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Consider the standard vector potential for an infinite solenoid of radius $R$, $$\vec{A}(r) = \frac{\Phi}{2\pi} \left(\frac{\theta(R-r)}{R^2}r + \frac{\theta(r-R)}{r} \right) \hat{\varphi}.$$ Here, $\theta(\cdot)$ is the Heaviside theta function, $\Phi$ is the magnetic flux and $(r,\varphi)$ are the polar coordinates. Suppose now we do the following gauge transformation $$\vec{A} \to \vec{A} - \vec{\nabla} \Lambda$$ with $$\Lambda = \frac{\Phi}{2\pi} \varphi.$$ This would give rise to a new vector potential $$\vec{A}'(r) = \frac{\Phi}{2\pi} \left(\frac{r}{R^2} - \frac{1}{r} \right) \theta(R-r) \hat{\varphi},$$ which clearly invalidates the (gauge-invariant) requirement that $$\oint \vec{A}' \cdot d \vec{r} = \Phi,$$ since $\vec{A}'$ completely vanishes outside the solenoid.

  1. Why is it that $\Lambda$ seems to be an ill-defined gauge function? What conditions does it not satisfy? I assume it has something to do with multivaluedness of the angle $\varphi$?
  2. I have seen a lot of research papers (e.g. concerning anyons) in which $\Lambda$ is used. Is there a way to compensate $\Lambda$'s "ill-defined"-ness by some other means? For example, is there a way to obtain the correct magnetic flux from the new vector potential $\vec{A}'$?
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    $\begingroup$ How are you computing $\nabla \Lambda$ here? Your $\Lambda$ is not even continuous at $\phi=0=2\pi$, let alone differentiable! $\endgroup$ – ACuriousMind Mar 22 '17 at 12:20
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Indeed $\Lambda$ is ill-defined in the plane because of its multivaluedness: Any point (except $r=0$) has multiple values of the cordinate $\phi$ differing by multiples of $2\pi$. Alternatively, you can restrict the range of $\phi$ to e.g. $(0,2\pi)$ to make it single-valued, but then you have to exclude a ray from the origin from your plane, say $\left\{\phi=0\right\}$.

If you like more complicated expressions with somewhat less rigour, you can even extend $\phi$ to the range $[0,2\pi)$ to cover the full plane. But then the derivative includes a $\delta$ function, as $$\nabla \Lambda = \frac{\Phi}{2\pi} \,\frac{1}{r} \hat{\phi} \cdot \left(1-2\pi\delta(\phi)\right)\,,$$ as there is a jump of $2\pi$ in your thus-defined $\phi$ coordinate at the line $\left\{\phi=0\right\}$. Then you get the correct integral, as $$\oint_{r=\text{const.}} \nabla\Lambda\, \text{d}r =\frac{\Phi}{2\pi} \,\frac{1}{r} \int_{-\epsilon}^{2\pi-\epsilon} \left(1-2\pi\delta(\phi) \right)\text{d}\phi=0 \,.$$ But of course now your new vector potential has a delta function line going out to infinity, a bit similar to a Dirac string.

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