Past exam of the theoretical physics II course of the University of Cambridge / 2017 / 4 / d / Solution

For $m\ge 0$, expand $e^{i\rho \cos \theta }=\exp [i\rho (e^{i\theta }+e^{-i\theta })/2]$. The lowest power producing the Fourier harmonic $e^{im\theta }$ is $(i\rho /2{)}^m/m!$. Meanwhile $J_m(\rho )\sim (\rho /2{)}^m/m!$. Matching these lowest powers gives $a_m=i^m$.

For negative order use $J_{-m}=(-1{)}^m{J}_m$ and the equality of the positive and negative Fourier coefficients of this even function of $\theta$. This gives $a_{-m}=(-1{)}^m{i}^m=i^{-m}$. Therefore the Jacobi-Anger expansion isThe cosine form has multiplicity one for $m=0$ and two for every positive $m$. Those multiplicities are essential for matching the incident plane wave.