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  • $\begingroup$ No, that is not correct. With the choice of partitioning of the Hilbert space I have specified in the problem, the two subsystems are unequivocally unentangled. I cite, for example, the use of the orbital cut in the entanglement spectrum of QH states (go look it up) - in there, IQH states are unentangled because they have the form $|\psi\rangle = \prod_i c_i^\dagger |0\rangle$. $\endgroup$ Commented Jun 3, 2013 at 17:39
  • $\begingroup$ Next, $c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger}|0 \rangle_1 \otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right)$ is wrong. The multiparticle Fock state vacuum is built up from the tensor product of individual vacuums $\prod_i |0\rangle_i$. Then a Fock state representation arises from acting the creation operators on this: $c_a^\dagger c_b^\dagger |0\rangle = |0\rangle_i \cdots c_a^\dagger |0\rangle_a \otimes \cdots c_b^\dagger |0\rangle_b \cdots|0\rangle_j$ $\endgroup$ Commented Jun 3, 2013 at 17:43
  • $\begingroup$ The anticommutativty arises because of the canonical anticommutation relations $\{ c_i, c_j^\dagger \} = \delta_{ij}$. This then results in $c_a^\dagger c_b^\dagger |0\rangle = -c_b^\dagger c_a^\dagger |0\rangle$. the whole point of the Fock state number representation is that it treats the particles immediately as indistinguishable and does not have an unphysical label (the positions, $x_1, x_2, \cdots$) for each particle like a first quantized notation (wavefunction) would. The first quantized notation is really cumbersome and awkward, because what it does $\endgroup$ Commented Jun 3, 2013 at 17:47
  • $\begingroup$ is to say that each particle is distinguishable with position labels $x_1, x_2, \cdots$, but then it enforces the rule that they're actually indistinguishable and should be odd under a swap of labels, leading to the Slater determinant / antisymmetrization scheme. So, -1. $\endgroup$ Commented Jun 3, 2013 at 17:49
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    $\begingroup$ I've elaborated that comment in my answer. $\endgroup$ Commented Sep 20, 2013 at 19:56