When finding a potential vector for the $\vec{B}$ field I understand that we have certain freedom because if $\nabla \times \vec{A}=\vec{B}$ then $\vec{A'} = \vec{A} + \nabla \psi$ also satisfies $\nabla \times \vec{A'}=\vec{B}$
What I don't understand is why that gives us the freedom to choose $\nabla \cdot \vec{A}=0$, when you can only choose any scalar function $\psi$.
I thought that maybe it has something to do with the Helmholtz theorem but I got nowhere.
Thank you, in advance
The proof is like this. Suppose you have some vector potential $\mathbf{A}$, not necessarily satisfying your gauge condition. Now choose some $\psi$ such that
$$\nabla^2 \psi = - \nabla \cdot \mathbf{A}$$
This is Poisson's equation for $\psi$, and it always has a solution (which is unique if you specify boundary conditions). Now, if we define
$$\mathbf{A}' = \mathbf{A} + \nabla\psi$$
Then it holds that
$$\nabla\cdot\mathbf{A}' = 0$$
i.e., we've found an equivalent vector potential that satisfies the gauge condition.
