 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2016/paper-1/2/d/solution

Use

ϵ_{x yz} = + 1

and the supplied commutators. Antisymmetry also gives

[K_{a}, L_{b}] = ϵ_{ab c} K_{c}

. Expanding the combinations,

[L_{a}^{\pm}, L_{b}^{\pm}] = \frac{1}{4} ([L_{a}, L_{b}] + [K_{a}, K_{b}] \pm [L_{a}, K_{b}] \pm [K_{a}, L_{b}]) = ϵ_{ab c} L_{c}^{\pm} .

The mixed commutator cancels:

[L_{a}^{+}, L_{b}^{-}] = \frac{1}{4} ([L_{a}, L_{b}] - [L_{a}, K_{b}] + [K_{a}, L_{b}] - [K_{a}, K_{b}]) = 0.

Thus

[L_{a}^{+}, L_{b}^{+}] = ϵ_{ab c} L_{c}^{+}, [L_{a}^{-}, L_{b}^{-}] = ϵ_{ab c} L_{c}^{-}, [L_{a}^{+}, L_{b}^{-}] = 0.

There are two mutually commuting three-generator rotation algebras. As real compact Lie algebras,

so (4) ≅ su (2) \oplus su (2) .

- i σ_{a} /2

obeys the same commutation convention in each

su (2)

factor. This is an algebra statement. Globally, the simply connected group is

Spin (4) ≅ S U (2) \times S U (2)

, and

SO (4) ≅ \frac{S U ( 2 ) \times S U ( 2 )}{{( I , I ) , ( - I , - I )}} .

Dropping that central quotient would incorrectly identify the global groups.

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