Since $M^2=nM$, direct multiplication gives the <matrix inverse>
$$
\boxed{(I_n+pM)^{-1}=I_n-\frac p{1+np}M,}
$$
provided $1+np\ne0$, as required by invertibility of the <K-matrix>. For $q=e_J$,
$$
K^{-1}e_J=e_J-\frac p{1+np}\mathbf1.
$$
The attached gauge-flux vector is therefore
$$
\Phi=2\pi\hbar e_J
-\frac{2\pi\hbar p}{1+np}\mathbf1.
$$
Relative to the $K=I_n$ electron in part a, every one of the $n$ gauge fields consequently carries the additional flux
$$
\boxed{\Delta\Phi_I=-\frac{2\pi\hbar p}{1+np}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 342 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-342/1/b/solution