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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?

niobium
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    Possible interpretation: given the initial profile you find the one at later times by performing a convolution with the modes of the SE (i.e. the waves that satisfy the dispersion relation $\omega=\omega(k)$) https://en.wikipedia.org/wiki/Convolution , https://en.wikipedia.org/wiki/Wave_packet , https://en.wikipedia.org/wiki/Dispersion_relation – Quillo Apr 30 '22 at 13:07

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

mike stone
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  • 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? – niobium May 08 '22 at 19:54
  • 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 $\psi(x,t)= \sum_n c_n \psi_n(x)e^{-iE_nt/\hbar}$. – mike stone May 08 '22 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) – niobium May 09 '22 at 09:14
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    Yes. It's just basic spectral theory. Remember that the eigenfunctions of a self-adjoint operator form a complete set. – mike stone May 09 '22 at 12:21