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Magnetic Moments
Consider a particle of electric charge
and speed
performing a circular orbit of radius
in the
-
plane. The charge is equivalent to a current loop of radius
in the
-
plane carrying current
. The magnetic moment
of
the loop is of magnitude
and is directed along the
-axis.
Thus, we can write
  |
(458) |
where
and
are the vector position and velocity of the particle,
respectively. However, we know that
, where
is the vector momentum of the particle, and
is its mass. We also know that
, where
is the orbital angular momentum.
It follows that
  |
(459) |
Using the usual analogy between classical and quantum mechanics, we
expect the above relation to also hold between the quantum mechanical operators,
and
, 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
  |
(460) |
where
is called the gyromagnetic ratio. For an electron this ratio
is found to be
 |
(461) |
The factor 2 is correctly predicted by Dirac's relativistic theory of the electron (see Chapter 11).
The small correction
, derived originally by Schwinger, is due to
quantum field effects. We shall ignore this correction in the following,
so
  |
(462) |
for an electron (here,
).
Next: Spin Precession
Up: Spin Angular Momentum
Previous: Rotation Operators in Spin
Richard Fitzpatrick
2013-04-08