 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/6/c/solution

P = a^{†} b^{†}

and

Q = ab

[Q, P] = N_{a} + N_{b} + 1 = 2 K_{3}

, while

[K_{3}, P] = P

and

[K_{3}, Q] = - Q

. Substituting

K_{1} = (P + Q) /2

and

K_{2} = (P - Q) / (2 i)

yields

[K_{1}, K_{2}] = - i K_{3}, [K_{2}, K_{3}] = i K_{1}, [K_{3}, K_{1}] = i K_{2} .

Therefore

[i K_{1}, i K_{2}] = i K_{3}, [i K_{2}, i K_{3}] = - i K_{1}, [i K_{3}, i K_{1}] = - i K_{2},

S U (2)

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