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

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!

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?

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

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 $$A = \sqrt{b}$$.

• Where is the $1/sqrt(2pi)$? And thank you very much, I'm going over it. – user707335 Oct 7 '19 at 16:17
• How did you compute the integral to get $1/b$? for the normalization? – user707335 Oct 7 '19 at 16:18
• And how did you compute the integral exp(-bu)cos(ku) from 0 to infinity? – user707335 Oct 7 '19 at 16:28
• And how does exp(-iku)=cos(ku)? – user707335 Oct 7 '19 at 16:31
• @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. – eyeballfrog Oct 7 '19 at 16:51