5.5: Magnetic Moments

Consider a particle of electric charge $v$ performing a circular orbit of radius $r$ in the $y$ plane. The charge is equivalent to a current loop of radius $r$ in the $y$ plane carrying current $\mu$ of the loop is of magnitude $z$ -axis. Thus, we can write

$= \frac{q}{2}\, {\bf x} \times {\bf v},$ $\ref{458}$

where ${\bf x}$ and ${\bf p} = {\bf v} /m$ , where $m$ is its mass. We also know that ${\bf L}$ is the orbital angular momentum. It follows that

$= \frac{q}{2\,m} \,{\bf L}.$ $\ref{459}$

Using the usual analogy between classical and quantum mechanics, we expect the above relation to also hold between the quantum mechanical operators, ${\bf L}$ , which represent magnetic moment and orbital angular momentum, respectively. This is indeed found to the the case.

Spin angular momentum also gives rise to a contribution to the magnetic moment of a charged particle. In fact, relativistic quantum mechanics predicts that a charged particle possessing spin must also possess a corresponding magnetic moment (this was first demonstrated by Dirac--see Chapter 11). We can write

$= \frac{q}{2\,m} \left({\bf L} + g \,{\bf S}\right),$ $\ref{460}$

where $g$ is called the gyromagnetic ratio. For an electron this ratio is found to be

$1/(2\pi\, 137)$, derived originally by Schwinger, is due to quantum field effects. We shall ignore this correction in the following, so $\simeq - \frac{e}{2\,m_e} \left({\bf L} + 2 \,{\bf S}\right)$ $\ref{462}$

for an electron (here, $e> 0$).

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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