How to find wavepacket time dependence from the $k$-wavefunction?

Question

I am trying to code the time dependence of a gaussian wavepacket using the Fourier transform techniques. I began with constructing a wavepacket (real parts only at the moment) at $t=0$ by multiplying the expression for a plane wave $cos(kx)$ by a gaussian envelope $e^frac{-ax^2}{2}$, giving $$psi(x) = cos(kx)e^frac{-ax^2}{2}.$$

To get the momentum wavefunction, $phi(k)$, I have simply performed the Fourier transform of $psi(x)$ giving another gaussian wavepacket which desmonstrates the uncertainty relation between $x$ and $k$.

From my understanding, $phi(k)$ is needed to obtain the time-dependent position wavefunction, $psi(x,t)$ through the expression:
$$psi(x,t) = int_{-infty}^{infty}phi(k)cos(kx-omega t) dk$$

Again, this is only focusing on the real components, I am not trying to show the complex behaviour.

I'm ot sure how to go about obtaining $psi(x,t)$. Is another Fourier transform required or do I just need to perform the integral for every value of $x$ to construct the wavepacket at the next time step?

ErickShock · Answer

Use separation of variables in Schrödinger's equation. The separation constant will be the energy $E$ of the particle
$$
E = frac{ihbar}{T(t)} frac{dT}{dt}.
$$
The solution will be an exponential, namely $e^{-iEt/hbar}$ (you may take the real part at the end of the calculation). Just keep in mind that $omega$ will have a dependence on $k$:
$$
omega= E/hbar = frac{hbar^2 k^2/2m}{hbar} = frac{hbar k^2}{2m}.
$$
Then you'll have to take the inverse transform of $phi(k)T(t)$ to get $psi(x,t)$.

How to find wavepacket time dependence from the $k$-wavefunction?

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