 Solution

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.

For $A=(-y{B}_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot{x}=\Omega y$, where $\Omega =e{B}_0/(mc)$. Since $p_y=m\dot{y}$, its Hamilton equation gives ${\dot{p}}_y=-(e/c)B_0\dot{x}=-m{\Omega }^{2}y$, henceFor nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E^{\prime }/(m{\Omega }^2)}$ and $\sqrt{2m{E}^{\prime }}$. Its enclosed phase-space area is $2\pi E^{\prime }/|\Omega |$, so the action variable isThe absolute value keeps the action positive if the signed charge or magnetic field is negative.

When the field varies slowly along the trajectory, with $|d\Omega /dt|\ll {\Omega }^2$ and no zero-frequency crossing, adiabatic invariance of the action makes $I$ approximately constant. Thus $E^{\prime }\simeq I|\Omega (z,t)|$ and the slow motion has effective HamiltonianThe effective potential is the second term. Direct averaging checks its force: the exact equation is $m\ddot{z}=-m\Omega {\partial }_z\Omega y^2$, while the oscillator has $⟨y^2⟩=E^{\prime }/(m{\Omega }^2)$. Hence $m\ddot{z}=-I{\partial }_z|\Omega |$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent effective potential need not conserve total energy.