I asked in this thread Time-dependet Schrödinger equation how to solve the Time-dependent Schrödinger equation. One of JamalS' recommendations was the Fourier transform, which is why I want to quote his example:


As an example, consider the case $V(x,t)=\delta(t)$, in which case the Schrödinger equation becomes,

$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \delta(t)\psi$$

We can take the Fourier transform with respect to $t$, rather than $x$, to enter angular frequency space:

$$-\hbar\omega \, \Psi(\omega,x)=-\frac{\hbar^2}{2m}\Psi''(\omega,x) + \psi(0,x)$$

which, if the initial conditions are known, is a potentially simple second order differential equation, which one can then apply the inverse Fourier transform to the solution.

Now, my question would be: What are meaningful initial conditions for this ODE? I mean, what you probably want to look at is how a wavefunction $\Psi(t=0,x)$ propagates in time? So how do you set up meaningful initial conditions for this Fourier-transformed Schrödinger equation? You don't need to refer to this particular ODE(with this potential). My question is rather: When you solve this ODE, what are appropriate initial/boundary conditions for this Fourier transformed ODE, cause this is were my imagination fails.

If anything is unclear, please let me know.

  • $\begingroup$ Just for your information, I plucked that example from thin air because it was convenient, so don't expect a physical interpretation. $\endgroup$
    – JamalS
    Jun 20, 2014 at 12:54

1 Answer 1


Working in the frequency space helps simplify the differential equation you need to solve. Now it should be possible to find a bunch of solutions to the new differential equation. However, in the end what you want to solve is still the time-dependent one. So you need to come back to the initial or boundary conditions of the original time-dependent equation to fix the uncertainty. To be more specific, you can try to build the time-dependent wave function with those solutions you obtained. Certainly there will be unknown coefficients to be determined at the last step.

  • 1
    $\begingroup$ so you are saying: Fourier-Transform Schrödinger equation-> Solve it-> Transform back-> Adjust initial/boundary conditions? $\endgroup$
    – Xin Wang
    Jun 21, 2014 at 11:23
  • 2
    $\begingroup$ Yes, that's what I meant. $\endgroup$
    – Pu Zhang
    Jun 21, 2014 at 11:35

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