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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?

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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 $$ \hbar \omega(k) = \frac{\hbar^2}{2m}k^2. $$

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  • $\begingroup$ My teacher says something like that: "Since the Schroedinger Equation is first order in time, if you know the wave function at one time, and you solve it, you get the wave function at any time". Do you know what he means by that? $\endgroup$
    – niobium
    Commented May 8, 2022 at 19:54
  • $\begingroup$ 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 $\psi(x,t)= \sum_n c_n \psi_n(x)e^{-iE_nt/\hbar}$. $\endgroup$
    – mike stone
    Commented May 8, 2022 at 20:43
  • $\begingroup$ 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) $\endgroup$
    – niobium
    Commented May 9, 2022 at 9:14
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    $\begingroup$ Yes. It's just basic spectral theory. Remember that the eigenfunctions of a self-adjoint operator form a complete set. $\endgroup$
    – mike stone
    Commented May 9, 2022 at 12:21

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