Uniqeness of Coulomb gauge Say I have some magnetic vector potential $A$ which is not in Coulomb gauge, meaning $\nabla \cdot A \neq 0$. I can set it to Coulomb gauge by adding some scalar potential function $\nabla \phi$ (and that's okay because the rotor - the magnetic force, will stay the same).
My questions is whether there is only one unique potential function we can add to $A$ to make it in Coulomb's gauge. It doesn't seem right to me that there is only one way to do it, is that so? If I'm wrong, why?
 A: You can find the answer just by computing the gauge transformation. You want to look for a escalar field $\phi$ whose gradient, added to the vector potential $\vec{A}$ makes a new vector potential, say $\vec{A}'$ which satisfies the Coulomb gauge:
$$\nabla \cdot \vec{A}'=0.$$
Now, plug in the expression of $\vec{A}'$
$$\nabla \cdot \vec{A}'=\nabla \cdot \left( \vec{A}+\nabla\phi\right)=\nabla\cdot \vec{A}+\nabla\cdot (\nabla\phi),$$
for the divergence operator is linear. Now, you can identify the laplacian ($\nabla^2\equiv\Delta$) of $\phi$ in the right-hand side of the equation. According to the Coulomb gauge, this quantity equals zero, then
$$\nabla \cdot \vec{A}'=\nabla \cdot \vec{A}+\Delta\phi=0
\Rightarrow \Delta\phi=-\nabla\cdot \vec{A}$$
So as you expected, the choice of $\phi$ isn't unique, and will depend on the form of the magnetic vector potential. You can look further on the resolution of this problem as it is a very well-known problem called Laplace-Poisson's problem. You can also verify that the result I posted is true, for example try to do problem 12.3  of Griffith's Introduction to Electrodynamics.
