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

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?

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 $$\hbar \omega(k) = \frac{\hbar^2}{2m}k^2.$$
• He means that if $\psi(x,t=0)= \sum_i c_n \psi_n(x)$ with $\psi_n$ an energy eigenfunction with energy $E_n$, then $\psi(x,t)= \sum_n c_n \psi_n(x)e^{-iE_nt/\hbar}$. Commented May 8, 2022 at 20:43
• And do you know why it is possible? (the same applies for the continuous case: if $\Psi(x,0)= \frac{1}{\sqrt{2\pi \hbar}} \int dp\Phi(p)e^{ipx/\hbar}$ then $\Phi(x,t)= \frac{1}{\sqrt{2\pi \hbar}} \int dp\Phi(p)e^{i(px/\hbar-\omega(p)t)}$ isn't it) Commented May 9, 2022 at 9:14
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