Consider a classical theory described by a Lagrangian $\mathscr{L}$ under the constraint $C=0$. We may make use of the Lagrange multipliers method and write the following, $$\mathscr{L}\mapsto\mathscr{L}+\lambda C, \tag{1}$$ as a modified Lagrangian from which we can derive our equation of motion.
Classical electrodynamics can be described by the following Lagrangian: $$\mathscr{L}_{ED}=-\frac{1}{4} F^{\mu \nu} F_{\mu \nu} \tag{2}$$ In $\mathscr{L}_{ED}$, we have gauge symmetry. The symmetry allows one to fix a gauge, the Lorenz gauge ($\partial_\mu A^\mu=0$) is one such example. One may think of the Lorenz gauge as a constraint on our system and we can make use of the Lagrange multipliers method (as in eq.(1)) to write, $$\mathscr{L}_{ED}\mapsto -\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{2\xi}(\partial_\mu A^\mu)^2 \tag{3}.$$ Somewhat sneakily, we have used $(\partial_\mu A^\mu)^2=0$ instead of $\partial_\mu A^\mu=0$ as our constraint; we have also rewritten the Lagrange multiplier in eq.(1) as $\lambda\equiv1/2\xi$.
Now, the Lagrange multiplier cannot be a priori fixed. The Lagrange multiplier must be found by solving the Euler Lagrange equations under the constraint.
My question is the following: Why can we choose the Feynman gauge $\xi=1$ (or the Landau gauge $\xi=0$ for that matter)? The $\xi$ term essentially is the Lagrange multiplier, so why can we just fix it to be 1 or 0?
Comment: I suspect that this has something to do with the Lorenz gauge not fully fixing the gauge, but I am certainly not confident in that statement.
