In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freedom
$A_\mu$ has 4 components
$A_0$ is non-dynamical (-1)
$\mathcal{L}_\text{Maxwell}$ enjoys gauge symmetry and should be removed (-1)
So in total, we have $4 - 1 - 1 = 2$ dofs.
However, in a quantization procedure as discussed in David Tong's Quantum Field Theory lecture note (section 6.2.2 Lorenz Gauge), the Lagrangian is modified to $$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2\xi} (\partial_\mu A^\mu)^2 \ . $$
$A_\mu$ has 4 components.
$A_0$ is dynamical again
there is no gauge symmetry left.
Question: at this stage, can we claim that there are only 2 propagating dofs? (I feel we cannot.)
Of course, subsequently we introduced the Gupta-Bleuler condition $$ \partial^\mu A_\mu^{(+)} |\text{phy}\rangle = 0 $$ to make the observation (or rather, to decide) that only the two transverse polarisation are "physical". I feel, only at this stage, that we are in position to fix #dof to 2.
Please correct me if my understanding is incorrect (the counting of dofs has been fuzzy in my head, although I know what words to say to get the right number).
