2
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – d_b
    Oct 5 '21 at 3:34
  • 2
    $\begingroup$ A pedestrian point: your "gauge transformation" changes magnetic field. $\endgroup$ Oct 5 '21 at 4:38
4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Oct 5 '21 at 5:54
  • $\begingroup$ 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? $\endgroup$ Oct 5 '21 at 7:08
  • 1
    $\begingroup$ @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. $\endgroup$
    – J. Murray
    Oct 5 '21 at 11:31
  • $\begingroup$ Thanks, all clear now! $\endgroup$ Oct 5 '21 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.