lagrangian formalism - Are Maxwell's equations "physical"?

The canonical Maxwell's equations are derivable from the Lagrangian

$${\cal L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$

by solving the Euler-Lagrange equations.

However: The Lagrangian above is invariant under the gauge transformation

$$A_\mu \to A_\mu - \partial_\mu \Lambda(x) $$

for some scalar fiend $\Lambda(x)$ that vanishes at infinity. This implies that there will be redundant degrees of freedom in our equations of motion (i.e. Maxwell's equations).

Therefore, as I understand gauge fixing, this implies that Maxwell's equations (without gauge fixing) can lead to unphysical predictions.

Question: Hence my question is simply are Maxwell's equations (the ones derived from $\cal{L}$ above) actually physical, in the sense they do not make unphysical predictions?

Example: The general solution to the equations of motion derived from $\cal{L}$ is given by

$$A_\mu(x) = \sum_{r=0}^3 \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(\epsilon^r_\mu(\mathbf{p}) a^r_\mathbf{p}e^{-ipx} + \epsilon^{*r}_\mu(\mathbf{p}) (a^r_\mathbf{p})^\dagger e^{ipx} \right)$$

where we have, at first, 4 polarization states for external photons.

My understanding: is that we can remove one of these degrees of freedom by realizing that $A_0$ is not dynamical, but to remove the other one we have to impose gauge invariance (cf. (2)). This seems to imply that unless we fix a gauge Maxwell's equations will predict a longitudinal polarization for the photon.

Answer

Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.

At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.

However, such a theory has a deep problem: the equation of motion $\partial_\mu F^{\mu\nu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.

Note that this isn't true if you add a mass term $m^2 A_\mu A^\mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.