# Very simple example of the way the Fourier transform is used in quantum mechanics?

According to a book I'm reading, the Fourier transform is widely used in quantum mechanics (QM). That came as a huge surprise to me. (Unfortunately, the book doesn't go on to give any simple examples of how it's used!) So can someone provide one please?

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Given that leftaroundabout and vonjd have addressed the fundamental place of the Fourier transform in the formalism, let me talk a little about an experimental application.

## What is the shape and size of a atomic nucleus?

From Rutherford we learned that the nucleus is rather a lot smaller than the atom as a whole. Now, electron microscopy can just about provide vague picture of a medium or large atom as a out-of-focus ball, but there is no hope of employing that technique to something orders of magnitude smaller.

What we do is scatter things off of the component parts of the nucleus. A nice reaction here is

$$e + A \to e + p + B$$

where $A$ represents that target nucleus and $B$ the remnant after we bounce a proton out. (This is what nuclear physicists call "quasi-elastic scattering".) Now, if (1) we are shooting a beam of electrons at a stationary target, (2) we have a precision measurement of the momenta of the incident and scattered electrons and the ejected proton, (3) we are willing to neglect excitation energy of the remnant nucleus, and (4) we assume that $p$ mostly did not interact with the remnant after being scattered, we know the momentum of the proton inside the nucleus at the time it was struck.

Collect enough statistics on this and we have sampled the proton momentum distribution of the nucleus.

Now, here's the fun part: you can show that the spacial distribution of protons in the nucleus is the Fourier transform of the momentum distribution.

And bingo, a measurement of the size of the nucleus.

Do it with a polarized target and you can get info on the shape as well.

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+1 F**********K! –  Pratik Deoghare Nov 12 '12 at 22:54

In quantum mechanics, the momentum and position wave functions are Fourier transform pairs, to within a factor of Planck's constant. With this constant properly taken into account, the inequality above becomes the statement of the Heisenberg uncertainty principle.

You can find more here:
http://en.wikipedia.org/wiki/Fourier_uncertainty_principle#Uncertainty_principle

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Apart from its physical meaning (canonically conjugated observables (energy and time, position and momentum...) are fourier transforms of one another), the fourier transform is an important tool in any field that deals with linear differential equations. In quantum mechanics, everything is linear, so it is quite often useful to perform a fourier transform to make solving the equations easier.

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