# Bounds of a Fourier Transform

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1. Oct 8, 2015

### NickCouture

If I have a wave function given to me in momentum space, bounded by constants, and I have to find the wave function in position space, when taking the Fourier transform, what will be my bounds in position space?

2. Oct 8, 2015

### blue_leaf77

What do you mean by "bounded by constants"?

3. Oct 8, 2015

### NickCouture

p is found from -ϒ+p_0 to ϒ+p_0 where ϒ and p_0 are positive constants.

4. Oct 8, 2015

### blue_leaf77

So, your wavefunction is strictly bound in momentum space. But what is the exact shape of it in between those boundaries?

5. Oct 8, 2015

### NickCouture

It is equal to another constant, C, between those bounds

6. Oct 8, 2015

### blue_leaf77

Which means the momentum space wavefunction forms a rectangle of height C and width 2Y. And you want to calculate its position space version, do you know where you should start from? Or have you even got your result?

7. Oct 8, 2015

### NickCouture

I've normalized the wave function in momentum space and I've started taking the Fourier transform of the normalized function over the same bounds given above. It does not look like I'm going in the right direction.

8. Oct 8, 2015

### blue_leaf77

What function do you get in position space? Is it even bound at all?

9. Oct 8, 2015

### NickCouture

It gives me 1/√(2ϒ)

10. Oct 8, 2015

### blue_leaf77

Is that the wavefunction in position space you have calculated? If yes, then you have certainly made a mistake. It will be helpful if you can provide your work so that we can find where you have made the mistake.

11. Oct 8, 2015

### NickCouture

Oh I'm sorry that was my answer for momentum space. In position space I haven't found an answer yet.

12. Oct 8, 2015

### blue_leaf77

You start from
$$\psi(x) = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty}\phi(p) e^{ipx/\hbar} dp = \frac{1}{\sqrt{2\pi \hbar}} \int_{p_0-Y}^{p_0+Y} \frac{1}{\sqrt{2Y}} e^{ipx/\hbar} dp$$
How will you execute the next step?

13. Oct 8, 2015

### NickCouture

Yes that's what I have, I then compute the integral?

14. Oct 8, 2015

### blue_leaf77

Yes of course. In the end you should find that the wavefunction in position space is unbounded.

15. Oct 8, 2015

### NickCouture

would it be useful to change my exponential into the form cos(theta)+isin(theta)?

16. Oct 8, 2015

### blue_leaf77

It is easier to stay in exponential form, do you know how to integrate an exponential function?

17. Oct 8, 2015

### NickCouture

Yes. Thank you for your help!