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NickCouture
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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?
p is found from -ϒ+p_0 to ϒ+p_0 where ϒ and p_0 are positive constants.blue_leaf77 said:What do you mean by "bounded by constants"?
It is equal to another constant, C, between those boundsblue_leaf77 said:So, your wavefunction is strictly bound in momentum space. But what is the exact shape of it in between those boundaries?
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 said: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?
It gives me 1/√(2ϒ)blue_leaf77 said:What function do you get in position space? Is it even bound at all?
Oh I'm sorry that was my answer for momentum space. In position space I haven't found an answer yet.blue_leaf77 said: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.
Yes that's what I have, I then compute the integral?blue_leaf77 said: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 of course. In the end you should find that the wavefunction in position space is unbounded.NickCouture said:Yes that's what I have, I then compute the integral?
would it be useful to change my exponential into the form cos(theta)+isin(theta)?blue_leaf77 said:Yes of course. In the end you should find that the wavefunction in position space is unbounded.
Yes. Thank you for your help!blue_leaf77 said:It is easier to stay in exponential form, do you know how to integrate an exponential function?
The Fourier Transform is a mathematical operation that decomposes a function into its constituent frequencies, allowing us to analyze the frequency components of a signal or function.
The Bounds of the Fourier Transform refer to the range of frequencies that can be represented by the transform. The lower bound is typically 0, representing the DC or constant component, while the upper bound is determined by the sampling rate or maximum frequency present in the signal.
The Nyquist Frequency is the maximum frequency that can be accurately represented by the Fourier Transform. It is equal to half of the sampling rate and is important because any frequencies above this limit will be aliased, or folded back into the frequency range below the Nyquist Frequency, causing errors in the analysis.
The Bounds of the Fourier Transform determine the range of frequencies that can be accurately analyzed and processed. If the frequencies of interest are outside of this range, they may be distorted or lost in the analysis, leading to errors in signal processing.
The Bounds of the Fourier Transform are directly related to the time-domain signal through the sampling rate. A higher sampling rate results in a wider frequency range and a more accurate representation of the time-domain signal in the frequency domain.