# Do we fix divergence of the vector potential $A$, because $\nabla \cdot \nabla \psi \ne 0$?

Because $\nabla \times \nabla \psi = 0$, we can transform the vector potential $A \longmapsto A + \nabla \psi$, without changing the magnetic field. Is the reason we specify $\nabla \cdot A$ in the gauge theory, because $\nabla \cdot \nabla\psi\ne0$? Thus, we always can find a function $\psi$, such that $\nabla \cdot A$ equals to whatever we want.

A bit more elaborated:

Let's say I found $A$ and $\phi$, and $\nabla \cdot A \ne 0$. In principle, nothing stops me from finding a function $\psi$, such that $\nabla \cdot (A + \nabla \psi)=0$ (Coulomb gauge). But now my scalar potential will be modified. Then from Gauss' law: $\nabla \cdot( \nabla\phi+\frac{\partial \nabla\psi}{\partial t}- \frac{\partial }{\partial t}(A + \nabla \psi))=\nabla \cdot( \nabla\phi+\frac{\partial \nabla\psi}{\partial t})$. Thus, the new 4-potential in the Coulomb gauge $(\phi,A)\longmapsto (\phi+\frac{\partial \nabla\psi}{\partial t},A + \nabla\psi)$. So, the freedom of choosing the divergence stems from the freedom of choice of $\nabla \phi$.

• I think you are confusing different gauge choices. The potentials are $(\phi,{\bf A})$ and the choice of $\psi$ has effect also on $\phi$. Different gauge choices have different effects on these four potentials.
– Jon
Mar 11, 2018 at 11:25

Gauge freedom is the freedom to transform $\mathbf{A} \to \mathbf{A}' = \mathbf{A} + \nabla \psi$ and $\phi \to \phi' = \phi - \partial \psi/\partial t$ for any scalar function $\psi$. (Note the slight difference from your expressions.) If we want to demand that $\nabla \cdot \mathbf{A}' = 0$ (Coloumb), this boils down to being able to find a function $\psi$ such that $\nabla^2 \psi = -\nabla \cdot \mathbf{A}$. Since we know that there is always a function $\psi$ that satisfies this equation, the condition $\nabla \cdot \mathbf{A}' = 0$ is always realizable via a gauge transformation.
That said, there are plenty of other gauge conditions out there that we could impose that do not lead to $\nabla \cdot \mathbf{A}' = 0$. Other common choices include $\nabla \cdot \mathbf{A}' + \partial \phi'/\partial t= 0$ (Lorenz gauge), $\phi' = 0$ (temporal gauge), and $\mathbf{A}' \cdot \hat{n} = 0$ for some unit vector $\hat{n}$ (axial gauge). All of these are realizable, in the sense that given an arbitrary $\mathbf{A}$ and $\phi$, there always exists a function $\psi$ such that the transformed potentials satisfy the desired condition.