For one generation, the <Standard Model representation> content under $SU(2)_L\times SU(3)_C$ is
$$
Q_L=(u_L,d_L):(\mathbf2,\mathbf3),\quad
u_R:(\mathbf1,\mathbf3),\quad d_R:(\mathbf1,\mathbf3),
$$
$$
L_L=(\nu_{eL},e_L):(\mathbf2,\mathbf1),\quad
e_R:(\mathbf1,\mathbf1),\quad
H:(\mathbf2,\mathbf1),
$$
with 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 $n_s=0$. Therefore
$$
\boxed{\beta_3=\frac{11}{3}(3)-\frac13(6+6)=7.}
$$
For $SU(2)_L$, 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
$$
\boxed{\beta_2=\frac{11}{3}(2)-\frac13(12)-\frac16=\frac{19}{6}.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-305/4/e/solution