Quantum Chromodynamics

Derivation. Derives Phenomenon 102.1. Write the strong Hamiltonian as \(H=H_{0}+H_{1}\), with \(H_{0}\) \(\SU(3)\)-symmetric and \(H_{1}\) the symmetry-breaking piece. The breaking comes from the quark mass term of Section 102.3.3, which for \(m_{u}=m_{d}\neq m_{s}\) splits into a piece proportional to the identity — absorbed into \(H_{0}\) — and a piece proportional to \(\lambda_{8}\) of Equation (14.65). So \(H_{1}\) transforms as the eighth component of an octet. Treat it in first-order perturbation theory (Approximation Methods), so that the mass shift of a state \(\ket{B}\) is \(\ev{H_{1}}{B}\).

By the Wigner–Eckart theorem for \(\SU(3)\) — the same statement as for \(\SU(2)\), and proved in the same way from Schur's lemma Theorem 5.158 — the diagonal matrix elements of an octet operator inside an octet are determined by two reduced matrix elements, because \(\vect{8}\otimes\vect{8}\) contains \(\vect{8}\) twice (once symmetrically, through \(d_{abc}\) of Equation (14.70), and once antisymmetrically, through \(f_{abc}\) of Equation (14.69)). Two reduced matrix elements mean two constants, and the only \(\SU(3)\) scalars that can be built from \(T_{8}\) and the state's own quantum numbers, to first order, are \(Y\) and the isospin Casimir \(I(I+1)\) corrected by the trace term \(Y^{2}/4\). Hence

with \(M_{0}\), \(a\) and \(b\) the same for the whole multiplet.

For the octet, \((Y,I)\) is \((1,\tfrac{1}{2})\) for \(N\), \((0,0)\) for \(\Lambda\), \((0,1)\) for \(\Sigma\) and \((-1,\tfrac{1}{2})\) for \(\Xi\), so that \(I(I+1)-Y^{2}/4\) equals \(\tfrac{1}{2}\), \(0\), \(2\), \(\tfrac{1}{2}\) respectively, and

The three unknowns are eliminated from the four masses in one way: \(M_{N}+M_{\Xi}=2M_{0}+b\) and \(3M_{\Lambda}+M_{\Sigma}=4M_{0}+2b\), whence Equation (102.3). Numerically, with the isospin averages \(M_{N}c^{2}=938.92\,\mathrm{MeV}\), \(M_{\Lambda}c^{2}=1115.68\,\mathrm{MeV}\), \(M_{\Sigma}c^{2}=1193.15\,\mathrm{MeV}\) and \(M_{\Xi}c^{2}=1318.29\,\mathrm{MeV}\) [Navas:2024],

a discrepancy of \(25.8\,\mathrm{MeV}\), that is \(0.57\,\mathrm{\%}\) of either side, against a spread of nearly \(400\,\mathrm{MeV}\) among the masses being related. For the decuplet the states satisfy \(I=1+Y/2\), so that \(I(I+1)-Y^{2}/4=2+\tfrac{3}{2}Y\) and Equation (102.4) collapses to a linear function of \(Y\): consecutive isospin multiplets are equally spaced in mass. That is what Section 102.1.2 exploits.

Derivation. Derives Phenomenon 102.3. Apply Equation (102.4) to the decuplet, where it reduces to \(M=\text{const}+\left(a+\tfrac{3}{2}b\right)Y\) as shown above. The measured spacings between consecutive isospin multiplets of the decuplet are, with modern isospin-averaged masses [Navas:2024],

equal to within \(5\,\mathrm{\%}\), which is the accuracy first-order perturbation theory in a breaking of this size deserves. With the masses available in 1962 the first two spacings were about \(145\,\mathrm{MeV}\)\(/c^{2}\) each, and extrapolating one further step placed the missing state near \(1680\,\mathrm{MeV}\)\(/c^{2}\).

The quantum numbers follow with no freedom at all. The tenth weight of the decuplet sits at \(Y=-2\), \(I=I_{3}=0\), so Equation (102.1) gives \(Q=-e\), and \(Y=B+S\) with \(B=1\) gives \(S=-3\). The lifetime argument is kinematic: the lightest \(S=-3\) combination reachable by a strong decay, which cannot change strangeness, is \(\Xi K\), whose threshold is \(\left(1315+494\right)\mathrm{MeV}/c^{2} =1809\,\mathrm{MeV}/c^{2}\) [Navas:2024] — above the predicted mass. Every strong channel is therefore closed, the state must decay through the weak interaction of Weak Interactions with \(\Delta S=1\), and its lifetime must be of the order of \(10^{-10}\,\mathrm{s}\) rather than the \(10^{-23}\,\mathrm{s}\) of its nine partners. That is the difference between a resonance visible only as a bump in a mass distribution and a particle that travels centimetres and is photographed.

Derives Proposition 102.6. The first decomposition is Theorem 14.57, proved there both by splitting \(\operatorname{End}(\C^{3})\) into its trace and traceless parts and by counting weights. For the second, tensor Equation (14.81) once more with \(\vect{3}\), or count directly: \(\C^{3}\otimes\C^{3}\otimes\C^{3}\) has dimension \(27\) and decomposes under the symmetric group \(S_{3}\), which commutes with the \(\SU(3)\) action, into a totally symmetric part of dimension \(\binom{5}{3}=10\), a totally antisymmetric part of dimension \(\binom{3}{3}=1\), and two copies of a mixed-symmetry part of dimension \(8\) each; \(10+1+8+8=27\). Each symmetry class is \(\SU(3)\)-invariant, and the four pieces are the representations named. Baryon number is additive with \(B=\tfrac{1}{3}\) per quark, so \(q\bar{q}\) has \(B=0\) and \(qqq\) has \(B=1\), matching mesons and baryons respectively, and Equation (102.1) is reproduced because \(Q\) and \(Y\) are additive by construction.

Derivation. Derives Phenomenon 102.7. A Dirac particle of charge \(Q_{q}e\) and mass \(m_{q}\), with \(Q_{q}\) the dimensionless charge in units of the elementary charge, has magnetic moment \(\mu_{q}=Q_{q}e\hbar/(2m_{q})\), of SI dimension \(\mathrm{J}/\mathrm{T}\); this is the \(g=2\) result of the Dirac equation (The Dirac Equation), whose small radiative correction is the subject of Section 100.8.2 and is irrelevant at the accuracy in question here. Write \(\hat{\mu}:=e\hbar/(2m_{q})\) for the common scale, with \(m_{u}=m_{d}=:m_{q}\) by isospin symmetry, so that \(\mu_{u}=+\tfrac{2}{3}\hat{\mu}\) and \(\mu_{d}=-\tfrac{1}{3}\hat{\mu}\).

The proton is \(uud\) with total spin \(\tfrac{1}{2}\) and no orbital angular momentum. The spin-flavour wavefunction that is symmetric under exchange of all three quarks — required by Section 102.2.1, and there is exactly one such state with \(J=\tfrac{1}{2}\) — gives, on projecting the magnetic moment operator \(\sum_{i}\mu_{i}\sigma_{z}^{(i)}\) onto it,

The combinatorial factors \(4\) and \(-1\) are the statement that in the symmetric \(J=\tfrac{1}{2}\) state the like pair (\(uu\) in the proton) is coupled to spin \(1\) with weight \(\tfrac{2}{3}\) and the odd quark carries the remaining spin with weight \(-\tfrac{1}{3}\). Substituting, \(\mu_{p}=\hat{\mu}\) and \(\mu_{n}=-\tfrac{2}{3}\hat{\mu}\), whence Equation (102.10). The unknown \(m_{q}\) has cancelled, which is why the ratio is a prediction and the individual moments are not.

Feeding the measured \(\mu_{p}=2.792847\,\mu_{N}\) [Mohr:2025], with the nuclear magneton \(\mu_{N}=5.0507837393\times 10^{-27}\,\mathrm{J}/\mathrm{T}\) [Mohr:2025], back into \(\mu_{p}=e\hbar/(2m_{q})\) gives

almost exactly one third of the nucleon mass. A model in which the nucleon is three objects of mass \(m_{N}/3\) therefore reproduces both the moment ratio and the moment scale. That the number \(336\,\mathrm{MeV}\)\(/c^{2}\) has nothing to do with the \(u\) and \(d\) masses in the Lagrangian, which are \(2.16\,\mathrm{MeV}\)\(/c^{2}\) and \(4.70\,\mathrm{MeV}\)\(/c^{2}\) [Navas:2024], is the first appearance in this chapter of the fact developed in Section 102.8.2: the mass of ordinary matter is not the mass of its quarks.

Derives Theorem 102.12. The centre of \(\SU(3)\) is the group of the three matrices \(\omega^{r}\identity\), \(r=0,1,2\), with \(\omega=\ee^{2\pi\ii/3}\); these are unitary, unimodular (\(\det(\omega\identity)=\omega^{3}=1\)) and commute with everything, and no other multiple of the identity is unimodular. On the fundamental \(\vect{3}\) the element \(\omega\identity\) acts as multiplication by \(\omega\), and on \(\bar{\vect{3}}\) as multiplication by \(\omega^{-1}\). On a tensor product of \(m\) copies of \(\vect{3}\) and \(n\) of \(\bar{\vect{3}}\) it therefore acts as \(\omega^{m-n}\), and on the trivial representation it must act as \(1\). Hence \(\omega^{m-n}=1\), i.e. \(3\mid m-n\). This number, \(m-n \bmod 3\), is the triality, and it is a representation-theoretic invariant: no equivalence can change it, so the obstruction is absolute and not a matter of which coupling scheme is used.

For the multiplicities: the number of singlets in \(V\otimes W\) is \(\dim\operatorname{Hom}_{G}(V^{*},W)\), which by Schur's lemma Theorem 5.158 is \(1\) if \(W\) is equivalent to \(V^{*}\) and \(0\) otherwise. Since \(\bar{\vect{3}}\) is inequivalent to \(\vect{3}\) (Proposition 14.55), \(\vect{3}\otimes\vect{3}\) has no singlet, while \(\vect{3}\otimes\bar{\vect{3}}\) has exactly one — as Theorem 14.57 exhibits explicitly, the trace part of a \(3\times3\) matrix. For three quarks, the decomposition Equation (102.9) contains \(\vect{1}\) once, and the invariant realizing it is \(\epsilon_{ijk}\): the unique (up to scale) totally antisymmetric invariant tensor of \(\SL(3,\C)\), hence of \(\SU(3)\).

Derivation. Derives Phenomenon 102.15. At leading order both processes proceed through a single virtual photon, so they differ only in the charge of the produced pair and in how many distinct pairs can be produced. The denominator is Equation (100.32), \(\sigma(e^{+}e^{-}\to\mu^{+}\mu^{-}) =4\pi\alpha^{2}(\hbar c)^{2}/3E_{\mathrm{cm}}^{2}\), computed in Section 100.3.1. For a pointlike spin-\(\tfrac{1}{2}\) particle of electric charge \(Q_{f}e\) — here and throughout, \(Q_{f}\) is the dimensionless charge of flavour \(f\) in units of \(e\) — created well above its threshold, the same computation applies with the single change \(e^{2}\to Q_{f}e^{2}\) at the production vertex, so the cross-section is that of Equation (100.32) times \(Q_{f}^{2}\); and the rates for distinguishable final states add incoherently, so a quark flavour existing in \(N_{c}\) colours contributes \(N_{c}\) times, the electromagnetic vertex being blind to the colour label. Dividing by the muon channel, for which \(Q=-1\) and there is one internal state,

the sum running over the flavours light enough to be produced. With \(Q_{u}=Q_{c}=+\tfrac{2}{3}\) and \(Q_{d}=Q_{s}=Q_{b}=-\tfrac{1}{3}\) this gives \(R=\tfrac{2}{3}N_{c}\) for \(u,d,s\), \(R=\tfrac{10}{9}N_{c}\) once charm opens and \(R=\tfrac{11}{9}N_{c}\) once bottom opens — that is \(2\), \(10/3\) and \(11/3\) for \(N_{c}=3\), which is what is measured.

Confinement does not spoil the count. The quarks produced at the photon vertex cannot emerge as free particles, but whatever they do afterwards they end as hadrons with unit probability, so the hadronic total is settled at the moment of pair creation and the long-distance physics cancels in the ratio. This is the same inclusiveness argument that makes Theorem 100.58 work, applied to colour instead of to soft photons.

The residual excess of the data over the integers above is calculable. Gluon emission from the produced pair multiplies Equation (102.15) by \(1+\alpha_{s}(E_{\mathrm{cm}})/\pi+\dots\), the first term being the same \(\mathcal{O}(\alpha_{s})\) real-plus-virtual sum that gives the three-jet rate of Section 102.6.2; at \(E_{\mathrm{cm}}=5\,\mathrm{GeV}\), where \(\alpha_{s}\approx0.2\), this is a \(6\,\mathrm{\%}\) excess, and it is seen. So \(R\) is simultaneously a colour counter at the \(10\,\mathrm{\%}\) level and, once the count is granted, one of the determinations of \(\alpha_{s}\) collected in Section 102.4.2.

Derives Proposition 102.16. The whole content is the colour factor of the triangle graph. The axial current whose divergence carries the anomaly is the third isospin component, \(j^{\mu5}_{3} =\tfrac{1}{2}(\bar{u}\gamma^{\mu}\gamma^{5}u -\bar{d}\gamma^{\mu}\gamma^{5}d)\), and each quark in the loop is weighted by the square of its electric charge and summed over colours, the electromagnetic vertices being colour-blind. The sum is

the relative minus sign coming from the isospin structure of the current. The amplitude is proportional to Equation (102.17) and the width to its square, which is Equation (102.16); the remaining factors — the anomaly coefficient itself, the reduction of the amplitude to the pion decay constant through the partial conservation of the axial current, and the two-body phase space — are derived in Section 106.7.2. The reason this count is sharper than \(R\) is that it is a low-energy theorem: the anomaly receives no corrections at higher order in \(\alpha_{s}\) (Theorem 106.89), so the only uncertainty is the measured \(F_{\pi}\), entering squared.

Derives Theorem 102.19. Under Equation (102.20) the derivative picks up an inhomogeneous term, \(\pp_{\mu}q\mapsto U\pp_{\mu}q +(\pp_{\mu}U)q\), so \(\Lag_{0}\) is not invariant. Postulate a matrix field \(\mathcal{A}_{\mu}\) valued in the Lie algebra and define \(D_{\mu}\) by Equation (102.21). Demanding \(D_{\mu}q\mapsto U(D_{\mu}q)\) gives

for all \(q\), whence \(\ii\mathcal{A}'_{\mu}U=\ii U\mathcal{A}_{\mu}-(\pp_{\mu}U)\) and, multiplying on the right by \(U^{\dagger}\), Equation (102.22). Conversely Equation (102.22) makes \(\bar{q}\gamma^{\mu}D_{\mu}q\) invariant by the same computation read backwards, so the condition is necessary as well as sufficient.

For the infinitesimal form put \(U=\identity+\ii\theta^{a}T_{a}\) and keep first order. The first term of Equation (102.22) gives \(\mathcal{A}_{\mu}+\ii\theta^{b}\comm{T_{b}}{\mathcal{A}_{\mu}}\), and with \(\comm{T_{b}}{T_{c}}=\ii f_{bca}T_{a}\) of Equation (14.67) together with the total antisymmetry of \(f\) this is \(\mathcal{A}_{\mu}-(g_{s}/\hbar)f^{abc}\theta^{b}A^{c}_{\mu}T_{a}\). The second term gives \(\ii(\ii\pp_{\mu}\theta^{a}T_{a})=-\pp_{\mu}\theta^{a}T_{a}\). Dividing by \(g_{s}/\hbar\) yields Equation (102.23). For one abelian generator the second term of Equation (102.23) is absent and the first reproduces Equation (100.8) with \(\Lambda=-\hbar\theta/g_{s}\).

Derives Proposition 102.21. Compute the commutator on a quark field:

the derivative terms acting on the quark cancelling between the two orderings. Factoring \(\ii g_{s}/\hbar\) and using \(f_{bca}=f_{abc}\) gives Equation (102.24). Since \(D_{\mu}\mapsto UD_{\mu}U^{\dagger}\) by construction — Equation (102.22) is exactly the statement that \(D_{\mu}q\) transforms like \(q\) — the commutator of two such objects transforms as \(U\comm{D_{\mu}}{D_{\nu}}U^{\dagger}\), which is Equation (102.25). Finally \(F^{a}F^{a}\propto\tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu})\) by Equation (14.66), and a trace of a product of two adjoint objects is invariant by cyclicity — it is the Killing form of Proposition 14.53, which is what makes this the only quadratic invariant available.

Derives Proposition 102.23. Write \(F^{a}_{\mu\nu}=G^{a}_{\mu\nu} -(g_{s}/\hbar)f^{abc}A^{b}_{\mu}A^{c}_{\nu}\) with \(G^{a}_{\mu\nu}=\pp_{\mu}A^{a}_{\nu}-\pp_{\nu}A^{a}_{\mu}\) and square, the cross term appearing twice by the symmetry of the contraction. Multiplying by \(-1/4\mu_{0}\) gives Equation (102.28). For an abelian group all \(f^{abc}\) vanish and only the first line survives, which is Equation (100.12); the normalization \(1/4\mu_{0}\) is fixed by that limit. Dimensionally \(F^{2}/\mu_{0}\) carries \(\mathrm{T}^{2}\,\mathrm{A}^{2}/\mathrm{N} =\mathrm{J}/\mathrm{m}^{3}\), as in Equation (100.12).

Derives Theorem 102.25. This is Theorem 106.52, proved there; only the group-theoretic input is quoted here. The variation of the gauge condition under Equation (102.23) is \(\delta G^{a}=-(\hbar/g_{s})\pp^{\mu} \left(\delta^{ab}\pp_{\mu}+ (g_{s}/\hbar) f^{acb}A^{c}_{\mu}\right) \theta^{b}\), whose kernel is the adjoint covariant derivative. The \(f^{acb}\) term is absent for an abelian group, which is Corollary 106.53 and the reason the photon needed no ghosts in Section 100.1.3.

Derives Proposition 102.28. The gauge sector is fixed by the group: the number of gauge fields is \(\dim\mathfrak{su}(3)=8\) by Proposition 14.50, and the cubic and quartic couplings of Equation (102.28) carry the structure constants Equation (14.69), which are numbers, multiplied by powers of the same \(g_{s}\) that appears in the quark vertex. The quark sector contributes one mass per flavour. That the coupling is the same for every flavour is the non-abelian version of Proposition 100.6: \(D_{\mu}\) depends on the matter field only through its colour representation, and every quark is a \(\vect{3}\). Renormalizability (Theorem 102.26) is what guarantees the count is not enlarged order by order, and the Slavnov–Taylor identities are what keep the three gauge vertices tied to one number.

Derives Proposition 102.29. Rules 1–3 are Proposition 100.14 with the substitution \(q\gamma^{\mu}\to g_{s}\gamma^{\mu}T_{a}\) forced by Equation (102.21), together with the observation that the free parts of Equation (102.29) are diagonal in colour; rules 4–6 are read off the interaction terms of Equation (102.28) and of Theorem 102.25 by the same expansion of \(\exp(\ii S/\hbar)\). For Equation (102.31), \(T_{a}T_{a}\) commutes with every generator and is therefore a multiple of the identity on the irreducible \(\vect{3}\); Equation (14.77) evaluates the multiple as \(4/3\), and Equation (14.78) does the same for the adjoint, where the generators are \(-\ii f^{abc}\).

Derives Lemma 102.31. In a medium of permittivity \(\varepsilon\) and permeability \(\mu\) the phase velocity of light is \(\left(\varepsilon\mu\right)^{-1/2}\). The vacuum, polarized or not, is Lorentz invariant, and by the postulates of Lorentz Transformations light propagates in it at \(c\) in every frame; a medium with \(\varepsilon\mu\neq c^{-2}\) would define a rest frame. Hence Equation (102.32). The second form follows from \(\varepsilon_{0}\mu_{0}c^{2}=1\), and the statement about signs is immediate. Finally, the coupling measured at a scale is \(\alpha_{s}=g_{s}^{2}/ \left(4\pi\varepsilon_{\mathrm{eff}}\hbar c\right)\) by Equation (102.19), so \(\varepsilon_{\mathrm{eff}}<\varepsilon_{0}\) means a larger effective coupling at long distance and a smaller one at short distance, which is asymptotic freedom.

Derives Lemma 102.32. For a spinless particle the transverse motion is a harmonic oscillator of frequency \(\omega_{c}=\abs{q}c^{2}B/E\), whose quantization gives \(E^{2}=c^{2}p_{z}^{2}+(2n+1)\abs{q}\hbar c^{2}B\) — the relativistic Landau problem, obtained by squaring the wave operator so that \(\vect{p}_{\perp}^{2}\to(2n+1)\abs{q}\hbar B\). A magnetic moment \(\vect{\mu}=g\left(q/2m\right)\vect{S}\) adds an interaction energy \(-\vect{\mu}\cdot\vect{B}\), which in the squared equation appears as \(-g\abs{q}\hbar c^{2}Bs_{z}\); putting \(g=2\) gives Equation (102.33). That \(g=2\) is the Dirac value for a spin-\(\tfrac{1}{2}\) particle (The Dirac Equation), and its small radiative correction (Theorem 100.41) is of higher order in the coupling and irrelevant to a one-loop coefficient. That \(g=2\) also for the charged gluons is a property of Equation (102.28): expanding \(-F^{a}_{\mu\nu}F^{a\,\mu\nu}/4\mu_{0}\) to second order about a background field produces, besides the covariant Laplacian, exactly the term \(-2\ii\mathcal{F}_{\mu\nu}\) acting on the vector index of the fluctuation, and a coefficient \(2\) in front of the field-strength coupling to the spin generator is the statement \(g=2\). The degeneracy \(\abs{q}B/(2\pi\hbar)\) per unit area is the standard one: one state per flux quantum \(2\pi\hbar/\abs{q}\).

Derives Theorem 102.33. Put the vacuum in a uniform field \(B\) pointing along the chosen Cartan direction in colour space, so that every field component has a definite charge \(x_{i}g_{s}\) and the problem is the abelian one of Lemma 102.32. The vacuum energy density is the sum of the zero-point energies of all modes,

the factor \((-1)^{2s}\) being the sign of a fermionic zero-point energy, \(-\tfrac{1}{2}\hbar\omega\) per mode instead of \(+\tfrac{1}{2}\hbar\omega\).

Cut the longitudinal momentum off at \(\abs{p_{z}}\leq P\) and write \(\varepsilon_{n}^{2}=A\left(2n+1+\kappa\right)\) with \(A=\abs{q}\hbar c^{2}B\). Then

the first two terms being independent of \(B\) and of \(n\) in a way that contributes no \(B^{2}\ln B\), and the third being the one that matters. Split \(\ln\varepsilon_{n}^{2}=\ln A+\ln(2n+1+\kappa)\): the second piece carries no \(B\) and produces a term proportional to \(B^{2}\) with no logarithm, i.e. a renormalization of the field energy and not a running. Keeping the first piece,

The sum over Landau levels diverges and is defined by analytic continuation: writing \(2n+1+\kappa=2(n+x)\) with \(x=\left(1+\kappa\right)/2\), and using the Hurwitz zeta function \(\zeta(\sigma,x)=\sum_{n\geq0}(n+x)^{-\sigma}\) continued to \(\sigma=-1\), where \(\zeta(-1,x)=-\tfrac{1}{2}\left(x^{2}-x+\tfrac{1}{6}\right)\),

using \(\kappa=-2s_{z}\). The two terms of Equation (102.39) are the two magnetic responses: the \(1\) comes from the \(2n+1\) of the orbital motion and the \(-12s_{z}^{2}\) from the \(\kappa\) of the moment.

Substituting \(A_{i}=\abs{q_{i}}\hbar c^{2}B\) into Equation (102.38) and writing \(L_{i}=\ln\left[4c^{2}P^{2}/(\abs{q_{i}}\hbar c^{2}B)\right]\),

Comparing with the field energy density \(u=B^{2}/2\mu_{\mathrm{eff}}\) and using Equation (102.32) together with \(q_{i}^{2}=x_{i}^{2}g_{s}^{2}\) and \(g_{s}^{2}/(\varepsilon_{0}\hbar c)=4\pi\alpha_{s}\) from Equation (102.19),

The species-dependence of \(L_{i}\) is a \(B\)-independent additive constant under the logarithm and is therefore a choice of subtraction scheme, not physics: only the coefficient of \(\ln B\) is common to all schemes, which is the standard statement that a one-loop beta-function coefficient is scheme-independent (Proposition 100.55). Identifying the energy scale probed by the field, \(E^{2}\sim g_{s}\hbar c^{2}B\) — the field sets the transverse momentum of the modes — and writing \(\Lambda_{E}=2cP\), all \(L_{i}\) become \(\ln(\Lambda_{E}^{2}/E^{2})\) up to constants, which is Equation (102.34).

Finally \(\alpha_{s}(E) =g_{s}^{2}/\left(4\pi\varepsilon_{\mathrm{eff}}\hbar c\right)\), so \(\alpha_{s}(E)^{-1}=\alpha_{s}(\Lambda_{E})^{-1} +\left(S/24\pi\right)\ln(\Lambda_{E}^{2}/E^{2})\). Differentiating with respect to \(\ln E\) gives \(-\alpha_{s}^{-2}\,\dd\alpha_{s}/\dd\ln E=-S/12\pi\), which is Equation (102.35).

Derives Corollary 102.34. The one-particle states are the fermion and the antifermion, each of charge magnitude \(e\), so \(\sum_{i}x_{i}^{2}=2\). The statistics factor is \((-1)^{2s}=-1\) and the spin sum over \(s_{z}=\pm\tfrac{1}{2}\) is \(2\left(1-12\cdot\tfrac{1}{4}\right)=-4\). Hence \(S=2\cdot(-1)\cdot(-4) =8\) and \(\dd\alpha/\dd\ln E=8\alpha^{2}/12\pi=2\alpha^{2}/3\pi\). The sign is positive — QED screens — and by Equation (102.39) the reason is arithmetical: for \(s_{z}=\pm\tfrac{1}{2}\) the paramagnetic term \(12s_{z}^{2}=3\) beats the diamagnetic \(1\), so the electron vacuum is in fact paramagnetic, and it is the Fermi statistics — the minus sign of a fermion loop — that turns the result back into screening.

Derives Theorem 102.35. Apply Theorem 102.33 twice.

Quarks. A quark flavour is a colour triplet; taking the Cartan direction to be \(T_{3}\), its three colour states carry charges \(x=+\tfrac{1}{2},-\tfrac{1}{2},0\), and each is a Dirac field with its own antiparticle. Hence \(\sum_{i}x_{i}^{2}=2\tr\left(T_{3}^{2}\right)=2T_{F}=1\) per flavour, the factor \(2\) counting antiparticles and \(T_{F}=\tfrac{1}{2}\) being Equation (14.66). With the spin factor \((-1)\times(-4)=+4\) from Corollary 102.34, quarks contribute \(S_{q}=4n_{f}\).

Gluons. The gluons form the adjoint \(\vect{8}\), whose charges under \(\ad_{T_{3}}\) are the first components of the roots Equation (14.80), namely \(\pm1\), \(\pm\tfrac{1}{2}\), \(\pm\tfrac{1}{2}\) and two zeros, so that \(\sum_{a}x_{a}^{2}=f^{3cd}f^{3cd}=C_{A}=3\) by Equation (102.31). Here the charged components already occur in \(\pm\) pairs — a gluon of charge \(+1\) is the antiparticle of the one with charge \(-1\) — so no further doubling is applied. A gluon is massless and has two physical polarizations, \(s_{z}=\pm1\); the unphysical ones are removed by the ghosts of Theorem 102.25, which is where that machinery earns its keep. The statistics factor is \((-1)^{2}=+1\) and the spin sum is

so \(S_{g}=3\times(-22)=-66\).

Adding, \(S=4n_{f}-66=-6\left(11-\tfrac{2}{3}n_{f}\right)=-6b_{0}\), and Equation (102.35) becomes Equation (102.42). Restoring the group factors in place of their numerical values, the quark term is \(-\tfrac{4}{3}T_{F}n_{f}\) and the gluon term \(+\tfrac{11}{3}C_{A}\), which is the form used for any gauge group. \(b_{0}>0\) requires \(n_{f}<33/2\), i.e. \(n_{f}\leq16\).

Derivation. Derives Phenomenon 102.39. This is dimensional transmutation, and Equation (102.45) is the whole of it. A statement of the form “\(\alpha_{s}=0.118\)” is meaningless without naming the scale at which it holds, because Equation (102.44) says the number changes with the scale; and a statement of the form “\(\alpha_{s}=0.118\) at \(E_{0}\)” contains one dimensionless number and one energy. The combination in Equation (102.45) is invariant under changing \(E_{0}\): differentiating \(\ln\Lambda_{\mathrm{QCD}}=\ln E_{0}-2\pi/ \left(b_{0}\alpha_{s}(E_{0})\right)\) with respect to \(\ln E_{0}\) gives \(1+\left(2\pi/b_{0}\alpha_{s}^{2}\right) \dd\alpha_{s}/\dd\ln E_{0}=1-1=0\) by Equation (102.42). So the theory has exactly one free parameter in its massless limit, and that parameter is an energy. The classical scale invariance of the massless Lagrangian is broken by the quantum theory — it is the trace anomaly, the same phenomenon in another guise — and \(\Lambda_{\mathrm{QCD}}\) is what the breaking produces. Numerically, taking \(E_{0}=m_{Z}c^{2}=91.188\,\mathrm{GeV}\) and \(\alpha_{s}=0.1180\) [Navas:2024] with \(n_{f}=5\), so that \(b_{0}=11-\tfrac{10}{3}=\tfrac{23}{3}\),

and at four loops in the \(\overline{\mathrm{MS}}\) scheme the same data give about \(0.21\,\mathrm{GeV}\) for five active flavours and about \(0.33\,\mathrm{GeV}\) for three [Navas:2024]. The factor of two between Equation (102.46) and the four-loop value is not a discrepancy: \(\Lambda\) is defined by a truncation of Equation (102.42) and by a scheme, and changes when either changes, whereas \(\alpha_{s}\) at a stated scale does not. That is why the Particle Data Group quotes \(\alpha_{s}(m_{Z}c^{2})\) and not \(\Lambda\). What is robust, and what Phenomenon 102.39 asserts, is the order of magnitude and the fact that a scale exists at all.

Derivation. Derives Phenomenon 102.41. The curve is Equation (102.44) with Equation (102.42), together with the rule that a quark decouples below its own mass — the decoupling theorem [Appelquist:1975] already used in Proposition 100.63 — so that \(n_{f}\) and hence \(b_{0}\) change at each flavour threshold and \(\alpha_{s}\) is matched continuously across it. Starting from Equation (102.47) and integrating downwards through the \(b\) threshold and upwards through the \(t\) threshold with the masses of Table 102.1 gives Table 102.2. The prediction is not one number but the shape of a curve over three decades, and the fact that determinations using entirely different physics land on it is the content of the phenomenon.

Derivation. Derives Phenomenon 102.45. Scaling. Suppose the photon, at large \(Q^{2}\), resolves a time short compared with the time over which the proton's constituents interact with one another, so that it scatters elastically off one of them and the rest are spectators. This is the impulse approximation, and Feynman's name for the constituents is partons [Feynman:1969]. Let the struck parton carry a fraction \(\xi\) of the proton's four-momentum, \(p=\xi P\), and let it remain on its mass shell after the collision, its mass being \(\xi M\) in the same approximation. Then

So the measured variable \(x\) is the momentum fraction of the struck constituent: this is what makes \(x\) physical rather than a convenient abbreviation. The cross-section is then the incoherent sum over constituents of pointlike elastic cross-sections, each of which is the Mott formula Equation (102.50) times the squared charge, and the only place a scale could enter — a form factor, i.e. a size — is absent by hypothesis. Hence

with \(q(x)\dd x\) the number of quarks of flavour \(q\) carrying momentum fraction in \(\dd x\): a function of \(x\) alone. Any structure of the constituents themselves would reintroduce a length and destroy the scaling, so scaling is the statement that the constituents are pointlike down to the resolution \(\hbar c/Qc\) reached — about \(10^{-16}\,\mathrm{m}\) at the original SLAC energies, a tenth of the proton radius \(8.4075\times 10^{-16}\,\mathrm{m}\) [Mohr:2025]. The contrast is with elastic scattering, whose cross-section carries the squared dipole form factor and therefore falls as \(Q^{-12}\): the same beam sees a soft object elastically and hard objects inelastically, which is precisely Rutherford's situation.

The Callan–Gross relation. Decompose the virtual photon into its three polarization states in the frame where it moves along \(+z\) and the struck parton along \(-z\). A massless spin-\(\tfrac{1}{2}\) parton coupling through the vector current \(\bar{q}\gamma^{\mu}q\) conserves helicity, because \(\gamma^{\mu}\) commutes with \(\gamma^{5}\) up to a sign that leaves the chiral projectors intact and, for a massless fermion, chirality is helicity. After absorbing the photon the parton moves along \(+z\); its helicity is unchanged, so its spin projection on the \(z\) axis has flipped, \(\Delta J_{z}=\pm1\). Only a transverse photon, \(\lambda=\pm1\), can supply that. The longitudinal photon, \(\lambda=0\), therefore cannot be absorbed at all:

Expressing the two cross-sections in terms of the structure functions, the transverse one is proportional to \(F_{1}\) and the longitudinal to the combination \(F_{L}:=F_{2}-2xF_{1}\) (up to terms of order \(M^{2}c^{4}/Q^{2}c^{2}\), negligible in the deep inelastic region), so Equation (102.55) is exactly Equation (102.52) [Callan:1969]. A spin-\(0\) parton gives the opposite conclusion by the same argument — it has no spin to flip, so only \(\lambda=0\) is absorbed, \(\sigma_{T}=0\) and \(F_{1}=0\) — and the data exclude it decisively: the measured ratio \(R_{L}=\sigma_{L}/\sigma_{T}\) is small, as QCD predicts; the experimental write-up belongs to Section 116.4.3.

Derives Proposition 102.46. The photon couples to a quark with strength \(Q_{q}e\), so Equation (102.54) weights each flavour by \(Q_{q}^{2}\). The charged weak current of Weak Interactions couples with a strength independent of electric charge, so the corresponding neutrino structure function is \(F_{2}^{\nu N}(x)=x\sum_{q}[q(x)+\bar{q}(x)]\) with unit weights. On an isoscalar target the \(u\) and \(d\) distributions enter symmetrically, and the ratio of the two sums is the mean of \((2/3)^{2}\) and \((1/3)^{2}\), which is Equation (102.56). Had the constituents carried integer charges the ratio would have been of order unity instead of \(0.28\). This is a measurement of the charges themselves, independent of the spectroscopy of Section 102.1.3, and it is why neutrino scattering was decisive.

Derivation. Derives Phenomenon 102.47. The momentum of the proton is shared among its constituents by definition, \(\int_{0}^{1}\dd x\,x\sum_{\text{all}}f_{i}(x)=1\), since \(x\) is the momentum fraction (Equation (102.53)). The measured left-hand side of Equation (102.57) is obtained from Equation (102.54) together with Equation (102.56), the two providing enough information to separate the charge weights from the distributions; the details of the extraction belong in Section 116.5.2. The deficit is a fact about the data and not an inference: whatever carries the missing half is electrically neutral and weakly inert. In QCD it is the gluon, which appears here for the first time as an experimental necessity rather than a theoretical convenience — and it appeared this way, in the momentum sum rule, before it was seen in the three-jet events of Section 102.6.2.

Derives Proposition 102.48. Two features are derived here and the third is quoted. The factor \(\dd k_{T}^{2}/k_{T}^{2}\) is the collinear singularity of a massless emitter: the propagator of the quark after emission has virtuality \(\left(p-k\right)^{2}\propto -k_{T}^{2}/(1-z)\), and the squared amplitude carries its inverse square against a two-body phase space \(\propto\dd k_{T}^{2}\), leaving \(\dd k_{T}^{2}/k_{T}^{2}\). It is logarithmic precisely because the emitter is massless, and it is the mass singularity of Theorem 100.60 in its QCD form. The factor \(1/(1-z)\) is the soft singularity: as \(z\to1\) the gluon energy goes to zero and the amplitude reduces to the eikonal current Equation (100.76), whose square gives exactly this pole, with \(C_{F}\) replacing the squared electric charge because the emission vertex is \(T_{a}\) and \(T_{a}T_{a}=C_{F}\identity\). The numerator \(1+z^{2}\) is the spin structure of the vertex, the two terms being the helicity-conserving and helicity-flip configurations of the emitting quark; it is computed from the same trace technology as Section 100.3.2 [Altarelli:1977] [Gribov:1972].

The plus prescription is not an extra assumption but a normalization forced by conservation of quark number. The number of quarks of a given flavour minus antiquarks is a conserved charge and cannot change with \(Q^{2}\), so \(\int_{0}^{1}\dd z\,P_{qq}(z)=0\); the unregulated \(1/(1-z)\) is not integrable, and the divergence is cancelled by the virtual correction, which lives entirely at \(z=1\). The plus prescription is exactly that cancellation written as a distribution: it subtracts, at \(z=1\), whatever the real emission supplies. Equivalently \([(1+z^{2})/(1-z)]_{+}=(1+z^{2})\left[1/(1-z)\right]_{+} +\tfrac{3}{2}\delta(1-z)\), and the \(\tfrac{3}{2}\) is the virtual term.

Derives Theorem 102.49. A parton of fraction \(x\) at resolution \(Q^{2}+\dd Q^{2}\) is either one that was already there at \(Q^{2}\), or one produced by an emission from a parton of larger fraction \(y>x\) that the coarser resolution could not separate. Multiplying Equation (102.58) by the density of emitters \(q_{i}(y)\), changing variables from \(z\) to \(x=zy\) at fixed \(y\) so that \(\dd z=\dd x/y\), and integrating \(\dd k_{T}^{2}/k_{T}^{2}\) over the shell between the two resolutions gives the first term of Equation (102.59). The second term is the same statement for a gluon converting to a quark–antiquark pair, with \(P_{qg}(z)=T_{F}\left[z^{2}+(1-z)^{2}\right]\). That the kernel depends on \(x\) and \(y\) only through \(x/y\) is a consequence of the collinear factorization: the emission probability knows only the momentum fraction taken, not the absolute momenta. Comparing with Equation (102.63) below identifies the integral as \(\left(P\star q\right)(x)\) with the Haar measure \(\dd y/y\) of Definition 17.97, which is the structure that Corollary 102.50 exploits.

Derives Corollary 102.50. Write the convolution of Theorem 102.49 in the form

which is Equation (17.113): both factors vanish for argument greater than one, so the range is \(x\leq y\leq1\) and the convolution stays supported on the unit interval. By Corollary 17.99, and ultimately by the Mellin convolution theorem Theorem 17.98, the moments of a multiplicative convolution are the products of the moments, Equation (17.115). Taking \(M_{N}\) of Equation (102.59) therefore turns the integral operator into multiplication by \(\gamma_{N}=M_{N}[P_{qq}]\), giving Equation (102.60). This is the whole reason the Mellin transform is the right instrument here: it diagonalises dilations, and Equation (102.59) is an equation about how momentum fractions dilate.

To solve, divide Equation (102.60) by Equation (102.42) rewritten as \(\dd\alpha_{s}/\dd\ln Q^{2}=-b_{0}\alpha_{s}^{2}/4\pi\) — the factor \(\tfrac{1}{2}\) relative to Equation (102.42) because \(\ln Q^{2}=2\ln Q\). Then \(\dd\ln M_{N}/\dd\alpha_{s}=-2\gamma_{N}/(b_{0}\alpha_{s})\), which integrates to \(M_{N}\propto\alpha_{s}^{-2\gamma_{N}/b_{0}}\) and hence to Equation (102.61).

For Equation (102.62), use the plus prescription:

the subtraction at \(z=1\) making the integrand finite. Split the numerator: \(\left(z^{N}-1\right)+\left(z^{N+2}-z^{2}\right)\), and write the second bracket as \(\left(z^{N+2}-1\right)-\left(z^{2}-1\right)\). With \(\int_{0}^{1}\left(z^{m}-1\right)/(1-z)\,\dd z=-H_{m}\) — immediate from \(\left(1-z^{m}\right)/(1-z)=1+z+\dots+z^{m-1}\) integrated term by term — this gives \(\gamma_{N}=C_{F}\left[-H_{N}-H_{N+2}+H_{2}\right]\) and \(H_{2}=\tfrac{3}{2}\), which is Equation (102.62). Checking: \(\gamma_{0} =C_{F}[\tfrac{3}{2}-0-\tfrac{3}{2}]=0\), as quark-number conservation demands, and \(\gamma_{1}=C_{F}[\tfrac{3}{2}-1-\tfrac{11}{6}] =-\tfrac{4}{3}C_{F}\).

Derives Proposition 102.55. Section 100.3.1 gives the spin-averaged squared amplitude for this process as \(\overline{\abs{\mathcal{M}}^{2}} =32\pi^{2}\alpha^{2}\hbar^{4}\left(t^{2}+u^{2}\right)/s^{2}\). For massless kinematics in the centre-of-mass frame, \(t=-\tfrac{1}{2}s\left(1-\cos\vartheta\right)\) and \(u=-\tfrac{1}{2}s\left(1+\cos\vartheta\right)\), so

Inserting this into the two-body phase space of Equation (100.40) and multiplying by \(Q_{f}^{2}\) for the charge and by \(N_{c}\) for the colour multiplicity gives Equation (102.67); integrating over angles returns Equation (100.32) times the same factors, which is Equation (102.15). The \(\sin^{2}\vartheta\) for scalars follows because a scalar pair produced by a vector current must be in a \(p\)-wave with \(J_{z}=0\) along the pair axis, and the corresponding Wigner function is \(\sin\vartheta\).

Derivation of the two-jet structure. Derives Phenomenon 102.56. Two facts make a jet. First, the parton pair is produced back to back with energy \(E_{\mathrm{cm}}/2\) each, and the angular distribution of the pair axis is Equation (102.67) — an observable that survives hadronization because the subsequent radiation is soft. Second, the radiation is soft and collinear: by Equation (102.58) the emission probability is concentrated at small \(k_{T}\), so the hadrons that materialize share the parton's direction to within an angle of order \(\Lambda_{\mathrm{QCD}}/E\). The transverse momentum inside a jet is therefore bounded by a hadronic scale while its longitudinal momentum grows with the beam energy, and the spray narrows as \(1/E\). Hanson and collaborators established this at SPEAR by measuring the sphericity tensor and showing that its principal axis is distributed as \(1+\cos^{2}\vartheta\) [Hanson:1975]: the axis of a spray of hadrons remembers the spin of a particle that was never observed.

Derivation of the singular structure. Derives Proposition 102.58. The two limits that dominate the rate are derivable from results already established, and they already distinguish a vector from a scalar.

Soft. As \(x_{3}\to0\) both \(x_{1}\) and \(x_{2}\) tend to \(1\), and the amplitude factorizes into the two-jet amplitude times the eikonal current Equation (100.76) of Section 100.6.1, with the electric charge replaced by \(g_{s}T_{a}\) and the colour factor \(T_{a}T_{a}=C_{F}\) of Equation (102.31). Squaring gives the double pole \(\left[(1-x_{1})(1-x_{2})\right]^{-1}\) of Equation (102.68) with numerator \(2\), which is \(x_{1}^{2}+x_{2}^{2}\) at \(x_{1}=x_{2}=1\). A scalar emitter has no such factorization: a soft scalar does not couple to the eikonal current at all, because the coupling to a fast fermion line is suppressed by the emitted energy, and \(x_{3}^{2}\to0\) in the same limit. The observed events pile up at the soft edge, and that alone excludes a scalar.

Collinear. As \(x_{1}\to1\) at fixed \(x_{2}\) the gluon becomes collinear with the antiquark, and Equation (102.68) reduces to \(\left(\alpha_{s}/2\pi\right)C_{F} \left(1+x_{2}^{2}\right)/\left[(1-x_{1})(1-x_{2})\right]\), whose \(x_{2}\) dependence is the splitting function Equation (102.58) with \(z=x_{2}\) — as it must be, since Proposition 102.48 and Equation (102.68) are two views of the same emission. That the same function \(C_{F}(1+z^{2}) /(1-z)\) governs scaling violation in Section 102.5.2 and the three-jet rate here is a consistency requirement of the theory and is tested by the agreement of the two determinations of \(\alpha_{s}\) in Section 102.4.2.

Derives Theorem 102.60. The area law itself is Theorem 106.72. The second sentence is the point at issue and is established by a counterexample. Compact \(\U(1)\) lattice gauge theory — lattice electrodynamics — obeys the same strong-coupling area law, by the same plaquette-tiling argument, since that argument uses only the group integration formula \(\int\dd U\,U=0\) and never the non-abelian structure. But continuum electrodynamics manifestly does not confine: the Coulomb potential of Theorem 100.25 is measured every day. The resolution is that compact \(\U(1)\) in four dimensions has a genuine phase transition at intermediate \(\beta\) separating a confining strong-coupling phase from a Coulomb weak-coupling phase, and the continuum limit is taken in the latter. Hence a strong-coupling area law establishes nothing about the continuum theory unless one also knows that no such transition intervenes between \(\beta\ll1\) and the continuum limit \(\beta\to\infty\) demanded by Remark 106.69. For \(\SU(3)\) the absence of a transition is a numerical finding, not a theorem.

Derivation of the two terms and a numerical test. Derives Phenomenon 102.62. The Coulomb term is one-gluon exchange between a colour source in \(\vect{3}\) and one in \(\bar{\vect{3}}\) coupled to a singlet. The colour factor is \(\avg{T_{a}\otimes T_{a}}_{\vect{1}} =\tfrac{1}{2}\left[C_{2}(\vect{1})-2C_{2}(\vect{3})\right] =-C_{F}\), using \(C_{2}(\vect{1})=0\) and Equation (14.77); the sign is attraction, and the magnitude is \(4/3\) times the electromagnetic case with \(\alpha\to\alpha_{s}\). Note that the same computation for a \(\vect{3}\otimes\vect{3}\) pair coupled to \(\bar{\vect{3}}\) gives \(-C_{F}/2\), attractive but weaker, and for the symmetric \(\vect{6}\) gives \(+C_{F}/4\), repulsive: colour singlets are the most tightly bound configurations available, which is Postulate 102.11 made plausible.

The linear term is not derived here — it is measured on the lattice and inferred from the spectrum — but it has a check that costs one page and is worth doing, because it connects Equation (102.69) to a completely independent body of data. Model the confining flux as a relativistic string of tension \(\sigma\) with massless ends rotating rigidly at the speed of light. If the string has total length \(2R\), the transverse speed at distance \(r\) from the centre is \(v=cr/R\), and

using \(\int_{0}^{1}u^{2}(1-u^{2})^{-1/2}\dd u=\pi/4\). Eliminating \(R\) and writing \(E=Mc^{2}\) and \(J=\hbar j\),

a linear Regge trajectory: spin proportional to squared mass, with a slope fixed by the string tension alone. The name records Regge's poles in complex angular momentum [Regge:1959]; that the observed hadronic trajectories are in fact linear is an empirical statement about the spectrum, tested on the \(\rho\) data below [Navas:2024]. Numerically, \(\sigma=1.442\times 10^{5}\,\mathrm{N}\) gives \(j=0.896\) for \(Mc^{2}=1\,\mathrm{GeV}\): by Equation (102.71) the slope of \(j\) in the squared rest energy \((Mc^{2})^{2}\) is \(1/(2\pi\sigma\hbar c)=3.49\times 10^{19}\,/\mathrm{J}^{2}\), that is \(0.896\,/\mathrm{GeV}^{2}\) in the customary units. The measured \(\rho\) trajectory — \(\rho(775)\) with \(j=1\), \(a_{2}(1320)\) with \(j=2\), \(\rho_{3}(1690)\) with \(j=3\) [Navas:2024] — has slope \(0.882\,/\mathrm{GeV}^{2}\) between the first two states and \(0.895\,/\mathrm{GeV}^{2}\) between the second and third. Agreement to about one percent, from a string tension measured in an entirely different way, and with no adjustable parameter: the linearity of Equation (102.69) and the linearity of the observed meson trajectories are the same fact.

Derivation. Derives Phenomenon 102.64. The arithmetic is immediate from Table 102.1: \(2m_{u}+m_{d}=9.02\,\mathrm{MeV}/c^{2}\) against \(938.272\,\mathrm{MeV}/c^{2}\), so \(99\,\mathrm{\%}\) of the mass is not quark mass. What has to be explained is not the deficit but the presence of a scale at all, and that is Phenomenon 102.39: in the limit \(m_{u},m_{d}\to0\) the Lagrangian Equation (102.29) contains no dimensionful parameter, yet the quantum theory generates \(\Lambda_{\mathrm{QCD}}\) through Equation (102.45), and every mass in the spectrum is a pure number times \(\Lambda_{\mathrm{QCD}}/c^{2}\). The proton mass is therefore predicted up to that one overall scale, which is fixed by measuring any single hadron mass; everything else is then a prediction with no freedom.

Turning that into numbers is a computation and not a derivation, and the distinction is worth keeping. The lattice programme [Duerr:2008] proceeds by: choosing the bare quark masses and \(\beta\); computing hadron correlators by Monte Carlo evaluation of Equation (106.81) with dynamical quarks; setting the scale by matching one measured quantity, so that \(a\) is known in metres; tuning the light quark masses by matching two more, typically \(m_{\pi}\) and \(m_{K}\); extrapolating to zero spacing, to infinite volume and, where necessary, to the physical light quark masses; and then predicting the remaining spectrum. Three inputs, and the masses of the nucleon, the \(\Delta\), the \(\Lambda\), the \(\Sigma\), the \(\Xi\) and the \(\Omega\) come out right to a few percent. That is the strongest single piece of evidence that Equation (102.29) is the correct theory of the strong interaction, and it is the reason this chapter can call confinement an observed regularity of a known Lagrangian rather than a mystery.

Derivation of the string-breaking mechanism. Derives Phenomenon 102.65. Take Equation (102.69) at face value and try to separate a quark from an antiquark. At separation \(r\) the stored energy is \(\sigma r\), and it grows without bound. But the vacuum is not empty: as soon as \(\sigma r\) exceeds the energy needed to create a light quark–antiquark pair, roughly twice the constituent mass Equation (102.12), it is energetically favourable for the flux tube to break, the new pair capping the two broken ends. The critical separation is

which is smaller than a hadron. So the attempt to isolate a quark produces, at a separation of less than one hadron radius, two hadrons instead: what would have been a free colour source is always screened by a newly created one before it can be separated. This is why the null results above are expected, and it is also why they cannot be strengthened into an observation of the mechanism: the mechanism predicts precisely that nothing anomalous is ever seen.

Derives Proposition 102.67. \(\gamma^{5}\) anticommutes with every \(\gamma^{\mu}\), so \(P_{L}\gamma^{\mu}=\gamma^{\mu}P_{R}\) and \(\bar{q}_{L}\gamma^{\mu}q_{R} =q^{\dagger}P_{L}\gamma^{0}\gamma^{\mu}P_{R}q =q^{\dagger}\gamma^{0}P_{R}\gamma^{\mu}P_{R}q=0\) because \(P_{R}\gamma^{\mu}P_{R}=\gamma^{\mu}P_{L}P_{R}=0\); the cross terms in the kinetic piece therefore vanish, which is the first half of Equation (102.73). For the mass term the same algebra keeps only the cross terms, since \(\bar{q}_{L}q_{L}=0\). The covariant derivative Equation (102.21) contains no \(\gamma^{5}\) and acts only on colour, so it commutes with the flavour rotations \(q_{L}\to U_{L}q_{L}\), \(q_{R}\to U_{R}q_{R}\); with the mass matrix set to zero nothing else connects the two chiralities, giving Equation (102.74). Splitting \(\U(n_{f})=\SU(n_{f})\times\U(1)\) on each side and recombining the two \(\U(1)\)s into vector (\(U_{L}=U_{R}\)) and axial (\(U_{L}=U_{R}^{\dagger}\)) combinations names the last two factors. The vector one is the conservation of baryon number. The axial one is not a symmetry of the quantum theory at all: the measure of the functional integral is not invariant under it, which is Theorem 106.87, and the classical current has an anomalous divergence proportional to the gluon topological density.

Derivation. Derives Phenomenon 102.68. If a symmetry generator \(Q\) annihilates the vacuum then it commutes with the Hamiltonian and maps energy eigenstates to degenerate energy eigenstates. The axial generators \(Q_{5}^{a}\) change parity, because the axial current is a pseudovector; so \(Q_{5}^{a}\ket{h}\) would be a state of the same mass and opposite parity to \(\ket{h}\), for every hadron. The splittings quoted are not small corrections to such a degeneracy: they are of order \(\Lambda_{\mathrm{QCD}}\), which is the largest scale available, whereas the explicit breaking by the quark masses is of order \(7\,\mathrm{MeV}\). Symmetry breaking by the Lagrangian cannot produce an effect a hundred times its own size. Hence the axial generators do not annihilate the vacuum, while the vector ones evidently do — the isospin multiplets are degenerate to the accuracy at which electromagnetic and \(m_{d}-m_{u}\) effects enter. The symmetry is spontaneously broken, in the pattern

Derives Theorem 102.70. Write \(v=\avg{\comm{Q_{a}}{\Phi(0)}}\neq0\) and insert a complete set of states between the operators:

Translation invariance gives \(\avg{0|j^{0}_{a}(x)|n}=\avg{0|j^{0}_{a}(0)|n} \ee^{-\ii p_{n}\cdot x/\hbar}\), so the integral over \(\dd^{3}x\) produces \((2\pi\hbar)^{3}\delta^{3}(\vect{p}_{n})\): only states of zero three-momentum contribute. For those the remaining \(x\) dependence is \(\ee^{-\ii E_{n}t/\hbar}\), and therefore

But \(Q_{a}\) is time independent, because \(\pp_{\mu}j^{\mu}_{a}=0\) and the spatial part integrates to a surface term; so \(v\) cannot depend on \(t\). Since \(v\neq0\), at least one term must survive, and a surviving term must have \(E_{n}=0\) at \(\vect{p}_{n}=0\). A state with zero energy at zero momentum is a massless particle, and by construction it has a non-vanishing matrix element with the current. Its quantum numbers are those of the current: for the axial current \(A^{\mu}_{a}\), which is a pseudovector, the state must be a pseudoscalar, since \(\avg{0|A^{\mu}_{a}|\pi_{b}}\) can only be proportional to \(p^{\mu}\) and \(p^{\mu}\) is the only vector available for a spinless state. Applying this to Equation (102.75): three generators are broken — \(\SU(2)_{L}\times\SU(2)_{R}\) has six and \(\SU(2)_{V}\) has three — so there are three massless pseudoscalars, an isotriplet.

Derivation. Derives Phenomenon 102.72. Theorem 102.70 supplies three massless pseudoscalars in the limit \(m_{u}=m_{d}=0\), and Phenomenon 102.68 identifies the broken symmetry as the one whose Goldstone bosons they are. The pions are those states, made massive only by the explicit breaking. It remains to find how their mass depends on the breaking, and the answer is the Gell-Mann–Oakes–Renner relation [GellMann:1968]:

where \(F_{\pi}\) is the pion decay constant of Equation (106.103), an energy, and every factor is written so that both sides carry (energy)\(^{4}\): \(\avg{\bar{q}q}\) carries \(/\mathrm{m}^{3}\) by Definition 102.69 and \(\left(\hbar c\right)^{3}\) converts it.

The derivation is a Ward identity. The divergence of the isotriplet axial current, exact in QCD, is \(\pp_{\mu}A^{\mu}_{a}=\left(m_{u}+m_{d}\right)c\,P_{a}\) with \(P_{a}=\ii\bar{q}\gamma^{5}\tau_{a}q\) the pseudoscalar density; it vanishes in the chiral limit, which is Proposition 102.67. Take its matrix element between the vacuum and a one-pion state. The left-hand side is fixed by the definition of the decay constant, \(\avg{0|A^{\mu}_{a}|\pi_{b}(p)}=\ii F_{\pi}p^{\mu}\delta_{ab}/c\), so that \(\avg{0|\pp_{\mu}A^{\mu}_{a}|\pi_{b}} =F_{\pi}\left(m_{\pi}c\right)^{2}\delta_{ab}/\hbar c\) on using \(p^{2}=m_{\pi}^{2}c^{2}\). The right-hand side is \(\left(m_{u}+m_{d}\right)c\avg{0|P_{a}|\pi_{b}}\), and the remaining matrix element is related to the condensate by the same chiral rotation that produced the Goldstone boson: acting with the broken charge on the pseudoscalar density returns the scalar density, so \(\avg{0|P_{a}|\pi_{b}}\) is proportional to \(\avg{\bar{q}q}/F_{\pi}\). Assembling the factors gives Equation (102.77). The structural point survives any bookkeeping: \(m_{\pi}^{2}\), not \(m_{\pi}\), is linear in the quark masses, because the quark mass enters the Lagrangian linearly and the pion mass enters the Lagrangian squared.

Derivation of the size of the jump. Derives Phenomenon 102.75. The energy density of an ideal ultrarelativistic gas is the standard blackbody result, generalized to \(g\) internal states and to fermions by Fermi–Dirac counting:

the factor \(7/8\) being the ratio of the Fermi to the Bose integral \(\int_{0}^{\infty}x^{3}\dd x/(\ee^{x}\pm1)\). Counting: a gluon has \(8\) colours and \(2\) polarizations, so \(g_{B}=16\); a quark flavour has \(3\) colours, \(2\) spins, and particle and antiparticle, so \(g_{F}=12\) per flavour and \(g_{F}=36\) for the three light flavours. Hence \(g_{\mathrm{eff}}=16+\tfrac{7}{8}\cdot36=47.5\). Below the transition the degrees of freedom are three pions, \(g_{\mathrm{eff}}=3\). The ratio is nearly sixteen, and that factor — not the precise value of \(T_{c}\) — is what the lattice sees as a steep rise in \(\varepsilon/T^{4}\).

Numerically at \(k_{B}T=155\,\mathrm{MeV}\), Equation (102.81) gives \(\varepsilon\approx1.2\,\mathrm{GeV}\) per cubic femtometre, that is

against about \(0.4\,\mathrm{GeV}\) per cubic femtometre inside a nucleon at rest. The lattice value near \(T_{c}\) is some tens of percent below the ideal-gas limit Equation (102.81), and the deficit does not go away quickly as the temperature rises: the medium is not a weakly coupled gas, which is exactly what the heavy-ion data below also say.

Derives Proposition 102.76. Power counting (Theorem 100.17) allows operators of mass dimension at most four. The gauge-invariant, Lorentz-invariant operators of dimension four that can be built from \(F^{a}_{\mu\nu}\) are \(F^{a}_{\mu\nu}F^{a\,\mu\nu}\), already in Equation (102.29), and \(\epsilon^{\mu\nu\rho\sigma}F^{a}_{\mu\nu}F^{a}_{\rho\sigma}\), which is Equation (102.83); from the quark fields they are the kinetic and mass terms. There is no other. The identification of Equation (102.83) with the topological density, its periodicity in \(\theta\), and its origin in the vacuum structure are Corollary 106.84 and Section 106.6.2. The statement that only \(\bar{\theta}\) is observable is Equation (105.111): it ties the gluon sector to the phases of the quark mass matrix, which come from a different sector of the Standard Model altogether.

Derivation. Derives Phenomenon 102.77. The step that converts a measured bound on a dipole moment into a bound on a Lagrangian parameter is the chiral estimate Equation (105.112), derived in Proposition 105.45: \(d_{n}\approx\bar{\theta}\times 1.6\times 10^{-37}\text{–}6.4\times 10^{-37}\,\mathrm{C}\,\mathrm{m}\), the parametric form following from the requirement that the answer vanish with the quark masses — because a massless quark makes \(\theta\) unobservable by the chiral rotation of Proposition 102.76 — and the coefficient from a chiral loop. Combining with the ultracold-neutron measurement [Abel:2020] gives \(\abs{\bar{\theta}}\lesssim10^{-10}\), which is Equation (105.113). The uncertainty of the estimate is a factor of a few, which is why the bound is quoted as an order of magnitude.