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Fourier Transform and Wave Function

by atomicpedals
Tags: fourier, function, transform, wave
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atomicpedals
#1
Apr28-12, 03:19 PM
P: 196
1. The problem statement, all variables and given/known data

a) Find the normalization constant N for the Gaussian wave packet [itex]\psi (x) = N e^{\frac{-(x-x_{0})^{2}}{2K^{2}}}[/itex]. b) Find the Fourier Transform and verify it is normalized.

2. The attempt at a solution

a) I think I've got
[tex]\psi (x) = N e^{\frac{-(x-x_{0})^{2}}{2K^{2}}}[/tex]
[tex]\int |N e^{\frac{-(x-x_{0})^{2}}{2K^{2}}}|^{2}dx = 1[/tex]
[tex]N^{2}=\frac{1}{\sqrt{\pi}K}[/tex]
[tex]N = \frac{1}{\pi^{1/4}\sqrt K}[/tex]
b) This is where the trouble starts...
[tex]\psi (x) = \frac{1}{\pi^{1/4}\sqrt K} e^{\frac{-(x-x_{0})^{2}}{2K^{2}}}[/tex]
[tex]F(\omega)=\frac{1}{\sqrt{2\pi}} \int \frac{1}{\pi^{1/4}\sqrt K} e^{\frac{-(x-x_{0})^{2}}{2K^{2}}} e^{i \omega x}dx[/tex]
I think I can pull the normalization constant out of the integrand and get
[tex]F(\omega)=\frac{1}{\sqrt{2\pi}}\frac{1}{\pi^{1/4}\sqrt K} \int e^{\frac{-(x-x_{0})^{2}}{2K^{2}}} e^{i \omega x}dx[/tex]
And the exponents should combine (here I'm not so sure)
[tex]F(\omega)=\frac{1}{\sqrt{2\pi}}\frac{1}{\pi^{1/4}\sqrt K} \int e^{\frac{-(x-x_{0})^{2}}{2K^{2}}+i \omega x}dx[/tex]
Assuming I've not gone horribly wrong earlier, evaluation of this integral stumps me. Any help and suggestions are much appreciated.
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atomicpedals
#2
Apr28-12, 03:53 PM
P: 196
If I crank through it, I think the result is

[tex]F(\omega)=\frac{1}{\pi^{1/4}\sqrt{1/K^{}2}\sqrt K}e^{\frac{-K^{2}k{2}}{2}+i x_{0} \omega}[/tex]
vela
#3
Apr28-12, 05:07 PM
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Thanks
PF Gold
P: 11,774
There should be an ##\omega^2## somewhere in the exponent. Your answer seems to have other typos as well. It would help to see what you actually did, but I think you have it under control.


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