I Fourier Integral of the Schrodinger Equation

[Neothilic]

TL;DR Summary

So I am confused on the steps to find out how you would get to having the second order differential operator to k^2 in the exponent.

 

[Charles Link]

The operator $\partial^2_x$ is to the left of the $dk$ integral. The only thing that is of importance here is the $e^{-ikx}$ term in the integrand. If the operator were by itself, (not in an exponential), I think you can see you get $-k^2 e^{-ikx}$ when it operates on this term. The effect of the $\partial^2_x$ operator is $-k^2$. The same thing applies when the operator is in an exponential.

 

[Neothilic]

The operator $\partial^2_x$ is to the left of the $dk$ integral. The only thing that is of importance here is the $e^{-ikx}$ term in the integrand. If the operator were by itself, (not in an exponential), I think you can see you get $-k^2 e^{-ikx}$ when it operates on this term. The effect of the $\partial^2_x$ operator is $-k^2$. The same thing applies when the operator is in an exponential.

I see how the effect of the $\partial^2_x$ operator is $-k^2$, but what do you mean The operator $\partial^2_x$ is to the left of the $dk$ integral? To show this effect of $\partial^2_x$ operator, could I apple the operator on $u_0(x)$? Then I could then say that the effect of the $\partial^2_x$ operator is $-k^2$.

 

[Charles Link]

Yes, you apply it to $u_o(x)$, and $u_o(x)$ is a Fourier integral in $dk$. The only term in the integrand with an $x$ is $e^{-ikx}$.
To clarify, you apply $e^{it \partial^2_x }$ to $u_o(x)$. If you do a Taylor expansion on $e^{it \partial^2_x }$ , you will get the various powers of $\partial^2_x$ which result in various powers of $-k^2$ as the result. Those then go back up in the exponential and become $e^{-it k^2}$.

 

[Neothilic]

Could I not just do $\partial^2_x u_0(x) = - \int e^{-ikx}k^2 \tilde u_0(k) \, dk$. Then I can say that the derivative produces a pre-factor of $(-ik)$. Then replace this pre-factor with the $\partial^2_x$ in the exponent?

 

[Charles Link]

I think you have the basic idea. You need to operate on $u_o(x)$ with $e^{it \partial^2_x }$ though, and the result is $e^{-it k^2}$.

 

[Neothilic]

I think you have the basic idea. You need to operate on $u_o(x)$ with $e^{it \partial^2_x }$ though, and the result is $e^{-it k^2}$.

Oh I see. Like this:
$e^{it \partial^2_x } (u_0(x)) = e^{it \partial^2_x } (\int e^{-ikx} \tilde u_0(k) \, dk)$
$e^{it \partial^2_x } (u_0(x)) = \int e^{it \partial^2_x }(e^{-ikx}) \tilde u_0(k) \, dk$
$e^{it \partial^2_x } (u_0(x)) = \int e^{kt \partial_x }(e^{-ikx}) \tilde u_0(k) \, dk$
$e^{it \partial^2_x } (u_0(x)) = \int e^{-ik^2t}e^{-ikx} \tilde u_0(k) \, dk$
$e^{it \partial^2_x } (u_0(x)) =e^{-ik^2t} \int e^{-ikx} \tilde u_0(k) \, dk$
Therefore, I can see the operator $\partial^2_x$ gives the result of $-k^2$.
Correct?

 

[Charles Link]

Yes. One minor correction though: In the last line, the $e^{-ik^2t}$ needs to stay inside the integral.