0
$\begingroup$

If the wavefunctions of two fermions interact, after which they separate can it be that the space parts of the wavefunctions become entangled, just like their spin functions will become entangled, no matter how far they are apart ("transcending" space)? After all, the position is a (quantum mechanical) degree of freedom (each position wavefunction is a continuos superposition of delta distributions with a continuous spectrum of weights for the distributions) and interaction can entangle these degrees of freedom.
This would imply that measuring the fermion positions (when their wavefunctions are far away from each other) of these once overlapping wavefunctions will show a entanglement (correlation) between the positions of the fermions (just like the spins show entanglement after the interaction), so these positions also "transcend" space.
So can it be that the interaction can induce entanglement between the continuous distribution of positions?
How can this entanglement show itself? Well, maybe the distance between the measured values (when far apart) of the position on the fermions stays constant in time, or instead of being constant, these measured distances vary in a certain defined way because their wavefunctions spread out. I.e. when one measures the (range of) position(s) of one particle, the (range of) position(s) of the other will be known too.

Have experiments been done to look for this or is it ruled out in the first place?

$\endgroup$

1 Answer 1

2
$\begingroup$

No, two overlapping non-interactive fields won't become entangled. To become entangled they would need to interact. The reason is, that to form an entangled state from the tensor product of two separable states, one needs a global unitary operation that operates on the entire tensor product and can thus produce correlation among the degrees of freedom of the two states. The evolution of two non-interactive fermions is described by local unitary operations that operate separately on the two fields and thus would not produce entanglement.

$\endgroup$

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.