Wave function Fourier transform with time I found the Fourier transform at $t=0$ for the wave function of a wave packet (and it's inverse Fourier transform) :
$$\Psi(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{ikx}dk$$
$$\Phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,0)e^{-ikx}dx$$
Now I want to know the same Fourier transform, but at a time $t\neq0$
What would be the formula ?
My guess is:
For the Fourier transform: $\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{i(kx-\omega t)}dk$
For the inverse Fourier transform: $\Phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,t)e^{i(-kx-\omega t)}dx$
But I am really not sure
 A: As you probably know, the Fourier transform of the wave function
$$\Psi(x,0)={1\over\sqrt{2\pi}}\int \phi(k)e^{ikx}dk$$
can be understood as a change of basis
$$|\Psi(0)\rangle=\int\Psi(x,0)|x\rangle dx=\int\phi(k)|k\rangle dk$$
with $\langle x|k\rangle={1\over\sqrt{2\pi}}e^{ikx}$. For a free particle, the quantum states $|k\rangle$ are eigenvectors of the Schrödinger Hamiltonian for the energies $E_k={\hbar^2k^2\over 2m}=\hbar\omega_k$. Note that I have emphasized the fact that $\omega_k$ is a function of $k$ and not an independent variable. The time evolution of the states $|k\rangle$ is therefore trivial:
$$\phi(k,t)=\phi(k)e^{-iE_kt/\hbar}=\phi(k)e^{-i\omega_kt}$$
and therefore
$$\Psi(x,t)={1\over\sqrt{2\pi}}\int \phi(k)e^{i(kx-\omega_kt)}dk$$
The Fourier transform of the wavefunction at time $t$ is
$${1\over\sqrt{2\pi}}\int \Psi(x,t)e^{-ikx}dx=\phi(k)e^{-i\omega_kt}$$
and is a function of $k$ and $t$.
You may also want to take the Fourier transform with respect to both the variable $x$ and $t$. The result will be in this case
$${1\over 2\pi}\int\int \Psi(x,t)e^{-i(kx-\omega t)}dxdt
   =\phi(k)\delta(\omega-\omega_k)$$
which is now a function of $k$ and $\omega$.
