A Plasma in a Magnetic Field

An important place where both magnetic terms come into play is in a plasma.There, many electrons are not bound to atoms and external Electric fields are screened out. Let's assume there is a constant (enough) B field in the z direction. We then have cylindrical symmetry and will work in the coordinates, \bgroup\color{black}$\rho$\egroup, \bgroup\color{black}$\phi$\egroup, and \bgroup\color{black}$z$\egroup.

\begin{displaymath}\bgroup\color{black}{-\hbar^2\over 2m_e}\nabla^2\psi+{eB\over...
..._z\psi+{e^2B^2\over 8m_ec^2}

The problem clearly has translational symmetry along the z direction and rotational symmetry around the z axis. Given the symmetry, we know that \bgroup\color{black}$L_z$\egroup and \bgroup\color{black}$p_z$\egroup commute with the Hamiltonian and will give constants of the motion. We therefore will be able to separate variables in the usual way.


In solving the equation in \bgroup\color{black}$\rho$\egroup we may reuse the Hydrogen solution ultimately get the energies

\begin{displaymath}\bgroup\color{black}E={eB\hbar\over m_ec}\left(n+{1+m+\vert m\vert\over 2}\right)+{\hbar^2k^2\over 2m}\egroup\end{displaymath}

and associated LaGuerre polynomials (as in Hydrogen) in \bgroup\color{black}$\rho^2$\egroup (instead of \bgroup\color{black}$r$\egroup).

The solution turns out to be simpler using the Hamiltonian written in terms of \bgroup\color{black}$\vec{A}$\egroup if we choose the right gauge by setting \bgroup\color{black}$\vec{A}=Bx\hat{y}$\egroup.

H & = & {1\over {2 m_e}}\left(\vec{p}+{e\over c}\vec{A}\right)...
...{2eB\over c}x p_y
+ \left(eB\over c\right)^2 x^2 + p^2_z\right)

This Hamiltonian does not depend on \bgroup\color{black}$y$\egroup or \bgroup\color{black}$z$\egroup and therefore has translational symmetry in both x and y so their conjugate momenta are conserved. We can use this symmetry to write the solution and reduce to a 1D equation in \bgroup\color{black}$v(x)$\egroup.


Then we actually can use our harmonic oscillator solution instead of hydrogen! The energies come out to be

\begin{displaymath}\bgroup\color{black}E_n={eB\hbar\over m_ec}\left(n+{1\over 2}\right)+{\hbar^2k^2\over 2m_e}.\egroup\end{displaymath}

Neglecting the free particle behavior in \bgroup\color{black}$z$\egroup, these are called the Landau Levels.This is an example of the equivalence of the two real problems we know how to solve.

Jim Branson 2013-04-22