 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2013/paper-1/5/b/solution

The bosonic commutator is important in

S_{+} S_{-} = ℏ^{2} n_{↑} (n_{↓} + 1)

and

S_{-} S_{+} = ℏ^{2} n_{↓} (n_{↑} + 1)

. Therefore

S^{2} = S_{z}^{2} + \frac{S _{+} S _{-} + S _{-} S _{+}}{2} = \frac{ℏ ^{2}}{4} [(n_{↑} + n_{↓})^{2} + 2 (n_{↑} + n_{↓})] .

On the

N

-particle sector,

S^{2} = ℏ^{2} \frac{N}{2} (\frac{N}{2} + 1), s = \frac{N}{2} .

For example

N = 0, 1, 2

gives spin

0, \frac{1}{2}, 1

. The total number operator can be retained in the same identity before restricting to a sector.

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