Plane Wave Decomposition of Electric Field I've tried to understand the decomposition of an HF electrical field in a series of plane waves.
$$\vec{E}(\vec{r}, t) = \int\int\int \hat{\vec{E}}(\vec{k}) \cdot\mathrm{e}^{\mathrm{i}(\vec{k}.\vec{r}-\omega t)} \mathrm{d}^3\vec k$$
The fourier component corresponding to $\vec{k} = 0$ does not represent any plane wave. But it oscillates with $\omega$. How can I imagine this component, since from Maxwell's equations there is no homogeneous and oscillating field possible?
 A: Your expression is a solution of the wave equation
$$\vec\Delta\vec E-{1\over c^2}{\partial^2\vec E\over\partial t^2}=0$$
of the electric field only if the pulsation $\omega$ is not a free parameter but a function of $\vec k$, i.e. $\omega=\omega(\vec k)$. When plugging the plane wave $e^{i(\vec k.\vec r-\omega t)}$ into the wave equation, one gets the so-called dispersion law
$$\omega^2=||\vec k||^2c^2$$
Therefore, the homogeneous solution $\vec k=0$ leads to $\omega(0)=0$ and, as you correctly stated, does not oscillate.
To answer your comment, introduce the 3D Fourier transform
$$\vec E(\vec r,t)=\int \hat{\vec E}(\vec k,t)e^{i\vec k.\vec r}d^3\vec k$$
Note that the Fourier coefficients are time-dependent. When pluging into the wave equation, one gets the differential equation
$$-k^2 \hat{\vec E}(\vec k,t)-{1\over c^2}{\partial^2\hat{\vec E}
  \over\partial t^2}=0$$
whose solutions are
$$\hat{\vec E}_\pm(\vec k,t)=\hat{\vec E}(\vec k,0)e^{\pm i\omega t}$$
where $\omega^2=k^2c^2$. $\vec E(\vec r,t)$ is recovered by an inverse Fourier transform. Alternatively, one can introduce the 4D Fourier transform
$$\vec E(\vec r,t)=\int\hat{\vec E}(\vec k,\omega)e^{i(\vec k.\vec r-\omega t)}d^3\vec k d\omega$$
In this case, the wave equation implies that the Fourier coefficients vanish unless the dispersion law is satisfied.
