As the title of the question suggest; how you could determine if a gauge fixing is a condition or a requirement. Let me explain.
Imagine you are working with Maxwell's Equations. By the definition of the fields:
$$
\mathbf E = -\nabla\phi – \frac{\partial \mathbf A}{\partial t}
$$
$$
\mathbf B = \nabla \times \mathbf A
$$
and by using the Maxwell's Equations, you arrive to the following expressions:
$$
\Box \mathbf A -\nabla \left ( \nabla \cdot \mathbf A + \frac{1}{c^2} \frac{\partial \phi}{\partial t} \right ) = -\mu_0\mathbf J
$$
$$
\frac{\rho}{\varepsilon_0} = -\nabla^2\phi – \frac{\partial}{\partial t} (\nabla \cdot\mathbf A).
$$
By imposing the Lorenz gauge fixing condition:
$$
\nabla\cdot\mathbf A = – \frac{1}{c^2}\frac{\partial \phi}{\partial t}
$$
We get:
$$
\Box \mathbf A = -\mu_0 \mathbf J \\ \; \\
\Box \phi = -\frac{\rho}{\varepsilon_0}.
$$
My question is, once our gauge is fix, and we get the simplified equations, does the gauge condition satisfies itself automatically, or after solving the resultant equations, you will have to go back to the gauge fixing equations and require them (by hand) to satisfy as well? How would you prove this?
Best Answer
This was stated in a comment, but I wish to make it really clear.
It is not sufficient to look for potentials merely satisfying $$ \Box \mathbf A = -\mu_0 \mathbf J \\ \; \\ \Box \phi = -\frac{\rho}{\varepsilon_0}. $$ You have to find potentials satisfying those equations and the gauge condition $$ \nabla\cdot\mathbf A = - \frac{1}{c^2}\frac{\partial \phi}{\partial t}. $$ Thus you have three equations to solve, not two.
Of course you can if you like replace the potentials with some other set related to the first by a gauge transformation. In that case your new potentials don't necessarily satisfy any of the above! An exception is when you replace a set of potentials by another (related by a gauge transformation) but which still satisfy the Lorenz gauge condition. In that case the new potentials will satisfy all the above, but I think they won't necessarily join smoothly to your first set. (I'm not sure of the correctness of that last clause but it seems reasonable).
Added comment
After pondering this a little it occurred to me that I can offer an easy example. In empty space the first two equations become wave equations and then there are plane wave solutions (among other solutions). If you take a plane wave solution for $\phi$ (at some non-zero frequency) and the solution ${\bf A}=0$ for $\bf A$ then the first two equations are satisfied but this is is not a valid solution overall because it does not satisfy the gauge condition.