The field operator is still a wave§
From the continuous‑mode expansion we have
E ^ ( x , t ) = i ∫ 0 ∞ d ω ℏ ω 4 π ϵ 0 c E ^ ( ω ) e i ( k x − ω t ) + h.c. , k = ω c . \hat{E}(x,t)=i\int_{0}^{\infty}\mathrm{d}\omega\,\sqrt{\frac{\hbar\omega}{4\pi\epsilon_0c}}\,\hat{E}(\omega)\,e^{i(kx-\omega t)}+\text{h.c.},
\qquad k = \frac{\omega}{c}. E ^ ( x , t ) = i ∫ 0 ∞ d ω 4 π ϵ 0 c ℏ ω E ^ ( ω ) e i ( k x − ω t ) + h.c. , k = c ω . Each term is proportional to the plane wave e i ( k x − ω t ) e^{i(kx-\omega t)} e i ( k x − ω t ) x x x t t t ( ∂ x 2 − 1 c 2 ∂ t 2 ) e i ( k x − ω t ) = 0 (\partial_x^2 - \frac{1}{c^2}\partial_t^2)e^{i(kx-\omega t)}=0 ( ∂ x 2 − c 2 1 ∂ t 2 ) e i ( k x − ω t ) = 0
Commutation in x , t x, t x , t §
[ E ^ ( x , t ) , E ^ ( x ′ , t ′ ) ] = i ∫ 0 ∞ d ω d ω ′ ℏ ω ω ′ 4 π ϵ 0 c [ E ^ ( ω ) , E ^ ( ω ′ ) ] e i ( k x − ω t ) e i ( k ′ x ′ − ω ′ t ′ ) + E ^ † E ^ † terms + E ^ E ^ † terms + E ^ † E ^ terms \left[\hat{E}(x,t), \hat{E}(x',t')\right]=i\int_0^\infty\mathrm{d}\omega\mathrm{d}\omega'\frac{\hbar\sqrt{\omega\omega'}}{4\pi\epsilon_0c}\left[\hat{E}(\omega),\hat{E}(\omega')\right]e^{i(kx-\omega t)}e^{i(k'x'-\omega't')} + \hat{E}^\dagger\hat{E}^\dagger \text{terms} + \hat{E}\hat{E}^\dagger \text{terms} + \hat{E}^\dagger\hat{E} \text{terms} [ E ^ ( x , t ) , E ^ ( x ′ , t ′ ) ] = i ∫ 0 ∞ d ω d ω ′ 4 π ϵ 0 c ℏ ω ω ′ [ E ^ ( ω ) , E ^ ( ω ′ ) ] e i ( k x − ω t ) e i ( k ′ x ′ − ω ′ t ′ ) + E ^ † E ^ † terms + E ^ E ^ † terms + E ^ † E ^ terms Then, using the known commutation relation, only [ E ^ , E ^ † ] \left[\hat{E},\hat{E}^\dagger\right] [ E ^ , E ^ † ]
[ E ^ ( x , t ) , E ^ ( x ′ , t ′ ) ] = i ∫ 0 ∞ d ω d ω ′ ℏ ω ω ′ 4 π ϵ 0 c ( δ ( ω − ω ′ ) e i ( k x − ω t ) e − i ( k ′ x ′ − ω ′ t ′ ) − δ ( ω − ω ′ ) e − i ( k x − ω t ) e i ( k ′ x ′ − ω ′ t ′ ) ) = i ∫ 0 ∞ d ω ℏ ω 4 π ϵ 0 c ( e i k ( Δ x − c Δ t ) − e − i k ( Δ x − c Δ t ) ) = − ℏ 2 π ϵ 0 c ∫ 0 ∞ d ω ω sin ( ω c ( Δ x − c Δ t ) ) = − i ℏ c 2 ϵ 0 δ ′ ( Δ x − c Δ t ) \begin{align*}
\left[\hat{E}(x,t), \hat{E}(x',t')\right]&=i\int_0^\infty\mathrm{d}\omega\mathrm{d}\omega'\frac{\hbar\sqrt{\omega\omega'}}{4\pi\epsilon_0c}\left(\delta(\omega-\omega')e^{i(kx-\omega t)}e^{-i(k'x'-\omega't')}-\delta(\omega-\omega')e^{-i(kx-\omega t)}e^{i(k'x'-\omega't')}\right) \\
&=i\int_0^\infty\mathrm{d}\omega\frac{\hbar\omega}{4\pi\epsilon_0c}\left(e^{ik(\Delta x-c\Delta t)}-e^{-ik(\Delta x-c\Delta t)}\right)\\
&=-\frac{\hbar}{2\pi\epsilon_0c}\int_0^\infty\mathrm{d}\omega\,\omega\sin\left(\frac{\omega}{c}(\Delta x - c\Delta t)\right)\\
&=-\frac{i\hbar c}{2\epsilon_0}\delta'(\Delta x - c\Delta t)
\end{align*} [ E ^ ( x , t ) , E ^ ( x ′ , t ′ ) ] = i ∫ 0 ∞ d ω d ω ′ 4 π ϵ 0 c ℏ ω ω ′ ( δ ( ω − ω ′ ) e i ( k x − ω t ) e − i ( k ′ x ′ − ω ′ t ′ ) − δ ( ω − ω ′ ) e − i ( k x − ω t ) e i ( k ′ x ′ − ω ′ t ′ ) ) = i ∫ 0 ∞ d ω 4 π ϵ 0 c ℏ ω ( e ik ( Δ x − c Δ t ) − e − ik ( Δ x − c Δ t ) ) = − 2 π ϵ 0 c ℏ ∫ 0 ∞ d ω ω sin ( c ω ( Δ x − c Δ t ) ) = − 2 ϵ 0 i ℏ c δ ′ ( Δ x − c Δ t ) This is right chiral as we didn't use the full form from the start, but anyway the left chiral is a symmetry to the right chiral. So easily the full form is
− i ℏ c 2 ϵ 0 ( δ ′ ( Δ x − c Δ t ) − δ ′ ( Δ x + c Δ t ) ) -\frac{i\hbar c}{2\epsilon_0}\left(\delta'(\Delta x - c\Delta t)-\delta'(\Delta x + c\Delta t)\right) − 2 ϵ 0 i ℏ c ( δ ′ ( Δ x − c Δ t ) − δ ′ ( Δ x + c Δ t ) )