Gauge invariance of magnetic vector potential

Question

If we go by transformation for magnetic vector potential $\mathbf{A} \rightarrow\tilde{\mathbf{A}}=\mathbf{A}-\nabla\psi$, as well as $\varphi \rightarrow \tilde{\varphi}=\varphi+\frac{\partial\psi}{\partial t}$ for scalar potential, and if we consider that $\nabla\times\mathbf{B}=\mu_0\mathbf{j}+\frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}$, we have that

$$\Delta \mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}-\nabla\left(\nabla\mathbf{A}+\frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2}\right)=-\mu_0\mathbf{j},$$

which is equivalent to

$$\Delta \mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}=-\mu_0\mathbf{j},$$

if we take in Lorenz gauge in consideration.

Now, if we replace $\mathbf{A}$ with $\tilde{\mathbf{A}}$, I need to prove that

$$\Delta\left(\mathbf{A}-\nabla\psi\right)-\frac{1}{c^2}\frac{\partial^2\textrm{ }}{\partial t^2} \left(\mathbf{A}-\nabla\psi\right)=-\mu_0\mathbf{j},$$

or, more precisely, that

$$\Delta\left(\mathbf{A}-\nabla\psi\right)-\frac{1}{c^2}\frac{\partial^2\textrm{ }}{\partial t^2} \left(\mathbf{A}-\nabla\psi\right)=\Delta \mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2},$$

but I'm completely stuck. Have I made any mistakes so far?

Any help (in standard vector notation, if possible) would be appreciated.

Answer 1

By assuming the Lorenz gauge you also have restricted $\psi$ to obey $$\square \psi = 0 \,.$$ $\square = \partial_\mu\partial^\mu=\partial_t^2- \Delta$ up to a minus sign depending on your Minkowski metric convention. This solves your problem. By the way, I recommend to use covariant notation.