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Hamiltonian of spin 1/2 in tangential magnetic field

  1. May 27, 2010 #1
    I have this article in which I saw that for a spin 1/2 particle confined to move along a ring positiond in a magnetic field with a z and \varphi

    The Hamiltonian is given by: (in second attacment)
    What I do not understand is how do you get the last term in the Hamiltonian.

    Any help?

    Thanks in advance

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  2. jcsd
  3. May 27, 2010 #2
    It comes from the standard spin-orbit coupling between the particle's magnetic moment, usually written as [tex]-\mathbf{\mu}_0\cdot \mathbf{B}[/tex]. You can find a treatment on this in any book on quantum mechanics. Here, [tex]\mathbf{\mu}_0[/tex] is the magnetic moment of the spin-1/2 particle (not equal to the [tex]\mu[/tex] used in your text)

    Now, to obtain the form they use you use the fact that you are dealing with spin-1/2. In that case the magnetic moment [tex]\mathbf{\mu}_0[/tex] can be written as [tex]-\mu \mathbf{S}/\hbar[/tex], where [tex]\mu[/tex] is, again, called the magnetic moment (confusing!). Furthermore, [tex]\mathbf{S}[/tex] is the spin operator which for spin-1/2 particles can be represented by the pauli matrices:

    [tex]\mathbf{S} = (S_x,S_y,S_z) = \frac{\hbar}{2}(\sigma_x,\sigma_y,\sigma_z)[/tex]

    Plugging this into the spin-orbit coupling reproduces the term from the article, up to some constants. But you can absorb these all into the prefactor by redefining [tex]\mu[/tex] (since this is just some numerical value anyway). Hope this helps!
  4. May 28, 2010 #3
    Thank you for your reply. By the way it was mentioned in the article that h bar was taken to be 1
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