Magnetic Moments

Consider a particle of electric charge $q$ and speed $v$ performing a circular orbit of radius $r$ in the $x$ -$y$ plane. The charge is equivalent to a current loop of radius $r$ , lying in the $x$ -$y$ plane, and carrying a current $I=q\,v/(2\pi\, r)$ . The magnetic moment $\mu$ of the loop is of magnitude $\pi\, r^{\,2}\, I$ and is directed along the $z$ -axis (the direction is given by a right-hand rule with the fingers of the right-hand circulating in the same direction as the current) [49]. Thus, we can write

$$\mbox{\boldmath$\mu$} = \frac{q}{2}\, {\bf x} \times {\bf v},$$ (5.42)

where ${\bf x}$ 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 particle's vector momentum, and $m$ its mass. We also know that ${\bf L} = {\bf x}\times{\bf p}$ , where ${\bf L}$ is the orbital angular momentum. It follows that

$$\mbox{\boldmath$\mu$} = \frac{q}{2\,m} \,{\bf L}.$$ (5.43)

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

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 [30,32,9]. We can write

$$\mbox{\boldmath$\mu$} = \frac{q}{2\,m} \left[{\bf L} + {\rm sgn}(q)\,g \,{\bf S}\right],$$ (5.44)

where the parameter $g$ is called the $g$ -factor. For an electron this factor is found to be

$$g_e = -2\left( 1 + \frac{\alpha}{2\pi}\right),$$ (5.45)

where

$$\alpha= \frac{e^{\,2}}{4\pi \,\epsilon_0\,\hbar \,c}\simeq \frac{1}{137}$$ (5.46)

is the fine structure constant. The factor $-2$ is correctly predicted by Dirac's famous relativistic theory of the electron [30,32]. (See Chapter 11.) The small correction $1/(2\pi\, 137)$ is due to quantum field effects [101]. We shall ignore this correction in the following, so

$$\mbox{\boldmath$\mu$} \simeq - \frac{e}{2\,m_e} \left({\bf L} + 2 \,{\bf S}\right)$$ (5.47)

for an electron. (Here, $e>0$ is the magnitude of the electron charge, and $m_e$ the electron mass.)