5.5: Magnetic Moments
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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 μ of the loop is of magnitude z -axis. Thus, we can write
=q2x×v, ???where x and p=v/m , where m is its mass. We also know that L is the orbital angular momentum. It follows that
=q2mL. ???Using the usual analogy between classical and quantum mechanics, we expect the above relation to also hold between the quantum mechanical operators, 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
=q2m(L+gS), ???where g is called the gyromagnetic ratio. For an electron this ratio is found to be
1/(2π137), derived originally by Schwinger, is due to quantum field effects. We shall ignore this correction in the following, so ≃−e2me(L+2S) ???for an electron (here, e>0).
Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)