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  • B

Two electrons in the same orbital is clearly an entangled quantum state since it is not a tensor product: $$|\psi\rangle=\frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes|\downarrow\rangle-|\downarrow\rangle \otimes|\uparrow\rangle)$$

  • A

Two fermions in the same orbital can be described by fermionic creation operators a†↑ and a†↓, which increase the occupation numbers: $$|\psi\rangle= a_{\uparrow}^{\dagger} a_{\downarrow}^{\dagger}|0\rangle \otimes|0\rangle=\left|1_{\uparrow}\right\rangle \otimes\left|1_{\downarrow}\right\rangle$$

The resulting singlet state is clearly a tensor product and is thus not entangled according to A


I already have reviewed the entangled states and separable states

  1. But I just wonder What is the basic origin of their confusion ? Are these two states are the same state just in two different basis?
  2. Where is B’s entanglement in A’s picture? Why B looks like entangled state and A not?
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  • $\begingroup$ Why would A not be $\frac{1}{\sqrt{2}}(a^\dagger_\uparrow a^\dagger_\downarrow - a^\dagger_\downarrow a^\dagger_\uparrow)|0\rangle \otimes |0\rangle$? $\endgroup$ – probably_someone Jul 1 at 7:48
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A and B are the same state: simply the second identity in B is false. The fermionic creation operators add particles in the antisymmetric subspace of the Fock space. The vector you wrote in B is not invariant up to phase (sign) under interchange of the two electrons so that it does not respect the general principle of indistinguishable particle. Furthermore $|0\rangle \otimes |0\rangle$ has to be replaced for $|vacuum\rangle$ which is in common for all electrons. Have a look at https://en.m.wikipedia.org/wiki/Fock_space (section Definition).

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  • $\begingroup$ So I can understand B in Hilbert space, and A in Fock space. Actually they are the same state,just different space representation? $\endgroup$ – Zeo Jul 1 at 14:15
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    $\begingroup$ The Hilbert space of identical Fermions is a (fermionic) Fock space. Properly speaking all the description is in the 2-particle Fock subspace whose explicit expression is A. $\endgroup$ – Valter Moretti Jul 1 at 14:19
  • $\begingroup$ Thank you, now I finally get the relationship $\endgroup$ – Zeo Jul 1 at 14:46
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Simply put, A and B are different states, representing different physical situations. One of them is entangled and the other is not.

In state A, electron 1 and electron 2 both have definite spins. The first electron, when measured, will always be spin-up, and the second electron, when measured, will always be spin-down.

In state B, electron 1 and electron 2 have indefinite spins. The first electron, when measured, will be spin-up in half of the measurements (on average) and spin-down in the other half. Likewise, the second electron will be spin-up in half of the measurements (on average) and spin-down in the other half, in such a way that each individual measurement of both electrons' spins will have one electron spin-up and the other electron spin-down.

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  • $\begingroup$ So I can understand that B is two mutually entangled particles can not tell which is which, and for A is to let one of the particles spin up the other spin down?Where is B’s entanglement in A’s picture? Why B looks like entangled state and A not $\endgroup$ – Zeo Jul 1 at 11:46
  • $\begingroup$ @Zeo B and A are not the same state. They're not equivalent. A doesn't look like an entangled state because it isn't an entangled state. $\endgroup$ – probably_someone Jul 1 at 11:56

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