4 ms·

"Spooky action" indeed

lisper 12y ago
It actually isn't spooky at all once you understand it properly (but it is almost never explained properly, especially in the popular press). Measurement and entanglement are really the same phenomenon. The process of "measuring" photon A was actually begun when it became entangled with photon B (and vice versa). We call it a "measurement" when a large number of particles (like a measurement apparatus or a brain) become mutually entangled with each other. See: http://www.flownet.com/ron/QM.pdf http://www.flownet.com/ron/QM.pdf for a more detailed explanation.
- ClayFerguson 12y ago
  lisper, actually measurement is the act of collapsing the wave, and entanglement only exists before the wave function collapse. It's possible to never measure something, and it could be entangled forever.
  - waqf 12y ago
    You didn't read lisper's link, did you?
    - ClayFerguson 12y ago
      Yes I read his comment. Both entanglement and measurement apply to individual particles, so the opposite of what he said is true.
      - lisper 12y ago
        You didn't read waqf's comment, did you? You may have read my comment but you clearly didn't read my link. > actually measurement is the act of collapsing the wave, and entanglement only exists before the wave function collapse. It's possible to never measure something, and it could be entangled forever. Entangled particles continue to be entangled even after they are "measured". If this were not true, faster-than-light communication would be possible. To understand why, read the paper. The only way to "undo" an entanglement is to time-reverse the process that created the entangled pair to begin with, i.e. to bring the members of the entangled pair back together.
        ClayFerguson 12y ago
        You are correct, I didn't read your link. Why? Because every word of what you said in your comment was wrong. So why would I click the link?
        lisper 12y ago
        I dunno, because you might learn something?
trhway 12y ago
it is actually pretty spooky setup. May be there is a Nobel Prize lurking there, yet looking at the Figure 2 of their arxiv paper http://arxiv.org/ftp/arxiv/papers/1401/1401.4318.pdf http://arxiv.org/ftp/arxiv/papers/1401/1401.4318.pdf one can't wonder why for example the beam L4'-L5 isn't filtered out - if this beam does carry information about the image then we would have FTL communication possible. Another question is why L3-L3' red beam - the image info carrying beam - is fed into NL2 crystal thus providing a chance to physically affect formation of NL2-D5-L6 yellow beam, ie. to exchange information. Some of the red beam photons entering NL2 would be consumed and re-emitted by the NL2 crystal's atoms (and the location of these atoms will be affected by the red beam internal density ie. by the image of the object "O") and thus would physically affect the yellow beam generation, i.e. passing some image info into the yellow beam. Thus no entanglement or other "spooky" QM features necessary for the NL2-D5-L6 yellow beam to become carrying some image info.
- bkcooper 12y ago
  if this beam does carry information about the image then we would have FTL communication possible. See Michael Nielsen's answer elsewhere in the comments for why this is not the case. Another question is why L3-L3' red beam - the image info carrying beam - is fed into NL2 crystal thus providing a chance to physically affect formation of NL2-D5-L6 yellow beam, ie. to exchange information. The interference requires us to not know which path the photon took way back at BS1, in that diagram. The probability of pair formation in the nonlinear crystals is quite low, so with very high likelihood, only one of the crystals will be responsible for producing a pair of photons (including the yellow signal photon.) Suppose you did not route the imaging photon through NL2. Then observing an output photon at D3 would tell you that the pair production happened at NL2 (since there is no way for the imaging photon to get there.) This will destroy the interference pattern of the "yellow" photons. That is why the beam is sent through NL2.
  - trhway 12y ago
    >This will destroy the interference pattern of the "yellow" photons. That is why the beam is sent through NL2 why do you need interference pattern in situation when the imaging beam is sent through NL2? Sending the imaging beam through NL2 causes it to transfer the image info into the yellow beam formed in NL2 (the red beam heats the crystal and excites the crystal's atoms which affects yellow beam formation) and thus makes all the discussion of interference and entanglement moot and unnecessary. The other yellow beam can be just discarded as the yellow beam from NL2 alone is enough to produce the picture.

It actually isn't spooky at all once you understand it properly (but it is almost never explained properly, especially in the popular press). Measurement and entanglement are really the same phenomenon. The process of "measuring" photon A was actually begun when it became entangled with photon B (and vice versa). We call it a "measurement" when a large number of particles (like a measurement apparatus or a brain) become mutually entangled with each other. See: http://www.flownet.com/ron/QM.pdf http://www.flownet.com/ron/QM.pdf for a more detailed explanation.

ClayFerguson 12y ago
lisper, actually measurement is the act of collapsing the wave, and entanglement only exists before the wave function collapse. It's possible to never measure something, and it could be entangled forever.
- waqf 12y ago
  You didn't read lisper's link, did you?
  - ClayFerguson 12y ago
    Yes I read his comment. Both entanglement and measurement apply to individual particles, so the opposite of what he said is true.
    - lisper 12y ago
      You didn't read waqf's comment, did you? You may have read my comment but you clearly didn't read my link. > actually measurement is the act of collapsing the wave, and entanglement only exists before the wave function collapse. It's possible to never measure something, and it could be entangled forever. Entangled particles continue to be entangled even after they are "measured". If this were not true, faster-than-light communication would be possible. To understand why, read the paper. The only way to "undo" an entanglement is to time-reverse the process that created the entangled pair to begin with, i.e. to bring the members of the entangled pair back together.
      - ClayFerguson 12y ago
        You are correct, I didn't read your link. Why? Because every word of what you said in your comment was wrong. So why would I click the link?
        lisper 12y ago
        I dunno, because you might learn something?

trhway 12y ago

it is actually pretty spooky setup. May be there is a Nobel Prize lurking there, yet looking at the Figure 2 of their arxiv paper http://arxiv.org/ftp/arxiv/papers/1401/1401.4318.pdf http://arxiv.org/ftp/arxiv/papers/1401/1401.4318.pdf one can't wonder why for example the beam L4'-L5 isn't filtered out - if this beam does carry information about the image then we would have FTL communication possible. Another question is why L3-L3' red beam - the image info carrying beam - is fed into NL2 crystal thus providing a chance to physically affect formation of NL2-D5-L6 yellow beam, ie. to exchange information. Some of the red beam photons entering NL2 would be consumed and re-emitted by the NL2 crystal's atoms (and the location of these atoms will be affected by the red beam internal density ie. by the image of the object "O") and thus would physically affect the yellow beam generation, i.e. passing some image info into the yellow beam. Thus no entanglement or other "spooky" QM features necessary for the NL2-D5-L6 yellow beam to become carrying some image info.

bkcooper 12y ago
if this beam does carry information about the image then we would have FTL communication possible. See Michael Nielsen's answer elsewhere in the comments for why this is not the case. Another question is why L3-L3' red beam - the image info carrying beam - is fed into NL2 crystal thus providing a chance to physically affect formation of NL2-D5-L6 yellow beam, ie. to exchange information. The interference requires us to not know which path the photon took way back at BS1, in that diagram. The probability of pair formation in the nonlinear crystals is quite low, so with very high likelihood, only one of the crystals will be responsible for producing a pair of photons (including the yellow signal photon.) Suppose you did not route the imaging photon through NL2. Then observing an output photon at D3 would tell you that the pair production happened at NL2 (since there is no way for the imaging photon to get there.) This will destroy the interference pattern of the "yellow" photons. That is why the beam is sent through NL2.
- trhway 12y ago
  >This will destroy the interference pattern of the "yellow" photons. That is why the beam is sent through NL2 why do you need interference pattern in situation when the imaging beam is sent through NL2? Sending the imaging beam through NL2 causes it to transfer the image info into the yellow beam formed in NL2 (the red beam heats the crystal and excites the crystal's atoms which affects yellow beam formation) and thus makes all the discussion of interference and entanglement moot and unnecessary. The other yellow beam can be just discarded as the yellow beam from NL2 alone is enough to produce the picture.