Solving for Φ(k) in Quantum Fourier Transform with ψ(x,0)=e^(-λ*absvalue(x))

[chris2020]

Homework Statement

Assume ψ(x,0)=e^(-λ*absvalue(x)) for x ± infinity, find Φ(k)

Homework Equations

Φ(k)=1/√(2π)* ∫e^{(-λ*absvalue(x))}e^(-i*k*x)dx,-inf, inf[/B]

The Attempt at a Solution

, my thought was Convert the absolute value to ± x depending on what of the number line was being integrated.[/B]

U=i*k*x
du/(i*k)=dx

1/√(2π)*∫e^{-λ*√(u2/(i*k)2)}*e^(-u)du,-inf,inf

Now fixing abs value

1/((2π)*(i*k))*∫e^λ/(i*k)*ue^(-u),du,-inf,o

the integrand for one half of the number line looks like:

E^{(u*(λ/(ik)-1)}

For which i get: after limits are taken for that half of the integral

(1/((λ/ik)-1))

Then similar integral for other half

Is this the right track or am i totally off?

 

[TSny]

Hello and welcome to PF!

Your work looks OK. I don't think you need to make the substitution u = ikx. Looks like a typo in one place where you left out the square root for the $2 \pi$ factor.

If you feel more comfortable with working with real functions, write $e^{-ikx} = \cos (kx) - i \sin (kx)$. You can then check to see if the resulting integrands are even or odd functions over the interval $-\infty < x < \infty$.

 

[chris2020]

yea i was thinking that route also but i forgot about about the even or odd shortcut

Thanks!