Stationary states in quantum mechanics I know that the general one-dimensional Schroedinger equation is given by:
$$-\frac{\hbar^2}{2m} \frac{\partial^2\Psi(x,t)}{\partial x^2} + U(x)\Psi(x,t) = i\hbar \frac{\partial \Psi(x,t)}{\partial t} $$
The source I am using mentions: "If the potential energy function is nonzero, these sinusoidal waves do not satisfy the Schroedinger equation". Why? I thought the reason why the general equation included $U(x)$ was exactly to address this problem.
Then it continues by: "However, we can still write the wave function for state of definite energy $E$ in the following form" $$\Psi(x,t) = \psi(x)e^{iEt/\hbar} $$
I understand that this comes from $$ \Psi(x,t) = Ae^{i(kx - wt)} $$
But why a state of definite energy must have potential energy non-zero?
 A: The form $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ does not "come" from $Ae^{i(kx-\omega t)}$ but thee from basic ansatz $\Psi(x,t)=\psi(x)\Phi(t)$ used to solve partial differential equations.
The form $\Psi(x,t)=\psi(x)\Phi(t)$ is used to convert the partial differential equation to a pair of ordinary differential equations connected by a separation constant.
Inserting $\Psi(x,t)=\psi(x)\Phi(t)$ into the time-dependent Schrodinger equation produces a result easily rearranged to
\begin{align}
\frac{1}{\psi(x)}\left(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+U(x)\psi(x)\right)=\frac{1}{\Phi(t)}\left(i\hbar \frac{d\Phi(t)}{dt}\right)=E
\end{align}
with $E$ the separation constant (to be identified with the energy). It is possible to solve the $\Phi$ equation immediately as it is independent of the potential:
$$
\Phi(t)=e^{-iEt/\hbar}\, ,
$$
from which $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ follows, with $\psi(x)$ a solution to the time-independent Schrodinger equation
$$
-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+U(x)\psi(x)=E\psi(x)\, .
\tag{1}
$$
The form $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ thus holds for arbitrary potential $U(x)$ provided $\psi(x)$ satisfies (1).
