Introduction represents just one possible choice (out of infinitely many) where the vector field is referred to the origin. Since for this problem there is no preferred origin the wave functions will reflect this choice and be gauge-dependent. For this reason we highlighted “[*from zero*]{}” in italics above. An alternative would have been to use $$\mathbf{A}_S=\frac{B}{2}[(x-x_0)\boldsymbol{\hat{y}}-(y-y_0)\boldsymbol{\hat{x}}],
        \label{gauge_as}$$ and then the wave functions would have been different, even though this represents the same problem, i.e. a charged particle in free space in the presence of a uniform magnetic field $B$ (see, for example, Ref. \[\] for a discussion of this point).

Returning to the particular gauge choice given in Eq. (\[gauge\_as\]) with $x_0=y_0=0$, the values of $r_{\ell}$ also reveal that the amount of magnetic flux $\Phi_B$ associated with each wave function in the LLL is quantized in units of the magnetic flux quantum $\Phi_0={h}/{e}$. This is easily seen: consider the magnetic flux $\Phi_B$ between a circle of radius $r_{\ell}$ and that of (smaller) radius $r_{\ell + 1}$ (recall that $\ell \leq 0$, so that $r_{\ell} > r_{\ell +1}$ ): $$\Phi_B=B\pi (r_{\ell}^2-r_{\ell+1}^2) = 2 \pi B \ell_B^2(|\ell| - |\ell+1|) = 2 \pi B\frac{\hbar}{eB}=\Phi_0.
    \label{flux_quantum}$$ Thus, there is exactly a single quantum of 