Physical meaning of the photon propagator using gauge fixing term and path integral

I am considering the following generating functional in terms of functional integral for electromagnetic gauge fields
begin{equation}
Z[J] = intmathcal{D}A_{mu} exp{left(iint d^4x left(-frac{1}{4}F_{mu nu}F^{mu nu} – frac{1}{2alpha}(partial_{mu}A^{mu})^{2} + J_{mu}A^{mu} right) right)},
end{equation}
where I have used the gauge-fixing term in the second term and the source term as the third term, and I have normalized the path integral measure such that $Z[J=0]=1$. It is usually described in books that we can calculate
begin{equation}
langle 0|That{A}_{mu}(x)hat{A}_{nu}(y)|0rangle
end{equation}
where $mu$ and $nu$ range from $0sim 3$, through
begin{equation}
frac{delta }{idelta J^{mu}(x)} frac{delta }{idelta J^{mu}(y)} Z[J]
end{equation}
and then set $J^{mu}=0$. However, I am a bit confused about the physical meaning of $langle 0|That{A}_{mu}(x)hat{A}_{nu}(y)|0rangle $, and how to calculate it from operator formalism, mentioned in 1 below.

In particular, when we are considering the canonical quantization of $A_{mu}$ for the Lagrangian $-frac{1}{4}F_{mu nu}F^{mu nu}$ without the gauge fixing term, we know that we can not quantize all the four $A_{mu}$. We need to, say pick Coulomb gauge together with $A_{0}=0$ and quantize the spatial components $A_{i}$ only, though we still can only have two physical degrees of freedom. It then makes sense to me that we can calculate
begin{equation}
langle 0|That{A}_{i}(x)hat{A}_{j}(y)|0rangle
end{equation}
where $i$ and $j$ range from $1 sim 3$ in this canonical quantization method.

My questions are then:

1. Can I view $langle 0|That{A}_{mu}(x)hat{A}_{nu}(y)|0rangle $ mentioned above as the propagator if we canonically quantized the theory with Lagrangian $-frac{1}{4}F_{mu nu}F^{mu nu} – frac{1}{2alpha}(partial_{mu}A^{mu})^{2}$? If we start from this Lagrangian with gauge fixing term, we will be able to canonically quantized the theory for all four components of $A_{mu}$.

2. I understand that, since now $langle 0|That{A}_{mu}(x)hat{A}_{nu}(y)|0rangle $ depends on the gauge we choose, it can’t be physical. But we can use it to calculate something that is physical, namely gauge invariant, for example the transition probability. So can I understand this as saying that: OK, we want to calculate the transition amplitude from a photon state to another, and we can use canonical quantization, which can make the calculation very difficult, but we can also use this Lagrangian with gauge fixing term that gives a covariant and $alpha$-dependent propagator $langle 0|That{A}_{mu}(x)hat{A}_{nu}(y)|0rangle $. We will also yield nice, covariant Feynman rule. And we will arrive the same final answer if we were to proceed using canonical quantization?

3. If I go back to the canonical quantization method and calculate $langle 0|That{A}_{i}(x)hat{A}_{j}(y)|0rangle $ where $i$ and $j$ range from $1 sim 3$, is this $langle 0|That{A}_{i}(x)hat{A}_{j}(y)|0rangle $ physical? I think since we need to pick a gauge to really calculate it, it can not be physical? But just as I mentioned in 2, we can use $langle 0|That{A}_{i}(x)hat{A}_{j}(y)|0rangle $ to calculate physical stuff, like transition probability?

4. A another maybe not that related question (but I just want to double check my understanding), it is always stated in books, or lectures, that we will view $1/alpha$ as a Lagrange multiplier. However I think $1/alpha$ is not a Lagrange multiplier, for if we want it to be a Lagrange multiplier, it should be dynamical, namely $1/alpha(x)$. And when we do a variational calculation on it, we yield the constraint equation. However we are just having $alpha$ as a constant. Then why I always see people saying that $1/alpha$ is a Lagrange multiplier? Is there any reason to make this (seemingly false) statement?