![]() ![]() ![]() I'll spare you the details, but for this definition of time-frequency area the family of signals gives you the minimum. That means instead of the uncertainty product between F and T we measure the minimal uncertainty product of any two conjugate variables spanned by F and T. ![]() The notion of time frequency area can however be generalised to measure the area of shapes that are not aligned with the time and frequency axis. If you want minimal $\Delta F \cdot \Delta T$ then $\gamma$ from above must vanish. The signals I gave above minimise the time-frequency area, but not the time-frequency uncertainty product. Please go ahead an check out the derivate of the above mentioned Fourier transform. Since the Fourier transform is simply a superposition of different sine-waves (a cosine-wave is a sine-wave, which is shifted by pi/2) this is an example of your problem. These symplectomorphisms preserve the Heisenberg uncertainty relation.Įdit: Let me make this more precise and also in fact more correct. The Fourier transform of a rectangle is the sinc function. The Heisenberg Uncertainty Principle is atheoremabout Fourier transforms. This family is generated by quadratic symplectomorphisms in time-frequency from a single Gabor atom. 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.Īnd bingo, a measurement of the size of the nucleus.ĭo it with a polarized target and you can get info on the shape as well.It is important to define the time and frequency widths $\Delta_t$ and $\Delta_\right)^2 + i \left(\phi + \omega_0 (t-t_0) + \gamma (t-t_0)^2\right)\right) $$ (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.Ĭollect enough statistics on this and we have sampled the proton momentum distribution of the nucleus. Where $A$ represents that target nucleus and $B$ the remnant after we bounce a proton out. If we try to make a particle with a definite momentum. If we try to localize a particle to a very small region of space, its momentum becomes uncertain. What we do is scatter things off of the component parts of the nucleus. For the Gaussian wave packet, we can rigorously read the RMS width of the probability distribution as was done at the end of the section on the Fourier Transform of a Gaussian. 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 is the shape and size of a atomic nucleus?įrom Rutherford we learned that the nucleus is rather a lot smaller than the atom as a whole. 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. This is the Heisenberg Uncertainty Principle expressed with respect to the Fourier transforms. ![]()
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