0
$\begingroup$

What is the Fourier Transform of the spatial portion of $Ψ(x,t)=A\exp(-b|x-2|)\exp(-iwt)$?

I tried applying the regular exponential Fourier transform, but not getting it.

Do you just bring out the $exp(iwt)$? If so, then how do you integrate the $exp(-b|x-2|)exp(-ikx)$ left inside from negative to positive infinity?

Any help will be much appreciated. Thank you very much!

$\endgroup$
0
$\begingroup$

It's linear, so the time portion just comes along for the ride. We can use the table here to note that $$\mathcal{F}\left\{e^{-a|x|}\right\}=\frac{2a}{a^2+4\pi^2\xi^2}. $$ This is assuming the unitary, ordinary frequency type of FT. You can use a different column if you're using a different convention. Together with the shifting theorems, that should get you all the way there. Can you continue?

$\endgroup$
4
  • $\begingroup$ I have not a clue what that... squiggly E^2 is. Sorry, could you clarify? Thank you! $\endgroup$ – user707335 Oct 7 '19 at 16:04
  • $\begingroup$ The $\xi$ is the transformed variable of the Fourier Transform. It's the equivalent of $s$ for a Laplace Transform. $\endgroup$ – Adrian Keister Oct 7 '19 at 16:05
  • $\begingroup$ Oh, thank you, good to know. $\endgroup$ – user707335 Oct 7 '19 at 16:13
  • $\begingroup$ @ArthurKarapetov it's the greek letter "xi" $\endgroup$ – user438666 Oct 7 '19 at 16:25
0
$\begingroup$

Going straight to the definition of the Fourier transform, we have \begin{multline} \mathcal{F}[\Psi](k) = \int_{-\infty}^\infty Ae^{-b|x-2|}e^{iwt}e^{-ikx}dx = Ae^{iwt}\int_{-\infty}^\infty e^{-b|x-2|}e^{-ikx}dx \\= Ae^{iwt}e^{-2ik}\int_{-\infty}^\infty e^{-b|u|}e^{-iku}du = 2Ae^{i(wt-2k)}\int_{0}^\infty e^{-bu}\cos(ku)du = \frac{2Abe^{i(wt-2k)}}{b^2+k^2} \end{multline} where the substitution $u = x-2$ was used and the parity of the integrand was used to simplify the last integral.

Also, this wave function is normalizable. The normalization integral is \begin{multline} \int_{-\infty}^\infty \Psi^*\Psi dx = \int_{-\infty}^\infty \left[A^*e^{-b|x-2|}e^{-iwt}\right]\left[Ae^{-b|x-2|}e^{iwt}\right] dx \\= |A|^2\int_{-\infty}^\infty e^{-2b|x-2|}dx = |A|^2\int_{-\infty}^\infty e^{-2b|u|}du=\frac{|A|^2}{b}= 1, \end{multline} with the same substitution and parity considerations were used here. So $A = \sqrt{b}$.

$\endgroup$
6
  • $\begingroup$ Where is the $1/sqrt(2pi)$? And thank you very much, I'm going over it. $\endgroup$ – user707335 Oct 7 '19 at 16:17
  • $\begingroup$ How did you compute the integral to get $1/b$? for the normalization? $\endgroup$ – user707335 Oct 7 '19 at 16:18
  • $\begingroup$ And how did you compute the integral exp(-bu)cos(ku) from 0 to infinity? $\endgroup$ – user707335 Oct 7 '19 at 16:28
  • $\begingroup$ And how does exp(-iku)=cos(ku)? $\endgroup$ – user707335 Oct 7 '19 at 16:31
  • $\begingroup$ @ArthurKarapetov Note the symmetry of the integral. $e^{-b|u|}\cos(ku)$ is an even function, so its integral over the whole real line is equal to twice the integral over the positive reals. $e^{-b|u|}\sin(ku)$ is an odd function, so its integral over the whole real line is zero. Similarly for the normalization integral, though I skipped over the same $u =x-2$ substitution. I'll make these more explicit in the answer. $\endgroup$ – eyeballfrog Oct 7 '19 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy