# Gauge choice in Aharonov-Bohm effect

In p.385 of Griffiths QM the vector potential $$\textbf{A} = \frac{\Phi}{2\pi r}\hat{\phi}$$ is chosen for the region outside a long solenoid. However, couldn't we also have chosen a vector potential that is a multiple of this, namely $$\textbf{A} = \alpha \frac{\Phi}{2\pi r} \hat{\phi}$$ where $$\alpha$$ is some constant? The two are related by a gauge transformation: $$$$\alpha\frac{\Phi}{2\pi r}\hat{\phi} = \frac{\Phi}{2\pi r} \hat{\phi}+\nabla\bigg((\alpha-1)\frac{\Phi}{2\pi}\phi\bigg).\tag{1}$$$$ When I solve the TISE with this new gauge I get that the energy levels are: $$$$E_n = \frac{\hbar^2}{2mb^2}\bigg(n-\alpha \frac{\Phi}{\Phi_0}\bigg)^2\tag{2}$$$$ which is different from what Griffiths even if the magnetic flux is quantized. How is it possible that the ground state depends on the gauge choice?

• Your gauge-transformed vector potential won't be continuous at the radius of the solenoid. I am curious if this has something to do with your issue.
– d_b
Oct 5 '21 at 3:34
• A pedestrian point: your "gauge transformation" changes magnetic field. Oct 5 '21 at 4:38

## 1 Answer

Your proposed gauge transformation fails because $$\phi$$ is not a continuous function. The reason a gauge transformation $$A_\mu \mapsto A_\mu + \partial_\mu \alpha$$ is invisible from the perspective of the Aharanov-Bohm effect is because the AB phase difference takes the form $$\delta \varphi = \oint A_\mu \mathrm dx^\mu\underbrace{ \longmapsto}_{\text{gauge transformation}} \oint A_\mu \mathrm dx^\mu + \underbrace{\oint (\partial_\mu \alpha)\mathrm dx^\mu}_{=0\text{ if }\alpha\text{ differentiable}} = \delta\varphi$$

Your proposed gauge transformation uses an $$\alpha$$ which has a branch cut somewhere in the plane. As a result, its closed-loop integral will generically not be equal to zero. But this is not a valid gauge transformation. This can be seen even more clearly by noting that the transformation $$A \mapsto \alpha A$$ does not preserve the values of the electric and magnetic fields (i.e. it scales them by a factor of $$\alpha$$) which is an immediate indication that it isn't a gauge transformation.

Recall that the Faraday tensor $$F$$ - which contains the physical degrees of freedom in electromagnetism - is obtained from the 4-potential $$A$$ via $$F = \mathrm dA \iff F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$ while the (vacuum) Maxwell equations take the form $$\partial_{[\alpha}F_{\beta\gamma]}=0 \qquad \partial_\mu F^{\mu\nu}=0$$

A gauge transformation $$A \mapsto A + \mathrm d\alpha$$ does not change the Maxwell equations$$^\dagger$$ because $$F \mapsto F + \mathrm d^2\alpha$$, and $$\mathrm d^2 \alpha$$ is identically zero for all twice continuously differentiable functions $$\alpha$$; this is what a gauge transformation is. If $$\alpha$$ does not possess this smoothness property, then this isn't a gauge transformation.

$$^\dagger$$In fact, electromagnetic gauge transformations preserve not just the Maxwell equations but the Faraday tensor itself. In more complicated (non-Abelian) gauge theories this is not true - the equations of motion are preserved, but the non-Abelian analogue of $$F$$ is not itself gauge-invariant.

• Hmm, interesting. So how do we know if we should choose one potential or the other, should the continuous vector potential be chosen on physical grounds? Oct 5 '21 at 5:54
• Also, the integral will involve a delta function, but that contribution to the gradient will point in the radial direction, so it cancels out when taking the path integral right? Oct 5 '21 at 7:08
• @KoutaDagnino I intended my "delta function" comment to be a loose intuitive way to see that what you propose is not a gauge transformation. However, I've replaced it with what I think is a better argument. The vector potential certainly should be continuous because it is defined as an auxiliary field which, when differentiated, gives us $F$ (or equivalently $\vec E$ and $\vec B$). This clearly indicates that $A$ should at least be differentiable. Oct 5 '21 at 11:31
• Thanks, all clear now! Oct 5 '21 at 14:33