Gauge choice in Aharonov-Bohm effect

Question

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: \begin{equation} \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} \end{equation} When I solve the TISE with this new gauge I get that the energy levels are: \begin{equation} E_n = \frac{\hbar^2}{2mb^2}\bigg(n-\alpha \frac{\Phi}{\Phi_0}\bigg)^2\tag{2} \end{equation} 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?

Answer 1

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.

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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.