I would like to make the link between the Gupta-Bleuler Lagrangian and the Lorenz Gauge for Electromagnetism because everything is not clear to me.
I am looking for a simple explanation without too many references if possible.
So, from what I have understood :
We have $$\mathcal{L}_0=-\frac 14 F_{\mu \nu} F^{\mu \nu}$$
The problem with this Lagrangian is that we have $\Pi_0=0$: the momentum associated to $A_0$ is $0$.
Also, the field equations are:
$$ \Box A_{\mu}-\partial_\mu \partial^{\nu} A_\nu=0$$
And there is no propagator associated with this equation (no Green function).
Two problems here. The momentum associated to $A_0$ is $0$ and there is no propagator associated with the equation of motion.
The "trick" we do is to add a term $\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$ to the Lagrangian.
When we do it, the Lagrangian is now $$\mathcal{L}=-\frac 14 F_{\mu \nu} F^{\mu \nu}+\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$$
Then, we can show if we take $\xi=1$, and do a few integrations by parts, that this new Lagrangian is equivalent to $\mathcal{L}=\partial_\mu A_\nu \partial^\mu A^\nu$ and thus, will give the equations of motion :
$$ \Box A_{\mu} = 0$$
Then, we can quantize everything as the Klein Gordon field.
$$A_\mu=\int d\widetilde{k}\left(A_\mu(k)^+ e^{ikx} + A_\mu(k)^- e^{-ikx}\right) $$
And we implement the Lorenz Gauge by saying $\partial_\mu A^{\mu, +} |\psi_\text{Phys} \rangle=0$ where $|\psi_\text{Phys} \rangle$ are "physical" states of our Hilbert space. (we define them like this).
My questions:
I don't understand the global spirit of this. For me we modified the dynamics because we modified the Lagrangian. I think everything relies upon the fact that we work with an equivalent Lagrangian as the one we started with but I don't find where this is true? Why can we modify our Lagrangian like this?
I would like the most simple answer even if it is not very rigorous, I just would like to get the average Idea of the thing.