Residual gauge fixing in Lorenz gauge

Question

Background

Let $A^{\mu}$ be a 4-potential that satisfies the Lorenz condition $$\partial_\mu A^\mu =0$$ We can make a gauge transformation $$A_\mu \to A'_\mu=A_\mu + \partial_\mu \Lambda$$ such that $$ \partial_\mu A'^\mu =\partial_{\mu}\left(A^{\mu}+\partial^{\mu}\Lambda\right)=0 $$ and still stay in the Lorenz gauge. Since $\partial_{\mu}A^{\mu}=0$, this condition for $\Lambda$ is \begin{equation} \partial_{\mu}\partial^{\mu}\Lambda=0\tag{1} \end{equation} This possibility to do some gauge transformations while having fixed the Lorenz condition is called residual gauge freedom.

Claim

We want to remove this residual gauge freedom by fixing another condition on $A_{\mu}$. My book claims that \begin{align} \begin{aligned}A_{3}(0,\vec{x}) & =0\\ \partial_{t}A_{3}(0,\vec{x}) & =0 \end{aligned} \tag{2} \end{align} is such a condition.

Proof

The solution to (1) is uniquely determined by its initial conditions \begin{align*} \Lambda(0,\vec{x}) & =\Phi_{1}(\vec{x})\\ \partial_{t}\Lambda(0,\vec{x}) & =\Phi_{2}(\vec{x}) \end{align*} so it is sufficient to prove that the conditions (2) imposed on the transformed 4-potential uniquely determine $\Phi_{1}$ and $\Phi_{2}$. Since $$ A'_{3}=A_{3}+\partial_{3}\Lambda $$ they read \begin{alignat*}{2} A_{3}(0,\vec{x})+\partial_{3}\Lambda(0,\vec{x}) & = & A_{3}(0,\vec{x})+\partial_{3}\Phi_{1}(\vec{x}) & =0\\ \partial_{t}A_{3}(0,\vec{x})+\partial_{t}\partial_{3}\Lambda(0,\vec{x}) & = & \partial_{t}A_{3}(0,\vec{x})+\partial_{t}\partial_{3}\Phi_{2}(\vec{x}) & =0 \end{alignat*} So we can choose an antiderivative wrt $x^{3}$ of the given functions $A_{3}(0,\vec{x})$ and $\partial_{t}A_{3}(0,\vec{x})$ to find the $\Lambda$ we're searching for.

Question

It seems to me that this antiderivative is not uniquely determined, as you can add to it an arbitrary function of $x^{1}$ and $x^{2}$. So $\Lambda$ is not uniquely determined either. Is then the residual gauge not removed? Or am I missing something?

Answer 1

You are right that you can add a function of $\lambda(x,y)$ to your gauge parameter. But note that it should satisfy the Lorenz gauge which implies that $$(\partial_x^2+\partial_y^2)\lambda=0$$ whose general solutions consists of arbitrary holomorphic and antiholomorphic functions on the plane $$\lambda=f(z)+g(\bar z), \qquad z=x+i y,\quad \bar z =x-iy$$ The point is that this solution diverges somewhere. Expanding $f=\sum_n c_n (z-z_0)^n$ we see that the function diverges either at $z=z_0$ or at $z\to \infty$ except for the constant function.

Imposing some boundary conditions, say $A_\mu\sim\cal{O}(1/r)$ on the gauge field at $r=\sqrt{x^+y^2+z^2} \to \infty$ and requiring smoothness of the gauge field in the bulk of spacetime excludes the additional freedom you mentioned.

There is however a nice story about asymptotic symmetries surviving gauge fixing and boundary conditions which is not necessary for your question here.