# Quantum particle on a ring with magnetic flux through it: Can we gauge away the magnetic field?

Consider a quantum particle on a ring and a non-zero homogeneous magnetic field perpendicular to the disk that the ring defines and is non-zero only in the inside of the perimeter of the ring. Let $\vec{B}=B_0 \hat{z}$ and the flux through the ring be $\Phi$.
For the vector potential, we can choose (in cylindrical coordinates) $\vec{A}=\dfrac{\Phi}{2\pi}\dfrac{1}{\rho}\hat{\phi}$.

If I try to perform a gauge-transformation $\vec{A'}=\vec{A}-\vec{\nabla}f$ to gauge-away the magnetic field by going to a new gauge where $\vec{A'}=\vec{0}$, I find $f=\dfrac{\Phi}{2\pi}\phi$. So, since I have found a function $f$ to do this, it seems as if I have successfully gauged-away the magnetic field, which is physically impossible!

What is happening here?
I suspect that something is not right because $f$ is multi-valued at $x=0$ (which corresponds to $\phi=0, 2\pi, ..)$. If this is the case, how do I fix this and get $\vec{A'}\neq0$? Is there a systematic way to treat cases such as this, i.e. to find a proper $f$ that would deal with this problem and give the correct magnetic field?

• Sounds like you answered your own question; this isn’t right because your $f$ is multivalued. Jan 11, 2018 at 18:31
• @knzhou I have edited the question to make more clear what I am asking. Jan 11, 2018 at 18:35
• I'm still not sure what you're asking. What do you mean by 'give the correct magnetic field'? You already know what it is; it's $B_0 \hat{z}$. Jan 11, 2018 at 18:48
• @knzhou through the above gauge transformation, I found an $f$ such that $\vec{A'}=0$ which gives $\vec{B}=0$, which is clearly wrong. So, the question is, how to see where exactly the whole thing goes wrong and what is the systematic and correct way to perform gauge transformations so that the above problem does not show up? Jan 11, 2018 at 18:52

Describing the problem as "because $f$ is multi-valued" is one way to describe the problem, but it's not actually the root of the problem. The root of the problem is the singularity at $r=0$. In order to qualify as a gauge transformation, the function needs to obey $$\nabla\times \nabla f(\mathbf{x},t) = 0$$ everywhere.
It looks like this function does if you just take the derivatives, but that's misleading. $-\frac{1}{4\pi}\nabla^2 r^{-1}$ looks like it vanishes by the same critereon when it is, in fact, the Dirac delta function. Same thing happens here. Just like how it is possible to show the existence of the delta function in the divergence of the gradient case using the divergence theorem, an application of Stoke's theorem will show you that $$\oint_{\gamma} \nabla f \cdot \operatorname{d}\vec{s} \neq 0$$ whenever the origin is enclosed by the loop, $\gamma$, proving that the gauge transformation is invalid.
• @TheQuantumMan You mean "exclude"? Or do you want to make a gauge transformation with these functions, anyway? If the former, it's easiest to just stick with continuous functions. It may be possible to allow discontinuities depending on the topology of the discontinuity, but I don't know definitively. If the latter, that would take modifying what we mean by a gauge transformation to keep $\vec{E}$ and $\vec{B}$ invariant under the wider class of functions. Jan 11, 2018 at 19:05
• @TheQuantumMan I don't know, and I don't know if anyone has bothered to work it out. I suspect that you can allow discontinuities as long as they don't have an edge to them. It might be possible to allow edges in the discontinuities if you modify the gauge transformation equations in such a way that they keep $\vec{E}$ and $\vec{B}$ fixed, even for this wider class of functions, but I haven't looked into it. Jan 11, 2018 at 19:12