Schrödinger equation obtain $ψ(x,t)$ from $ψ(x,0)$

Question

In this answer of the post "Wave packet expression and Fourier transforms" it is said that for the S.E. we have this property:

• If we start with an initial profile $ψ(x,0)=e^{ikx}$, then the solution to our wave equation is $ψ(x,t)=e^{i(kx−ω_kt)}$, where $ω_k$ is a constant that may depend on $k$.

I would like if someone can explain to know how we can obtain $ψ(x,t)$ from $ψ(x,0)$ in (or with the help of) the S.E.

EDIT: My first attempt was to compose the $ψ(x)$ function with an $f$ function defined as follow $f(u,t) = \frac{u}{k}-\frac{\omega t}{k}$ but I don't think it's possible in math to compose a function of one variable with another function of two variables, is it?

Answer 1

Just plug $\psi(x,t)= e^{i(k x-\omega t)}$ into your translation-invariant wave equation and read off what $\omega(k)$ has to be to satisfy it.

For example the free Schrodinger equation $$ i\hbar \frac{\partial}{\partial t}\psi= -\frac{\hbar^2 }{2m}\frac{\partial^2 \psi}{\partial x^2} $$ gives $$ \hbar \omega e^{i(k x-\omega t)}= \frac{\hbar^2}{2m}k^2 e^{i(k x-\omega t)}, $$ so $$ \hbar \omega(k) = \frac{\hbar^2}{2m}k^2. $$