 Solution

The canonical momenta are $p_x=m\dot{x}$, $p_y=m\dot{y}+qBx$, $p_z=m\dot{z}$. The Legendre transform in mechanics givesHamilton's equations areThus $\ddot{x}=\Omega \dot{y}$, $\ddot{y}=-\Omega \dot{x}-qE/m$, and $\ddot{z}=0$, where $\Omega =qB/m$. The signs also follow from $\mathbf{E}=-E{\mathbf{e}}_y$ and $\mathbf{B}=B{\mathbf{e}}_z$.

For $Bq\ne 0$, put $v_d=-E/B$, $a=v_{x0}-v_d$, $b=v_{y0}$. The motion isDifferentiation verifies both equations and all initial conditions. The centre of the circular motion has the electric drift velocity $-E{\mathbf{e}}_x/B$. For $B=0$, take $x=x_0+v_{x0}t$, $y=y_0+v_{y0}t-qE{t}^2/(2m)$ and the same $z$. For $q=0$, the particle moves freely.