An \bgroup\color{black}$\ell=1$\egroup System in a Magnetic Field*

We will derive the Hamiltonian terms added when an atom is put in a magnetic field in section 20. For now, we can be satisfied with the classical explanation that the circulating current associated with nonzero angular momentum generates a magnetic moment, as does a classical current loop. This magnetic moment has the same interaction as in classical EM,

\begin{displaymath}\bgroup\color{black}H=-\vec{\mu}\cdot \vec{B}.\egroup\end{displaymath}

For the orbital angular momentum in a normal atom, the magnetic moment is

\begin{displaymath}\bgroup\color{black}\vec{\mu}={-e\over 2mc}\vec{L}.\egroup\end{displaymath}

For the electron mass, in normal atoms, the magnitude of \bgroup\color{black}$\vec{\mu}$\egroup is one Bohr magneton,

\begin{displaymath}\bgroup\color{black}\mu_B={e\hbar\over 2m_ec}.\egroup\end{displaymath}

If we choose the direction of \bgroup\color{black}$B$\egroup to be the \bgroup\color{black}$z$\egroup direction, then the magnetic moment term in the Hamiltonian becomes

\begin{displaymath}\bgroup\color{black}H={\mu_B B\over\hbar} L_z
=\mu_BB\left(\matrix{1&0&0\cr 0&0&0\cr 0&0&-1}\right).\egroup\end{displaymath}

So the eigenstates of this magnetic interaction are the eigenstates of \bgroup\color{black}$L_z$\egroup and the energy eigenvalues are \bgroup\color{black}$+\mu_BB$\egroup, \bgroup\color{black}$0$\egroup, and \bgroup\color{black}$-\mu_BB$\egroup.

* Example: The energy eigenstates of an $\ell =1$ system in a B-field.*
* Example: Time development of a state in a B field.*

Jim Branson 2013-04-22