1, $$ as well as $$\label{poisson2}
\{R_x,P_y\} = - \frac{1}{e B} \{P_y,P_y\} = 0, \
\{R_y,P_x\} = \frac{1}{e B} \{P_x,P_x\} = 0.$$ While eqs.(\[poisson1\]) and (\[poisson2\]) look like the usual Poisson brackets of position and momentum, one should not forget that $R_x$ and $R_y$ are just multiples of $P_y$ and $P_x$, and should hence not be mistaken as independent variables. In particular, one also obtains the relation $$\{R_x,R_y\} = \frac{1}{e B}.$$ Hence, just like the two generators of translations, the $x$- and $y$-components of the Runge-Lenz vector do not have a vanishing Poisson bracket. At the quantum level, this will imply that the $x$- and $y$-components of the center of a cyclotron circle are not simultaneously measurable with absolute precision.

Another conserved quantity is the radius $r$ of the circular cyclotron orbit which can be expressed as $$r^2 = (x - R_x)^2 + (y - R_y)^2 = \frac{1}{M^2 \omega^2} 
\left(p_y + e B x\right)^2 + \frac{p_x^2}{M^2 \omega^2} = 
\frac{2 H}{M \omega^2}.$$ Since $r^2$ is proportional to the energy, it obviously is indeed conserved.

Semi-classical Treatment
------------------------

Next, we consider the same problem semi-classically, i.e. by using Bohr-Sommerfeld quantization, which, in this case, is equivalent to the quantization of angular momentum, 