The QCD Lagrangian and the Colour Charges · Elementary Particle Physics

The photon carries no charge and the gluon carries two. Every difference between this chapter and the last one descends from that single fact.

🎯 Why this matters

A carrier that feels its own force makes the field equations non-linear, so none of Chapter 5’s methods survive the crossing intact. Everything difficult about the strong interaction begins there.

Two sections of evidence, and now the theory. §5.1 built QED by demanding that a phase be choosable independently at every point, and the photon came out forced rather than added. This section runs the identical argument with a bigger group — and one thing goes differently, which is the whole of QCD.

From U(1) to SU(3)

The gauge group of QCD is SU(3): the 3 × 3 unitary matrices of determinant 1. Three colour charges , called red, green and blue, and the symmetry is exact — unlike isospin or flavour SU(3), which are accidents of small quark masses (§4.6), colour is not approximately conserved. It is conserved.

The Lagrangian looks like QED’s with the objects promoted:

	QED — U(1)	QCD — SU(3)
gauge potential	A^μ, a real-valued field	G^μ, a 3 × 3 Hermitian matrix field
field tensor	F^{μν} = ∂^μA^ν − ∂^νA^μ	G^{μν} = ∂^μG^ν − ∂^νG^μ − i g_s [G^μ, G^ν]
charge	q_f, one number per particle	a colour, in the 3 or the 3̄
coupling	α = e²/4πε₀ħc	α_s = g_s²/4π, Eq. (6.30)
interaction term	−q_f ψ̄γ^μψ A_μ	−g_s ψ̄γ^μ ψ G_μ, with the colour index summed
mediators	1 photon, uncharged	8 gluons, each carrying colour
do mediators self-couple?	no	yes — and everything follows from that

Only the second row is a genuinely new kind of object. Every other line is a promotion of a number to a matrix, which is bookkeeping. The commutator −i g_s [G^μ, G^ν] has no QED counterpart at all, because numbers commute and matrices do not.

💡 What this really says — one commutator, and everything that follows from it

In QED the field tensor is a curl, $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , and it is linear in the field. The free-field term $F_{\mu\nu}F^{\mu\nu}$ is therefore quadratic — two photons and no more. A quadratic term describes propagation, not interaction. Photons do not touch each other.

In QCD the tensor carries an extra piece, $-ig_s[G^\mu, G^\nu]$ , which is quadratic in the field. Square it and you get terms with three and four gluon fields and no quark anywhere. Those are interaction vertices. The gluon field is a source of itself.

Everything distinctive about the strong interaction is downstream of that:

gluons carry colour, so emitting one changes a quark’s colour, where emitting a photon leaves an electron’s charge alone;
gluon loops contribute to the running with the opposite sign to quark loops, which is asymptotic freedom (§6.5);
the colour field between two quarks collapses into a tube instead of spreading, which is confinement;
there should exist bound states of pure glue — glueballs — with no quarks at all. They are expected, and hard to identify, because they mix with ordinary $q\bar q$ mesons of the same quantum numbers.

the non-abelian difference · two colour rotations, in both orders

Pick any two generators. The panel multiplies them both ways round and subtracts.

with

λ₁

0	1	0
1	0	0
0	0	0

λ₂

0	-1i	0
1i	0	0
0	0	0

[λ₁, λ₂]

2i	0	0
0	-2i	0
0	0	0

‖[λ_a, λ_b]‖ in SU(3)

2.828

the same thing in U(1)

0.000

Non-zero. Two colour rotations applied in the other order do not give the same result, so the field tensor cannot be the plain curl it is in QED. It acquires a term −i g_s [G_μ, G_ν], which contains the gluon field with no quark in sight — the gluon field is its own source. Everything that separates QCD from QED comes out of this one box being non-zero: gluons carry colour, gluons couple to gluons, and α_s runs the other way.

⚙️ Engineer’s bridge — order of operations is the entire content

The widget above is doing something an engineer does daily: checking whether two operations commute.

Rotations in 3D do not commute — roll then pitch is not pitch then roll, which is why quaternions and rotation matrices exist and why “just add the angles” is a bug. Matrix multiplication does not commute, which is why A*B and B*A are different lines of code. Two writes to overlapping memory do not commute, which is the entire subject of memory ordering.

U(1) is the abelian case: gauge transformations are complex phases, phases are numbers, numbers commute, and $[\,\cdot,\cdot\,]$ is identically zero for every choice. There is nothing to check.

SU(3) is the non-abelian case, and “non-abelian” is simply the statement that the box in the widget is not zero. The physical consequence is that the mediator carries the charge it mediates — which in the memory analogy is the difference between a bus that carries transactions and a bus that is a transaction.

Note the honest wrinkle the widget makes you find: not every pair fails to commute. $\lambda_3$ and $\lambda_8$ commute perfectly happily — they are the two diagonal generators, and that pair is exactly the Cartan subalgebra that gave §4.6 its two-dimensional weight diagrams. A group is non-abelian if some pair fails, not if all do.

Where it breaks: non-commutativity in software has no consequences beyond the result you computed — do the operations in the wrong order and you get a wrong number, not a different machine. Here the failure to commute is dynamical: it puts a term in the Lagrangian, and that term is a gluon–gluon vertex. The non-abelian structure does not merely change the answer, it changes what particles exist and what they can do — asymptotic freedom, confinement and the three-jet events of §6.1 are all that one commutator. Order-of-operations is the right way in; it stops being the right way once you ask what the disagreement produces.

Eight gluons, and why not nine

Combining a colour with an anticolour gives nine states, and they split, Eq. (6.31):

\htmlClass{t-l}{3 \otimes \bar 3} = \htmlClass{t-e}{8} \oplus \htmlClass{t-s}{1}

(6.31)

Why there are eight gluons and not nine. Nine colour–anticolour combinations exist; they split into a set of eight that transform into one another and one that transforms into nothing.

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.

which is exactly the decomposition §4.6 used for the meson nonet — the same group, the same algebra, a completely different physical meaning.

Fig. 6.15/6.16 — the gluon octet, 3 ⊗ 3̄ = 8 ⊕ 1 (Eq. 6.33) · J^P = 1⁻

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

The colour weight diagram, with colour isospin across and colour hypercharge up — the same lattice §4.6 used for flavour, now for a symmetry that is exact rather than approximate. Six gluons carry a colour and a different anticolour and form three particle–antiparticle pairs: RḠ with GR̄, RB̄ with BR̄, GB̄ with BḠ. The two at the centre, g₇ = (RR̄ − GḠ)/√2 and g₈ = (RR̄ + GḠ − 2BB̄)/√6, are their own antiparticles — the analogues of the π⁰ and the η₈. The ninth combination, the singlet, would also sit at the centre and does not exist; the worked box below is why. Unlike the meson nonet there is no octet–singlet mixing here, because colour SU(3) is unbroken.

🔢 Worked example — there is no ninth gluon, and the book’s reason is not quite the reason

The book says of the singlet $g_0 = \tfrac{1}{\sqrt3}(R\bar R + G\bar G + B\bar B)$ : “the colour charges neutralize each other. As a result it does not interact with the quarks.”

The conclusion is right and the argument as stated does not carry it — the photon is electrically neutral too, and it interacts with electrons perfectly well. Being uncharged does not stop a mediator from mediating. Two better reasons:

1. Group theory. Gauge bosons live on the generators of the gauge group. The generators of SU(3) are the traceless Hermitian 3 × 3 matrices, and there are exactly eight of them. The singlet is proportional to the identity, whose trace is 3, so it is not a generator of SU(3) at all — it generates a separate U(1). The gauge group is SU(3). There is no ninth gluon because there is no ninth generator.

2. What it would do if it existed. A singlet gluon couples equally to R, G and B, i.e. to total colour. Sum any octet generator’s couplings over the three quarks of a colour-singlet baryon and you get zero — that is what makes the octet force short-ranged and confined inside hadrons. Do the same for the singlet and you get $\sqrt3$ , not zero.

So a ninth gluon would produce an unscreened, long-range strong force between ordinary hadrons — nuclei binding to each other across macroscopic distances. Nothing remotely like it exists, and that absence is an experimental fact rather than a definition.

the colour factors, and the missing ninth gluon

import numpy as np
s2, s3, s6 = np.sqrt(2), np.sqrt(3), np.sqrt(6)
R, G, B = 0, 1, 2

# The book's basis, Eq. (6.33), with the diagonal colour factors of Eq. (6.34).
# g1..g6 are off-diagonal -- one colour, a different anticolour -- and carry 1.
g7 = {R: 1/s2, G: -1/s2, B: 0.0}          # (RRbar - GGbar)/sqrt2
g8 = {R: 1/s6, G:  1/s6, B: -2/s6}        # (RRbar + GGbar - 2BBbar)/sqrt6

print("Eq. (6.35): a blue quark scattering off a blue quark.")
print("  g7 has no blue component, so only g8 can mediate it:")
print(f"    (1/2)(-2/sqrt6)(-2/sqrt6) = {0.5*g8[B]*g8[B]:.4f} = 1/3      book: 1/3")
print()
print("Eq. (6.36): a red quark scattering off a red quark.")
print("  now both g7 and g8 have a red component, so both contribute:")
print(f"    g7: (1/2)(1/sqrt2)^2 = {0.5*g7[R]**2:.4f}")
print(f"    g8: (1/2)(1/sqrt6)^2 = {0.5*g8[R]**2:.4f}")
print(f"    total = {0.5*(g7[R]**2 + g8[R]**2):.4f} = 1/3   -- identical, as symmetry demands")
print()
print("Why there is no ninth gluon.")
print("  1. Gluons sit on the GENERATORS, and SU(3)'s generators are traceless:")
print(f"       trace of g7 over R,G,B = {sum(g7.values()):+.4f}")
print(f"       trace of g8 over R,G,B = {sum(g8.values()):+.4f}")
g0 = {R: 1/s3, G: 1/s3, B: 1/s3}
print(f"       trace of the singlet   = {sum(g0.values()):+.4f}   <- not traceless")
print("     A non-traceless matrix generates a separate U(1), not SU(3).")
print()
print("  2. If it existed it would couple to TOTAL colour, which does not cancel")
print("     in a colour-singlet hadron -- so it would mediate a long-range strong")
print("     force between nuclei. No such force is observed.")

prints

Eq. (6.35): a blue quark scattering off a blue quark.
g7 has no blue component, so only g8 can mediate it:
  (1/2)(-2/sqrt6)(-2/sqrt6) = 0.3333 = 1/3      book: 1/3

Eq. (6.36): a red quark scattering off a red quark.
now both g7 and g8 have a red component, so both contribute:
  g7: (1/2)(1/sqrt2)^2 = 0.2500
  g8: (1/2)(1/sqrt6)^2 = 0.0833
  total = 0.3333 = 1/3   -- identical, as symmetry demands

Why there is no ninth gluon.
1. Gluons sit on the GENERATORS, and SU(3)'s generators are traceless:
     trace of g7 over R,G,B = +0.0000
     trace of g8 over R,G,B = +0.0000
     trace of the singlet   = +1.7321   <- not traceless
   A non-traceless matrix generates a separate U(1), not SU(3).

2. If it existed it would couple to TOTAL colour, which does not cancel
   in a colour-singlet hadron -- so it would mediate a long-range strong
   force between nuclei. No such force is observed.

The vertex, and how colour flows

The QCD vertex differs from the QED one in three ways, all visible in Fig. 6.16.

The incoming and outgoing quark can be different — same flavour, different colour. The gluon carries away the difference.
There is a colour factor $\lambda_\ell^{c_i c_j}$ alongside $\sqrt{\alpha_s}$ , depending on which gluon and which colours.
Colour lines are continuous through the diagram, which is the bookkeeping rule that makes these diagrams checkable by eye.

Fig. 6.16(c) — a blue quark scattering off a red quark

Click a vertex or an internal line.

The rule that makes these checkable: follow each colour line through the diagram and it must be continuous. Blue enters top-left, leaves along the gluon, and exits bottom-right. Compare Fig. 6.15, where the electron's charge is the same on both sides of the vertex because the photon carries none.

⚠️ Colour is not gauge invariant — and that is fine

A surprising and important asymmetry with QED, which the book states near the end of the section and which is easy to skim past.

The electric charge of an electron is gauge invariant. Change gauge and it is still −1.

The colour of a quark is not. A quark that is red in one gauge can be blue in another. There is no gauge-independent answer to “what colour is this quark?”

That sounds fatal until you notice what is invariant: a system that is a colour singlet in one gauge is a singlet in every gauge. And singlets are the only things that exist as free particles. So the gauge dependence of colour has no observable consequence — not because it is small, but because every observable is built from singlets.

The reason for the difference is the same commutator as ever. Take a small volume around an electron and integrate the charge density: only the electron contributes, because the photon is neutral. Do it around a quark and the gluons in the volume contribute too — and their contribution is gauge dependent.

🔑 If you remember only three things

Eight, because one combination carries nothing. The missing ninth would be a colourless carrier, and a colourless carrier would give the strong force infinite range.
A single quark’s colour is not observable. Unlike electric charge it is not gauge invariant, so only colourless combinations can ever be measured.
The same derivation, one bigger group. Nothing new is assumed here that Chapter 5 did not already assume — only the group is larger, and it does not commute.

Where this goes next

§6.4 uses these colour factors to compute why $q\bar q$ and $qqq$ bind and nothing else does — and gets the hyperfine splittings right in a way a U(1)-like force cannot.
§6.5 puts the gluon self-coupling into a loop and gets asymptotic freedom, the sign flip against §5.8.
§5.1 is the U(1) version of this argument, and the widget above runs in both modes for exactly that comparison.
§4.6 is where 3 ⊗ 3̄ = 8 ⊕ 1 was first used — for flavour, where the symmetry is badly broken, rather than for colour, where it is exact.

✅ Check yourself — SU(3), gluons, and the non-abelian difference

0/5 answered · 0 correct

1.The QCD field tensor carries an extra term −ig_s[G^μ, G^ν] that QED's does not. What does that one term cause?
2.In the widget, λ₃ and λ₈ commute exactly. Does that mean SU(3) is abelian in some corner?
3.The book says the singlet gluon 'does not interact with the quarks' because its colour charges neutralize. Why is that argument, as stated, insufficient?
4.Blue-on-blue scattering (Eq. 6.35) and red-on-red (Eq. 6.36) both give a colour factor of 1/3, but by different routes. What is the difference, and why must the answers agree?
5.A quark's colour is not gauge invariant — red in one gauge can be blue in another. Why is that not a catastrophe?