The field operator is still a wave

From the continuous‑mode expansion we have

\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}.

Each term is proportional to the plane wave $e^{i(kx-\omega t)}$ (or its complex conjugate), which is the only factor depending on $x$ and $t$ .
Plane waves satisfy the wave equation $(\partial_x^2 - \frac{1}{c^2}\partial_t^2)e^{i(kx-\omega t)}=0$ . Because the operator is a linear superposition of such solutions, it obeys the same wave equation. Thus the quantize electric field is indeed a wave – it respects Maxwell’s equations (free field, Coulomb gauge) as an operator identity.

Commutation in $x, t$ domain

\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}

Then, using the known commutation relation, only $\left[\hat{E},\hat{E}^\dagger\right]$ can be non-zero,

\begin{aligned} [\hat{E} (x, t), \hat{E} (x^{'}, t^{'})] & = i \int_{0}^{\infty} d ω d ω^{'} \frac{ℏ \sqrt{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^{'}} \end{aligned}

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

-\frac{i\hbar c}{2\epsilon_0}\left(\delta'(\Delta x - c\Delta t)-\delta'(\Delta x + c\Delta t)\right)

-

ω^{'}

t^{'}

)

)

= i \int_{0}^{\infty} d ω \frac{ℏ ω}{4 π ϵ_{0} c} (e^{i k (Δ x - c Δ t)} - e^{- i k (Δ x - c Δ t)})

= - \frac{ℏ}{2 π ϵ_{0} c} \int_{0}^{\infty} d ω ω \sin (\frac{ω}{c} (Δ x - c Δ t))

= - \frac{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*}

Advance Properties of Field Operators

The field operator is still a wave

Commutation in x,tx, tx,t domain

Commutation in $x, t$ domain