# Quantum ring and Dirac Quanitzation

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.

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.

• Thank you for your answer. I understand this part of the computation, however, I still think my comment about $g(\vec{r})$ not being well defined is still of concern. If it were, then you could arrive at a point on the other end of the solenoid by going either around the top or the bottom of the solenoid. If $g(\vec{r})$ were well defined the line integrals for both of these paths would have the same value and thus the closed path around the solenoid would integrate to 0, not the flux. – user33525 Sep 19 '14 at 7:01
• @BRayhaun : You have different homotopy classes for the closed paths, each one corresponding to some winding number. These are called homotopy classes, because, inside one class, one may go continuously from one closed path to another (without crossing the solenoid). In fact, here, the 1st homotopy group, the fundamental group is just $\mathbb Z$. Closed paths which are not enclosing the solenoid correspond to the $0$ th class or $0$ winding number. – Trimok Sep 19 '14 at 8:55
• @BRayhaun : So, $g(\vec r)$ is constant inside a homotopy class. The example I gave in the answer correspond to the $1$ st class or $1$ winding number. – Trimok Sep 19 '14 at 8:56