[H,L] = 0$, with the Runge-Lenz vector and the angular momentum operator given by $$\begin{aligned}
&&R_x = - \frac{P_y}{e B} = \frac{i {\partial}_y}{e B}, \ 
R_y = \frac{P_x}{e B} = y - \frac{i {\partial}_x}{e B}, 
\nonumber \\
&&L = x \left(- i {\partial}_y + \frac{e B x}{2}\right) - 
y \left(- i {\partial}_x + \frac{e B y}{2}\right).\end{aligned}$$ As in the classical case, the radius of the cyclotron orbit squared is given by $$r^2 = (x - R_x)^2 + (y - R_y)^2 = \left(x - \frac{i {\partial}_y}{e B}\right)^2
- \frac{ {\partial}_x^2}{e^2 B^2} = \frac{2 H}{M \omega^2},$$ and is thus again a conserved quantity. In particular, we can express the Hamiltonian as $$H = \frac{1}{2} M \omega^2 r^2.$$ Remarkably, although the two coordinates $R_x$ and $R_y$ of the center of the cyclotron circle are not simultaneously measurable, its radius $r$ has a definite value in an energy eigenstate.

As it should, under spatial rotations the Runge-Lenz vector $(R_x,R_y)$ indeed transforms as a vector, i.e. $$[L,R_x] = i R_y, \ [L,R_y] = - i R_x.$$ These relations suggest to introduce $$R_\pm = R_x \pm i R_y,$$ which implies $$[L,R_\pm] = \pm R_\pm.$$ Hence, $R_+$ and $R_-$ act as raising and lowering operators of angular 