What is the difference between entangled states and separable states or electron spin-singlets?

  • $\begingroup$ related on physics.SE: physics.stackexchange.com/q/489114/58382 $\endgroup$
    – glS
    Jul 3, 2019 at 8:51
  • $\begingroup$ You have completely changed the nature of the question, invalidating existing answers (ie mine). Generally it would be better to ask a new question. $\endgroup$
    – DaftWullie
    Jul 4, 2019 at 5:57

1 Answer 1


There could be a few things going on here. However, I think the fundamental issue is the tensor product that you write in the second expression. It implies an issue of identification. Your second expression is effectively saying:

  • there are two different positions at which I can create a fermion. Call them P1 and P2.

  • create a spin-up at P1 and a spin down at P2.

This is quite a different thing from "there are two spins, one up and one down, but I cannot distinguish which is which", corresponding to your first expression.

This may need some further elaboration, but I don't want to overcomplicate it at this point, and I'll be guided by comments.

If you want to write B using A's description, you need to have terms that look like both $a^\dagger_{\uparrow}a^\dagger_{\downarrow}$ and $a^\dagger_{\downarrow}a^\dagger_{\uparrow}$ because you don't know which is which (that's a very vague hand-wavy description). Now, since electrons are fermions, they are antisymmetric. That shows how we have to combine those two terms: $$ (a^\dagger_{\uparrow}a^\dagger_{\downarrow}-a^\dagger_{\downarrow}a^\dagger_{\uparrow})|0\rangle\otimes|0\rangle. $$ Hopefully you can now see how that matches with the B description.

  • 1
    $\begingroup$ It's not that A is artificial. There's very real scenarios in which either is valid, it's just that those scenarios are different. If you're talking about two electrons in an orbital, you can't tell them apart, and B applies. $\endgroup$
    – DaftWullie
    Jul 1, 2019 at 14:22

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