# Charged particle on a sphere with magnetic dipole at its center

#### ShayanJ

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 $\mu$ which is constrained to move on the surface of a sphere of radius R. There is a magnetic dipole with moment $\vec m=m \hat z$ at the centre of the sphere.
The vector potential produced by the magnetic dipole is $\vec A= \frac {\mu_0 m \sin\theta} {4\pi R^2}\hat \varphi$ and so the Hamiltonian is:

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

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

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

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

$T''+\cot\theta T'+\frac{2\mu R^2 E}{\hbar^2} T=0$

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:

$\frac{2\mu R^2 E}{\hbar^2}=n(n+1) \Rightarrow E=\frac{\hbar^2}{2\mu R^2}n(n+1)$

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 $A^2$ 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 !!!

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#### Greg Bernhardt

Admin
@ShayanJ did you find any more insight on this topic?

#### ShayanJ

Gold Member
Wow, that was years ago. Back then I had a habit of playing with such problems and gathering any little bit of interesting information I could get from them. There was no organized effort toward anything.
I can't quite figure out what my line of thought was. But I'm not sure this is related to the Aharonov-Bohm effect. For that to be true, the $\vec B$ field should have no effect on the particle.
Anyway, even if I was right, its just about a non-typical configuration for the Aharonov-Bohm effect. You can come up with so many crazy configurations for so many physical phenomena and study the effect of that particular configuration on that phenomenon. In this particular case, the effect seems to be that the Aharonov-Bohm effect is only manifesting itself at the second order of approximation.
Actually I miss those days when I just played with such problems, maybe I think about it again.

• Greg Bernhardt

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