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According to the paper: A. S. Parkins and H. J. Kimble, Phys. Rev. A 61, 52104 (2000). http://pra.aps.org/abstract/PRA/v61/i5/e052104

You can entangle position and momenta of two atoms by using entangled light. Atoms A and B are located at sites A and B. You get entangled light by parametric-down-conversion and send one photon to atom A and the corresponding entangled photon to atom B. You keep on sending entangled pairs until atom A and B come to a steady-state and they will be entangled with each other. Essentially what is happening is that "entanglement is transferred from a pair of quantum-correlated light fields to a pair of trapped atoms in a process of quantum state exchange."

enter image description here The atom is trapped in the x direction. The atom is stuck in a harmonic potential. The entanglement is only in x direction.

position are anti-correlated between two atoms: (q1 = -q2) and momentum are correlated (p1 = p2)

You can measure the position and momenta of each particle using homodyne measurement techniques. So, if I were to measure the position of atom A very very precisely, then its momentum would be very uncertain. Its distribution would be so smeared out that it'd be approaching a uniform distribution of momentum, if position is a sharp delta function. At this point, there would exist a probability that the momentum is very large, and thus its kinetic energy would be very large too! KE = p^2/2m. (Need a checking here) Would there be a non-negligible probability that the energy of the atom would be so large it would escape the harmonic oscillator trap?!

Since the two atoms are entangled, p1 = p2, thus atom B would also have correlated momentum which translates to correlated kinetic energy. Atom B would also have enough energy to escape the harmonic trap as well.

If everything I have stated is correct, haven't I found a method to communicate via translational entanglement, because I can control the outcome of the measurement? (Unlike spin...) Let's say we have an ensemble of these entangled atoms. A scientist at Lab A can send a binary message by choosing to measure the position of an atom at Lab A to high precision thereby causing the atom at Lab B to escape the harmonic trap. Not measuring would keep the atoms trapped. The presence or absence would correspond to a 1 or a 0. Since there would exist chances that the momentum would not be large enough to cause an escape, then we could have batches of entangled atoms that would represent one bit. Very costly, but you can interact instantaneously over large distances...

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    $\begingroup$ Your description is very vague, to say the least. If you add some actual quantum state descriptions that specifically show the exact form of entanglement then I will try to answer your question. And I don't have time to read that paper. $\endgroup$
    – A.O.Tell
    Oct 18, 2012 at 16:01
  • $\begingroup$ Entanglement is given by equation (9) in this paper: prola.aps.org/abstract/PR/v47/i10/p777_1 Also, the teleportation of quantum states using translational entanglement is located here: pra.aps.org/abstract/PRA/v49/i2/p1473_1 More references to the nature of entanglement: (I recommend looking at this paper) arxiv.org/abs/quant-ph/9909021 $\endgroup$
    – QEntanglement
    Oct 18, 2012 at 22:43
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    $\begingroup$ Your references are all about entanglement in general, not about what you're describing here. If you don't provide a more quantitative and mathematical description of your proposal then I doubt anyone will care enough to try to guess what you could possibly mean $\endgroup$
    – A.O.Tell
    Oct 22, 2012 at 14:40
  • $\begingroup$ Saying "measure the position of atom A very very precisely" means very little. $\endgroup$
    – bla
    Oct 25, 2012 at 8:29
  • $\begingroup$ How do you propose the scientist at Lab B will measure if the atom B is still on the harmonic trap or not? $\endgroup$
    – user56771
    Oct 29, 2012 at 14:05

1 Answer 1

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It has already been proved a long time ago that quantum entanglement cannot be used to transmit information faster than light, by any mean. I'm sure this is no exception (you control the next outcome of the measurement in spin entanglement too), but I'm not sure of how to prove it.

Here is my guess: Once the atoms are entangled, you can't touch them without breaking the entanglement. You have the two atoms in the same state, with low knowledge about position and momentum, then you separate them and they keep being entangled. Should you measure one's position very precisely then you break the entanglement, and you know atom B's position very precisely as well, but this doesn't associate with momentum uncertainty. I think I remember from my lessons of quantum metrology that since you are using two atoms to extract your measurements, you can bypass Heisenberg uncertainty principle, knowing both position and momentum of the atom (before measurement, of course) with great precision. You can do the same thing with lasers measuring number of photons and phase (which are conjugated variables).

Don't you have an (ex-)advisor to ask this to? He/She could definitely help you out with this. Otherwise you can send an email to Fabio Sciarrino (you can google him), he's a very friendly researcher on experimental quantum information theory I had the pleasure of studying with for some time. He definitely knows this.

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  • $\begingroup$ Yes, once you measure it, they stop being entangled. However, I'm suggesting that you can control the position measurement because you choose to measure the position very precisely! I know that it has been established that you can't do it with spin or discrete variables like that because you CAN't control the outcome. However, I'm suggesting this scheme does. This scheme would also mean that you have limited amount of messages to send. $\endgroup$
    – QEntanglement
    Oct 29, 2012 at 1:58
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    $\begingroup$ I edited my post to make it more sense. $\endgroup$
    – Ferdinando Randisi
    Oct 29, 2012 at 13:22

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