the Hamiltonian as a separable product, $\Psi=e^{i\ell\phi}\psi(r)$, where $\hbar \ell$ is the same $L_z$ eigenvalue mentioned earlier and $\ell$ must be an integer. Substituting this into the Schrödinger Equation and introducing the wave vector $k=\frac{\sqrt{2mE}}{\hbar}$ results in $$\frac{d^2\psi}{dr^2}(r)+\frac{1}{r}\frac{d\psi}{dr}(r)+\left(k^2-\frac{l^2}{r^2}\right)\psi(r)=0.
\label{bess_eqn}$$ Equation (\[bess\_eqn\]) is the Bessel equation, with solutions given by $$\psi(r)=AJ_\ell(kr)+BN_\ell(kr)
    \label{bess_fn}$$ where $A$ and $B$ are arbitrary constants, and $J_{\ell}(r)$ and $N_{\ell}(r)$ are the $\ell^{\rm th}$ order Bessel and Neumman functions, respectively. Because the Neumman functions are non-normalizable at the origin, we discard this solution, and apply the boundary condition: $AJ_{\ell}(ka)=0$. The solution is given by $ka=\beta_{n,\ell}$, where $\beta_{n,\ell}$ is the $n$th zero of the $\ell$th order Bessel function. We thus obtain the eigenvalues, $$E_{n,\ell}=\frac{\hbar^2\beta_{n,\ell}^2}{2ma^2},
    \label{bess_eigen}$$ where $n$ is a positive integer that enumerates the zeros for the $\ell^{\rm th}$ order Bessel function. Notice the lack of degeneracy for all $n,\ell$. Bessel functions of the same order are both normalizable and orthogonal for different zeros over $r \in [0,a]$ for all integer $\ell$,[@Weisstein1] $$\int_{0}^aJ_\ell\left(\beta_{n,\ell}\frac{r}{a}\right)J_\ell\left(\beta_{n',\ell}\frac{r}{a}\right)rdr = \delta_{n,n'} \frac{a^2}{2}J_{\ell+1}^2\left(\beta_{n,\ell}\right).
\label{ortho}$$ Because of the orthogonality of $e^{i\ell \phi}$, we need not worry about the orthogonality relations for Bessel functions of different order. These solutions imply that a suitable basis set for the “empty” 