infinite circular well are the functions $$\Psi_{n,\ell}(r,\phi)=\frac{e^{i\ell\phi}}{\sqrt{2\pi}}\frac{\sqrt{2}}{a}\frac{J_\ell\left(\beta_{n,\ell}\frac{r}{a}\right)}{J_{\ell+1}(\beta_{n,\ell})}.
    \label{circ_basis}$$ Since $J_{-\ell}(r)=(-1)^\ell J_\ell (r)$ then wave functions with opposite angular momentum are degenerate, as expected in the absence of a magnetic field.

[Matrix Mechanics with Bessel Functions]{}
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Matrix mechanics techniques have been utilized for problems like this before, albeit for a study of disordered systems, with a different gauge and confining potential.[@ohtsuki1988numerical] Here, when we include a magnetic field with the Symmetric gauge, we obtain $$H = \frac{\mathbf{p}^2}{2m} + \frac{\omega_c}{2}L_z+\frac{1}{8}m\omega_c^2r^2.
        \label{ham_circ}$$ To solve the Schrödinger equation with this Hamiltonian we follow the usual procedure of expanding in a basis (see, for example, Refs. \[,\]). We use $$\ket{\Psi}=\sum_{n^\prime=1}^{\infty}\sum_{\ell^\prime= -\infty}^{\infty}c_{n^\prime,\ell^\prime} \ket{n^\prime,\ell^\prime}$$ where $\ket{n',\ell'}$ is the ket corresponding to the circular well basis state, Eq. (\[circ\_basis\]). Acting on this state with the Hamiltonian, Eq. (\[ham\_circ\]), and taking the inner product with an arbitrary basis state $\ket{n,\ell}$ results in the matrix equation: $$\sum_{n^\prime=1}^{\infty}\sum_{\ell^\prime=-\infty}^{\infty} H_{n \ell, n^\prime \ell^\prime} c_{n^\prime,\ell^\prime} =E_{n,\ell}c_{n,\ell}
    \label{matrix1}$$ where $$H_{n \ell, n^\prime,\ell^\prime}=\bra{n,\ell}H\ket{n^\prime,\ell^\prime},
\label{hmatrix}$$ and, using Eq. (\[ham\_circ\]), $$H_{n \ell, n^\prime\ell^\prime}= \delta_{n,n^\prime}\delta_{\ell,\ell^\prime}E^{(0)}_{n,l}+\frac{1}{2}\omega_c\bra{n,\ell}L_z\ket{n^\prime,\ell^\prime}+\frac{1}{8}m\omega_c^2\bra{n,\ell}r^2\ket{n^\prime,\ell^\prime}.
  \label{hmatrix2}$$ Here, $E_{n,l}^{(0)}$ is simply the infinite circular well eigenenergy Eq. (\[bess\_eigen\]) and $\delta_{ij}$ is the standard Kronecker delta function. It should now be apparent that the entire matrix is diagonal in 