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Heisenberg's uncertainty principle is well-studied and has become a bit of a pop science phenomenon due to its widespread implications in quantum mechanics. (Though interpretations are often misrepresented.) The Heisenberg uncertainty principle can be arrived at in a couple of ways:

  1. By considering the Cauchy-Schwarz inequality, noting that equality only holds for eigenfunctions of the Fourier transform, knowing a priori what the eigenfunctions of the Fourier transform are and then minimizing the uncertainty product over all eigenfunctions. (The minimizer being the Gaussian of course.)

  2. Considering commutators of the position and momentum operators.

Many integral transforms that I have come across have uncertainty products associated to them. In the case of the Fourier transform it's not so hard to see since scaling in one domain causes an inverse scaling in the conjugate domain so that the more localized a function is, the more spread out its Fourier transform is. My question is: is this an accident? Or is there something intrinsic about integral transforms on the whole that somehow naturally encodes uncertainty products?

Cameron L. Williams
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1 Answers1

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Once i almost had a fight with one of my professors on mathematics because i maintained that this is not accidental.

The point is to have an understanding of what the fourier transform means for a signal or system and then how this relates to momentum (at least in QM).

Since fourier transforms (or integral representations in other domains) are used heavily in real-world applications, it is important.

The intuition is that a signal (or measurement) cannot be localized (arbitrarily and at the same point) in both domains. Roughly because the information for a point at one domain depends on information of multiple points of the other (this is what a linear integral transform does). Mathematically this is expressed as an inequality (upper bound) of the uncertainties of both points and the correlation between the measurements (also called fourier uncertainty principle).

Furthermore one can say that the two measurements cannot be represented exactly on the the same basis (or that the measurements do not commute).

Nikos M.
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  • I agree entirely and in the Fourier case, it's manifest but I guess I was speaking in much more generality than just the Fourier case. The argument to real world phenomena is a little impervious in the general case. However it does serve as a good guide to go by. – Cameron L. Williams Jun 28 '14 at 04:29
  • @CameronWilliams, of course there is variety of integral transforms. However in the case of fourier (or others, eg short-time fourier transform), this is very clear. There are other transforms that use different kernels, e.g Wigner distribution etc.., yet what they do is interchange localization in one domain for the other. – Nikos M. Jun 28 '14 at 04:44
  • @CameronWilliams, much in this discussion also depends on one's view (or affiliation) with specific interpretations of QM (which was at least part of my almost fight). Nevertheless it still a valid argument. (ps. i dont like the offical interpretation too much myself) – Nikos M. Jun 28 '14 at 04:47
  • @CameronWilliams, one point missing (or rather not fully elaborated) in the answer, is why (or how) momentum relates to fourier transform and why it is needed (in QM). This needs a much longer analysis. Shortly, fourier transform encodes information about the group structure and group characters (for example: http://people.cs.uchicago.edu/~laci/reu02/fourier.pdf, http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/charthy.pdf) – Nikos M. Jun 28 '14 at 04:55
  • Yeah I'm aware of dual groups and such but I think that you have cause and effect backwards. The general case was motivated by the Fourier transform on $\mathbb{R}$. – Cameron L. Williams Jul 13 '14 at 20:52
  • @CameronWilliams, i'm not sure i understand what you mean, cause and effect backwards, in any case check also this post about transforms in general and how they relate to one-another – Nikos M. Jul 14 '14 at 16:39
  • @CameronWilliams, im not sure i reverse "cause-effect", the full answer to this, is the reason (in QM) the momentum is the fourier transform of the position (which in clasical case is the legendre transform), previous comment gives an answer in this direction, a full answer requires a little more, but still i would not say i reverse cause or effect or anything of the like – Nikos M. Jul 14 '14 at 16:43