# Solving the Schrödinger equation for a free particle with Fourier transformations

So the differential equation looks as follows:

$$i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \Delta \psi$$

where $$\hbar, m > 0$$, $$\psi(t,x) \in \mathbb{C}$$, $$t > 0$$, $$x \in \mathbb{R}^3$$ and

$$\psi(0,x) = \exp(-|x|^2).$$

I think this can be solved elegantly using a spatial Fourier transform. I know that:

$$\hat{\frac{\partial \psi}{\partial t}} = \frac{\partial \hat{\psi}}{\partial t},$$

but how do I calculate

$$\hat{\Delta \psi} \; \;$$

and then use it to solve the PDE?

You shouldn't do a Fourier in the time coordinate just in the position coordinates (i.e. x y z). And you know by simple calculations that $$\hat{\Delta \psi}$$ is just $$-k^2\psi(t, \vec{k})$$ where $$\vec{k}$$ is (x, y, z) after the Fourier transform. And the by substituting this to the equation you get a simple ode for $$\hat{\psi}(t, \vec{k})$$ because $$\hat{\frac{\partial \psi}{\partial t}} = \frac{\partial \hat{\psi}}{\partial t},$$ when you transform in the position coordinates as you said.
• Thank you. Could you clarify on how to calculate $\hat{\Delta \psi}$? – maths_student Jan 26 at 11:16
• Just plug in the fourier transform df/dx and get -ik$\hat{f}$ and the other coordinates are the same. To get the laplcian just apply the derivative twice to get $(-ik)^2 = k^2$. – Joe Jan 26 at 16:14