Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the physical significance of a fourier transform?

I am interested in knowing exactly how it works when crossing over from momentum space to co ordinate space and also how we arrive at the uncertainity relations and how do we interpret them physically.

P.S.: Links to articles, papers or book recommendations are welcomed as long as it is relevant to the topic.

share|cite|improve this question
Dear @ramanujan_dirac: Just a minor point. Please make sure not to overuse the math-phys tag, which is meant for rigorous math-phys analysis only. In this question (v3) OP is mostly asking for physical intuition. – Qmechanic Jan 26 '13 at 17:16
up vote 5 down vote accepted

A very brief answer: The Fourier transform as used in quantum physics interprets a point in one set of three coordinates as a broad spectrum in another set of three coordinates, and vice-versa. This is the uncertainty relationship, since it means that when the object is point-like in one set of coordinates (space, for example) it become vast and diffuse in the other (momentum, for example).

To make that a bit more precise you need to stop thinking of Fourier transforms as flat sinusoidal graphs, and instead think of them as having complex values -- that is, as having phase values in a plane (two more axes really) that is orthogonal to the other three. In this form, an object that is stretched out along one axis -- say X -- looks more like a stretched-out slinky if you add the two complex coordinates and move along X. The various continuity equations of physics simply say that the coils of any such slinky like to stay as smooth and unkinked as possible. (The slinkies also rotate around their axis based on how massive the object is, but that's a different issue).

If such a slinky is infinitely long in X and perfectly smooth at some frequency of coiling, it represents a particle that whose wave function is infinitely long in X, and whose probability of finding it at any one spot is essentially zero. That's position uncertainty, the worst possible case of it.

But it's the frequency of coiling that's more important for uncertainty, because that frequency just happens to represent the momentum of the particle parallel to the X axis. If the coiling is perfectly regular, there's only one such frequency, and you could map that on frequency out (radio-dial style) into another triplet of axes labeled $p_x$, $p_y$, and $p_z$. That's momentum space, the other side of the Fourier coin.

In that space, the particle has a very precise location indeed, that being the location of that particular frequency along the momentum axis $p_x$.

But what if you try it the other way around? That is, what if you construct a similar helix using the same rules of adding a complex plane perpendicular to $p_x$ and creating a very long, very regular coil there. (There's a beautiful symmetry in physics here, because it turns out that momentum space has continuity and smoothness rules very much like those of regular XYZ space, despite some different meanings of them.)

Well, pretty much exactly the same thing happens, only in reverse: The long, precise coil in momentum space also has a precise frequency, which also has an interpretation over in XYZ space as a precise location. And again, having the coil stretched out in momentum space means that it's hard to find the particle there; it could with equal likelihood be almost anywhere along the coil, meaning it could have almost any momentum. So, in this case, being "uncertain" about the location of the particle in momentum space means that it has a very precise location in regular space.

The symmetry is simply gorgeous, and has many real implications in physics. Metals, for example, are substances in which pairs of electrons exist more in momentum space than in regular space, causing some of them to have very high momenta (energies) and all of them to be distributed rather oddly over a metal crystal. The high energy ones make the metal reflective, the movement (momenta) of the electrons make it conductive, and the delocalization of the electrons combine with charge cancellation to create rather remarkable properties such as tensile strength (it sticks together tightly!) and ductility (it's easier to bend when half the charges are always in motion).

But while the Fourier part of this symmetry between the XYZ and momentum spaces is exact, the meanings of frequencies in the two spaces is anything but. The biggest difference is that while the size of an object in XYZ space has no huge energy implications, size in momentum space has huge implications, because higher momentum means higher energy. So, the longer a slinky (more properly called a wave function) becomes in momentum space, the higher in energy the particle becomes.

And, since the length of the slinky in momentum space determines through the Fourier transform how precisely it can be positioned in regular XYZ space, extremely precise positions come at a cost -- a very high cost. The Stanford linear accelerator, for example, has to accelerate electrons to extremely high velocities (momentums) because it's the only way to make electron locations precise enough to probe the innards of particles such as neutrons and protons.

Oddly, no such cost is accrued when particles become lost in regular XYZ space. So, an electron wandering through the cosmos can in principle be represented by a very large slinky with a very precise frequency, if something (e.g. diffraction) encourages it to form that way. But even though the energy cost is low, such wave functions are unstable for a very different reason: Any kind of information-imparting interaction between them and other matter causes the wave function to be irrelevant (yes, I'm trying to avoid getting into schools of quantum doctrine with that careful wording) and for all practical purpose a vastly smaller wave function. Just not too small, since only the energy of the interaction is available to alter the original diffuse form.

And again, to bring all that back home: It's the Fourier transform and its ability to say how well each frequency "matches" a slinky-form in either of the spaces that creates this interpretation back-and-forth between the two, and thereby creates quantum uncertainty.

share|cite|improve this answer
Though the FAQ of Stack Exchange discourages users from adding comments expressing thanks etc. I must say, this was just awesome! Thank you, very much! – noir1993 Jan 24 '13 at 7:20
It does?? Well, thanks back anyway. Hopefully the Powers That Be are OK with occasional one-liners from both sides! – Terry Bollinger Jan 24 '13 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.