# Can we explicitly solve the Hamilton–Jacobi equation for a particle in a uniform magnetic field?

HJE for nonrelativistic charged particle in an electromagnetic field is

$$\frac{1}{2m}\left(\nabla S - q\mathbf{A}\right)^2 + q\phi + \frac{\partial S}{\partial t} = 0.$$

For a uniform magnetic field $\mathbf{B} = B_0 \hat{\mathbf{z}}$ and a particular choice of gauge this becomes something like:

$$\left(\frac{\partial S}{\partial x}\right)^2 + \left(\frac{\partial S}{\partial y} - qB_0 x\right)^2 + \left(\frac{\partial S}{\partial z}\right)^2 = -2m \frac{\partial S}{\partial t}.$$

Can we solve this explicitly? It seems to me we can start by separating the variables $t, z$ by letting $S = f(x, y) + p_z z - Et$, giving

$$\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y} - qB_0 x\right)^2 = 2mE - p_z^2,$$

but I'm really bad at solving differential equations, so I don't know how to proceed. I also have no intuition for what the solution $S$ is supposed to look like, even though I already know how a particle moves in a uniform magnetic field, so I don't have any idea how to guess a form for $S$.

• Is S the action? If so you might try reading the original equation as ${1\over 2m}({\bf{p}}-q{\bf{A}})+q\phi+\dot{S}$ If $\dot S$ is $dS\over{dt}$ here I would be very careful about calling that ${\partial S\over \partial t}$ – IntuitivePhysics Sep 4 '14 at 0:26

Hints:

1. Since there is no explicit time dependence in the Landau problem, we can use Hamilton's characteristic function $W$ rather than Hamilton's principal function $$\tag{1} S~=~W - Et.$$ Thus $$\tag{2} \left(\frac{\partial W}{\partial x}\right)^2 + \left(\frac{\partial W}{\partial y} - qB_0 x\right)^2 + \left(\frac{\partial W}{\partial z}\right)^2 = 2m E .$$

2. The two variables $y$ and $z$ are cyclic variables, so the corresponding momenta $p_y$ and $p_z$ are conserved, and Hamilton's characteristic function becomes $$\tag{3}W(x,y,z)~=~ w(x) +p_y y +p_z z.$$ Thus the first-order PDE (2) reduces to a first-order ODE $$\tag{4} \left(\frac{dw}{d x}\right)^2 + \left(p_y - qB_0 x\right)^2 + p_z^2 = 2m E ,$$ which has a well-known explicit solution.

• Thanks, solved it. And got the trajectory, eventually. It was a lot more work than I expected! – Brian Bi Oct 23 '14 at 21:20

1) Follow the procedure of paragraph $3.3$ (A Charged Particle in a Magnetic Field) page $77$ of this paper, by specializing your case with $k=\lambda=0$, so that $y=r$

2) You finally get the formula $(46)$ at the top of the page $80$, and you may take:

$S = f(y)+ \gamma\theta - \alpha t$

So you have a differential equation for $f$, and you may get the result, from W.A. for instance : here

Remark : The example is given in $2$ dimensions, so with $3$ dimensions, you may simply take $p_z$= Cte and $\alpha \to \alpha - \dfrac{p_z^2}{2m}$