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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 2. 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, 0, 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.


next up previous
Next: Spin Up: Operators Matrices and Spin Previous: Rotation operators for
James Branson
2001-09-17