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Do they really split a single photon or just absorb one and generate 3 new one? I assumed that photon is an elementary particle.

JadeNB 7y ago
> Do they really split a single photon or just absorb one and generate 3 new one? I think that there is no meaningful difference between these two descriptions. (IANAP.)
- lisper 7y ago
  You have it exactly right. Photons are fungible. Splitting a photon is just like splitting a bitcoin. It makes no sense to say whether the output coins are "the same" as the input coins.
  - y04nn 7y ago
    If you produce 2 entangled photons, if you "split" it into 3, do the the 3 new photons still entangled to the first one?
    - lisper 7y ago
      I presume you meant to ask: If you produce 2 entangled photons, and then split one of them again, do you end up with three mutually-entangled photons? The answer to that is: yes. The next obvious question then is: what then is the big deal about producing three mutually entangled photons? And the answer to that is that spontaneous parametric down conversion is very sensitive to the wavelength/frequency of the input light. You can't just throw in any old photon and split it. It has to be a very specific wavelength for the process to work, so as a practical matter it is nearly impossible to do the two-step process that you've asked about.
contravariant 7y ago
With the way quantum mechanics works it might not be possible to meaningfully distinguish between the two. Any attempt to verify which of the two is happening is likely to change the result.
athaht 7y ago
The photon is split into multiple photons with lower frequencies. Energy conserved
- lisper 7y ago
  > Energy conserved As well as linear and angular momentum. The latter is the most important quantities for quantum optics because that is what produces polarization correlations.
- eveningcoffee 7y ago
  This is what I presumed that the output is photons with lower frequencies but is it splitting in sense of really separating parts of the photon or just transfer to lower energy state that for conservation of energy produces two new photons while consuming the first one?
  - gus_massa 7y ago
    The photons have no inner parts. This is more similar to destroying the photon and creating three new photons. (I don't like "absorbing". They use probably a non linear crystal, and the transformation is probably not in a single spot. I'm not sure about the details, but probably the photon slowly disappear while the photon is traveling inside the crystal, and at the same time the other three photons slowly apear. This must be interpreted as probabilities of seen the original photons or the new three photons at the other end of the crystal. You can't have half a photon, only a 50% probability of seeing a whole photon. This explanation has too many weird things and handwaving ... just use "destroying" instead of "absorbing". More info https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion https://en.wikipedia.org/wiki/Spontaneous_parametric_down-co... )
analog31 7y ago
There are processes that do the latter, such as fluorescence. There are features of the two processes that let you figure out which is which. In fluorescence, there is a time delay that might be long enough to observe. Also, the generated photons will come out with energies characteristic of the material, rather than of the incoming photon. And there are differences related to polarization and the angular distributions of the photons.
bvinc 7y ago
There are many different ways of looking at quantum mechanics, so here is one way of answering your confusion. Probably the standard way of looking at quantum mechanics right now is quantum field theory. In quantum field theory, you give up the notion of "particles" and instead view them as vibrations in fields. Once you do that, it makes more sense that these "particles" can turn into other particles, be absorbed or emitted by other particles, or split. Here's a video of Sean Carroll talking about it, roughly 28:00-33:00. https://youtu.be/gEKSpZPByD0?t=1680 https://youtu.be/gEKSpZPByD0?t=1680
rpz 7y ago
I doubt the photon is truly fundamental. I've mentioned this recently in another thread but I'm pretty sure E=hf is hiding information and the units of plancks constant are wrong. I think the equation should be E=htf where h is in Joules/oscillation (quanta of energy) t is measure time and f is osillations/second. Check out Juliana H.J Mortensen's work. The upshot is that in this model, every oscillation of light carries the same energy regardless of frequency. This supports wave particle duality even more so than the current model, and takes the uncertainty out of the uncertainty principle.
- vermilingua 7y ago
  Forgive the layperson question, but: if all light carries the same energy regardless of frequency, would this experiment (where one photon is split into three with the same sum frequency) not violate conservation of energy?
  - rpz 7y ago
    The idea here is that what people generally regard as a "photon" today is actually one second's worth of oscillations of light at a particular frequency. Imagine each oscillation of light as the true elementary particle instead. There are plenty of those elementary particles within the classical "photon" that can be split out into smaller "photons" of a different frequency.
- fsh 7y ago
  Not even wrong
- rpz 7y ago
  Lol downvote me all you want folks. https://img1.wsimg.com/blobby/go/e266e5e7-739c-437c-89c3-eed60d3a9f3f/downloads/1cfqec4a8_58871.pdf https://img1.wsimg.com/blobby/go/e266e5e7-739c-437c-89c3-eed...

JadeNB 7y ago

> Do they really split a single photon or just absorb one and generate 3 new one? I think that there is no meaningful difference between these two descriptions. (IANAP.)

lisper 7y ago
You have it exactly right. Photons are fungible. Splitting a photon is just like splitting a bitcoin. It makes no sense to say whether the output coins are "the same" as the input coins.
- y04nn 7y ago
  If you produce 2 entangled photons, if you "split" it into 3, do the the 3 new photons still entangled to the first one?
  - lisper 7y ago
    I presume you meant to ask: If you produce 2 entangled photons, and then split one of them again, do you end up with three mutually-entangled photons? The answer to that is: yes. The next obvious question then is: what then is the big deal about producing three mutually entangled photons? And the answer to that is that spontaneous parametric down conversion is very sensitive to the wavelength/frequency of the input light. You can't just throw in any old photon and split it. It has to be a very specific wavelength for the process to work, so as a practical matter it is nearly impossible to do the two-step process that you've asked about.

contravariant 7y ago

With the way quantum mechanics works it might not be possible to meaningfully distinguish between the two. Any attempt to verify which of the two is happening is likely to change the result.

athaht 7y ago

The photon is split into multiple photons with lower frequencies. Energy conserved

lisper 7y ago
> Energy conserved As well as linear and angular momentum. The latter is the most important quantities for quantum optics because that is what produces polarization correlations.
eveningcoffee 7y ago
This is what I presumed that the output is photons with lower frequencies but is it splitting in sense of really separating parts of the photon or just transfer to lower energy state that for conservation of energy produces two new photons while consuming the first one?
- gus_massa 7y ago
  The photons have no inner parts. This is more similar to destroying the photon and creating three new photons. (I don't like "absorbing". They use probably a non linear crystal, and the transformation is probably not in a single spot. I'm not sure about the details, but probably the photon slowly disappear while the photon is traveling inside the crystal, and at the same time the other three photons slowly apear. This must be interpreted as probabilities of seen the original photons or the new three photons at the other end of the crystal. You can't have half a photon, only a 50% probability of seeing a whole photon. This explanation has too many weird things and handwaving ... just use "destroying" instead of "absorbing". More info https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion https://en.wikipedia.org/wiki/Spontaneous_parametric_down-co... )

analog31 7y ago

There are processes that do the latter, such as fluorescence. There are features of the two processes that let you figure out which is which. In fluorescence, there is a time delay that might be long enough to observe. Also, the generated photons will come out with energies characteristic of the material, rather than of the incoming photon. And there are differences related to polarization and the angular distributions of the photons.

bvinc 7y ago

There are many different ways of looking at quantum mechanics, so here is one way of answering your confusion. Probably the standard way of looking at quantum mechanics right now is quantum field theory. In quantum field theory, you give up the notion of "particles" and instead view them as vibrations in fields. Once you do that, it makes more sense that these "particles" can turn into other particles, be absorbed or emitted by other particles, or split. Here's a video of Sean Carroll talking about it, roughly 28:00-33:00. https://youtu.be/gEKSpZPByD0?t=1680 https://youtu.be/gEKSpZPByD0?t=1680

rpz 7y ago

I doubt the photon is truly fundamental. I've mentioned this recently in another thread but I'm pretty sure E=hf is hiding information and the units of plancks constant are wrong. I think the equation should be E=htf where h is in Joules/oscillation (quanta of energy) t is measure time and f is osillations/second. Check out Juliana H.J Mortensen's work. The upshot is that in this model, every oscillation of light carries the same energy regardless of frequency. This supports wave particle duality even more so than the current model, and takes the uncertainty out of the uncertainty principle.

vermilingua 7y ago
Forgive the layperson question, but: if all light carries the same energy regardless of frequency, would this experiment (where one photon is split into three with the same sum frequency) not violate conservation of energy?
- rpz 7y ago
  The idea here is that what people generally regard as a "photon" today is actually one second's worth of oscillations of light at a particular frequency. Imagine each oscillation of light as the true elementary particle instead. There are plenty of those elementary particles within the classical "photon" that can be split out into smaller "photons" of a different frequency.
fsh 7y ago
Not even wrong
rpz 7y ago
Lol downvote me all you want folks. https://img1.wsimg.com/blobby/go/e266e5e7-739c-437c-89c3-eed60d3a9f3f/downloads/1cfqec4a8_58871.pdf https://img1.wsimg.com/blobby/go/e266e5e7-739c-437c-89c3-eed...