Quantum Mechanics in an Electromagnetic Field

The classical Hamiltonian for a particle in an Electromagnetic field is

\begin{displaymath}\bgroup\color{black} H={1\over 2m}\left(\vec{p}+{e\over c}\vec{A}\right)^2-e\phi \egroup\end{displaymath}

where \bgroup\color{black}$e$\egroup is defined to be a positive number. This Hamiltonian gives the correct Lorentz force law. Note that the momentum operator will now include momentum in the field, not just the particle's momentum. As this Hamiltonian is written, \bgroup\color{black}$\vec{p}$\egroup is the variable conjugate to \bgroup\color{black}$\vec{r}$\egroup and is related to the velocity by \bgroup\color{black}$\vec{p}=m\vec{v}-{e\over c}\vec{A}$\egroup.

In Quantum Mechanics, the momentum operator is replaced in the same way to include the effects of magnetic fields and eventually radiation.

\begin{displaymath}\bgroup\color{black} \vec{p}\rightarrow \vec{p}+{e\over c}\vec{A} \egroup\end{displaymath}

Starting from the above Hamiltonian, we derive the Hamiltonian for a particle in a constant magnetic field.

\begin{displaymath}\bgroup\color{black}{-\hbar^2\over 2m}\nabla^2\psi+{e\over 2m...

This has the familiar effect of a magnetic moment parallel to the angular momentum vector, plus some additional terms which are very small for atoms in fields realizable in the laboratory.

So, for atoms, the dominant additional term is

\begin{displaymath}\bgroup\color{black}H_B={e\over 2mc}\vec{B}\cdot\vec{L}=-\vec{\mu}\cdot\vec{B},\egroup\end{displaymath}

where \bgroup\color{black}$\vec{\mu}=-{e\over 2mc}\vec{L}$\egroup. This is, effectively, the magnetic moment due to the electron's orbital angular momentum.

The other terms can be important if a state is spread over a region much larger than an atom. We work the example of a plasma in a constant magnetic field. A charged particle in the plasma has the following energy spectrum

\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}

which depends on 2 quantum numbers. \bgroup\color{black}$\hbar k$\egroup is the conserved momentum along the field direction which can take on any value. \bgroup\color{black}$n$\egroup is an integer dealing with the state in x and y. This problem can be simplified using a few different symmetry operators. We work it two different ways: in one it reduces to the radial equation for the Hydrogen atom; in the other it reduces to the Harmonic Oscillator equation, showing that these two problems we can solve are somehow equivalent.

Jim Branson 2013-04-22