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, $\rho$, $\phi$, and $z$.

$$\bgroup\color{black}{-\hbar^2\over 2m_e}\nabla^2\psi+{eB\over... ..._z\psi+{e^2B^2\over 8m_ec^2} (x^2+y^2)\psi=(E+e\phi)\psi\egroup$$

Given the symmetry, we know that $L_z$ and $p_z$ commute with the Hamiltonian and will give constants of the motion. We therefore will be able to separate variables in the usual way.

$$\bgroup\color{black}\psi(\vec{r})=u_{nmk}(\rho)e^{im\phi}e^{ikz}\egroup$$

In solving the equation in $\rho$ we may reuse the Hydrogen solution ultimately get the energies

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

and associated LaGuerre polynomials (as in Hydrogen) in $\rho^2$ (instead of $r$). The solution turns out to be simpler using the Hamiltonian written in terms of $\vec{A}$ if we choose the right gauge by setting $\vec{A}=Bx\hat{y}$. Then we actually can use our harmonic oscillator solution instead of hydrogen! The energies come out to be

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

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