next up previous
Next: Spin precession Up: Angular momentum Previous: Rotation operators in spin

Magnetic moments

Consider a particle of charge $q$ and velocity $v$ performing a circular orbit of radius $r$ in the $x$-$y$ plane. The charge is equivalent to a current loop of radius $r$ in the $x$-$y$ plane carrying current $I=q v/2\pi  r$. The magnetic moment $\mbox{\boldmath$\mu$}$ of the loop is of magnitude $\pi  r^2  I$ and is directed along the $z$-axis. Thus, we can write
\begin{displaymath}
\mbox{\boldmath$\mu$}= \frac{q}{2}  {\bf r} \times {\bf v},
\end{displaymath} (462)

where ${\bf r}$ and ${\bf v}$ are the vector position and velocity of the particle, respectively. However, we know that ${\bf p} = {\bf v} /m$, where ${\bf p}$ is the vector momentum of the particle, and $m$ is its mass. We also know that ${\bf L} = {\bf r}\times{\bf p}$, where ${\bf L}$ is the orbital angular momentum. It follows that
\begin{displaymath}
\mbox{\boldmath$\mu$}= \frac{q}{2 m}  {\bf L}.
\end{displaymath} (463)

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

Does spin angular momentum also give rise to a contribution to the magnetic moment of a charged particle? The answer is ``yes''. In fact, relativistic quantum mechanics actually predicts that a charged particle possessing spin should also possess a magnetic moment (this was first demonstrated by Dirac). We can write

\begin{displaymath}
\mbox{\boldmath$\mu$}= \frac{q}{2 m} \left({\bf L} + g  {\bf S}\right),
\end{displaymath} (464)

where $g$ is called the gyromagnetic ratio. For an electron this ratio is found to be
\begin{displaymath}
g_e = 2\left( 1 + \frac{1}{2\pi} \frac{e^2}{4\pi \epsilon_0 \hbar  c} \right).
\end{displaymath} (465)

The factor 2 is correctly predicted by Dirac's relativistic theory of the electron. The small correction $1/(2\pi  137)$, derived originally by Schwinger, is due to quantum field effects. We shall ignore this correction in the following, so
\begin{displaymath}
\mbox{\boldmath$\mu$}\simeq - \frac{e}{2 m_e} \left({\bf L} + 2  {\bf S}\right)
\end{displaymath} (466)

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


next up previous
Next: Spin precession Up: Angular momentum Previous: Rotation operators in spin
Richard Fitzpatrick 2006-02-16