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Consider a magnetic field $\mathbf B= \Phi\delta(x) \delta(y) \hat z$. The corresponding vector potential becomes $\mathbf A = \frac{\Phi}{2\pi r} \hat\theta$ in the cylindrical coordinates. Furthermore, we may write $\mathbf A$ as a gradient $\nabla \chi$ if we choose $\chi = \Phi\theta/2\pi$. Note that the singularity of $\chi$ is inevitable.

Now, consider a magnetic field $\mathbf B = c\delta(y)\,\hat z$, where $c$ is a constant. The corresponding vector potential is $\mathbf A = c \operatorname{sgn}y\,\hat x$ up to constant. Can this vector potential be written as a gradient of a singular scalar function?

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The corresponding vektorpotential is $\vec A=\frac{1}{2}c\,{\rm sgn}(y)\,\hat x$, beacause ${\rm sgn}(y)=2H(y)-1$ and therefor $\frac{d}{dy}{\rm sgn}(y)=2\delta(y)$. Consider the Helmholz decomposition theorem for some vektorfield $\vec F(\vec r)$: $$ \vec F(\vec r)=\vec\nabla\left[-\frac{1}{4\pi}\int dV' \frac{(\vec\nabla\cdot\vec F)(\vec r')}{|\vec r-\vec r'|}\right]+\vec\nabla\times\left[\frac{1}{4\pi}\int dV' \frac{(\vec\nabla\times\vec F)(\vec r')}{|\vec r-\vec r'|}\right]. $$ Now the divergence of $\vec A$ is $0$. So we get that it can only be written as a rotation. Although by this argument a constant function $\vec F(\vec r)=c\,\hat x$ wouldn't be able to be described as a gradient as well which is not true: $\vec F(\vec r)=\vec\nabla c\, x$. Maybe the Helmholz theorem is not able to halp here after all, as it just shows that $\vec A$ can be written as a rotation, while it does not show it can't be written as a gradient.

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This magnetic field is that of an infinitely long, infinitely thin current carrying solenoid along the z-direction, so this is about the Aharonov-Bohm effect. The vector potential cannot be written as the gradient of a single valued scalar function except at the origin.

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