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No matter what lens is put in the beam path of a Gaussian beam, it will always go through a waist of non-zero width.

Why not just a point? I know the maths, I'm wondering whether there is any physics that prevents it.

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  • $\begingroup$ There are no point like objects in physics, that concept only exists in mathematics. Everything naturally occurring always has a finite size, as far as we can tell experimentally. $\endgroup$ – CuriousOne Dec 21 '15 at 20:58
  • $\begingroup$ Although the electron seems to be awfully small... $\endgroup$ – Jon Custer Dec 21 '15 at 22:29
  • $\begingroup$ @JonCuster: The classical electron radius is about three times larger than the proton radius... what electrons don't have, within that radius, is the kind of parton-structure that one can find in protons, at least not up to the energy range, to which it has been measured with precision, so far. If you do allow for a bit of intellectual nonsense, then you can actually approximate "the real electron" with some order of the QED perturbation series of a "naked" point electron, and then you do get quite a bit of structure from the induced distribution of virtual photons and electrons. $\endgroup$ – CuriousOne Dec 21 '15 at 23:07
  • $\begingroup$ @CuriousOne - perhaps I was misremembering scattering cross section measurements of electrons which only result in a maximum upper bound that is still pretty darn small. Always open to corrections... $\endgroup$ – Jon Custer Dec 22 '15 at 0:17
  • $\begingroup$ @JonCuster: If we do electron-electron scattering at energies above 1MeV we get plenty of "structure", after all, one can make pretty much everything in the particle zoo with that. So, in some sense, there is even a Higgs in there etc., but that's because "in some sense" is intellectual nonsense, just like calculating the first or higher order perturbation series of the virtual particle cloud would be. Yes, it's all in there, but only because we are simply exciting higher mass particle states of the quantum field that happens to have "electron" as its stable, low energy, charged state. $\endgroup$ – CuriousOne Dec 22 '15 at 0:30
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While I concur that you may use the uncertainty principle to understand this, it isn't necessary. If you have a classical EM field that's governed by the nice wave equation derived from Maxwell's equations, then you can compute a diffraction integral that tells you that you must have a finite waist, even if the far-field divergence is very large.

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  • $\begingroup$ And, physically, why is that? If there is an answer.. $\endgroup$ – SuperCiocia Jan 3 '16 at 22:49
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Why not just a point?

Uncertanty principle I guess. In a point focus, the momentum would be defined with a finite $\Delta p < \infty$ (either I guess you can relate it to the lens numerical aperture in a purely geometrical picture where all rays wave-vectors coincide in the focus, or you take $\Delta p=0$ if you consider that the waist of a gaussian beam has a plane front). In any case, because $\Delta p \Delta x \ge$ constant, problems arise with a finite $\Delta p$ ,as the position would be perfectly defined ($\Delta x=0$).

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