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
$$
e^{i\theta_{ff}}=
e^{i\theta_{ee}}e^{i\theta_{mm}}e^{2i\theta_{em}}=-1.
$$
The <topological spin> obeys $e^{2\pi i s_f}=e^{i\theta_{ff}}$, so
$$
s_f=\frac12\pmod 1.
$$
This is precisely the two-dimensional spin-statistics relation for a fermion, so the result is consistent.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-342/1/b/solution