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Is it possible that if you have 2 ultraviolet lasers, that are invisible to the human eye, and if you aim their beams to intersect at some point, that the place of intersection will show a lower visible wavelength of light, caused by interference of the light freqency? Can any other form of heat or energy be generated at the intersection?

If so, can someone provide a link to the explanation? I am curious about this.

EDIT: I should add that the 2 lasers would have different light frequencies while they are both invisible to the human eye. So, I am wondering about interference between light freqencies.

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    $\begingroup$ interference causes change in amplitude, not change in freq.en.wikipedia.org/wiki/… notice that the intensity is only thing which is changing. $\endgroup$ – Vineet Menon Dec 11 '11 at 4:55
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    $\begingroup$ Interference changes the spatial distribution of intensity of the same frequency. $\endgroup$ – Vladimir Kalitvianski Dec 11 '11 at 11:40
  • $\begingroup$ Interference only affects like-frequencies as long as your measurement is time averaged. Instantaneous measurement would show interference between frequencies. $\endgroup$ – Colin K Dec 14 '11 at 1:34
  • $\begingroup$ If all you care about is the invisible part, and not actually ultraviolet, then en.wikipedia.org/wiki/Two-photon_excitation_microscopy is for you. $\endgroup$ – jcohen79 Oct 9 '12 at 5:13
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What you say is not possible with interference. Interference of light does not produce new colors of light. Light would have to scatter inelastically off of some molecules to produce a down-shifted frequency (Raman scattering or some type of wave-mixing phenomena). The point is, the light should interact with matter to change its frequency.

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    $\begingroup$ Why not? $\endgroup$ – Peter Shor Dec 11 '11 at 5:18
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    $\begingroup$ Heterodyning and Homodyning techniques need a mixing element. In electronics, a simple diode arrangement acts as a mixer whereas in optics, a beam-splitter cube acts as a mixer. Photons do not interact with each other. This is fundamental. You need an interaction to change the energy of a photon. (Also, interference is not observed with completely uncorrelated light). $\endgroup$ – Antillar Maximus Dec 11 '11 at 15:17
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    $\begingroup$ I have never heard of a glass beam-splitter which can act as a mixer. This would make nonlinear optics substantially easier if it were true. I think you are confusing the beam-splitters role (giving two beams parallel k-vectors) and the mixing process which requires a material with a nonlinear polarizability. $\endgroup$ – user2963 Dec 11 '11 at 15:55
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    $\begingroup$ "Photons do not interact with each other." I'll caution you that this is true in classical optics, but not in a fully quantum treatment. That said, light-on-light effects are very weak for ordinary light intensities. $\endgroup$ – dmckee --- ex-moderator kitten Dec 11 '11 at 17:09
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    $\begingroup$ @AntillarMaximus You don't need a direct interaction...they interact at the one-loop level. $\endgroup$ – dmckee --- ex-moderator kitten Dec 11 '11 at 22:52
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You'd have more luck using two infrared lasers - if you shone two infrared lasers at your eye (or at a camera) with frequencies that are half that of a visible photon, a small percentage of them would undergo two-photon absorption; two photons each with half the energy needed would be absorbed by a sensor element at the same time, causing the sensor to detect the equivalent of a visible-light photon.

The intensities would have to be very high for such an effect to be visible; I haven't done the calculations, but I'm confident the infrared lasers would completely cook your eye before you managed to see anything.

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If you use the appropriate non-linear media, and both beams have a high enough intensity (probably requiring pulsed lasers), you may be able to have a visible effect at the point inside the media where the beams intersect. Such nonlinear optics are pretty difficult to get working in practice, but not theoretically impossible.

My understanding is that pretty much all optically-pumped lasers convert (typically incoherent) high-energy (short wavelength) photons into laser light with somewhat lower-energy (longer wavelength) photons.

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The basic reason the answer is NO, as everybody says, is because light, an electromagnetic wave, is an emergent phenomenon from innumerable individual photons. Thus your question boils down to :

is there a photon photon interaction? can a photon scatter off a photon?

The answer is : yes, there exists a two photon interaction, but with such a tiny probability that it is improbable to see a collective effect unless at very high energies .

photon photon

A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

Note that there are four electromagnetic vertices here, and the electromagnetic coupling constant that ends up as squared in the calculations is (1/137)^1/2, four times. This is a very small number and ensures the observations: that electromagnetic beams do not interact in vacuum except at very high energies when the crossection rises.

There are proposals for gamma gamma colliders, not exactly visible electromagnetic radiation as an output, but lots of elementary particles.

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  • $\begingroup$ Nice answer, however in all of these things - people like to say no, and that is generally true at least for a time until we find out what we do not know. Like meta materials and cloaking the things we know now that once were said to not be impossible. $\endgroup$ – Ken Jul 22 '20 at 5:59

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