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

Solution

For one generation, the Standard Model representation content under

S U (2)_{L} \times S U (3)_{C}

Q_{L} = (u_{L}, d_{L}) : (2, 3), u_{R} : (1, 3), d_{R} : (1, 3),

L_{L} = (ν_{e L}, e_{L}) : (2, 1), e_{R} : (1, 1), H : (2, 1),

S U (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

β_{3} = \frac{11}{3} (3) - \frac{1}{3} (6 + 6) = 7.

For

S U (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

β_{2} = \frac{11}{3} (2) - \frac{1}{3} (12) - \frac{1}{6} = \frac{19}{6} .

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