= Solution
Use signature $(+---)$ and take the covariant spatial components $A_i$ as canonical coordinates. The <canonical quantization of the electromagnetic field> begins with the canonical momenta
$$
\Pi^\mu=\frac{\partial\mathcal L}{\partial(\partial_0A_\mu)}=-F^{0\mu},\qquad\Pi^0=0,\qquad\Pi^i=F_{0i}=\dot A_i-\partial_iA_0.
$$
Thus the <primary momentum constraint of the electromagnetic potential> is $\Pi^0=0$: $A_0$ has no independent velocity. The <Hamiltonian> obtained by the <Legendre transform in mechanics>, up to a boundary term, is
$$
H=\int d^3x\left[\frac12\Pi^i\Pi^i+\frac14F_{ij}F_{ij}-A_0\partial_i\Pi^i\right].
$$
Preserving the primary constraint requires the <Gauss law constraint in gauge theory>, $\partial_i\Pi^i=0$. It also follows by varying $A_0$. These are two <first-class constraints>; they generate the gauge freedom and remove two <canonical pairs> from the four potential components. The reduced phase space has four dimensions per spatial mode, hence \b[two propagating <photon> degrees of freedom].
Impose <Coulomb gauge>, $\partial_iA_i=0$. With no charges, <Gauss's law> then gives $\nabla^2A_0=0$; vanishing boundary conditions set $A_0=0$. This is <radiation gauge>. The remaining components are transverse and obey the massless <wave equation>. Let $\epsilon_i^{(r)}(\mathbf k)$, $r=1,2$, be orthonormal transverse <polarization vectors>. Their <photon polarization completeness relation> is
$$
k_i\epsilon_i^{(r)}=0,\qquad\sum_{r=1}^2\epsilon_i^{(r)}\epsilon_j^{(r)*}=P^T_{ij}(\mathbf k):=\delta_{ij}-\frac{k_ik_j}{|\mathbf k|^2}.
$$
The <canonical transverse photon field> is the Hermitian operator
$$
A_i^T(x)=\sum_{r=1}^2\int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_{\mathbf k}}}\left[\epsilon_i^{(r)}a_r(\mathbf k)e^{-ik\cdot x}+\epsilon_i^{(r)*}a_r^\dagger(\mathbf k)e^{ik\cdot x}\right],\qquad\omega_{\mathbf k}=|\mathbf k|,
$$
where
$$
[a_r(\mathbf k),a_s^\dagger(\mathbf k')]=(2\pi)^3\delta_{rs}\delta^3(\mathbf k-\mathbf k'),\qquad[a_r,a_s]=[a_r^\dagger,a_s^\dagger]=0.
$$
The field and its <conjugate momentum> have the <transverse equal-time commutator>
$$
[A_i^T(t,\mathbf x),\Pi^{Tj}(t,\mathbf y)]=i\delta^T_{ij}(\mathbf x-\mathbf y),\qquad\delta^T_{ij}(\mathbf x)=\int\frac{d^3k}{(2\pi)^3}P^T_{ij}(\mathbf k)e^{i\mathbf k\cdot\mathbf x}.
$$
This is the quantized reduced bracket, or equivalently the <Dirac bracket> after imposing the constraints and gauge conditions. The normal-ordered <Hamiltonian> is $\sum_r\int d^3k\,\omega_{\mathbf k}a_r^\dagger a_r/(2\pi)^3$. Its excitations are <photons>; circular combinations of the two transverse polarizations have <helicity> $+1$ and $-1$. The scalar and longitudinal potential components do not create additional physical <photons>.
The <Feynman propagator> is the vacuum expectation of a <time-ordered product>. The mode expansion directly gives the <radiation-gauge photon propagator>, with $z=x-y$:
$$
\begin{aligned}
D^F_{ij}(z)&=\int\frac{d^3k}{(2\pi)^3}\frac{P^T_{ij}(\mathbf k)}{2\omega_{\mathbf k}}e^{i\mathbf k\cdot\mathbf z}\left[\theta(z^0)e^{-i\omega_{\mathbf k}z^0}+\theta(-z^0)e^{i\omega_{\mathbf k}z^0}\right]\\
&=\int\frac{d^4k}{(2\pi)^4}\frac{iP^T_{ij}(\mathbf k)}{k^2+i0}e^{-ik\cdot z}.
\end{aligned}
$$
The first expression comes from the creation-annihilation commutator; the second is its contour-integral representation. The positive-energy pole lies below the real axis and the negative-energy pole above it. In this reduced free-field description, the temporal operator is zero. A <photon propagator> must specify its gauge; the spatial transverse propagator is not the same tensor as the covariant four-potential propagator.
For the commonly used covariant form, add the <gauge fixing> term $-(\partial_\mu A^\mu)^2/(2\xi)$. The resulting Fourier-space kinetic operator is
$$
K^{\mu\nu}(k)=-k^2\eta^{\mu\nu}+(1-\xi^{-1})k^\mu k^\nu.
$$
The <inversion of the gauge-fixed Maxwell kinetic operator> gives $K^{\mu\nu}D^F_{\nu\rho}=i\delta^\mu{}_{\rho}$. In <Feynman gauge>, $\xi=1$, the <photon propagator> is
$$
\boxed{D^F_{\mu\nu}(x-y)=\int\frac{d^4k}{(2\pi)^4}\frac{-i\eta_{\mu\nu}}{k^2+i0}e^{-ik\cdot(x-y)}.}
$$
Equivalently, $\Box D^F_{\mu\nu}=i\eta_{\mu\nu}\delta^4(x-y)$ with the vacuum pole prescription. For general $\xi$, the momentum-space numerator is $\eta_{\mu\nu}-(1-\xi)k_\mu k_\nu/(k^2+i0)$.
A covariant canonical realization uses four polarization oscillators with $[a_r,a_s^\dagger]=-(2\pi)^3\eta_{rs}\delta^3(\mathbf k-\mathbf k')$. The resulting indefinite <inner product> is auxiliary. In <Gupta-Bleuler quantization>, impose $(\partial_\mu A^\mu)^{(+)}|\mathrm{phys}\rangle=0$ and take the <Gupta-Bleuler null-state quotient>. The scalar-longitudinal combination is thereby removed from the physical state space, leaving the same two transverse <photon> states. Thus the four-component Feynman-gauge numerator does not imply four physical polarization states.
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