Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 6 / d

Table of contents
- Solution d

Solution

The pair products are

PQ = N_{a} N_{b}

and

QP = (N_{a} + 1) (N_{b} + 1)

. Hence

K_{1}^{2} + K_{2}^{2} = \frac{1}{2} (PQ + QP) = N_{a} N_{b} + \frac{1}{2} (N_{a} + N_{b} + 1),

whereas

K_{3} = (N_{a} + N_{b} + 1) /2

. Subtraction gives

C = \frac{( N _{a} - N _{b} ) ^{2} - 1}{4} .

The number difference commutes with pair creation, pair annihilation and

K_{3}

, since each pair operation changes both occupations equally. Its boson-pair Casimir operator

C

therefore commutes with every generator. In a fixed-difference sector

N_{a} - N_{b} = d

, the Casimir eigenvalue is

(d^{2} - 1) /4

, including

- 1/4

for equal occupations.

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