Bounds of a Fourier Transform

  • #1
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?
 

Answers and Replies

  • #2
blue_leaf77
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What do you mean by "bounded by constants"?
 
  • #3
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.
 
  • #4
blue_leaf77
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So, your wavefunction is strictly bound in momentum space. But what is the exact shape of it in between those boundaries?
 
  • #5
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
 
  • #6
blue_leaf77
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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
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.
 
  • #8
blue_leaf77
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What function do you get in position space? Is it even bound at all?
 
  • #9
What function do you get in position space? Is it even bound at all?
It gives me 1/√(2ϒ)
 
  • #10
blue_leaf77
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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
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.
 
  • #12
blue_leaf77
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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
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?
 
  • #14
blue_leaf77
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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.
 
  • #15
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)?
 
  • #16
blue_leaf77
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It is easier to stay in exponential form, do you know how to integrate an exponential function?
 
  • #17
It is easier to stay in exponential form, do you know how to integrate an exponential function?
Yes. Thank you for your help!
 

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