= Solution
Use the free <Maxwell Lagrangian> $\mathcal L=-F_{\mu\nu}F^{\mu\nu}/4$ in mostly-minus signature. In terms of the spatial <electromagnetic four-potential>, define $\mathbf E=-\dot{\mathbf A}-\boldsymbol\nabla A_0$ and $\mathbf B=\boldsymbol\nabla\times\mathbf A$. Thus $\mathcal L=(\mathbf E^2-\mathbf B^2)/2$. Taking the Cartesian components of $\mathbf A$ as coordinates gives the <canonical momenta>
$$
\boxed{\Pi^0=0,\qquad \Pi_i=\frac{\partial\mathcal L}{\partial\dot A_i}=\dot A_i+\partial_iA_0=-E_i.}
$$
This sign convention treats $A_i$ here as Cartesian spatial-vector labels; it is not a claim that covariant and contravariant four-vector components coincide.
The temporal momentum vanishes because no $\dot A_0$ occurs. This is the <primary momentum constraint of the electromagnetic potential>, so one cannot treat all four components as independent unconstrained oscillators. The <Legendre transformation> for the spatial components gives $\Pi_i\dot A_i-\mathcal L=(\boldsymbol\Pi^2+\mathbf B^2)/2-\boldsymbol\Pi\cdot\boldsymbol\nabla A_0$. After integration by parts, the <Hamiltonian> is
$$
\boxed{H=\int d^3x\left[\frac12(\boldsymbol\Pi^2+\mathbf B^2)+A_0\boldsymbol\nabla\cdot\boldsymbol\Pi\right].}
$$
Surface terms vanish for decaying fields or the corresponding regulated boundary conditions. The total constrained <Hamiltonian> also contains an arbitrary multiplier of $\Pi^0$. Preserving $\Pi^0=0$ under time evolution requires
$$
\dot\Pi^0=-\frac{\delta H}{\delta A_0}=-\boldsymbol\nabla\cdot\boldsymbol\Pi=0,
$$
which is the <Gauss law constraint in gauge theory>. Both constraints are <first-class constraints>; they express the <gauge invariance> $\mathbf A\mapsto\mathbf A+\boldsymbol\nabla\alpha$, $A_0\mapsto A_0-\dot\alpha$. The electric and magnetic fields are unchanged, so different gauge potentials need not describe different physical states.
One consistent <canonical quantization of the electromagnetic field> first reduces this redundancy. Choose <Coulomb gauge> $\boldsymbol\nabla\cdot\mathbf A=0$. In the absence of charges, the <Gauss law constraint in gauge theory> gives $\nabla^2A_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf A$. The remaining variables are the transverse pairs $\mathbf A^T,\boldsymbol\Pi^T=\dot{\mathbf A}^T$. Their <gauge reduction of the Maxwell Hamiltonian> is
$$
\boxed{H_T=\frac12\int d^3x\left[(\boldsymbol\Pi^T)^2+(\boldsymbol\nabla\times\mathbf A^T)^2\right].}
$$
The two transverse coordinates for each nonzero momentum are the physical <degrees of freedom>. Treating the constraints as operator identities while retaining unconstrained componentwise commutators would be inconsistent: differentiating $[A_i^T,\Pi_j^T]=i\delta_{ij}\delta^3$ would give a nonzero right side even though $\partial_iA_i^T=0$.
The reduced <Dirac bracket>, promoted to a <canonical commutation relation>, instead gives the <transverse equal-time commutator>:
$$
\boxed{[A_i^T(t,\mathbf x),\Pi_j^T(t,\mathbf y)]=i\delta^T_{ij}(\mathbf x-\mathbf y),\qquad
[A_i^T,A_j^T]=[\Pi_i^T,\Pi_j^T]=0,}
$$
where
$$
\delta^T_{ij}(\mathbf r)=\int\frac{d^3k}{(2\pi)^3}e^{i\mathbf k\cdot\mathbf r}
\left(\delta_{ij}-\frac{k_i k_j}{|\mathbf k|^2}\right).
$$
Indeed the spatial <transverse projector of a vector field> $P^T_{ij}=\delta_{ij}-k_ik_j/|\mathbf k|^2$ is symmetric, has $P^T\mathbf k=0$ and $(P^T)^2=P^T$. Projecting both entries of the unconstrained canonical bracket onto the transverse subspace therefore produces $P^T\delta^3$, equivalently the constrained <Dirac bracket>. The spatial zero mode is excluded by the decay condition; in a finite-volume regulator it must be treated separately.
Choose two orthonormal transverse <photon polarization vectors> $\boldsymbol\epsilon_r(\mathbf k)$, so that
$$
\mathbf k\cdot\boldsymbol\epsilon_r=0,\qquad
\boldsymbol\epsilon_r^*\cdot\boldsymbol\epsilon_s=\delta_{rs},\qquad
\sum_{r=1}^2\epsilon_i^{(r)}\epsilon_j^{(r)*}=P^T_{ij}.
$$
The <canonical transverse photon field> has mode expansion
$$
A_i^T(x)=\sum_{r=1}^2\int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_{\mathbf k}}}
\left[\epsilon_i^{(r)}(\mathbf k)a_r(\mathbf k)e^{-ik\cdot x}
+\epsilon_i^{(r)*}(\mathbf k)a_r^\dagger(\mathbf k)e^{ik\cdot x}\right],
\qquad \omega_{\mathbf k}=|\mathbf k|.
$$
Differentiate in time to obtain $\Pi_i^T$: the annihilation coefficient acquires $-i\omega_{\mathbf k}$ and the creation coefficient $+i\omega_{\mathbf k}$. The oscillator <canonical commutation relations> are
$$
\boxed{[a_r(\mathbf k),a_s^\dagger(\mathbf q)]=(2\pi)^3\delta_{rs}\delta^3(\mathbf k-\mathbf q),\qquad
[a_r,a_s]=[a_r^\dagger,a_s^\dagger]=0.}
$$
Substitution into $[A_i^T,\Pi_j^T]$ gives $i/2$ times the sum of the two Fourier kernels with phases $\pm i\mathbf k\cdot(\mathbf x-\mathbf y)$. The polarization completeness relation and the evenness of $P^T(\mathbf k)$ make that sum exactly $i\delta^T_{ij}$. This derives, rather than presumes, the correspondence between the oscillator and field commutators.
Substitute the mode expansions into $H_T$. Spatial integration sets paired momenta equal or opposite. The $aa$ and $a^\dagger a^\dagger$ terms from the electric and magnetic energies cancel because $\omega_{\mathbf k}^2=|\mathbf k|^2$ and $\mathbf k\cdot\boldsymbol\epsilon_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by <normal ordering> gives
$$
\boxed{:\!H_T\!:=\sum_{r=1}^2\int\frac{d^3k}{(2\pi)^3}\omega_{\mathbf k}
a_r^\dagger(\mathbf k)a_r(\mathbf k).}
$$
The corresponding <momentum operator> is $\mathbf P=\sum_r\int d^3k\,\mathbf k\,a_r^\dagger a_r/(2\pi)^3$. These expressions imply $[H_T,a_r^\dagger(\mathbf k)]=\omega_{\mathbf k}a_r^\dagger(\mathbf k)$ and $[\mathbf P,a_r^\dagger(\mathbf k)]=\mathbf k a_r^\dagger(\mathbf k)$.
Define the <Fock vacuum> by $a_r(\mathbf k)|0\rangle=0$. Applying $a_r^\dagger(\mathbf k)$ creates a \b[<photon> of energy $|\mathbf k|$, momentum $\mathbf k$, and one of two transverse polarizations]. Its mass is zero because $k^2=0$. The relativistically normalized state is $|\mathbf k,r\rangle=\sqrt{2\omega_{\mathbf k}}a_r^\dagger(\mathbf k)|0\rangle$, with inner product $2\omega_{\mathbf k}(2\pi)^3\delta_{rs}\delta^3(\mathbf k-\mathbf q)$. Smearing these momentum eigenstates gives normalizable wave packets. Repeated commuting <bosonic creation operators> construct the symmetric multiphoton <bosonic Fock space>. Circular combinations of the two transverse <photon polarization vectors> transform with opposite phases under rotations about $\mathbf k$, giving the two physical <helicities> $\boxed{h=\pm1}$.
The same physical states can be obtained while retaining manifest Lorentz covariance through <Gupta-Bleuler quantization>. Add the <Feynman gauge> term $-(\partial_\mu A^\mu)^2/2$ and quantize four auxiliary oscillator polarizations. Their algebra has an indefinite metric, including a negative-norm temporal excitation. Impose $(\partial_\mu A^\mu)^{(+)}|\mathrm{phys}\rangle=0$ and take the <Gupta-Bleuler null-state quotient>. Temporal and longitudinal excitations then do not survive as independent physical states. The result agrees with the positive, two-polarization space obtained by <Coulomb gauge> reduction; covariant gauge fixing is a different implementation of the same physical constraint, not four physical photon states.
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