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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?

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1 Answer 1

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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.

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  • $\begingroup$ Thank you. Could you clarify on how to calculate $\hat{\Delta \psi}$? $\endgroup$ Jan 26, 2020 at 11:16
  • $\begingroup$ 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 $(-ik)^2 = k^2$. $\endgroup$
    – Joe
    Jan 26, 2020 at 16:14
  • $\begingroup$ In addition after you plug in df/dx do integration by parts to get what I said $\endgroup$
    – Joe
    Jan 26, 2020 at 16:47

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