Bounds of a Fourier Transform

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?
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
What do you mean by "bounded by constants"?
 
What do you mean by "bounded by constants"?
p is found from -ϒ+p_0 to ϒ+p_0 where ϒ and p_0 are positive constants.
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
So, your wavefunction is strictly bound in momentum space. But what is the exact shape of it in between those boundaries?
 
So, your wavefunction is strictly bound in momentum space. But what is the exact shape of it in between those boundaries?
It is equal to another constant, C, between those bounds
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
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?
 
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?
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.
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
What function do you get in position space? Is it even bound at all?
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
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.
 
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.
Oh I'm sorry that was my answer for momentum space. In position space I haven't found an answer yet.
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
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?
 
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?
Yes that's what I have, I then compute the integral?
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
Yes that's what I have, I then compute the integral?
Yes of course. In the end you should find that the wavefunction in position space is unbounded.
 
Yes of course. In the end you should find that the wavefunction in position space is unbounded.
would it be useful to change my exponential into the form cos(theta)+isin(theta)?
 

blue_leaf77

Science Advisor
Homework Helper
2,629
784
It is easier to stay in exponential form, do you know how to integrate an exponential function?
 

Related Threads for: Bounds of a Fourier Transform

  • Posted
Replies
4
Views
1K
  • Posted
Replies
1
Views
425
  • Posted
Replies
1
Views
475
Replies
3
Views
2K
Replies
1
Views
498

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top