I Fourier transform and locality/uncertainty

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The discussion centers on the relationship between locality and uncertainty in Fourier transforms, particularly how the uncertainty principle relates to time duration and bandwidth. It highlights that while one can increase uncertainty in measurements, a lower bound exists that prevents simultaneous precise measurement of position and momentum. The examples of a sine wave and delta function illustrate the trade-off between localization and frequency spread. Participants express confusion about whether the lower bound applies to uncertainties or the magnitudes of bandwidth and duration. The conversation suggests a need for clarity on the mathematical properties of Fourier transforms versus their physical implications.
bznm
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Could you explain a bit about the relationship between locality and uncertainty in Fourier pairs?
Many pages talk about uncertainty principle stating that the precision at which we can measure time duration of signal cannot unlimitedly grow without affecting precision on bandwidth.
Many other pages talk about the lower bound on time duration and bandwidth product themselves, not uncertainties about them.
What's exactly the link between them?

Furthermore, does the uncertainty principle forbid signal with unlimited time and bandwidth? I am talking about effective duration and effective bandwidth.
 
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It may be useful to look at two extreme examples.

First consider a simple sin wave. The wave extends through all space so there is no locality. The Fourier transform of a sin wave is a delta function, so we have exactly one frequency and hence exactly one value of momentum.

On the other hand, if we started with a delta function, then we have a particle localized at a single point. The Fourier transform of a delta function is a sin wave, so we have an infinite spread of frequencies and hence an infinite spread of momentum values.
 
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NFuller said:
It may be useful to look at two extreme examples.

First consider a simple sin wave. The wave extends through all space so there is no locality. The Fourier transform of a sin wave is a delta function, so we have exactly one frequency and hence exactly one value of momentum.

On the other hand, if we started with a delta function, then we have a particle localized at a single point. The Fourier transform of a delta function is a sin wave, so we have an infinite spread of frequencies and hence an infinite spread of momentum values.
thanks! however this example only justifies that one spread or the other has to be non infinite...
in particular I'd like to have a somehow intuitive explanation of the fact that uncertainty principle gives a lower bound and not an upper bound..
 
The point of these examples is to show that when one value is exactly known as a delta function, the other value consists of an infinite spread of values.
bznm said:
I'd like to have a somehow intuitive explanation of the fact that uncertainty principle gives a lower bound and not an upper bound..
You are always allowed to have as large of an uncertainty as you want, just use less accurate equipment. There is no physical justification to have an upper-bound on the uncertainty of any measurement. What is surprising is that a lower bound exists and no matter how good your measurement device is, you can never exactly measure position and momentum simultaneously. This is a purely mathematical consequence of defining the position of a particle using a wave function.

If you're looking for a simple proof of the formula, check out this page http://www.phys.ufl.edu/courses/phy4604/fall13/uncertaintyproof.pdf
 
NFuller said:
The point of these examples is to show that when one value is exactly known as a delta function, the other value consists of an infinite spread of values.

You are always allowed to have as large of an uncertainty as you want, just use less accurate equipment. There is no physical justification to have an upper-bound on the uncertainty of any measurement. What is surprising is that a lower bound exists and no matter how good your measurement device is, you can never exactly measure position and momentum simultaneously. This is a purely mathematical consequence of defining the position of a particle using a wave function.

If you're looking for a simple proof of the formula, check out this page http://www.phys.ufl.edu/courses/phy4604/fall13/uncertaintyproof.pdf
why is physics implied? aren't we talking about mathematical properties of an integral transform? this is something I can't understand.
Then, I still don't get if the lower bound is on uncertainties about bandwidth, duration or if it is about the magnitude itself of this quantities.
 
bznm said:
why is physics implied?
Because you posted this in the General physics sub-forum. Also when referencing locality in your original post, it sounds like you are referring to the position of a particle. If you are asking this from more of a signal analysis perspective, then you should post this in the mathematics sub-forums.
 
NFuller said:
Because you posted this in the General physics sub-forum. Also when referencing locality in your original post, it sounds like you are referring to the position of a particle. If you are asking this from more of a signal analysis perspective, then you should post this in the mathematics sub-forums.
You're right, that was my fault.

@MODERATORS, please: could you move the question?
 

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