# Homework Help: Fourier transform question (pretty simple, i think)

1. Oct 8, 2006

### holden

ok, i have a wave packet which is defined between (-pi/(2b)) and (pi/(2b)) as cos(bx), and it's zero everywhere else. here's what i've done so far:

i normalized and solved for b, getting pi/2. so now i'm thinking i should calculate the a0, an, and bn series, and add, right? and here is where i'm stuck. how do i go about doing this? what are the limits of integration? and do i just add the results together to get the final fourier transform? any and all attempts at help is much appreciated. thanks a lot!

2. Oct 8, 2006

### quasar987

huh? I understand you want to write your wave packet as a Fourier integral, like so,

$$F(x)=\int_{-\infty}^{+\infty}G(\omega)e^{i\omega x}d\omega$$

??

All you need to do is calculate $G(\omega)$ by doing the fourier transform of F:

$$G(\omega)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(x)e^{-i\omega x}dx=\frac{1}{2\pi}\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos(bx)e^{-i\omega x}dx$$

3. Oct 8, 2006

### holden

i'm sorry, i guess i should've been more clear. i need to write it in k space.. as in, i need an expression for phi(k) (sorry, can't get LaTeX to display it)

Last edited: Oct 8, 2006
4. Oct 8, 2006

### holden

soo.. i don't need to normalize and solve for b?

5. Oct 8, 2006

### quasar987

Ok, from the top. This is how it goes... You have a wave packet F(x,t) propagating in the positive x direction at speed v. In the spirit of Fourier series, you're hoping to write your wave packet as a "sum" of progressive waves of carefully chosen amplitudes: $\phi(k)e^{k(x-vt)}$ ($\phi(k)$ represents the amplitude of the wave of wave number k. It is the analogue in Fourier integrals of the coefficients $c_n$ in Fourier series):

$$F(x,t)=\int_{-\infty}^{+\infty}\phi(k)e^{ik(x-vt)}dk$$

But what is $\phi(k)$? You know that at t=0, the wave has the form $F(x,0)=\cos(bx)$ for $-\pi/2b<x<\pi/2b$ and 0 elsewhere. In this case, your wave packet becomes

$$F(x,0)=F(x)=\int_{-\infty}^{+\infty}\phi(k)e^{ikx}dk$$

This is a form that is suitable for a Fourier transform. According to Fourier's inversion theorem then,

$$\phi(k)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(x)e^{-ikx}dx$$

$$\phi(k)=\frac{1}{2\pi}\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos(bx)e^{-ikx}dx$$

6. Oct 8, 2006

### quasar987

In post #4, you're talking about normalization, this suggest that F(x,t) is intended as a wave function in the quantum mechanical sense? In that case, yes, b cannot have just about any value. For F to be a wave function satisfying the probabilistic interpretation of QM, $FF^*$ needs to have the properties of a probability density function. One of these is that the "sum" of the probabilities be 1. In this case,

$$1=\int_{-\infty}^{+\infty}FF^*dx$$

$$1=\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos^2(bx)dx$$

What condition does that set on b?

Last edited: Oct 8, 2006
7. Oct 8, 2006

### holden

Think I got it :) I got b as $$\frac{1}{\pi}$$ and my final equation in k space as $$\phi(k) = \frac {\pi}{\frac{\pi^2}{4}+k^2}e^{2ik}$$ .

Thanks a lot for the help!

8. Oct 8, 2006

### quasar987

I got b=pi/2, but I could be wrong. Still it's worth double checking your work I think.

9. Oct 8, 2006

### holden

er, you're right. That's what I got too. I was looking at something else. Thanks so much for the help, it's truly appreciated.