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Why isn't it possible to focus a laser beam to an infinitely small point in space? I am familiar with the shape of a gaussian beam, but why can't my $w_0$ be equal to zero?

  • Related/possible duplicates: https://physics.stackexchange.com/q/234996/50583, https://physics.stackexchange.com/q/421584/50583, https://physics.stackexchange.com/q/140949/50583 – ACuriousMind Mar 14 '19 at 20:07
  • I am sorry, but non of these answer my question. – Timo_BLN Mar 14 '19 at 20:08
  • The answer is diffraction. – Gilbert Mar 14 '19 at 20:16
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    It's amazing how the two current answers to this question, by Thomas Fritsch and @ACuriousMind, answer the question with the same underlying principle approached from two entirely different sides. Thomas Fritsch explains it from the mathematical model, with the result that you'd need a wavelength of $0$ to have a beam with a width of $0$, and ACuriousMind states that at the most extreme case, you'd have only a single wave, which logically would be bounded in width by wavelength and amplitude, thus requiring the same wavelength of $0$. This is why I love physics. – TheEnvironmentalist Mar 14 '19 at 20:42

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An ideal Gaußian beam is diffraction-limited - its wavefront inevitably spreads out due to Huygens' principle. An infinitesimally small beam would diffract infinitely strongly, i.e. not resemble a beam at all: By Huygens' principle a single point (i.e. a "beam source" with zero radius) as a source simply emits a single, spherical wave, not a beam.

ACuriousMind
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  • This does not explain why you cannot focus to a point though. – Paul Childs Mar 14 '19 at 22:05
  • @PaulChilds I linked three other questions about focusing to a point vs. conservation of etendue in the comments to the question, and the OP said they didn't answer their question, so I didn't see the point of explaining that yet again. – ACuriousMind Mar 14 '19 at 22:15
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According to Wikipedia:Gaussian beam the beam waist ($w_0$) and the total angular spread of the beam ($\Theta$) are related by

$$ \pi w_0 \tan \frac \Theta 2 = \lambda $$

From this formula you see: The wave-nature of light (via its wavelength $\lambda$) is responsible for the non-zero waist.

That means you would have a beam waist $w_0 = 0$ only for these cases:

  • wavelength $\lambda = 0$ (so that there is no diffraction)
  • total angular spread $\Theta = 180°$ (meaning we have a spherical wave from a point source)
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The Gaussian beam is not a precise solution of the Maxwell equation, it can only derived in the paraxial approximation, and this approximation is not applicable for small $w_0$.

akhmeteli
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Ok, I'm not happy with the answers so far so here's my 2 cents.

Yes $\omega_0$ can be zero. If you have a perfectly spherical mirror with a single particle lasing source inside, then by classical optics it will focus light to a infinitely narrow point at the centre.

The real reason then why it is not truly infinitely narrow (manufacturing defects aside) is that even if we are to ignore entanglement re: the photon and assume it to not ne governed by the exclusion principle, the emitter will be, and cannot be isolated to a infinitely narrow non-moving position.

  • Thank you, that was the answer I was looking for! – Timo_BLN Mar 14 '19 at 22:17
  • That answer is not correct. It is possible to cause a single atom in a crystal to emit light, and to know that the atom is localized to within a fraction of a nanometer, but it is not possible to focus the emitted light back to a region the same size. – S. McGrew Mar 15 '19 at 01:35
  • How does that explain why it is incorrect? Uncertainty in location and momentum cause respective spatial and temporal decoherences. These will govern how tightly you can focus light. – Paul Childs Mar 15 '19 at 01:54
  • This answer is wrong. You can't focus light into a single point, even if you try to describe it as a classical wave. There is no classical solution to the wave equation with an abitrary small focus-size. – Quantumwhisp Mar 30 '21 at 18:53
  • @Quantumwhisp please tell me then what is the classical solution for ω_0 in my example. – Paul Childs Mar 31 '21 at 00:04
  • Just to understand you right - you talk about a spherical wave solution? The spherical mirror you talk about covers all solid angles around the "focus" point? – Quantumwhisp Mar 31 '21 at 07:34
  • Yes. Otherwise any edge would break symmetry. – Paul Childs Apr 01 '21 at 00:55