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An asymptotically free theory has a running coupling that tends to zero at arbitrarily high energy. Its negative weak-coupling beta function makes short-distance perturbation theory increasingly accurate.

In Quantum chromodynamics, this means that a short-distance probe can resolve quarks with a small QCD coupling, even though isolated quarks are absent from the observed hadron spectrum. This made asymptotic freedom a bridge between the approximately free constituents of the parton model and an interacting quantum field theory. Asymptotic freedom supplies control at large momentum transfer; it does not itself prove confinement.^[ref]

Use the four-dimensional QCD coupling convention ${\alpha }_s=g_s^2/(4\pi )$. For the SU(3) gauge group and $n_f$ active quark flavors in the fundamental representation, the one-loop order beta function isHere $n_f$ counts quark flavors, not the three values of color charge. The chain rule converts the first equation to the last; changing from $\ln \mu$ to $\ln {\mu }^2$ supplies a factor of two. For ${\beta }_0>0$, the negative sign makes $g_s=0$ an attractive high-energy renormalization-group fixed point.^[ref]

For ${\beta }_0>0$, set the renormalization scale $\mu =Q$, where $Q$ is a positive energy scale. The one-loop running of the strong coupling then givesThe strong-coupling scale $\Lambda$ has the same units of measurement as $Q$, so the logarithm has a dimensionless argument. This is a fixed-$n_f$ approximation on the positive branch $Q>\Lambda$; controlled perturbative quantum field theory requires a small QCD coupling and suitably small omitted corrections, not merely a positive denominator. Heavy quark masses require matching the QCD scale across a quark threshold, changing $n_f$ and the associated $\Lambda$. Both the renormalization scheme and the loop order matter when quoting a numerical $\Lambda$.

A uniform incoherent initial mixture of two independently prepared fundamental color charges requires an average over their $N^2$ initial possibilities, together with a sum over unobserved final color charges. This differs from preparing a coherent color singlet. For quark-antiquark annihilation through one vector in the adjoint representation, $c_s=T_{ji}^a{T}_{kl}^a$ gives $N^{-2}\sum |c_s{|}^2=(N^2-1)/(4N^2)$. For the exchange contraction $c_t=T_{ki}^a{T}_{jl}^a$, $N^{-2}\sum c_s{c}_t^{\ast }=-(N^2-1)/(4N^3)$ by the SU(N) fundamental completeness relation. The negative overlap reverses the sign of the corresponding Abelian interference term.

The group-theoretic tensor multiplying the kinematic part of a Feynman diagram involving color charge. It is obtained by contracting internal indices among gauge generators, invariant tensors and structure constants of a Lie algebra. For instance, one gluon exchange on a single quark line contributes ${\sum }_a{T}^a{T}^a=C_{F}1$; averaging or summing the squared tensor over external colour labels gives the corresponding color average in a squared scattering amplitude.

A quark-gluon Feynman vertex whose initial and final quark have the same chosen color charge basis label. Such vertices exist: the diagonal Gell-Mann matrices act nontrivially on the red basis vector. Their gluons are traceless colour-octet combinations, not colour singlets. A ninth singlet generator proportional to the identity is absent from SU(3).

For distinct incoming and outgoing quark flavours, tree-level $q\bar{q}\to Q\bar{Q}$ has one gluon annihilation graph. Averaging the initial colours gives $2/9$, and averaging initial spins gives $1/4$. The outgoing fermion spin sum must use distinct particle and antiparticle projectors. For final mass $M$, $s+t+u=2M^2$ and the displayed result includes all outgoing colour and spin sums. Its amplitude squared remains finite at threshold even though the cross section vanishes through the final-state velocity.

A right-handed neutrino is a singlet of the entire Standard Model gauge group. Its same-chirality bilinear with the field obtained by charge conjugation therefore admits the displayed renormalizable Majorana mass term, in addition to a Higgs-induced Dirac mass term. It violates lepton number by two units. Analogous terms for electrically charged Standard Model fermions carry nonzero electric charge; quarks also have nontrivial color charge. The presence of a right-handed field permits mass but does not force a nonzero mass parameter.

A lepton is a Standard Model fermion without color charge. The charged leptons have electric charge $-1$ and their corresponding neutrinos have charge zero. There are three generations; their antiparticles carry opposite electric charge and lepton number. Charged leptons and neutrinos have lepton number $+1$, with the corresponding antiparticles carrying $-1$.

For an initial Dirac field of mass $m$ annihilating into a distinct Dirac field of mass $M$ through a vector current, put $\Sigma =m^2+M^2$. With Mandelstam variables satisfying $s+t+u=2\Sigma$, the spin average of the two contracted vector currents is $2F$, where $F=(\Sigma -t{)}^2+(\Sigma -u{)}^2+2s\Sigma =t^2+u^2+4s\Sigma -2{\Sigma }^2$. The gamma-matrix trace proof uses $\mathrm{tr}[(/p_2-m){\gamma }^{\mu }(/p_1+m){\gamma }^{\nu }]=4[p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-(s/2)g^{\mu \nu }]$ and the analogous final tensor. Couplings, the mediator denominator and color charge factors multiply this common kinematic expression.

The two electromagnetic tree-level Feynman diagrams for $q\gamma \to q\gamma$ exchange a quark in the $s$ and $u$ channels. Their sum obeys both photon Ward identities. Averaging over the initial quark spin, color charge and two physical incoming photon polarization vectors gives $-2e^4{Q}_q^4(s/u+u/s)$ after final-state sums. The color average is one because the interaction preserves the single open quark colour index. The interference trace is proportional to $(p_1+p_2)\cdot (p_1-p_4)=(s+t+u)/2=0$ for massless external states.

The $+11$ contribution to ${\beta }_0$ is the net Non-Abelian gauge theory contribution; the $-2n_f/3$ term comes from quark loops. Thus quarks screen color charge, opposing asymptotic freedom. It is the net gluon contribution that produces antiscreening: the inferred QCD coupling decreases toward shorter distances. Unlike the electrically neutral photon, a gluon carries color charge and has self-interactions. Describing this as vacuum polarization must not imply that all vacuum contributions have the screening sign of quantum electrodynamics.^[ref]

Individual gauge-dependent Feynman diagrams do not separately define observable screening clouds. A proof of the coefficient sums the gauge fixing, Faddeev-Popov ghost and matter contributions and extracts the charge renormalization; the final one-loop order coefficient is independent of the gauge choice. The beta function is negative near zero for integer $0\le n_f\le 16$. This is a statement about weak coupling, not a determination of the low-energy particle spectrum.^[ref]

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

There are two one-loop Feynman diagrams: the two orders of the labelled photons around a top quark fermion loop, with one Yukawa interaction insertion for the outgoing Higgs boson. They are added, rather than identified by a Feynman-diagram symmetry factor. Each has the same color charge trace $N_c=3$.

In the first routing, traverse the fermion loop from the $p_2$ insertion to the $p_1$ insertion with momentum $l$, then to the Higgs boson insertion with momentum $l+p_1$, then back to the $p_2$ insertion with momentum $l-p_2$. Conservation at the Higgs boson insertion gives $p_h=p_1+p_2$. The second routing exchanges $(p_1,\mu )$ and $(p_2,\nu )$: its three internal momenta are $l$, $l+p_2$ and $l-p_1$. Reversing the orientation is another way to represent this second ordering. Two vector insertions give no Furry theorem cancellation.

Both $s$-channel annihilation and $t$-channel gluon exchange are now allowed. Their spin sum factors are the Bhabha scattering factors with $e\mapsto g$. Their color charge tensors, with the external indices of Solution, areEach squared term has $K_{ss}=K_{tt}=(N_c^2-1)/(4N_c^2)$. The cross term instead has the color averageTo evaluate it, use the SU(N) fundamental completeness relationThis follows by completing the orthogonal basis of traceless Hermitian matrices with the identity. Since $\mathrm{tr}{T}^b=0$, it gives ${\sum }_a{T}^a{T}^b{T}^a=-T^b/(2N_c)$, and henceMultiplying the two squared Bhabha scattering terms by $K_{ss}$ and its cross term by $K_{st}$ givesFor the SU(3) gauge group, this isThe negative color charge overlap reverses the sign of the QED interference. Since $st<0$, this cross contribution is positive in the physical region. No additional spinor calculation is needed; all the new information is in the SU(N) fundamental completeness relation.

The two Bhabha scattering tree-level Feynman diagrams are the two panels of Solution with a photon mediator: $s$-channel annihilation with internal momentum $p_1+p_2$, and $t$-channel exchange with internal momentum $p_1-p_3$. Each interaction vertex preserves the electron species. There is no independent $u$-channel Feynman diagram, since the outgoing electron and Positron are distinguishable.

Removing the color charge factor from Solution and replacing $g$ by $e$ givesCrossing symmetry interchanges $s$ and $t$ for the exchange Feynman diagram, givingSquaring the sum of the two scattering amplitudes also produces the interference term $2\mathrm{Re}\overline{{\mathcal{M}}_s{\mathcal{M}}_t^{\ast }}$. The given third term must be this cross term: it has both propagator denominators $st$, whereas a squared individual Feynman diagram has $s^2$ or $t^2$. With the correct relative fermionic sign, its value is $4e^4{u}^2/(st)$. ThusThe coefficient of the last displayed fraction is positive, but $s>0$ and $t<0$ in the physical scattering region, so this interference is destructive. The relative sign originates in the two possible external fermion contractions; changing an external-state phase changes both scattering amplitudes together.^[ref]

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

Continue with $\eta =-g$ and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors areContracting givesLet $u=2M^2-s-t$. The on-shell products are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yieldsThe result is already averaged over incoming colours and spins and summed over outgoing ones. For $M\to 0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.

An additional check uses the centre-of-mass angle and velocity ${\beta }_t=\sqrt{1-4M^2/s}$:It tends to $4g_s^4/9$ at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t{)}^2+(M^2-u{)}^2-2M^{2}s]/(9s^2)$, displaying the incorrect mass contribution explicitly.

The different-flavour quark annihilation into a massive pair has only an $s$-channel gluon graph at tree level when the two quark flavours differ. Its amplitude contains one incoming and one outgoing colour current, with common magnitude factor $g_s^2/s$. On-shell current conservation removes the gauge-parameter-dependent longitudinal part of the gluon propagator. To display the Dirac signs transparently, evaluate spin traces with the auxiliary mostly-minus metric $\eta =-g$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$; the invariants $s,t$ are then the usual positive-energy Mandelstam invariants, numerically identical to those defined in the paper.

The incoming colour average and outgoing colour sum giveAveraging over the two spins of each incident particle supplies an additional $1/4$. The physical spin sums are $\sum u\bar{u}=/p+M$ and $\sum v\bar{v}=/p-M$. ThereforeThe coefficient is equivalently $8{\pi }^2{\alpha }_s^2/(9s^2)$, with ${\alpha }_s=g_s^2/(4\pi )$.

The two top projectors cannot both contain $-M$. The printed expression has that error. It is not repaired by changing the metric convention consistently: a particle and antiparticle still require the appropriate distinct projectors. In the auxiliary convention, its outgoing tensor would obey $q_{\mu }{H}_{\mathrm{printed}}^{\mu \alpha }=8M^2{q}^{\alpha }$, rather than zero for $q=p_1+p_2$. At threshold it gives a zero contracted trace, while the correctly averaged vector-current production amplitude squared has a nonzero limit $4g_s^4/9$. The phase-space cross section still vanishes at threshold because the final velocity tends to zero. The colour and spin prefactor above is consequently the prefactor for the corrected trace; no overall constant can make the misprinted massive trace the physical answer for all kinematics.

The exchanged particle is now a gluon, and the incoming up quark mass replaces $m_e$. DefineThe Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. Averaging over the $N_c^2$ initial color charge choices and summing the final color charge choices givesReplace the factor $N_c{e}^4{Q}_e^2{Q}_t^2$ from part (ii) by $g_s^4{C}_g$. ThereforeKeeping $m_u$ is allowed; if the usual negligible up quark mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial quark color charge states are averaged, whereas final ones are summed.

Let $\Sigma =m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude isThere are two initial spin angular momentum states for each particle, so the spin average is $1/4$. The final color charge sum gives ${\sum }_{c,d}|{\delta }_{cd}{|}^2=N_c$. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities giveHere $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting givesSince $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma -t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma -u$, this is the massive vector-current annihilation kernelThus the absolutely normalized result isAfter taking out the displayed prefactor, $\boxed{A=1,B=4,C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

There are two tree-level Feynman diagrams, with intermediate photon and gluon respectively. Their color charge tensors are ${\delta }_{ji}{\delta }_{kl}$ and ${\sum }_A{T}_{ji}^A{T}_{kl}^A$. The pure photon factor isTheir interference vanishes after the color charge sum because of photon-gluon colour orthogonality:Thus there is no term proportional to $e^2{g}_s^2$, and with the $F_u$ of part (iii),Here $Q_u=Q_t=2/3$. The absence of interference concerns the specified averaged and summed result; a fixed choice of external color charge need not have zero interference.