1
$\begingroup$

Most entanglement examples I see have two particles being entangled over the same observable (e.g. spin 1/2). But, can two particles become entangled with respect to two non-commuting observables? For example could the position of one particle be correlated through entanglement with the momentum of another particle?

Note: My thinking is this: I don't see any reason why not. After all, the fact that position and momentum are non-commuting is a HUP thing that shows up after collapse. But entanglement is entirely in the quantum realm. I guess the real issue is whether the two particles could become entangled over these two non-commuting observables at all. A complication would be that the entangled wave function would have to the eigenvectors of both particles' wave functions.

$\endgroup$
2
  • 1
    $\begingroup$ What does "'entangled over an observable" mean? They are entangled or they aren't. $\endgroup$
    – WillO
    Apr 10 at 3:05
  • $\begingroup$ To pick up on @WillO ‘s point: either the state is separable or it is not. This does not depend on observables or their eigenvectors. $\endgroup$ Apr 10 at 3:20

1 Answer 1

0
$\begingroup$

The position and momentum of two particles which don't interact do commute, since these operators act only on their respective systems and are therefore independent. The Heisenberg uncertainty principle applies only to the position and momentum of the same particle/system. Similar, you could imagine an entangled system of two particles whose spin measured along the z-axis and spin measured along the x-axis are correlated. However, I'm not aware of a way to prepare two particles whose position and momentum are entangled in a lab.

$\endgroup$
2
  • 1
    $\begingroup$ The latter can be easily prepared: Just entangle them both along the z-axis and rotate one of them. $\endgroup$ May 1, 2020 at 20:31
  • $\begingroup$ Thanks! I meant the position-momentum state. $\endgroup$
    – anticharm
    May 1, 2020 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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