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Charged particle on a sphere with magnetic dipole at its center

  1. Aug 4, 2014 #1

    ShayanJ

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    Gold Member

    I just wanted something to play with so I thought of this problem and solved it but now I have questions about it.
    Consider a charged particle with charge q and mass [itex] \mu [/itex] which is constrained to move on the surface of a sphere of radius R. There is a magnetic dipole with moment [itex] \vec m=m \hat z [/itex] at the centre of the sphere.
    The vector potential produced by the magnetic dipole is [itex] \vec A= \frac {\mu_0 m \sin\theta} {4\pi R^2}\hat \varphi [/itex] and so the Hamiltonian is:

    [itex] \hat H=\frac{1}{2\mu}\left( -\hbar^2 \nabla^2-\frac{\mu_0 q m}{4\pi R^3}\frac{\partial}{\partial \varphi}+\frac{\mu_0^2q^2m^2\sin^2\theta}{16\pi^2R^4}\right) [/itex].

    I assumed [itex] q^2 m^2 [/itex] to be small and so the [itex] A^2 [/itex] term to be negligible.
    So the time independent Schrodinger equation is:

    [itex]
    \left( \frac{\partial^2}{\partial \theta^2}+\cot\theta \frac{\partial}{\partial \theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}+ \frac {\mu_0 q m}{4 \pi R \hbar^2}\frac{\partial}{\partial \varphi} \right) \psi=-\frac{2\mu R^2}{\hbar^2}E \psi
    [/itex]

    Then I assumed [itex] \psi=T(\theta) P(\varphi) [/itex]. It turns out that [itex] P(\varphi)=const [/itex] otherwise the equation is not separable.This is also supported by symmetry considerations.
    So the equation becomes:

    [itex]
    T''+\cot\theta T'+\frac{2\mu R^2 E}{\hbar^2} T=0
    [/itex]

    Which is Legendre's differential equation. But the coefficient of T should be of the form n(n+1) so that the solutions are well-behaved which gives us the energy spectrum:

    [itex]
    \frac{2\mu R^2 E}{\hbar^2}=n(n+1) \Rightarrow E=\frac{\hbar^2}{2\mu R^2}n(n+1)
    [/itex]

    Now the questions:

    1- Can it be an example of Aharonov-Bohm effect. Because we can somehow isolate the magnetic dipole so that its magnetic field is zero on the surface of the sphere. How is that isolation?

    2-As you can see, the energies are independent of the charge of the particle. I know that by adding the [itex] A^2 [/itex] term as a perturbation, we'll have q dependence but it is still strange to me. Can it be somehow explained?

    I have a candidate answer to these solution but I don't know how much its correct. Looks like the geometry of the problem is pushing Aharonov-Bohm effect away, at least pushing it to the correction terms. Interesting.

    Thanks

    P.S.
    Looks like I know the answer to the questions. So there is no question!!!
    Consider it as my first publication so :biggrin:!!!
     
    Last edited: Aug 4, 2014
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