# 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

• This question can not be answered until you specify the hamiltonian of the system. For different hamiltonians evolution of wave functions and their Fourier transform is different. By the way, I think your guess is correct, but only in case of bound states from discrete spectrum with defined eigenenergies (omegas in your formula) Commented Sep 1, 2022 at 8:35

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 $$\omega$$.

• Is the bra-ket notation compulsory to study QM? My teacher (not really one, because it's a MOOC...) hasn't used so far the bra-ket notation (he didn't speak about it) and I'm almost at the end of his Introduction to QM course Commented Sep 1, 2022 at 15:32
• It is not compulsory. Sorry, I could have avoid it. You just have to know that plane waves are solutions of Schrödinger equation for the energies $\hbar\omega_k$. Now, time-dependent Schrödinger equation implies $i\hbar{\partial\over\partial t}\phi(k,t)=\hbar\omega_k\phi(k,t)$ so $\phi(k,t)=\phi(k,0)e^{-i\omega_kt}$. You can come back to my answer. Commented Sep 1, 2022 at 16:21