# Can the vector potential be written as $\mathbf A = \nabla \chi$ for some singular function $\chi$?

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?

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.