Colour, the Lagrangian, and the running coupling
The β-function, asymptotic freedom, Λ_QCD — with the one-loop running solved.
Chapter 1 built the parton model with no underlying theory: point constituents, probabilities, sum rules. This chapter supplies the theory those partons live in. Three things have to be established: that quarks carry a hidden threefold charge (colour), that the field theory of that charge — QCD — has a coupling which weakens at short distances (asymptotic freedom, the licence for every perturbative calculation in this course), and that the same coupling grows toward a scale $\Lambda_{\rm QCD}$ where perturbation theory dies and confinement takes over. Our fits sit deliberately close to that edge: the input scale $Q_0 = 1.65$ GeV of our pipeline is a place where $\alpha_s \approx 0.33$ — small enough to compute, large enough that every order matters.
Why colour must exist — three measurements
1. The $\Delta^{++}$ statistics crisis. The $\Delta^{++}$ baryon is $uuu$ with spin $3/2$. In its ground state the spatial wavefunction is symmetric ($L=0$), the spin state $|{\uparrow\uparrow\uparrow}\rangle$ is symmetric, and the flavour content $uuu$ is trivially symmetric. Three identical fermions in a totally symmetric state — Pauli forbids it outright. The rescue: give each quark a hidden index $a = 1,2,3$ and antisymmetrize in it,
The $\epsilon_{abc}$ needs exactly three values of the index — colour, with $N_c = 3$, and the requirement that observed hadrons be colour singlets (the $\epsilon$ is the singlet combination of three triplets).
2. The $R$-ratio counts colours directly. In $e^+e^- \to$ hadrons the virtual photon produces a $q\bar q$ pair; each colour is a distinct final state, so the cross-section is $N_c$ times the muon-pair one per flavour:
With $N_c = 3$: $u,d,s$ active gives $R = 3(\tfrac49 + \tfrac19 + \tfrac19) = 2$; opening charm adds $3\cdot\tfrac49$ to give $\tfrac{10}{3} \approx 3.33$; opening bottom gives $\tfrac{11}{3} \approx 3.67$. Without colour the same numbers would be $\tfrac23$, $\tfrac{10}{9}$, $\tfrac{11}{9}$ — excluded by data at a glance (worked example 2 below adds the QCD correction and compares to measurements).
3. $\pi^0 \to \gamma\gamma$. The decay proceeds through a quark triangle loop (the axial anomaly), and the amplitude is proportional to $N_c\,(e_u^2 - e_d^2) = N_c/3$. The anomaly prediction for the rate, $\Gamma \propto (N_c/3)^2$, gives $7.75$ eV for $N_c = 3$ against the measured $7.8 \pm 0.1$ eV; $N_c = 1$ would miss by a factor of nine. Three independent experiments, one integer: $N_c = 3$.
The QCD Lagrangian — and why gluons interact with themselves
Promote the colour index to a local SU(3) gauge symmetry and the Lagrangian is forced, term by term:
with the covariant derivative and field strength
Here $f$ runs over the six quark flavours, $i,j = 1\ldots 3$ are colour indices, $a = 1\ldots 8$ labels the gluons, and the $t^a$ are the SU(3) generators with $[t^a, t^b] = i f^{abc} t^c$. The structure constants $f^{abc}$ are the whole difference from QED. In QED, $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is linear in the photon field, so $F^2$ is quadratic: photons are free of each other. In QCD the extra $g_s f^{abc} A^b A^c$ term makes $F^2$ contain cubic ($g_s\, f^{abc}\, \partial A\, A A$) and quartic ($g_s^2\, f^{abc} f^{ade} A A A A$) pieces — three- and four-gluon vertices are not optional; gauge invariance under a non-abelian group demands them. Gluons carry colour charge and radiate gluons. Every distinctive feature of QCD — asymptotic freedom below, the $P_{gg}$ splitting kernel of §2.3, the gluon-driven small-$x$ growth of §2.4 — flows from that one term.
Three colour constants will label everything downstream, so fix them now: $\mathrm{tr}(t^a t^b) = T_R\,\delta^{ab}$ with $T_R = \tfrac12$; $(t^a t^a)_{ij} = C_F\,\delta_{ij}$ with $C_F = \tfrac{N_c^2-1}{2N_c} = \tfrac43$ (quark charge-squared); $f^{acd}f^{bcd} = C_A\,\delta^{ab}$ with $C_A = N_c = 3$ (gluon charge-squared).
Renormalization and the running coupling
Loop corrections in QCD are ultraviolet divergent; renormalization absorbs the divergences into the definition of the coupling, at the unavoidable price of introducing a reference scale $\mu$. The coupling $\alpha_s(\mu^2) = g_s^2(\mu^2)/4\pi$ becomes scale-dependent, and physical quantities are $\mu$-independent only because explicit logarithms $\ln(Q^2/\mu^2)$ in the perturbative coefficients compensate the running — the same compensation mechanism we will meet again for $\mu_F$ in §2.2. The running itself is governed by the renormalization-group equation
and everything hangs on the sign of $\beta_0$: positive means $\alpha_s$ decreases as $\mu$ grows.
Write $\beta_0 = \frac{1}{4\pi}\left(\frac{11}{3}C_A - \frac{4}{3}T_R\, n_f\right)$ and take the two terms separately.
The quark term is QED reborn. A quark loop in the gluon propagator is exactly the electron loop in the photon propagator with $e_q \to g_s t^a$: virtual $q\bar q$ pairs polarize, antiparticles migrate toward a colour charge, and the charge seen from far away is reduced — screening. Screening makes the coupling grow at short distance (as in QED), i.e. it contributes negatively to $\beta_0$: $-\tfrac{4}{3}T_R$ per flavour $= -\tfrac{2}{3} n_f$ in total.
The gluon term is new physics. The cleanest accounting (Nielsen–Hughes) computes the vacuum's response to a background chromomagnetic field. Each charged fluctuation contributes to the magnetic susceptibility in two ways: its orbital motion (Landau levels) is diamagnetic and screens, contributing $-\tfrac13$ per degree of freedom; its spin aligns with the field, is paramagnetic, and antiscreens, contributing $(2s)^2$. For the gluon, $s = 1$ in the adjoint representation:
$$ \underbrace{(2s)^2\, C_A}_{\text{spin, paramagnetic}} \;-\; \underbrace{\tfrac13\, C_A}_{\text{orbital, diamagnetic}} \;=\; 12 - 1 \;=\; 11 \quad (\text{for } C_A = 3) $$
The famous 11 is really $12 - 1$: a small QED-like screening piece, overwhelmed by the spin coupling of charged, spin-1 gluons. (For quarks the same accounting with $s = \tfrac12$, colour factor $T_R$, and the extra minus sign of a fermion loop reproduces $-\tfrac23 n_f$.) A paramagnetic medium has permeability $\mu > 1$; Lorentz invariance of the vacuum forces $\epsilon\,\mu = 1$, so $\epsilon < 1$ — the vacuum antiscreens colour charge. With $n_f \le 16$ the gluons win, $\beta_0 > 0$, and QCD is asymptotically free. In our world $n_f \le 6$: never close.
The one-loop equation is separable:
$$ \frac{d\alpha_s}{\alpha_s^2} = -\beta_0\, d\ln\mu^2 \;\;\Longrightarrow\;\; \frac{1}{\alpha_s(\mu^2)} - \frac{1}{\alpha_s(\mu_0^2)} = \beta_0 \ln\frac{\mu^2}{\mu_0^2} $$
$1/\alpha_s$ runs linearly in $\ln\mu^2$ — the most useful form for arithmetic (worked example 1). The integration constant is one number; trade it for a scale by defining $\Lambda^2$ through $1/\alpha_s(\mu_0^2) \equiv \beta_0 \ln(\mu_0^2/\Lambda^2)$. Then
$$ \alpha_s(\mu^2) \;=\; \frac{1}{\beta_0 \ln\!\big(\mu^2/\Lambda^2\big)} $$
Two readings. Up: $\alpha_s \to 0$ logarithmically as $\mu \to \infty$ — asymptotic freedom, and the reason DIS at large $Q^2$ is computable at all. Down: $\alpha_s$ blows up as $\mu^2 \to \Lambda^2$ (the Landau pole — really a signal that perturbation theory has quit, not a physical infinity). Note what happened conceptually: classical QCD with massless quarks has no parameter with dimensions, yet quantization produced a mass scale $\Lambda$. This is dimensional transmutation — the dimensionless $g_s$ traded for a scale.
$\beta_0$ depends on $n_f$, the number of quark flavours light enough to circulate in loops at the scale considered. Crossing a heavy-quark mass $m_h$, one switches theories: $n_f \to n_f + 1$. At one loop the matching condition is simply continuity, $\alpha_s^{(n_f+1)}(m_h^2) = \alpha_s^{(n_f)}(m_h^2)$ (beyond one loop there are small discontinuities). Continuity of $\alpha_s$ with a changed slope $\beta_0$ forces $\Lambda$ to jump: each $n_f$ regime has its own $\Lambda^{(n_f)}$, fixed by matching. With $\alpha_s(M_Z^2) = 0.118$ at one loop (worked example 1): $\Lambda^{(5)} \approx 88$ MeV, $\Lambda^{(4)} \approx 121$ MeV, $\Lambda^{(3)} \approx 146$ MeV. So "$\Lambda_{\rm QCD}$" is not one number — quoting it requires stating the order and $n_f$ — but all versions say the same thing: QCD's intrinsic scale is a couple of hundred MeV, i.e. $1/\Lambda \sim 1$ fm, the size of a hadron. That is no coincidence: it is the explanation.
Massless QCD has no dimensionful input, yet it generates $\Lambda \sim 200$ MeV by pure quantum mechanics. Every light-hadron property is then a fixed pure number times a power of $\Lambda$: the proton mass is $\approx 4.7\,\Lambda^{(3)}$-ish, the string tension $\approx (2\Lambda)^2$, hadron radii $\sim 1/\Lambda$. Since $u,d$ quark masses contribute only a percent-level correction, about 99% of the proton's mass — and of yours — is transmuted gluon field energy, traceable to the sign of $\beta_0$. This is also why our PDFs must be fitted, not computed: they are properties of the $\alpha_s \sim 1$ regime, on the wrong side of $\Lambda$ from perturbation theory.
Confinement — the other end of the running
Run the coupling downward instead. Around $Q \sim \Lambda$, $\alpha_s \sim 1$, gluon self-coupling dominates, and the chromo-field between a separating $q\bar q$ pair no longer spreads like a Coulomb field: it collapses into a flux tube of roughly constant cross-section. Constant field profile means constant energy per length — a linearly rising potential $V(r) \approx \sigma r$ with string tension $\sigma \approx (0.44\ \text{GeV})^2 \approx 0.9$ GeV/fm (measured from quarkonium spectra and lattice QCD). Pull the pair to $r \approx 1.5$ fm and the stored $\sim\!1.4$ GeV exceeds the cost of creating a light $q\bar q$ pair: the string snaps into two hadrons. That is why no free quark has ever been seen, and why DIS final states are jets of hadrons rather than a bare quark in a detector. For this course the practical import is the boundary: above $Q \sim 1$–$2$ GeV, quasi-free partons and perturbation theory (this chapter onward); below, strings and bound states, parametrized — never computed — by our PDFs at $Q_0$. The $Q^2$ and $W^2$ cuts of §1.1, and the $\mathcal{O}(\Lambda^2/Q^2)$ power corrections of §2.2, are all fences along this same line.
Worked example 1 — running $\alpha_s$ from $M_Z$ down to $Q_0 = 1.65$ GeV
Given: $\alpha_s(M_Z^2) = 0.118$ at $M_Z = 91.19$ GeV (our pipeline's input), thresholds at $m_b = 4.92$ GeV and $m_c = 1.51$ GeV (the values pinned by theory id 40000000). Use $1/\alpha_s(\mu^2) = 1/\alpha_s(\mu_0^2) + \beta_0 \ln(\mu^2/\mu_0^2)$ piecewise.
Leg 1, $n_f = 5$ ($M_Z \to m_b$): $\beta_0^{(5)} = \tfrac{33-10}{12\pi} = \tfrac{23}{12\pi} = 0.6101$. $\;\ln(m_b^2/M_Z^2) = 2\ln(4.92/91.19) = -5.839$. $\;1/\alpha_s(m_b^2) = 8.4746 - 0.6101 \times 5.839 = 8.4746 - 3.5626 = 4.912$, so $\alpha_s(m_b^2) = 0.2036$.
Leg 2, $n_f = 4$ ($m_b \to Q_0$): switch $\beta_0^{(4)} = \tfrac{25}{12\pi} = 0.6631$ ($\alpha_s$ continuous across the threshold at this order). $\;\ln(Q_0^2/m_b^2) = 2\ln(1.65/4.92) = -2.185$. $\;1/\alpha_s(Q_0^2) = 4.912 - 0.6631 \times 2.185 = 3.463$, so $\boxed{\alpha_s(Q_0^2) \approx 0.289}$. No charm switch was needed: $Q_0 = 1.65 > m_c = 1.51$ — the input scale is chosen to sit just above the charm threshold precisely so that fits parametrize an $n_f = 4$ proton (the "perturbative charm" convention, §2.4).
Extract the $\Lambda$'s: $\ln(M_Z^2/\Lambda_5^2) = 1/(\beta_0^{(5)}\alpha_s(M_Z^2)) = 13.89 \Rightarrow \Lambda_5 = 88$ MeV. Matching at $m_b$: $\ln(m_b^2/\Lambda_4^2) = 4.912/0.6631 = 7.407 \Rightarrow \Lambda_4 = 121$ MeV. Continuing to $m_c$ and matching: $\Lambda_3 \approx 146$ MeV.
Honesty check: the full NNLO running that EKO performs inside our pipeline gives $\alpha_s(1.65\ \text{GeV}) \approx 0.33$ (anchor: the world average $\alpha_s(m_\tau = 1.78\ \text{GeV}) \approx 0.31$). One loop undershoots by $\sim\!13\%$ — at $M_Z$ the orders agree to a percent, but near $Q_0$ the expansion is strained. This is exactly why the theory id pins the order (NNLO) along with $\alpha_s(M_Z)$: at our scales, "which loop" is a leading systematic.
Worked example 2 — the $R$-ratio against data
Below charm ($\sqrt{s} \approx 2.5$ GeV, $u,d,s$ active): parton level $R = 2$. QCD correction: $R = R_0\,(1 + \alpha_s/\pi + \ldots)$; with $\alpha_s(2.5\ \text{GeV}) \approx 0.27$, $R \approx 2 \times 1.086 = 2.17$. Measured (away from resonances): $\approx 2.2$. ✓
Above charm ($\sqrt{s} \approx 7$ GeV): $R_0 = \tfrac{10}{3} = 3.33$; with $\alpha_s \approx 0.20$, $R \approx 3.33 \times 1.064 = 3.55$. Measured: $\approx 3.6$. ✓
Above bottom ($\sqrt{s} = 34$ GeV, PETRA): $R_0 = \tfrac{11}{3} = 3.67$; with $\alpha_s \approx 0.14$, $R \approx 3.67 \times 1.045 = 3.83$. Measured: $\approx 3.9$. ✓
Without colour the three predictions would be $0.72$, $1.18$, $1.28$ — wrong by a factor of three everywhere. Two lessons: $N_c = 3$ is data, not doctrine; and the small mismatches ($2.17$ vs $2.2$, $3.83$ vs $3.9$) are themselves the $\mathcal{O}(\alpha_s)$ correction being seen — historically one of the first quantitative confirmations that $\alpha_s$ runs.