Quantum ring and Dirac Quanitzation

Question

I had a small question. If you look at the energy eigenvalue problem for a particle restricted to a ring, you get $$E_n = \frac{\hbar^2n^2}{2mR^2}.$$ If you then put a solenoid inside the ring, then the values shift to $$E_n = \frac{\hbar^2}{2mR^2}\left( n^2 + \left(\frac{q\Phi}{2\pi\hbar}\right)^2 - \frac{nq\Phi}{\pi\hbar}\right)$$ according to Griffiths. I thought it should be possible to derive Dirac's quantization condition on the charge of magnetic monopoles by demanding that the shift in energy be 0, but this gives me (using $\Phi = 4\pi g$) $$qg = n$$ in units of $\hbar=1$, $not$ $qg = n/2$. Is this a kosher way to derive the Dirac quantization condition? If not, where does my thinking fail?

The reason I tried to derive the quantization condition this way is because I was having trouble actually understanding the Aharonov-Bohm effect in the context of electron interference. In Griffiths explanation, he uses a function $$g(\vec{r}) = \frac{q}{\hbar}\int_{\mathfrak{O}}^{\vec{r}}\vec{A}\cdot d\vec{l}$$ and claims that it is well-defined because $\vec{\nabla}\times\vec{A} = 0$ and so the integral is path independent. He uses $g$ to essentially say that one can solve the problem in the absence of the solenoid, and then tack on a phase factor $e^{ig}$ to that solution to find the wave function which is the solution to the problem when the solenoid is present.

However, I argue that $g$ is not well defined, because the region in which $\vec{\nabla}\times\vec{A} = 0$ is not simply-connected. In fact, is not the entire point that the beams arrive with different phases precisely because they took different paths, i.e. $g$ is path dependent? I would really appreciate some help in understanding this subtle point.

Answer 1

It is only a remark about the interrogations contained in the second part of your question.

You are mixing closed paths and open paths.

For any closed path $C$ around the solenoid (taking $c = \hbar=1$ units), the phase difference is $\Delta \phi_C = q \, \Phi_B$, and this is independent of the choosen closed path (homotopy invariance).

Separating the closed path in two open paths $C_1 = A \to I_1 \to B$, and $C_2 = B \to I_2 \to A$, one has : $\Delta \phi_C = \Delta \phi_{C_1} + \Delta \phi_{C_2} = q \, \Phi_B$

If we want to compute the phase difference between $C_1$ and $C'_2 = A \to I_2 \to B$, we have :

$\delta \phi_i = \Delta \phi_{C_1} - \Delta \phi_{C'_2} = \Delta \phi_{C_1} + \Delta \phi_{C_2} = q \, \Phi_B$

So, there is no contradiction between the homotopy invariance for closed paths, and the phase difference between open paths having the same origin and the same end.