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Paramagnetic term of the hamiltonian

  1. Dec 27, 2009 #1
    The Hamiltonian for particle in an EM field is

    H = 1/2m (p - qA)^2 + q phi

    If we take the cross-terms, which corresponds to the paramagnetic term, we have
    H para = -q/2m * (p.A + A.p )
    = iqh/2m * (\nabla .A + A.\nabla)

    What I do not understand is how this simplifies into [tex] iqh/m * A.\nabla [/tex]?

    assuming that [tex] \nabla .A = 0[/tex] (i.e. Coulomb gauge). Why does the factor of 1/2 disappears? I'm only a first year undergraduate and I'm learning this on my own. I will appreciate it if you give a fuller answer. Thank you.
  2. jcsd
  3. Dec 31, 2009 #2


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    Science Advisor

    In QM, H is assumed to act on a wave function \psi.
    This means that del.A really means del.(A \psi)=(del.A)\psi + (A.del)\psi, so the (A.del) comes in twice.
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