Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was talking my professor about entanglement swapping between light and matter and it is briefly described here:

You start out with a crystal capable of doing parametric down conversion of incoming photons. When they go in, they undergo a physical process which produces two entangled photons that come out. At the same time, you get two atoms of the same type, say two Hydrogen atoms that are trapped in a harmonic oscillator potential in the, let's say x direction. Then, you send each entangled photon pair to each of the atoms along the x direction and have them interact. You keep on sending entangled photon pairs until the two atoms come to a steady state where they oscillate in a synchronous fashion. You essentially transfer the momentum of the photon to the atoms and do the opposite of laser cooling. The atoms are entangled in position and momenta. When one is measured at x1 = 1, the other is at x2 = -1. Their momenta are equal. p1 = p2.

So, if we are to imagine two people. These two people would be the Hydrogen atoms. The photons that bounce off each of the two people are like the entangled photons. With their vision, absorption of photons, the two people can come into sync. When two people dance mirror images of each other, this can be viewed as if they are in "opposite" positions but since they are moving at the same velocity, they share equal momenta. You can't take the analogy too far, but would this be good?

share|cite|improve this question
I'm afraid you've just missed the entry date for this year's "Dance your PhD" contest – EnergyNumbers Oct 5 '12 at 10:40

Entanglement is more than correlation, it is correlation in amplitude, not just in probability. The classical examples can only correlate probability. Correlation in amplitude leads to violations of Bell's inequality, which shows there is no good local classical analog.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.