Posted by MRH on June 16, 2014, 12:24 pm, in reply to "Derivation of paramagnetic susceptibility"
I am fairly certain that everything is okay with the Hamiltonian and subsequent discussion as it stands. My reason for this is that L = r ^ p always. The canonical momentum p = mv + qA has the same commutation relations with r as did p = mv. Hence L has the same commutation relations as the case when there is no electromagnetic field, hence it is the same operator as before in terms of eigenvalues etc. If you are not convinced by that, note that A(r) so [A(r) , r]= 0. Therefore [p, r ] = ihbar as before. There is a note in P.A.M Dirac's book on QM that the canonical momentum p to coordinate q is ihbar d/dq + f(q). Since L = r ^ p, one may derive the zeeman term as presented. Hope I haven't missed the point of you question 
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