q̄ | 3 ⊗ 3̄ | 1 | −4/3 α_s | attractive | a meson |
| q q̄ | 3 ⊗ 3̄ | 8 | +1/6 α_s | repulsive | — |
| q q | 3 ⊗ 3 | 3̄ | −2/3 α_s | attractive | the pair inside a baryon |
| q q | 3 ⊗ 3 | 6 | +1/3 α_s | repulsive | — |
Exactly two channels attract, and they are exactly the two that exist as hadrons. Nothing forbids a qq pair in the 6 or a qq̄ pair in the octet — they simply do not bind, because the sign came out wrong. The −4/3 and −2/3 are the book's Eqs. (6.39) and (6.43).
the test · hyperfine splittings, Eq. (6.51)
SU(3) gets both signs right and the ratio too. 8/3 against 2 predicts 1.333; the measurement gives 1.355 — under 2 % apart, from pure group theory with no free parameter beyond the common Kα_s. This is the sharpest quantitative evidence in the chapter that the colour force really is SU(3), and it comes from mass splittings rather than from any scattering experiment.
The whole of §6.4 in one table and one comparison. Click a row to select it; then toggle the hyperfine panel between the real SU(3) factors and a hypothetical U(1)-like colour force.
which colour channels bind, and which do not
from fractions import Fraction as F
# quadratic Casimirs of the SU(3) representations that appear
C = {'1': F(0), '3': F(4,3), '3bar': F(4,3), '6': F(10,3), '8': F(3)}
print("colour factor of a two-body channel = (C_total - C_a - C_b)/2")
print()
rows = [('q qbar', '3 x 3bar', '1', '3', '3bar', 'a MESON'),
('q qbar', '3 x 3bar', '8', '3', '3bar', ''),
('q q', '3 x 3', '3bar', '3', '3', 'the pair inside a BARYON'),
('q q', '3 x 3', '6', '3', '3', '')]
for pair, prod, tot, a, b, note in rows:
f = (C[tot] - C[a] - C[b]) / 2
print(f" {pair:7s} {prod:9s} -> {tot:5s}: {str(f):>6s} a_s "
f"{'ATTRACTIVE' if f < 0 else 'repulsive '} {note}")
print()
print("Exactly two channels attract, and they are exactly the two that exist.")
print("The factors -4/3 and -2/3 are the book's Eqs. (6.39) and (6.43).")
print("Nothing FORBIDS a qq pair in the 6 or a qqbar pair in the octet --")
print("they simply do not bind, because the sign came out wrong.")colour factor of a two-body channel = (C_total - C_a - C_b)/2 q qbar 3 x 3bar -> 1 : -4/3 a_s ATTRACTIVE a MESON q qbar 3 x 3bar -> 8 : 1/6 a_s repulsive q q 3 x 3 -> 3bar : -2/3 a_s ATTRACTIVE the pair inside a BARYON q q 3 x 3 -> 6 : 1/3 a_s repulsive Exactly two channels attract, and they are exactly the two that exist. The factors -4/3 and -2/3 are the book's Eqs. (6.39) and (6.43). Nothing FORBIDS a qq pair in the 6 or a qqbar pair in the octet -- they simply do not bind, because the sign came out wrong.
The test: hyperfine splittings
A colour factor is not directly observable. But it multiplies a spin–spin interaction, and that shows up as a mass difference — the same hyperfine structure that splits hydrogen’s 21 cm line, scaled up by the strength of the colour force.
Bettini p. 247. The QED version, Eq. (6.44), is the same with electric charges in place of κ — it is what gives hydrogen its 21 cm line.
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 ordering a U(1)-like force gets backwards
Spin algebra first, identical either way. For two spin-½, : that is for and for . For three, Eq. (6.49) gives : for and for .
Now pretend colour behaves like electric charge. A meson is a quark and an antiquark, so ; two quarks in a baryon have the same sign, so . Then Eqs. (6.48) and (6.50):
The first is right — vector mesons are heavier than pseudoscalars. The second says the decuplet should be lighter than the octet. The Δ(1232) outweighs the proton by 294 MeV. It is not lighter.
And there is a second failure: the baryon splitting would be times the meson one. Measured, it is times.
Now use the real colour factors, Eq. (6.51):
The hyperfine splittings with the real colour factors in place. Two sign flips — one from spin, one from colour — and both orderings come out right, where the charge-like model above got the second one backwards.
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.
Both positive — both orderings right. And the ratio is a prediction with no free parameter, since cancels: .
the ratio, which is a parameter-free prediction
spin2 = lambda J: J*(J+1) - 1.5 # 2<s1.s2>, two spin-1/2
spin3 = lambda J: J*(J+1) - 2.25 # Sigma, three spin-1/2
meson_spin = -(spin2(1) - spin2(0)) # the -2 of Eq. (6.48)
baryon_spin = -(spin3(1.5) - spin3(0.5)) # the -3 of Eq. (6.50)
print(f"spin factors: mesons {meson_spin:+.0f}K, baryons {baryon_spin:+.0f}K")
for label, cf_m, cf_b in [("colour like electric charge", -1, +1),
("the real SU(3) factors ", -4/3, -2/3)]:
dm, db = meson_spin*cf_m, baryon_spin*cf_b
print(f"\n{label}")
print(f" m(3S1) - m(1S0) = {dm:+.3f} -> vector {'heavier' if dm>0 else 'LIGHTER'}"
f" {'(right)' if dm>0 else '(WRONG)'}")
print(f" m(10) - m(8) = {db:+.3f} -> decuplet {'heavier' if db>0 else 'LIGHTER'}"
f" {'(right)' if db>0 else '(WRONG)'}")
print(f" ratio meson/baryon = {dm/db:+.3f}")
dK, dD = 891.67 - 493.677, 1232.0 - 938.272 # K*(892) - K, Delta(1232) - p
print(f"\nmeasured: m(K*) - m(K) = {dK:.0f} MeV, m(Delta) - m(p) = {dD:.0f} MeV")
print(f" ratio = {dK/dD:.3f} against the SU(3) prediction 4/3 = {4/3:.3f}")
print(f" agreement to {abs(dK/dD - 4/3)/(4/3)*100:.1f}%, with K a_s cancelling out")spin factors: mesons -2K, baryons -3K
colour like electric charge
m(3S1) - m(1S0) = +2.000 -> vector heavier (right)
m(10) - m(8) = -3.000 -> decuplet LIGHTER (WRONG)
ratio meson/baryon = -0.667
the real SU(3) factors
m(3S1) - m(1S0) = +2.667 -> vector heavier (right)
m(10) - m(8) = +2.000 -> decuplet heavier (right)
ratio meson/baryon = +1.333
measured: m(K*) - m(K) = 398 MeV, m(Delta) - m(p) = 294 MeV
ratio = 1.355 against the SU(3) prediction 4/3 = 1.333
agreement to 1.6%, with K a_s cancelling out ⚙️ Engineer’s bridge — one sign right is not a test; two signs and a ratio is
Notice the structure of the argument, because it is a template worth stealing.
A U(1)-like colour force gets the meson ordering right. If that were the only comparison available you would conclude the model works. It is only when a second prediction is put beside it — the decuplet–octet ordering — that the model fails, and it fails in a direction no amount of parameter-tuning can rescue, because the sign of a prediction is not adjustable.
Then the ratio closes it. and are unknown, but they are common to both splittings, so the ratio is a bare number: 4/3 against a measured 1.355.
This is the same discipline as validating a model against a held-out set rather than the one you fitted to. One agreement is consistency; two agreements with a shared unknown cancelled out is a test. The measured ratio came from mass spectroscopy — no scattering experiment, no collider — and it pins the gauge group of the strong interaction to within two per cent.
Where it breaks: a parameter-free ratio tests only what the ratio is sensitive to, and the cancellation that makes it clean is the same cancellation that makes it narrow. and divide out — so the 4/3 says nothing about whether has the right value, whether the potential has the right shape, or whether the non-relativistic treatment is justified for the light quarks it is applied to. It pins the group to two per cent and validates nothing else. Two signs and a ratio is a strong test of one structural claim, and it is easy to read it as a test of the theory.
| prediction | if colour were like charge | with SU(3) colour factors | measured |
|---|---|---|---|
| which qq̄ channel binds | any — charges just add | the singlet only, −4/3 α_s | only singlets exist |
| which qq channel binds | none — like charges repel | the antisymmetric 3̄, −2/3 α_s | baryons exist |
| vector vs pseudoscalar meson | +2K — vector heavier ✓ | +8/3 α_s K — vector heavier ✓ | K* − K = +398 MeV |
| decuplet vs octet baryon | −3K — decuplet lighter ✗ | +2 α_s K — decuplet heavier ✓ | Δ − p = +294 MeV |
| ratio of the two splittings | 2/3 = 0.67 | 4/3 = 1.333 | 1.355 |
The fourth row is where a U(1)-like colour force dies, and the fifth is where SU(3) earns its place. Note that the last column contains no scattering experiment at all — this is the gauge group of the strong interaction determined from a table of hadron masses.
Supplied — the book computes the two attractive factors diagram by diagram and never asks about the other two, which is the question a reader actually has. Drawn from the Casimir shortcut above, with , , , . The channels that bind are exactly the channels that are colourless, and nothing in the calculation was told to arrange that — it falls out of being most negative when . The 2 : 1 ratio of the two attractive factors is what the hyperfine test above turns into a parameter-free prediction of 4/3 for the splitting ratio, measured at 1.355.
🔑 If you remember only three things
-
Attraction alone is not the rule. Two quarks in the right channel do attract; it takes a third before the object is colourless and the binding closes.
-
The prediction is a ratio, which is why it survives not knowing the strength. An overall constant nobody can compute cancels between the two cases being compared.
-
Colour neutrality is stronger than electric neutrality. A neutral atom still attracts another one; a colour singlet has no long-range field left to do it with.
Where this goes next
- §6.5 turns these same colour factors into the running of α_s, and the gluon self-coupling flips its sign against §5.8’s α.
- §6.7 explains why the proton weighs a hundred times its quarks, using the antiscreening clouds these factors describe.
- §4.8 is where colour was required by exchange symmetry; this section is where it is quantitatively confirmed by mass splittings.
- §4.6 has the same tensor products, for flavour rather than colour — a symmetry that is broken where this one is exact.
✅ Check yourself — colour factors and why hadrons are what they are
0/5 answered · 0 correct
1.Of the four two-body colour channels in the widget, exactly two have a negative colour factor. What does that explain?
2.Two quarks in the antisymmetric combination (RB − BR) attract a green quark exactly as an antigreen charge would. Why is that impossible in electromagnetism?
3.A U(1)-like colour force predicts the vector mesons heavier than the pseudoscalars, which is correct. Why is that not evidence for it?
4.The SU(3) prediction for the ratio of the two splittings is 4/3, against a measured 1.355. Why is a ratio a stronger test than either splitting alone?
5.Why does the singlet in 3 ⊗ 3 ⊗ 3 come only from the 3̄ ⊗ 3 piece, never from 6 ⊗ 3?