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|n=0> photon number state

  1. Sep 18, 2011 #1
    Can you send a light pulse consisting of |n=0> , e.e. vacuum state, and what is the physical difference between sending zero photons and sending no photons?
     
  2. jcsd
  3. Sep 18, 2011 #2

    Ken G

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    What makes you think there is a difference? I'm not aware of any. You can't "not send" the vacuum, it's already there.
     
  4. Oct 14, 2011 #3
    All articles about quantum information exchange talk at length about the possibility of one party (an adversary) intercepting a dim laser pulse and substituting the signal with a vacuum state. So does ""substituting a vacuum state" mean that the adversary will modify the signal and resend it, or does it mean that he just puts a reflecting mirror on the way of the pulse and shuts down the communication channel?
    Am I correct to assume that since the vacuum state has energy, propagation of this state through space is not the same as the absence of any signaling?
     
  5. Oct 14, 2011 #4

    Ken G

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    I can't see that any current theory of physics can distinguish sending a vacuum state from not sending anything at all. Replacing a dim laser with a vacuum state sounds like blocking the laser to me, but maybe there is some distinction that quantum field theory can make. Perhaps someone more of an expert in QFT can say if there's something I'm overlooking, but it sounds like pretty speculative science to me.
     
  6. Oct 14, 2011 #5
    Should it be possible to have a photon-number resolving detector (n'th outcome = n photons) and then to have the pulse blocked for any n not equal to zero? Would such an apparatus provide for an experimental realization of "vacuum state substitution"?

    A separate question. Does the vacuum state mean the vanishing of all observables (energy, polarization) as well as phase exp(i*delta)? That is, will the admixture of |n=0> state in a red laser pulse be physically different from the same admixture of |n=0> state in a green laser pulse?
     
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