Schrödinger equation solutions confusion In class we've been learning about the Schrödinger equation and its solutions. We mentioned that one of the simplest solutions for a free particle are simply plane waves $\psi(x,t)=e^{i(\omega t-k x)}$, but then it was said that those solutions are invalid since they are not normalisable, and that every functions for which a fourier transform exists is a valid solution.
My questions:


*

*Is a plane wave a valid solution to the Schrödinger equation or not, since it is trivial to show that it does solve the S. Equation?

*Are only functions for which a fourier transform exists solution or is that not a necessary condition?
 A: A plane wave is obviously a solution to the Schrödinger equation, but it cannot be normalised and hence cannot represent a particle. 
The condition for $\psi(\boldsymbol{x},t)$ to be normalisable, i.e.
$
\int \mathrm{d}\boldsymbol{x}\,|\psi(\boldsymbol{x},t)|^2 < \infty,
$
is a sufficient condition for the existence of the Fourier transform. However, it is not necessary. For example, the $\delta$ function is not normalisable but does possess a Fourier transform (in fact every tempered distribution does), contradicting what you were told. Thus arguments based on the existence of the Fourier transform should be avoided.
The Fourier transform of any normalisable wave function $\psi(\boldsymbol{x},t)$ ist just a decomposition of $\psi(\boldsymbol{x},t)$ into a spectrum of plane waves. Thus, while each of these waves cannot be normalised, their combination can. Certain problems become much simpler when expressed in the wave domain, i.e. when Fourier transforming the Schrödinger equation (and the wave function). This is why plane waves are important wave functions to consider.
