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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 $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$ and $\mathbf{B}=\nabla \times \mathbf{A}$. Thus $\mathcal{L}=({\mathbf{E}}^2-{\mathbf{B}}^2)/2$. Taking the Cartesian components of $\mathbf{A}$ as coordinates gives the canonical momentaThis 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 ${\Pi }_i{\dot{A}}_i-\mathcal{L}=({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2-\mathbf{\Pi }\cdot \nabla A_0$. After integration by parts, the Hamiltonian isSurface 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 requireswhich is the Gauss law constraint in gauge theory. Both constraints are first-class constraints; they express the gauge invariance $\mathbf{A}\mapsto \mathbf{A}+\nabla \alpha$, $A_0\mapsto A_0-\dot{\alpha }$. 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 $\nabla \cdot \mathbf{A}=0$. In the absence of charges, the Gauss law constraint in gauge theory gives ${\nabla }^2{A}_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf{A}$. The remaining variables are the transverse pairs ${\mathbf{A}}^T,{\mathbf{\Pi }}^T={\dot{\mathbf{A}}}^T$. Their gauge reduction of the Maxwell Hamiltonian isThe 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 ${\partial }_i{A}_i^T=0$.

The reduced Dirac bracket, promoted to a canonical commutation relation, instead gives the transverse equal-time commutator:whereIndeed the spatial transverse projector of a vector field $P_{ij}^T={\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$ 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 ${\mathbf{ϵ}}_r(\mathbf{k})$, so thatThe canonical transverse photon field has mode expansionDifferentiate 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 areSubstitution 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 $i{\delta }_{ij}^T$. 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 $a^{†}{a}^{†}$ terms from the electric and magnetic energies cancel because ${\omega }_{\mathbf{k}}^2=|\mathbf{k}{|}^2$ and $\mathbf{k}\cdot {\mathbf{ϵ}}_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by normal ordering givesThe corresponding momentum operator is $\mathbf{P}={\sum }_r\int d^{3}k\mathbf{k}{a}_r^{†}{a}_r/(2\pi {)}^3$. These expressions imply $[H_T,a_r^{†}(\mathbf{k})]={\omega }_{\mathbf{k}}{a}_r^{†}(\mathbf{k})$ and $[\mathbf{P},a_r^{†}(\mathbf{k})]=\mathbf{k}{a}_r^{†}(\mathbf{k})$.

Define the Fock vacuum by $a_r(\mathbf{k})|0⟩=0$. Applying $a_r^{†}(\mathbf{k})$ creates a 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 $|\mathbf{k},r⟩=\sqrt{2{\omega }_{\mathbf{k}}}{a}_r^{†}(\mathbf{k})|0⟩$, 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 $\boxed{h=\pm 1}$.

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 $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ 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.