Fourier transform and locality/uncertainty

In summary, the relationship between locality and uncertainty in Fourier pairs is that the uncertainty principle sets a lower bound on the precision at which we can simultaneously measure the time duration and bandwidth of a signal, which is linked to the position and momentum of a particle. This lower bound is a mathematical consequence of using a wave function to define the position of a particle. This principle also applies to signal analysis, with the lower bound being on the uncertainties of bandwidth and duration. However, there is no upper bound on the uncertainty, and it is possible to have a larger uncertainty by using less accurate equipment. The concept of locality is relevant to the position of a particle in this context. This discussion is applicable to both physics and mathematics, but for a more focused
  • #1
bznm
184
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes olgerm
  • #3
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..
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 

Related to Fourier transform and locality/uncertainty

1. What is a Fourier transform?

A Fourier transform is a mathematical operation that converts a function from its original domain (often time or space) to a representation in the frequency domain. This allows for the analysis of a signal's frequency components and their relative strengths.

2. How is the Fourier transform related to locality?

The Fourier transform is related to locality through the Heisenberg uncertainty principle. This principle states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. The Fourier transform allows us to analyze a signal's frequency components at a specific point in time, but this comes at the cost of losing information about its exact location in time.

3. How is the Fourier transform related to uncertainty?

The Fourier transform is related to uncertainty through the Heisenberg uncertainty principle. This principle states that the more precisely we know the frequency of a signal, the less precisely we can know its duration, and vice versa. This is because a signal's duration and frequency are inversely related, and the Fourier transform allows us to analyze a signal's frequency components, but this comes at the cost of losing information about its duration.

4. What is the difference between the Fourier transform and the inverse Fourier transform?

The Fourier transform converts a signal from its original domain (often time or space) to the frequency domain, while the inverse Fourier transform converts it back to its original domain. In other words, the Fourier transform decomposes a signal into its frequency components, while the inverse Fourier transform synthesizes a signal from its frequency components.

5. How is the Fourier transform used in signal processing?

The Fourier transform is a fundamental tool in signal processing, used for tasks such as filtering, spectral analysis, and data compression. It allows for the identification of specific frequency components in a signal and the removal of unwanted noise or interference. It is also used in the design and analysis of communication systems, image processing, and many other applications in engineering and physics.

Similar threads

  • Other Physics Topics
Replies
1
Views
1K
  • Quantum Physics
Replies
7
Views
885
  • Biology and Medical
Replies
6
Views
446
Replies
3
Views
1K
  • Atomic and Condensed Matter
Replies
1
Views
902
  • Quantum Physics
Replies
3
Views
283
  • Linear and Abstract Algebra
2
Replies
43
Views
5K
Replies
11
Views
870
Replies
29
Views
6K
Back
Top