Question 8.9: (Farina de Souza & Gandelman, 1990) Obtain the solution ......

With B = B\widehat{z} and

A = \frac{1}{2}B × r = \frac{B}{2}\left(−y\widehat{x} + x\widehat{y}\right),   (8.152)

the Hamiltonian is

H =\frac{1}{2m}\left(p −\frac{e}{c}A\right)^{2}=\frac{1}{2m}\left[\left(p_{x} −\frac{eB}{2c}y\right)^{2}+\left(p_{y} −\frac{eB}{2c}x\right)^{2}+p^{2}_{z}\right] .  (8.153)

For the purpose of an algebraic computation of the pertinent Poisson brackets, it is convenient to express the Hamiltonian in the following form:

H =\frac{1}{2m}\left(p^{2}_{x}+p^{2}_{y}+p^{2}_{z}\right)+\frac{\omega}{2}\left(yp_{x}-xp_{y}\right)+\frac{m\omega^{2}}{8}\left(x^{2}+y^{2}\right), \omega= \frac{eB}{mc}.   (8.154)

For the z-coordinate we have:

\mathcal{L}_{H}z = \left\{z,H\right\} =\frac{p_{z}}{m}=v_{z}; \mathcal{L}^{2}_{H}z =\left\{\left\{z,H\right\}, H \right\} = \frac{1}{m}\left\{p_{z},H\right\}=0.   (8.155)

It follows that \mathcal{L}^{n}_{H}z =0 for all n ≥ 2 and, therefore,

z = z_{0} + v_{0z}t . (8.156)

The particle moves with constant velocity in the z-direction.

As to the x-coordinate, we have:

\mathcal{L}_{H}x = \left\{x,H\right\} =\frac{p_{x}}{m}+\frac{eB}{2mc}y = \frac{1}{m}\left(p_{x}+\frac{eB}{2c}y\right)=v_{x};   (8.157)

\mathcal{L}^{2}_{H}x =\left\{\frac{1}{m}\left(p_{x}+\frac{eB}{2c}y\right),H\right\} = \frac{\omega}{m}\left(p_{y}-\frac{eB}{2c}x\right)=\omega v_{y};  (8.158)

\mathcal{L}^{3}_{H}x =\left\{\frac{\omega}{m}\left(p_{y}-\frac{eB}{2c}x\right),H\right\} =-\omega^{2}v_{x};   (8.159)

\mathcal{L}^{4}_{H}x =-\frac{\omega^{2}}{m}\left\{p_{x}+\frac{eB}{2c}y,H\right\} =-\omega^{3} \frac{1}{m}\left(p_{y}-\frac{eB}{2c}x\right)=-\omega^{3} v_{y}.  (8.160)

By induction,

\mathcal{L}^{2n}_{H}x =\left(-1\right)^{n+1}\omega^{2n-1} v_{y},    \mathcal{L}^{2n+1}_{H}x =\left(-1\right)^{n}\omega^{2n} v_{x}.  (8.161)

Therefore,

x\left(t\right)=x_{0}+ t \mathcal{L}_{H}x\mid _{0} + \frac{t^{2}}{2!}\mathcal{L}^{2}_{H}x\mid _{0} + \frac{t^{3}}{3!}\mathcal{L}^{3}_{H}x\mid _{0} + \frac{t^{4}}{4!}\mathcal{L}^{4}_{H}x\mid _{0}+ · · ·

=x_{0}+ t v_{x0} + \frac{\omega t^{2}}{2!}v_{y0}- \frac{\omega^{2} t^{3}}{3!}v_{x0}- \frac{\omega^{3} t^{4}}{4!}v_{y0}+ · · ·

=x_{0}+ \frac{v_{x0}}{\omega}\left[\omega t – \frac{\left(\omega t\right)^{3}}{3!}+ · · ·\right] + \frac{v_{y0}}{\omega}- \frac{v_{y0}}{\omega}\left[1 – \frac{\left(\omega t\right)^{2}}{2!}+\frac{\left(\omega t\right)^{4}}{4!} – · · ·\right]

=x_{0}+ \frac{v_{y0}}{\omega} + \frac{v_{x0}}{\omega} \sin \omega t – \frac{v_{y0}}{\omega} \cos \omega t .   (8.162)

In analogous fashion one obtains

y\left(t\right)=y_{0}-\frac{v_{x0}}{\omega}+ \frac{v_{y0}}{\omega} \sin \omega t+ \frac{v_{x0}}{\omega} \cos \omega t .   (8.163)

A simple calculation gives

\left(x – x_{0} – \frac{v_{y0}}{\omega}\right)^{2} + \left(y – y_{0} – \frac{v_{x0}}{\omega}\right)^{2}= \frac{v^{2}_{x0}+ v^{2}_{y0}}{\omega^{2}},   (8.164)

which shows that the projection of the motion on the xy-plane is a circle. In general, the particle’s trajectory is a helix.