Quick question on choosing a gauge (E.g. Lorenz gauge) I have been quite confused when I read about choosing a gauge. For  example we have the gauge transformation 
$$
A_\mu\longrightarrow A_{\mu\prime}= A_\mu+\partial_\mu\alpha,
$$
and we can choose any $\alpha(x)$. This I understand. What I don't understand is what follows, for example 

in the Lorenz gauge we choose $\alpha$ such that $\partial^\mu A_\mu=0$. 

Is it supposed to be $\partial^\mu A_\mu=0$ or $\partial^\mu A_\mu^\prime=0$ (in the latter case then $\partial^2\alpha=-\partial^\mu A_\mu$)? If it's the first one, then how could we choose $\alpha$ or what $\alpha$ do we choose s.t. we get the Lorenz gauge? 
 A: It's supposed to be the latter, indeed. Let's go through the entire argument for completeness.
So if we start with the Maxwell equations (in units that nobody else uses but I like, with $\dot X = c^{-1}\partial_t X$),$$\begin{align}
\nabla\cdot E &= c\rho, &\nabla\times E &= -\dot B,\\
\nabla\cdot B &= 0, & \nabla\times B&=J + \dot E,
\end{align}$$and we observe that we can automatically satisfy the third equation by choosing some $A$ such that $B = \nabla \times A$, this means that the second equation is $\nabla \times (E + \dot A) = 0$ which can be automatically satisfied by choosing some $\phi$ such that $E + \dot A = -\nabla \phi.$ 
These two new entities $A,\phi$ are not directly physical as they are integrals of the fields and thus come with, in some sense, constants of integration. This choice of constant of integration is the "gauge freedom" of the entities. Taking them in the order they came, we can see that $B$ is kept identical even under a mapping $A \mapsto A + \nabla \sigma$ for any $\sigma,$  but we can see that this mapping would by itself have an impact on $E$, namely $E\mapsto E -\nabla\dot\sigma.$ However there is a very easy solution to this which is to simultaneously map $\phi\mapsto \phi-\dot\sigma,$ which has no further effect on $B$ and so both fields are preserved.
Now if we define $\lambda = \dot\phi + \nabla\cdot A,$ the first and fourth equations take the form,
$$\begin{align}\Box\phi &= c\rho + \dot\lambda,\\
\Box A &= J - \nabla\lambda.\end{align}$$
The simultaneous gauge mapping above maps $\lambda \mapsto \lambda - \Box\sigma$ and so one can verify that the first and fourth equations actually do not vary their shape based on this gauge mapping: the added $-\Box\dot\sigma$ on the left hand side of the first equation cancels perfectly with the added $-\Box\dot\sigma$ on the right hand side, and the same for the fourth. So one might wonder whether this really makes a difference.
The answer is yes, because one can choose a functional form for $\lambda$ and enforce it. So for example the Coulomb gauge would really like to return the first equation to $-\nabla^2\phi = c\rho$ and it can do that precisely by choosing the functional form $\lambda = \dot\phi.$ "But what if I did not happen to choose $A,\phi$ satisfying this form?" you may ask. Simple: take whatever $A,\phi$ you chose and then derive the corresponding $\lambda$ and then solve the wave equation $\Box \sigma =\lambda - \dot \phi,$ and you will find that with this choice of $\sigma$ we obtain $\lambda \mapsto \dot \phi.$ 
The Lorentz gauge solves $\Box\sigma=\lambda$ for $\sigma$ and thus obtains the functional form $\lambda = 0$. It obtains this form for the new choice of $A^\mu$, answering your question.
However there is a reason to be sloppy and it is that one can simply postulate, from the start, that we happened to choose the correct $A,\phi$ to give a certain functional form for $\lambda.$ That is, the existence theorem for solutions to the wave equation $\Box \sigma = \tau$ gives us an implicit existence theorem for solutions to the two potential-form Maxwell equations satisfying a wide variety of different gauge choices for $\lambda.$ So once that happens it makes sense to just say "we chose $A_\mu$ such that $\partial^\mu A_\mu = 0.$"
