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

  • 2
    $\begingroup$ 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) $\endgroup$
    – goroshek
    Sep 1, 2022 at 8:35

1 Answer 1


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 $\omega$.

  • $\begingroup$ 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 $\endgroup$
    – niobium
    Sep 1, 2022 at 15:32
  • 1
    $\begingroup$ 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. $\endgroup$
    – Christophe
    Sep 1, 2022 at 16:21

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