Why is the Coulomb Gauge enough to fix extra degrees of freedom?

In classical electrodynamics, we have after the Coulomb gauge is applied:

$$\Delta U = -\frac{\rho}{\epsilon_0}$$

$$\Box \vec{A} = \mu_0 \vec{j}-\frac{1}{c^2} \vec{\nabla} \frac{\partial U}{\partial t}$$

If I take the first equation, I need two conditions to fix the potential (its value and its derivative on a given point for example).

So, why do we exactly say that imposing $$\vec{\nabla} \vec{A}=0$$ is enough to fix the gauge as we still have some degrees of freedom at the end?

(I would like a simple answer using what I am talking about here and not more complicated field theories if possible).

• If $\theta=\exp(-t^2)$, independent of $\vec x$, then the gauge transform $(U,\vec A)\to (U+\partial \theta/\partial t,\vec A\to\vec\nabla\theta)$ preserves the condition $\vec\nabla\cdot \vec A=0$. Clearly, the Coulomb gauge condition $\vec\nabla\cdot \vec A=0$ by itself doesn't fix the gauge completely. Maybe the source where you read this is tacitly assuming some additional condition(s), or maybe it was using a phrase like "fix the gauge" without intending to imply "fix the gauge completely". Can you provide more context? – Chiral Anomaly Jun 1 at 16:07
• @ChiralAnomaly, as per the Wikipedia article Gauge fixing, it is the boundary condition requiring all fields vanish at infinity the fixes the gauge function $\psi$: "Further gauge transformations that retain the Coulomb gauge condition might be made with gauge functions that satisfy $\nabla^2\psi = 0$, but as the only solution to this equation that vanishes at infinity (where all fields are required to vanish) is $\psi(\mathbf{r},t) = 0$, no gauge arbitrariness remains." – Alfred Centauri Jun 1 at 16:18
• @AlfredCentauri I would like to check What I understood from you : You mean that in addition of the gauge fixing we ask for a vanishing of the fields at infinity (why ?). So if we consider $A \rightarrow A - \nabla \psi$ then $\Delta \psi = 0$ bc of coulomb gauge. And the only $\psi$ verifying $\Delta \psi = 0$ such that it vanishes at infinity is $\psi=0$. Is that correct ? – StarBucK Jun 1 at 16:46
• StarBucK, as I understand it, if $\mathbf{A}$ satisfies the Coulomb gauge condition then $\mathbf{A}' = \mathbf{A} + \nabla\psi$ also satisfies it provided that $\nabla^2\psi = 0$. Then, imposing the boundary condition at infinity, if follows that $\psi(\mathbf{r},t) = 0$ and then $\mathbf{A}' = \mathbf{A}$ – Alfred Centauri Jun 1 at 22:24