two translations $P_x$ and $P_y$ do not commute, i.e. $$\{P_x,P_y\} = e B.$$ How can these standard symmetry considerations be related to an accidental symmetry due to a Runge-Lenz vector? The Runge-Lenz vector is familiar from the Kepler problem. It points from the center of force to the perihelion position, and is conserved because all bound classical orbits are closed. Similarly, the orbit of a charged particle in a constant magnetic field is a closed circle with a fixed center. Indeed, in this case the position of this center plays the role of the conserved Runge-Lenz vector and is given by $$\begin{aligned}
\label{RP}
&&R_x = x - \frac{v_y}{v} r = x - \frac{v_y}{\omega} = 
x - \frac{1}{M \omega}\left(p_y + e B x\right) = 
- \frac{p_y}{M \omega} = - \frac{P_y}{e B}, \nonumber \\
&&R_y = y + \frac{v_x}{v} r = y + \frac{v_x}{\omega} = 
y + \frac{p_x}{M \omega} = \frac{P_x}{e B}.\end{aligned}$$ Interestingly, the position $(R_x,R_y)$ of the center of the cyclotron circle is, at the same time, proportional to $(- P_y,P_x)$, i.e. it is orthogonal to the generators of spatial translations. Consequently, we can write $$\label{poisson1}
\{R_x,P_x\} = - \frac{1}{e B} \{P_y,P_x\} = 1, \
\{R_y,P_y\} = \frac{1}{e B} \{P_x,P_y\} = 