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Stimulated generation of superluminal light pulses via four-wave mixing (NIST)

  1. May 17, 2012 #1
    Concerning this paper:

    Stimulated generation of superluminal light pulses via four-wave mixing.

    I have two questions:

    1) How can we check that information cannot be propagated faster than light with this system?

    2) What is the meaning of the last sentense of this paper:


    Thanks
     
  2. jcsd
  3. May 17, 2012 #2
    1) That can actually be tricky conceptually. The understanding is that 'new' information is transmitted with points of non-analyticity. that is kinks and edges on a pulse. If you go in with a smooth pulse, as they did in the experiment, this would be the very front of the pulse, which is well hidden in the experimental noise.
    Other experiments have been done with 'kinks' above the noise floor, and (with quite some uncertainty) showed that the 'kinks' travel with c, wether they go through a fast-light medium or not.
    2) Loosely speaking, the apparent superluminal effect here is created by different parts of the pulse being scrambled by the medium, so that the pulse peak looks like it travels faster than c.
    I think what they mean is that quantum correlations with a reference pulse might be a means to measure how that exactly works.
     
  4. May 18, 2012 #3
    Thanks honeyg,
    Would you also have a comment about the "negative group velocity" mentionned in the paper?
     
  5. May 18, 2012 #4
    The group velocity is by definition the velocity of the pulse peak. In dispersive media this velocity can be changed dramatically, in slow light media, e.g., people could show (almost) stopped light, which means the group velocity goes down from c to 0. In fast light media the opposite happens, and for some range the group velocity becomes larger than c, and eventually is infinity. That happens when the (reshaped) pulse peak exits the medium at exactly the moment when the original pulse peak enters the medium. And one can go even beyond this, and then the observed pulse peak exits the medium before the original pulse peak enters. That's then a negative group velocity.
     
  6. May 21, 2012 #5
    Thanks again!

    I read this page:

    http://www.nist.gov/pml/div684/fast_light.cfm

    and could not understand the meaning of this part:

    How should I understand these 50ns, considering that it would take 0.05 ns for light to go through the cell (1.7 cm) in vacuum.
    I can't visualize what happens.
    How much time does the light take to go through the cell in this experiment?
     
  7. May 21, 2012 #6
  8. May 22, 2012 #7
    Thanks a lot honeyg.
    This is quite clear!

    Would you have a further comment about the 50 ns compared to the 0.5 ns length of the cell?
    Would it be because the the width of the input signal is very large compared to the cell?
     
  9. May 22, 2012 #8
    So here's the link to the original piece:
    http://arxiv.org/abs/1204.0810

    I guess you have a good point. The used pulses are 200 ns long which correspond to something like 70 m, so only a small fraction of the pulse is at a time in the cell. So what happens is, because the front of the input pulse has different frequency components than its tail, the output pulse is reshaped by the medium, and if you look just at the pulse peak, it apparently propagates 50 ns faster through the cell than if you leave the medium away. (btw the spin of the paper seems to be rather the second pulse that is generated in the process.)
     
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