Symmetry and Particle Physics, 4
Michaelmas Term 2008, Mathematical Tripos Part III Hugh Osborn Symmetry and Particle Physics, 4 1. The generators of the Lie algebra of SU(3) are Ei± for i = 1, 2, 3 and H1,H2 where [H1,H2] = 0, † [E1+,E2+] = E3+,[E1+,E3+] = [E2+,E3+] = 0 and Ei− = Ei+ . Ei±,Hi obey, for each i, the SU(2) algebra [E+,E−] = H, [H, E±] = ±2E± if H3 = H1 + H2. Also we have [H1,E2±] = ∓E2±, [H2,E1±] = ∓E1±. A state |ψi has weight [r1, r2] if H1|ψi = r1|ψi,H2|ψi = r2|ψi. |n1, n2ihw is a highest weight state with weight [n1, n2] satisfying Ei+|n1, n2ihw = 0. Let (l,k) i l−i k−i |ψi i = (E3−) (E1−) (E2−) |n1, n2ihw . Show that these all have weight [n1 + k − 2l, n2 − 2k + l]. If l ≥ k explain why, in order to construct a basis, it is necessary to restrict i = 0, . , k except if k > n2 when i = k − n2, . , k. For SU(2) generators E±,H obtain n n−1 [E+, (E−) ] = n(E−) (H − n + 1) . From the SU(3) commutators [E3+,E1−] = −E2+,[E3+,E2−] = E1+,[E2+,E3−] = E1− and [E2+,E1−] = [E1+,E2−] = 0 obtain also l−i l−i−1 k−i k−i−1 [E3+, (E1−) ] = − (l − i)(E1−) E2+ , [E3+, (E2−) ] = (k − i)(E2−) E1+ , i i−1 l−i [E2+, (E3−) ] = i E1−(E3−) , [E2+(E1−) ] = 0 . Hence obtain (l,k) (l−1,k−1) (l−1,k−1) E3+|ψi i = i(n1 + n2 − k − l + i + 1)|ψi−1 i − (l − i)(k − i)(n2 − k + i + 1)|ψi i , (l,k) (l−1,k−1) (l−1,k−1) E2+|ψi i = E1− i |ψi−1 i + (k − i)(n2 − k + i + 1)|ψi i .
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