Solid State Physics 2014
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Derivation of paramagnetic susceptibility

Posted by Toby on June 15, 2014, 3:25 pm

 I'm going over the derivation of the paramagnetic term in the hamiltonian that leads to the susceptibility in section 19.3 of the book. In equation 19.4 we change the term from p.(B times r) to B.(r times p) by a vector identity and then claim the (r times p) is the angular momentum, l. In classical mechanics, the hamiltonian for a particle in a B field is the minimal coupling: (p-qA)^2/2m. Working through with hamilton's equations we find that p, the conjugate momentum, is not just equal to mv, but has an additional term which we describe as the field momentum, so that in the end, p = mv + qA. My question is where does this extra part of the conjugate momentum go in the QM version? Is it still valid that (r times p) = hbar l even if p is not mv? I know I'm posting this at the 11th hour and I'm not expecting a response before the exam, but it would just be nice to know for my own satisfaction.
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 Message Thread Derivation of paramagnetic susceptibility - Toby June 15, 2014, 3:25 pm Re: Derivation of paramagnetic susceptibility - MRH June 16, 2014, 12:24 pm Re: Derivation of paramagnetic susceptibility - Toby June 16, 2014, 6:51 pm Re: Derivation of paramagnetic susceptibility - Steve Simon June 19, 2014, 9:05 am Re: Derivation of paramagnetic susceptibility - Steve Simon June 16, 2014, 12:30 am « Back to index