Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 1 / b

Table of contents
- Solution b

Solution

The composite is

f = e \times m

. Exchanging two composites exchanges the two

e

constituents and the two

m

constituents and produces one mutual winding between unlike constituents. The identical-particle phases are both

+ 1

, while the mutual contribution is

- 1

, hence

e^{i θ_{ff}} = e^{i θ_{ee}} e^{i θ_{mm}} e^{2 i θ_{e m}} = - 1.

The topological spin obeys

e^{2 πi s_{f}} = e^{i θ_{ff}}

, so

s_{f} = \frac{1}{2} (mod 1) .

This is precisely the two-dimensional spin-statistics relation for a fermion, so the result is consistent.

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