Suppose you have a pool table (with no pockets) with a ball in the middle and another ball in one corner. The ball in the corner is hit toward the middle ball. The balls collide and each ball ends up on opposite sides of the table. The 'x-position' of each ball along the walls at each end of the table is observed for one of the balls after they collide. Because of the conservation of momentum, if you observe the x-position of one ball after the collision, you know the x-position of the second ball because of the conservation of momentum (e.g. if I find one ball in one corner, I will know the other ball will be located at the opposite corner due to the conservation of momentum).

So my first thought was that the collision of the balls is like the balls being entangled such that when the experiment is over, I only need to measure the position of one ball to know the position of the. other without directly measuring it. Secondly, if a third ball lightly hits one of the other balls after the original collision, then the accuracy with which I can predict the position of the second ball after measuring the final position of the first ball will not be perfect, but there will still be a correlation, and the strength of the correlation will depend on how hard the third ball hit one of the first two after the original collision. To me, that seems anbalagous to decoherence.

Assume that you cannot observe the trajectories of the balls as they move on the table, you can only measure the final positions of the balls along the walls. My question is, is this example just equivalent to the 'put a left hand glove in one box and a right hand glove in another box' example that is often used to show how quantum entanglement does not work? Or does it make no sense whatsoever as an analogy for quantum entanglement?

  • $\begingroup$ I'd be happy to answer any questions you have about quantum entanglement to the best of my abilities. But the scenario you have presented is most definitely not an example of quantum entanglement in the first place, that I really don't know where to begin. Might you choose a different example (perhaps one not quite such an example of classical physics)? $\endgroup$ – the_photon May 9 at 4:42
  • $\begingroup$ measurement outcomes can be correlated without the system being entangled. Indeed, as long as you only use one measurement basis, any degree of (classical) correlation can be produced by non-entangled states. Does that answer the question? $\endgroup$ – glS May 9 at 22:07
  • $\begingroup$ Why on earth would it differ from gloves?? $\endgroup$ – Norbert Schuch May 10 at 18:46

Entanglement exists not only in all mathematical models used in physics, classical or quantum mechanical, but also in everyday situations.

My easy example: Two brothers , Dick and Harry. You hear that both are working abroad one in Paris the other in Cairo. If you meet Dick in Cairo you immediately know that Harry is in Paris. They are entangled by being brothers.

Entanglement , used in the quantum mechanical level, means that there exists a quantum mechanical mathematical model that relates two quantities , allowing to determine the total solution when some partial variables are known.

So your example is an example on the concept of entanglement in classical physics models.

In my opinion of course.

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    $\begingroup$ what definition of the word "entanglement" are you using here? Because you are just describing classical correlations. Quantum entanglement is a much stronger property. $\endgroup$ – glS May 9 at 12:39
  • $\begingroup$ @glS QE imo it is as strong as the mathematics used to describe the physical situation, which is all I am hopefully saying. $\endgroup$ – anna v May 9 at 13:24
  • $\begingroup$ you are saying that correlations are a type of entanglement. That is simply not the case, as two quantum systems can produce correlated outcomes without being entangled. $\endgroup$ – glS May 9 at 21:58
  • $\begingroup$ @glS it is the type of variables/constants used for defining entanglement. In the QM use they use quantum numbers. But also in my two brothers use you can think of "quantum numbers" one brother spin up, instead of name the other down. $\endgroup$ – anna v May 10 at 3:05
  • $\begingroup$ no. You can have that with a separable state, e.g. $|00\rangle\!\langle00|+|11\rangle\!\langle11|$. There is no entanglement there, only correlations. Entanglement entails, roughly speaking, correlations in multiple measurement bases $\endgroup$ – glS May 10 at 7:17

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