Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 4

Since ${\alpha }_i=g_i^2/(4\pi )$, the stated renormalization-group beta function impliesFor ${\beta }_i>0$, the running coupling decreases at high energy: the theory is asymptotically free and becomes strong in the infrared, as in Quantum chromodynamics. For ${\beta }_i<0$, it is infrared free but grows toward an ultraviolet Landau pole.

Integrating the renormalization-group beta function and defining $\Lambda$ by ${\alpha }_3^{-1}(\Lambda )=0$ givessoThe scale $\Lambda$ is the QCD scale in this one-loop approximation.

Direct integration between the two renormalization scales gives

Put $L=(2\pi {)}^{-1}\log (M_{\mathrm{GUT}}/m_Z)$. At the gauge coupling unification scale, write the common normalized coupling as ${\alpha }_U={\alpha }_2(M_{\mathrm{GUT}})={\alpha }_3(M_{\mathrm{GUT}})=(5/3){\alpha }_1(M_{\mathrm{GUT}})$. Running downward givesSubtracting the second equation from the third determinesEliminating $L$ from the first two equations yields

For one generation, the Standard Model representation content under $SU(2{)}_L\times SU(3{)}_C$ iswith no right-handed neutrino. For $SU(3{)}_C$, three generations give $n_L=6$ from the two left-handed quark flavors and $n_R=6$ from the two right-handed quark flavors, while $n_s=0$. ThereforeFor $SU(2{)}_L$, each generation supplies three colored quark doublets and one lepton doublet, so $n_L=12$, $n_R=0$, and the single complex Higgs doublet gives $n_s=1$. Hence