# Why are modes other than TEM not allowed for unguided waves?

I know the argument for why TEM waves are only allowed in not simply connected domains in waveguides. But when talking about unguided waves propagating in the vacuum, in many sources plane waves $$E(\mathbf{r}, t) = \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$$ with $$\mathbf{E}_0$$ constant (and analogously for the B field) are simply proposed as solutions to the wave equations.

From the derivation of TE and TM modes in waveguides (where $$\mathbf{E}_0$$ is a function of the position perpendicular to the propagation direction and does have a non-perpendicular component), it is not obvious to me why this approach is not allowed for unguided waves. Do we still obtain just TEM waves or is there any other reason why in free space, only transversal waves are always shown?

$$\lim_{R\to \infty} R\mathbf E \text{ is finite and}\\ \lim_{R\to \infty} \left( \mathbf {\hat r}\times\mathbf H+\sqrt{\frac{\epsilon_0}{\mu_0}} \mathbf E\right)=0 \tag{1}\label{1}$$ or
$$\lim_{R\to \infty} R\mathbf H \text{ is finite and}\\ \lim_{R\to \infty} \left( \mathbf {\hat r}\times \sqrt{\frac{\epsilon_0}{\mu_0}} \mathbf E-\mathbf H\right) =0 \tag{2}\label{2}$$ without which the integral equations describing the radiated fields do not converge ($$\mathbf {\hat r}$$ is the unit vector in the direction of the observation that is at a distance $$R$$ away from the origin). Here $$\eqref{1}$$ and $$\eqref{2}$$ show how a TE or TM wave, resp., turn asymptotically to TEM waves at large distance from the source. A very good description of this you may find here.