# Controlling the outcome of a quantum measurement through translational entanglement

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." 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...

• 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. Oct 18, 2012 at 16:01
• 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 Oct 18, 2012 at 22:43
• 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 Oct 22, 2012 at 14:40
• Saying "measure the position of atom A very very precisely" means very little.
– bla
Oct 25, 2012 at 8:29
• How do you propose the scientist at Lab B will measure if the atom B is still on the harmonic trap or not? Oct 29, 2012 at 14:05