Colour factors

Every calculation at a hadron collider begins with a bookkeeping nightmare. When two quarks scatter by exchanging a gluon, each quark carries one of three colours, each gluon carries a colour–anticolour label from eight independent combinations, and the amplitude must be summed over every way those labels thread through the diagram. For the clean single-diagram process $ud\to ud$ there are $3^4=81$ ways to colour the four quarks and $8$ exchanged gluons — $648$ distinct colour/gluon channels stacked on top of the ordinary spin and momentum algebra. Written longhand it looks hopeless. The colour factor is the statement that it is not. All $648$ channels collapse into a single number, $C(ik\to jl)=\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}$ , that multiplies a matrix element otherwise identical to its QED cousin $e^-q\to e^-q$ . The entire chromodynamic content of the diagram — the eight gluons, the $\mathrm{SU}(3)$ structure, the requirement that colour be conserved at every vertex — is compressed into one factor whose value is fixed completely by the algebra of the Gell-Mann matrices. Evaluate it and the $81$ colour combinations sort into four tidy classes with values $\tfrac13$ , $-\tfrac16$ , $\tfrac12$ and $0$ ; average over the initial colours a real cross section cannot see and a single universal number, $\tfrac29$ , drops out. This article shows where that number comes from, why it is the same for quarks and antiquarks, and how the same $\mathrm{SU}(3)$ Casimirs govern jets, the heavy-quark potential, and every parton shower inside an LHC event generator.

Key takeaways

QCD scattering amplitudes come from their QED analogues by the replacement $-Q_q e^2\to g_S^2$ (equivalently $-Q_q\alpha\to\alpha_S$ ) and multiplication by a colour factor.
For quark–quark scattering the colour factor sums over the eight gluons through the Gell-Mann matrices: $C(ik\to jl)=\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}$ .
The completeness relation $\sum_a\lambda^a_{ij}\lambda^a_{kl}=2\delta_{il}\delta_{jk}-\tfrac23\delta_{ij}\delta_{kl}$ collapses it to the closed form $C=\tfrac12\delta_{jk}\delta_{il}-\tfrac16\delta_{ji}\delta_{lk}$ .
Individual configurations fall into four classes: $rr\to rr=\tfrac13$ , $rb\to rb=-\tfrac16$ , $rg\to gr=\tfrac12$ , and colour-changing $rb\to bg=0$ .
Averaging $|C|^2$ over the nine initial colour pairs gives the universal $\langle|C|^2\rangle=\tfrac29$ for single-gluon $qq$ , $q\bar q$ and $\bar q\bar q$ scattering; the same $\mathrm{SU}(3)$ Casimirs $C_F=\tfrac43$ , $C_A=3$ , $T_F=\tfrac12$ govern jets and potentials.
Key idea: all the bookkeeping over quark colours and the eight gluons reduces to one number — the colour factor — that multiplies an otherwise QED-like matrix element, with its value dictated entirely by the structure of the Gell-Mann matrices and $\mathrm{SU}(3)$ colour conservation.

The core insight

All the complicated bookkeeping over quark colours and the eight gluons reduces to one number — the colour factor — that multiplies an otherwise QED-like matrix element, with its value dictated entirely by the structure of the Gell-Mann matrices and the conservation of $\mathrm{SU}(3)$ colour at every vertex.

Where this lives today

Colour factors are hard-wired into every prediction at the Large Hadron Collider. The dominant $2\to2$ parton cross sections — $qq$ , $q\bar q$ , $qg$ , $gg$ scattering that make the QCD dijet spectrum — each carry their own colour-averaged factor, of which $\tfrac29$ for $qq$ is the simplest, and these feed the matrix elements that predict jet rates across ten orders of magnitude. In parton showers the DGLAP splitting functions are weighted by exactly the Casimirs of equation (6): $P_{q\to qg}\propto C_F=\tfrac43$ , $P_{g\to gg}\propto C_A=3$ , $P_{g\to q\bar q}\propto T_F=\tfrac12$ , so every event generator (PYTHIA, HERWIG, SHERPA) is threaded with them. Their ratios are measured — the broader spread of gluon jets confirms $C_A/C_F=\tfrac94$ , and LEP fits of $C_A=2.89\pm0.21$ , $C_F=1.30\pm0.09$ rule out alternative gauge groups. The colour-singlet factor $-\tfrac43$ is the coefficient of the static $q\bar q$ potential fit to quarkonium spectra on the lattice, while the singlet–octet difference underlies colour coherence and the "colour reconnection" ambiguities that limit precision top- and $W$ -mass measurements.

The physical picture

A quark and a quark scatter by tossing a gluon between them. Draw the diagram for $ud\to ud$ : the up quark runs along the top, the down quark along the bottom, a single gluon exchanged in the $t$ -channel. Where the QED version carried a colour-blind photon, the gluon reaches into each quark line and rotates its colour — the vertex factor is $\tfrac12 g_S\lambda^a$ , so the Gell-Mann matrix $\lambda^a$ turns incoming colour $i$ into outgoing $j$ , while the identical vertex at the other end turns $k$ into $l$ . Tie the two vertices together with the propagator's $\delta_{ab}$ , sum over the eight gluon types, and the whole colour content of the diagram is $\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}$ . The first figure is that diagram, with the colour indices made explicit.

Figure

Single-gluon t-channel exchange in ud → ud: the up quark colour i → j and the down quark colour k → l are rotated by the Gell-Mann matrix at each vertex, and summing over the eight gluons a leaves the single colour factor C = ¼ Σ λ(ji) λ(lk)

Now evaluate that object. The Gell-Mann matrices obey a completeness (Fierz) identity that does the sum over eight gluons once and for all, leaving only Kronecker deltas that ask a single question: which colours entered, and where did they leave? The answer sorts the $81$ colour combinations into four classes. All four colours equal gives $+\tfrac13$ ; two distinct colours passing straight through gives $-\tfrac16$ ; two distinct colours swapping gives $+\tfrac12$ ; and a diagram that would create net colour gives exactly $0$ , because colour is a conserved $\mathrm{SU}(3)$ charge. The second figure is the punchline a cross-section calculation actually uses — the four numbers and the universal average they collapse to.

Figure

The four classes of quark–quark colour factor and their values: colour-swap rg → gr gives +½, all-same rr → rr gives +⅓, distinct straight-through rb → rb gives −⅙, and colour-changing rb → bg gives 0; averaging C² over the nine initial colour pairs yields the universal 2/9 that enters every cross section

The core equations

The colour factor is defined by peeling the colour algebra off the QCD diagram. Each quark–gluon vertex contributes a Gell-Mann matrix element $\tfrac12\lambda^a$ , the gluon propagator ties the two gluon indices together with $\delta_{ab}$ , and the sum runs over all eight gluons. What remains is a pure number attached to the four external colours.

C(ik\to jl)\equiv\frac14\sum_{a=1}^{8}\lambda^a_{ji}\,\lambda^a_{lk} \qquad (1)

where: $i,k$ — incoming colours of the two quarks; $j,l$ — their outgoing colours; $\lambda^a$ — the eight $3\times3$ Gell-Mann matrices; the $\tfrac14$ is the product of the two vertex factors $\tfrac12$ . Reading: $\lambda^a_{ji}$ rotates colour $i\to j$ at one vertex, $\lambda^a_{lk}$ rotates $k\to l$ at the other, and the sum over $a$ is the sum over exchanged gluons. The reason a colour factor is all that separates QCD from QED is that the two theories share the same Dirac and propagator structure at this order. The QCD amplitude for $qq\to qq$ is obtained from the QED amplitude for $e^-q\to e^-q$ by swapping the coupling and attaching the colour factor.

\mathcal{M}_{\text{QCD}}=C(ik\to jl)\,\mathcal{M}_{\text{QED}}\Big|_{-Q_q e^2\,\to\, g_S^2}\qquad\Longleftrightarrow\qquad -Q_q\alpha\to\alpha_S \qquad (2)

where: $Q_q$ — quark electric charge in units of $e$ ; $e$ — electromagnetic coupling, $\alpha=e^2/4\pi$ ; $g_S$ — strong coupling, $\alpha_S=g_S^2/4\pi$ . Reading: the charge product $-Q_q e^2$ is replaced by $g_S^2$ and the amplitude multiplied by $C$ ; the kinematics (the $1/q^2$ propagator, the spinor traces) are untouched. To evaluate $C$ one never needs the eight matrices individually. The Gell-Mann matrices satisfy a completeness relation — the $\mathrm{SU}(3)$ Fierz identity — that performs the sum over gluons in one line and returns only Kronecker deltas.

\sum_{a=1}^{8}\lambda^a_{ij}\,\lambda^a_{kl}=2\,\delta_{il}\delta_{jk}-\frac23\,\delta_{ij}\delta_{kl} \qquad (3)

where: the indices run over the three colours $1,2,3$ ( $r,b,g$ ). Reading: the first term connects the colour entering one vertex to the colour leaving the other (colour flows through the gluon); the $-\tfrac23$ term, with coefficient $-\tfrac2N$ for $\mathrm{SU}(N)$ , subtracts the colour-singlet piece a physical gluon does not carry. This identity is the engine of every colour-factor evaluation. Feeding (3) into definition (1) gives a closed form with no sums left at all — just two deltas that read off the answer from the external colours.

C(ik\to jl)=\frac12\,\delta_{jk}\delta_{il}-\frac16\,\delta_{ji}\delta_{lk} \qquad (4)

where: $\delta_{jk}\delta_{il}$ is nonzero only when the two quarks exchange colours; $\delta_{ji}\delta_{lk}$ is nonzero only when each quark keeps its colour. Reading: swap ( $+\tfrac12$ ), keep-same ( $+\tfrac12-\tfrac16=\tfrac13$ when the kept colours coincide), keep-distinct ( $-\tfrac16$ ), and everything else ( $0$ ). The four classes of the figure fall straight out. A physical cross section cannot know the colours of the incoming quarks, so it averages over the nine equally likely initial colour pairs and sums over the final ones. The colour-summed, colour-averaged squared factor is the number that actually multiplies the cross section.

\langle|C|^2\rangle=\frac{1}{9}\sum_{i,k}\sum_{j,l}\big|C(ik\to jl)\big|^2=\frac{N^2-1}{4N^2}=\frac{C_F}{2N}=\frac29 \qquad (5)

where: $N=3$ colours; the sum uses the trace orthogonality $\mathrm{Tr}(\lambda^a\lambda^b)=2\delta^{ab}$ . Reading: the $9$ in the denominator is the average over the $3\times3$ initial colours; the result $\tfrac29$ is universal for single-gluon $qq$ , $q\bar q$ and $\bar q\bar q$ scattering, and it is what appears in $d\sigma/dq^2$ for $ud\to ud$ . The same $\mathrm{SU}(3)$ algebra produces three Casimir invariants that recur throughout QCD and set the colour factors of every splitting and every potential.

C_F=\frac{N^2-1}{2N}=\frac43,\qquad C_A=N=3,\qquad T_F=\frac12 \qquad (6)

where: $C_F=\sum_a(T^aT^a)$ with $T^a=\tfrac12\lambda^a$ is the quark Casimir ( $q\to qg$ ); $C_A$ is the adjoint Casimir (triple-gluon vertex $g\to gg$ ); $T_F$ normalises $\mathrm{Tr}(T^aT^b)=T_F\delta^{ab}$ ( $g\to q\bar q$ ). Reading: the ratio $C_A/C_F=\tfrac94$ is why gluon jets are broader than quark jets, and $C_F=\tfrac43$ is the coefficient of the colour-singlet static potential.

Derivation AExtracting the colour factor from the QCD Feynman rules

The goal is to show precisely why a whole field theory of coloured quarks and eight gluons differs from ordinary electrodynamics, at lowest order in $qq$ scattering, by nothing more than one multiplicative number. We compare the $t$ -channel amplitude for $ud\to ud$ term by term with the QED amplitude for $e^-q\to e^-q$ , isolating every factor that carries a colour index so the kinematic and spinor structure can be quotiented out cleanly.

Step 1. Write the two quark–gluon vertices. Each contributes $ig_S\tfrac12\lambda^a\gamma^\mu$ , so the up-quark vertex carries $\tfrac12\lambda^a_{ji}$ (colour $i\to j$ ) and the down-quark vertex carries $\tfrac12\lambda^b_{lk}$ (colour $k\to l$ ), where $a,b$ label the gluon at each end. Why: the colour matrix and the Dirac matrix factorise at a vertex, so the colour part $\tfrac12\lambda^a_{ji}$ can be handled entirely separately from the $\gamma^\mu$ that stays behind in the kinematics.

Step 2. Attach the gluon propagator, which in Feynman gauge is $-ig_{\mu\nu}\delta_{ab}/q^2$ . The colour Kronecker delta $\delta_{ab}$ forces the two gluon labels to be equal, collapsing the double sum over $a$ and $b$ into a single sum over one gluon index $a$ . Why: a gluon does not change type as it propagates, so the same $\mathrm{SU}(3)$ label appears at both ends, and this is exactly what ties the two vertices into a single colour object.

Step 3. Collect the colour-bearing factors: $\big(\tfrac12\lambda^a_{ji}\big)\big(\tfrac12\lambda^a_{lk}\big)$ summed over $a$ , which is $\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}\equiv C(ik\to jl)$ . Everything else — the two $\gamma^\mu$ , the propagator's $g_{\mu\nu}/q^2$ , the external spinors — is byte-for-byte the QED $e^-q$ amplitude with the photon replaced by the gluon. Why: this is the statement that the colour algebra decouples; the colour factor is the complete residue of "being QCD" at this order.

Step 4. Match the couplings. QED carries the charge product $Q_e Q_q e^2=-Q_q e^2$ (with $Q_e=-1$ ); QCD carries $g_S^2$ times the colour factor. Hence $\mathcal{M}_{\text{QCD}}=C\,\mathcal{M}_{\text{QED}}$ under $-Q_q e^2\to g_S^2$ . Why: this makes the whole toolbox of QED cross-section results reusable — one computes the strong process by scaling a known electromagnetic one.

Go deeper · Derivation B · 1 minThe completeness relation and the four colour classes

Derivation BThe completeness relation and the four colour classes

The aim is to turn the abstract sum $\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}$ into concrete numbers without ever multiplying out eight explicit matrices, by deploying the $\mathrm{SU}(3)$ completeness identity that does the gluon sum in closed form and then reading the resulting Kronecker deltas against each physically distinct colour configuration — exposing the four classes and why colour-non-conserving diagrams vanish identically.

Step 1. Quote the completeness relation, $\sum_a\lambda^a_{ij}\lambda^a_{kl}=2\delta_{il}\delta_{jk}-\tfrac23\delta_{ij}\delta_{kl}$ . It follows from the fact that the eight $\lambda^a$ together with the identity form a complete basis for $3\times3$ Hermitian matrices; the $-\tfrac23=-\tfrac2N$ coefficient removes the colour-singlet direction that the traceless generators do not span. Why: this identity is what makes the gluon octet "know" it is not a ninth colour-blind gluon — the subtraction is the physical absence of a singlet gluon.

Step 2. Substitute into (1) with the index pattern of $\lambda^a_{ji}\lambda^a_{lk}$ : this gives $C=\tfrac14\big(2\delta_{jk}\delta_{il}-\tfrac23\delta_{ji}\delta_{lk}\big)=\tfrac12\delta_{jk}\delta_{il}-\tfrac16\delta_{ji}\delta_{lk}$ . Why: the first delta pair fires when the two quarks swap colours, the second when each retains its own — two mutually exclusive questions about the colour flow.

Step 3. Evaluate the classes. All four colours equal, $rr\to rr$ : both terms fire, $C=\tfrac12-\tfrac16=\tfrac13$ . Two distinct, straight through, $rb\to rb$ : only the keep term, $C=-\tfrac16$ . Two distinct, swapped, $rg\to gr$ : only the swap term, $C=+\tfrac12$ . Why: these three exhaust every colour-conserving pattern, and their values are exactly the heights plotted in the second figure.

Step 4. Test a colour-changing case, $rb\to bg$ : the swap delta needs $j=k$ and $i=l$ , i.e. $b=b$ and $r=g$ — false; the keep delta needs $j=i$ and $l=k$ , i.e. $b=r$ — false. Both vanish, so $C=0$ . Why: a nonzero amplitude would create net colour out of nothing; the completeness relation enforces $\mathrm{SU}(3)$ charge conservation automatically, with no extra assumption.

Go deeper · Derivation C · 1 minAveraging over colour: the universal 2/9

Derivation CAveraging over colour: the universal 2/9

This derivation obtains the single number a measured $qq$ cross section carries — the squared colour factor summed over final colours and averaged over initial ones — with a trace trick that sidesteps enumerating the eighty-one configurations one by one. The payoff is not only $\tfrac29$ but the demonstration that it equals $C_F/2N$ , tying this scattering factor to the universal Casimir that governs the rest of QCD.

Step 1. Form $\sum_{ijkl}|C|^2=\tfrac{1}{16}\sum_{ijkl}\sum_{a,b}\lambda^a_{ji}\lambda^a_{lk}\,\lambda^{b*}_{ji}\lambda^{b*}_{lk}$ . Because the Gell-Mann matrices are Hermitian, $\lambda^{b*}_{ji}=\lambda^b_{ij}$ , so the up-line indices contract as $\sum_{ij}\lambda^a_{ji}\lambda^b_{ij}=\mathrm{Tr}(\lambda^b\lambda^a)$ . Why: Hermiticity converts the complex conjugate into an index transpose, which is what lets the free colour indices close into a trace.

Step 2. Use the orthogonality $\mathrm{Tr}(\lambda^a\lambda^b)=2\delta^{ab}$ on both quark lines. The up line gives $2\delta^{ab}$ and the down line gives $2\delta^{ab}$ , so $\sum_{ijkl}|C|^2=\tfrac{1}{16}\sum_{ab}(2\delta^{ab})(2\delta^{ab})=\tfrac{1}{16}\cdot4\sum_a 1=\tfrac{4}{16}\,(N^2-1)$ . Why: the two traces force $a=b$ , and the surviving sum simply counts the eight gluons, $N^2-1=8$ .

Step 3. Evaluate: $\sum_{ijkl}|C|^2=\tfrac{8}{4}=2$ . Divide by the nine initial colour pairs to average: $\langle|C|^2\rangle=\tfrac19\cdot2=\tfrac29$ . Why: only the initial colours are averaged (they are unknown and equally likely); the final colours are summed because a detector sums over them, so the normalisation is $1/9$ , not $1/81$ .

Step 4. Recognise the invariant form $\langle|C|^2\rangle=\tfrac{N^2-1}{4N^2}=\tfrac{C_F}{2N}$ , with $C_F=\tfrac43$ ; for $N=3$ this is $\tfrac{8}{36}=\tfrac29$ . Why: writing the answer through $C_F$ shows this scattering number is the same $\mathrm{SU}(3)$ Casimir that sets the $q\to qg$ splitting and the static potential — one algebra, many appearances.

Limiting cases

Case	Result	Meaning
All colours equal ( $rr\to rr$ )	$C=+\tfrac13$	both delta terms fire; the largest same-colour amplitude
Distinct, straight through ( $rb\to rb$ )	$C=-\tfrac16$	only the keep term; note the negative sign — an interference, not a probability
Distinct, colours swap ( $rg\to gr$ )	$C=+\tfrac12$	only the swap term; colour is handed across the gluon
Any net colour change ( $rb\to bg$ )	$C=0$	forbidden by $\mathrm{SU}(3)$ colour conservation at the vertices
Colour-averaged cross section	$\langle\|C\|^2\rangle=\tfrac29$	universal for $qq$ , $q\bar q$ , $\bar q\bar q$ single-gluon exchange
Formal $\mathrm{SU}(N)$ , large $N$	$\langle\|C\|^2\rangle\to\tfrac14$	singlet-subtraction $-\tfrac1N$ fades; 't Hooft planar limit
Colour-singlet vs octet $q\bar q$	$-\tfrac43$ vs $+\tfrac16$	attractive (binds a meson) vs repulsive (no bound state)

Common trapConfusing the individual factor with the averaged one

The single most common slip is to quote the wrong colour number. The individual colour factor for a definite colour configuration can be $+\tfrac13$ , $-\tfrac16$ , $+\tfrac12$ or $0$ — these are the entries of the amplitude for a specific choice of the eight colour indices, and some of them are negative because they are amplitudes, not probabilities. The number that multiplies a physical cross section is the colour-averaged square $\langle|C|^2\rangle=\tfrac29$ , obtained after averaging $|C|^2$ over the nine initial colour pairs and summing over final colours. You never insert $\tfrac13$ or $-\tfrac16$ into a rate; you insert $\tfrac29$ . A related trap: $\tfrac29$ belongs to the single $t$ -channel diagram — identical-quark processes such as $uu\to uu$ have an extra $u$ -channel diagram and its interference, with a different colour bookkeeping, and $q\bar q$ annihilation adds an $s$ -channel piece.

Worked examples

1 · Reading the four colour factors off the closed form (fully worked)

Using $C=\tfrac12\delta_{jk}\delta_{il}-\tfrac16\delta_{ji}\delta_{lk}$ with the colour labels $r=1,b=2,g=3$ , evaluate the colour factor for $rr\to rr$ , $rb\to rb$ , $rg\to gr$ and $rb\to bg$ . The index order is $C(ik\to jl)$ : incoming $i,k$ , outgoing $j,l$ .

Step 1. $rr\to rr$ : here $i=j=k=l=r=1$ . The swap term needs $j=k$ and $i=l$ : $r=r$ and $r=r$ , both true, contributing $+\tfrac12$ . The keep term needs $j=i$ and $l=k$ : $r=r$ and $r=r$ , both true, contributing $-\tfrac16$ . Sum: $C=\tfrac12-\tfrac16=\mathbf{+\tfrac13}$ .

Step 2. $rb\to rb$ : quark A is $i=r,j=r$ ; quark B is $k=b,l=b$ . Swap term needs $j=k$ ( $r=b$ ? no), so it vanishes. Keep term needs $j=i$ ( $r=r$ , yes) and $l=k$ ( $b=b$ , yes), giving $-\tfrac16$ . Sum: $C=\mathbf{-\tfrac16}$ .

Step 3. $rg\to gr$ : quark A is $i=r,j=g$ ; quark B is $k=g,l=r$ . Swap term needs $j=k$ ( $g=g$ , yes) and $i=l$ ( $r=r$ , yes), giving $+\tfrac12$ . Keep term needs $j=i$ ( $g=r$ ? no), vanishes. Sum: $C=\mathbf{+\tfrac12}$ .

Step 4. $rb\to bg$ : quark A is $i=r,j=b$ ; quark B is $k=b,l=g$ . Swap term needs $i=l$ ( $r=g$ ? no). Keep term needs $j=i$ ( $b=r$ ? no). Both vanish: $C=\mathbf{0}$ — the diagram would conjure a green quark from a red one with no source, which $\mathrm{SU}(3)$ forbids. The four classes give $\tfrac13,-\tfrac16,\tfrac12,0$ exactly as plotted.

2 · The colour-averaged factor and the ud → ud cross section

Confirm $\langle|C|^2\rangle=\tfrac29$ by direct enumeration of the classes, and write the single-gluon $ud\to ud$ differential cross section.

Step 1. Count the initial colour pairs: the up quark has $3$ colours and the down quark has $3$ , so there are $9$ equally likely initial states — the denominator of the average. For each we sum $|C|^2$ over final colours.

Step 2. Enumerate contributions. Same-colour pairs ( $rr,bb,gg$ ): $3$ of them, each only scatters into itself with $C=\tfrac13$ , giving $3\times(\tfrac13)^2=3\times\tfrac19=\tfrac13$ . Distinct-colour pairs ( $rb,rg,br,bg,gr,gb$ ): $6$ of them; each can go straight through ( $C=-\tfrac16$ ) or swap ( $C=+\tfrac12$ ), giving per pair $(\tfrac16)^2+(\tfrac12)^2=\tfrac1{36}+\tfrac9{36}=\tfrac{10}{36}=\tfrac{5}{18}$ , so $6\times\tfrac{5}{18}=\tfrac{30}{18}=\tfrac53$ .

Step 3. Total $\sum_{ijkl}|C|^2=\tfrac13+\tfrac53=\tfrac63=2$ , matching the trace result. Average over the $9$ initial states: $\langle|C|^2\rangle=\tfrac29$ . The universal colour factor is $\tfrac29$ .

Step 4. Insert it. The QED $e^-q\to e^-q$ result $d\sigma/dq^2\propto Q_q^2e^4/q^4$ becomes, under $-Q_q e^2\to g_S^2$ and $\times\langle|C|^2\rangle$ ,

\frac{d\sigma}{dq^2}(ud\to ud)=\frac{2}{9}\cdot\frac{2\pi\alpha_S^2}{q^4}\,\Big[\,\text{spin/kinematic factor}\,\Big].

The factor $\tfrac29$ is the entire colour content of the $ud\to ud$ rate — all $648$ colour/gluon channels are inside it.

3 · Singlet versus octet: the sign of the heavy-quark potential

A quark and antiquark can combine into a colour singlet ( $\mathbf{3}\otimes\bar{\mathbf{3}}=\mathbf{1}\oplus\mathbf{8}$ ) or a colour octet. Show the singlet is attractive and the octet repulsive, and give the potential coefficient.

Step 1. For $q\bar q$ the one-gluon-exchange colour factor is $\tfrac12\big[C_2(R)-2C_F\big]$ , where $C_2(R)$ is the Casimir of the combined colour state $R$ and $C_F=\tfrac43$ is the quark Casimir. This form follows from $T^a_q\!\cdot\!T^a_{\bar q}=\tfrac12[(T_q+T_{\bar q})^2-T_q^2-T_{\bar q}^2]$ .

Step 2. Singlet: $C_2(\mathbf{1})=0$ , so the factor is $\tfrac12(0-2\cdot\tfrac43)=-\tfrac43$ . Octet: $C_2(\mathbf{8})=C_A=3$ , so the factor is $\tfrac12(3-\tfrac83)=\tfrac12\cdot\tfrac13=+\tfrac16$ .

Step 3. Read the sign. A negative colour factor multiplying the (repulsive-by-default) Coulomb form flips it to attractive, so the singlet potential is $V(r)=-\tfrac43\,\dfrac{\alpha_S}{r}$ at short distance — binding. The octet coefficient $+\tfrac16$ is positive, hence repulsive: no bound state. The colour-singlet coefficient $-\tfrac43=-C_F$ is why mesons are colour singlets and why charmonium and bottomonium spectra are fit with $V=-\tfrac43\alpha_S/r+\sigma r$ .

Physics of the portrait

Colour factors are, in the end, statements about the algebra of eight $3\times3$ matrices — and those matrices, and the $\mathrm{SU}(3)$ that organises them, carry one name above all others. The story runs from a classification scheme for hadrons to the gauge theory of the strong force in barely a decade.

Origins · 1 minMurray Gell-Mann and how this was discovered

Murray Gell-Mann

1929–2019 · SU(3), quarks and the gauge theory of colour

Murray Gell-Mann was the architect of the $\mathrm{SU}(3)$ structure that every colour factor rests on. In 1961 he (independently of Yuval Ne'eman) proposed the "Eightfold Way", organising the hadron zoo into $\mathrm{SU}(3)$ flavour multiplets — a scheme whose prediction of the $\Omega^-$ was confirmed in 1964 and whose generators are the eight matrices that now bear his name. In 1964 he introduced quarks as the fundamental flavour triplet, a proposal so radical he hedged on whether they were physical. The missing ingredient was colour: to reconcile the symmetric $\Omega^-$ wavefunction with Fermi statistics a new threefold degree of freedom was needed, supplied through parastatistics by Greenberg and as an explicit $\mathrm{SU}(3)$ by Han and Nambu. Gell-Mann, with Harald Fritzsch and Heinrich Leutwyler, gauged that colour $\mathrm{SU}(3)$ in 1972–73 and named the theory quantum chromodynamics, coining "colour" for the charge and taking the eight gluons as the gauge bosons of an exact, unbroken symmetry. The colour factor is the direct fingerprint of that choice: $\tfrac13$ , $-\tfrac16$ , $\tfrac12$ , $0$ or the averaged $\tfrac29$ is nothing but the completeness algebra of the Gell-Mann matrices read off a Feynman diagram. He received the 1969 Nobel Prize.

History

1961

— Gell-Mann and Ne'eman independently propose the Eightfold Way, SU(3) flavour multiplets of hadrons

1964

— Gell-Mann and Zweig introduce quarks as the fundamental SU(3) triplet

1964

— Greenberg proposes parastatistics, the seed of a threefold colour degree of freedom

1965

— Han and Nambu introduce an explicit SU(3) colour with integer-charge quarks and gauge bosons

1973

— Fritzsch, Gell-Mann and Leutwyler formulate QCD as the gauge theory of colour SU(3)

1973

— Gross, Wilczek and Politzer discover asymptotic freedom, making QCD predictive at high energy

Going further

Texts. Thomson, Modern Particle Physics, §10.7 (colour factors) and §10.7.1 (processes with antiquarks), Eqs. 10.23–10.35 · Halzen & Martin, Quarks and Leptons, Ch. 2 and §14 (colour algebra and QCD) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §15–17 (non-abelian gauge theory, colour identities) · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 8–9 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 25–27 · Ellis, Stirling & Webber, QCD and Collider Physics, §1–3 (Casimirs and splitting functions). The key results are: the colour factor definition $C=\tfrac14\sum_a\lambda^a_{ji}\lambda^a_{lk}$ reduces to the closed form $C=\tfrac12\delta_{jk}\delta_{il}-\tfrac16\delta_{ji}\delta_{lk}$ via the Fierz identity; the four representative configurations give $\tfrac13,-\tfrac16,\tfrac12,0$ ; the colour sum $\sum_{ijkl}|C|^2=2$ and the average $\langle|C|^2\rangle=\tfrac29=C_F/2N$ ; and the singlet/octet $q\bar q$ factors $-\tfrac43$ and $+\tfrac16$ follow from $\tfrac12[C_2(R)-2C_F]$ . The completeness relation $\sum_a\lambda^a_{ij}\lambda^a_{kl}=2\delta_{il}\delta_{jk}-\tfrac23\delta_{ij}\delta_{kl}$ and the Casimirs $C_F=\tfrac43,C_A=3,T_F=\tfrac12$ are standard $\mathrm{SU}(3)$ results.

Prerequisites. Colour and QCD · QCD vertices · $\mathrm{SU}(3)$ and the Gell-Mann matrices · Electron–quark scattering. Leads to. Hadron collisions · The heavy-quark potential · QCD in $e^+e^-$ annihilation · Jets and fragmentation.

Ask the tutor: Ask me why $rb\to bg$ vanishes while $rg\to gr$ gives $\tfrac12$ , or to build the full $\langle|C|^2\rangle=\tfrac29$ average colour pair by colour pair.

Why is the colour factor a single number when there are 648 channels?

Because the sum over those channels is a linear-algebra operation that closes on itself. Each channel is a product of Gell-Mann matrix elements, and the completeness relation $\sum_a\lambda^a_{ij}\lambda^a_{kl}=2\delta_{il}\delta_{jk}-\tfrac23\delta_{ij}\delta_{kl}$ performs the entire sum over the eight gluons in one step, leaving only Kronecker deltas that depend on the external colours alone. So the whole eight-gluon apparatus condenses into one factor in front of a QED-like amplitude — $\mathrm{SU}(3)$ has already done the arithmetic, and the colour factor is the residue.

Why is the physical factor 2/9 and not 1/3 or 1/2?

Because $\tfrac13$ , $-\tfrac16$ and $\tfrac12$ are amplitudes for specific colour configurations, and no experiment prepares quarks in a definite colour. A cross section averages over the nine equally likely initial colour pairs and sums over final ones, so squaring and averaging gives $\langle|C|^2\rangle=\tfrac19\sum_{ijkl}|C|^2=\tfrac19\cdot2=\tfrac29$ . The individual factors, signs included, still matter — they interfere inside that sum — but the number you insert into a rate is always the averaged square $\tfrac29$ . Quoting a bare $\tfrac13$ in a cross section is the classic error.

Is the colour factor the same for antiquarks?

Yes, and it is a theorem, not a coincidence. For antiquarks the vertex carries $-\lambda^{a\,T}$ rather than $\lambda^a$ . Because the Gell-Mann matrices split into symmetric and antisymmetric types, $\lambda^{a\,T}=\pm\lambda^a$ , so the sign flips cancel in pairs when you form $\sum_a\lambda^a\lambda^a$ . The upshot: quark–antiquark and antiquark–antiquark scattering by single-gluon exchange carry the same individual factors and the same average $\tfrac29$ as quark–quark scattering. The $s$ -channel annihilation diagram in $q\bar q\to q\bar q$ has its own colour structure, but the $t$ -channel piece matches.

How do C_F, C_A and T_F relate to the scattering colour factor?

They are all outputs of the same $\mathrm{SU}(3)$ algebra applied to different colour flows. $C_F=\tfrac{N^2-1}{2N}=\tfrac43$ is $\sum_a(T^aT^a)$ on a quark line — it appears whenever a gluon is emitted from a quark, $q\to qg$ . $C_A=N=3$ comes from the gluon self-interaction $g\to gg$ , and $T_F=\tfrac12$ normalises the generator trace and governs $g\to q\bar q$ . The scattering average $\langle|C|^2\rangle=C_F/2N=\tfrac29$ is just $C_F$ in disguise: one set of Casimirs runs through splitting functions, potentials and cross sections alike.

Why is the colour-singlet potential attractive but the octet repulsive?

The one-gluon-exchange colour factor for $q\bar q$ is $\tfrac12[C_2(R)-2C_F]$ , where $R$ is the combined colour state. The singlet has $C_2=0$ , giving $-\tfrac43$ ; the octet has $C_2=C_A=3$ , giving $+\tfrac16$ . A negative colour factor flips the sign of the Coulomb-like exchange, so the singlet feels an attractive $V=-\tfrac43\,\alpha_S/r$ and can bind into a meson, while the octet's $+\tfrac16$ is repulsive and does not bind. This sign is the group-theoretic reason hadrons are colour singlets — only that channel is attractive enough, once confinement is added, to form observed states.

Does the −⅙ colour factor being negative mean a negative probability?

No. The individual colour factors are pieces of a quantum amplitude, and amplitudes carry any sign or phase; only the squared, summed, averaged quantity is a probability, and $\langle|C|^2\rangle=\tfrac29$ is safely positive. The $-\tfrac16$ appears for distinct colours passing straight through, and its sign encodes destructive interference relative to other colour paths — exactly the $-\tfrac1N$ singlet-subtraction term in the completeness relation. That minus sign matters inside the sum, but it never produces a negative rate.