The Hamiltonian for a particle of mass $m$, charge $e$ and position vector $\mathbf{q}=(x, y, z)$, moving in an electromagnetic field, is given by

$$H(\mathbf{p}, \mathbf{q}, t)=\frac{1}{2 m}\left(\mathbf{p}-\frac{e \mathbf{A}}{c}\right)^{2}$$

where $\mathbf{A}(\mathbf{q}, t)$ is the vector potential. Write down Hamilton's equations and use them to derive the equations of motion for the charged particle.

Show that, when $\mathbf{A}=\left(-y B_{0}(z, t), 0,0\right)$, there are solutions for which $p_{x}=0$ and for which the particle motion is such that

$$\frac{d^{2} y}{d t^{2}}=-\Omega^{2} y$$

where $\Omega=e B_{0} /(m c)$. Show in addition that the Hamiltonian may be written as

$$H=\frac{m}{2}\left(\frac{d z}{d t}\right)^{2}+E^{\prime}$$

where

$$E^{\prime}=\frac{m}{2}\left(\left(\frac{d y}{d t}\right)^{2}+\Omega^{2} y^{2}\right)$$

Assuming that $B_{0}$ is constant, find the action

$$I\left(E^{\prime}, B_{0}\right)=\frac{1}{2 \pi} \oint m\left(\frac{d y}{d t}\right) d y$$

associated with the $y$ motion.

It is now supposed that $B_{0}$ varies on a time-scale much longer than $\Omega^{-1}$ and thus is slowly varying. Show by applying the theory of adiabatic invariance that the motion in the $z$ direction takes place under an effective potential and give an expression for it.