Can we safely assume $\Psi(x,t) = \psi(x)e^{-i\omega t}$ always in QM? In the particle in a box, harmonic oscillator and in Hydrogen Atom, we can safely assume $$\Psi(x,t) = \psi(x)e^{-i\omega t}.$$ So why not make it a postulate to consider the wave function to be always in the form $$\psi(x)e^{-i\omega t}$$ We can still explain all the foundational experiments, so my question is why not make this amendment to the theory so that we can avoid a lot of nightmares and some peace of mind. Why unnecessarily consider so general thing when we can do away with simple things most of the time?
Oh my actual question is where do we run into problems (practical physical situations/experiments) if we consider such a thing? If we run into problem anywhere, I hope to make any other simple changes to fix them but without going back on this one.
 A: In a different vein from the other answers, an example of physical a problem in which the solution is not separable is one considered in Sakurai 2nd Edition, Chapter 2.1, Pgs. 70 and 71 where we have a spin magnetic moment in the presence of a magnetic field whose intensity can change with time I.e. $B(t)\neq B(0)$, or even more problematically when the magnitude and direction of the magnetic field change with time. In both cases the Hamiltonian is time dependent and so the solution is not separable.
A: It is not general. For both the harmonic oscillation and the hydrogen atom, we have a Hamitlonian $\hat{H} = -\frac{i\hbar}{2m}\nabla^2 + V$ with a time-independent potential, which implied that the eigenvalue equation $\hat{H}\Psi(x,t) = E\Psi(x,t)$ is separable, and we can therefore write it as a product of a time-independent factor and a space-independent factor.
In other words, $\Psi(x,t) = \Psi(x)e^{-i\omega t}$ is something we get for the energy eigenstates with a time-independent Hamiltonian. A general solution is a superposition of multiple energy eigenstates, and is not going to have that form.
A: Another, as of yet unmentioned, reason is for super-position. 
For separation of variables techniques, we find $\Psi(\mathbf{x},t)=\psi(\mathbf{x})f(t)$ to give
$$
\frac{i\hbar}{f(t)}\frac{df}{dt}=E=\left[\frac{\hbar^2}{2m}\nabla^2+V(\mathbf{x})\right]\psi(\mathbf{x})\tag{1}
$$
The term on the left side gives $f\propto\exp\left[iEt/\hbar\right]$, giving us
$$
\psi\left(\mathbf{x},t\right)=\psi(x)e^{iEt/\hbar}\tag{2}
$$
But this is a stationary state, but the particle that Equation (1) describes is not stationary. Thus, we have that the general solution is a linear super-position of states:
$$
\psi\left(\mathbf{x},t\right)=\sum_ic_i\psi_i(x)e^{iE_it/\hbar}
$$
Which cannot be separated into such a form as Equation (2).
A: Only when $V(\mathbf{x},t) = V(\mathbf{x})$, otherwise the time-dependent Schrödinger equation does not separate into time- and space-like parts.
Any explicit time-dependence in the problem and you cannot.
