Entangled or unentangled? I got a little puzzled when thinking about two entangled fermions.
Say that we have a Hilbert space in which we have two fermionic orbitals $a$ and $b$. Then the Hilbert space $H$'s dimension is just $4$, since it is spanned by
\begin{align}
\{ |0\rangle, c_a^\dagger |0\rangle, c_b^\dagger |0\rangle, c_a^\dagger c_b^\dagger |0\rangle\},
\end{align}
where $c_i^\dagger$ are the fermionic operators that create a fermion in orbital $i$.
Say we have a state $c_a^\dagger c_b^\dagger |0\rangle$. Then if I partition my Hilbert space into two by looking at the tensor product of the Hilbert spaces of each orbital, i.e. $H = H_a \otimes H_b$, then my state can be written as $c_a^\dagger |0\rangle_a \otimes c_b^\dagger |0\rangle_b$, from which it is obvious that this state is unentangled ($|0\rangle = |0\rangle_a \otimes |0\rangle_b$).
Now I was thinking about writing the state in first quantized i.e. a wavefunction. Let $\phi_a(r), \phi_b(r)$ be the wavefunctions of the orbitals $a$ and $b$. Then
\begin{align}
\psi(x_1,x_2) = \langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle = \phi_a(x_1) \phi_b(x_2) - \phi_a(x_1)\phi_b(x_2).
\end{align}
This is where I got confused. What object is $\psi(x_1,x_2)$, i.e. what Hilbert space does it belong to? What exactly are we doing when we do $\langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle$? We seem to be changing/expanding our Hilbert space by taking the position representation?
Written in this way, and assuming the same partition $H_a \otimes H_b$, the unentangled nature of the original state is no longer manifest. I'm not sure what the partition $H_a \otimes H_b$ even means in this context. Would that be saying $\psi(x_1, x_2) = \psi_a(x_1, x_2) \times \psi_b(x_1,x_2)$ where $\psi_i(x_1,x_2)$ is a linear combination of $\phi_i(x_1), \phi_i(x_2)$? This does not seem right to me.
Regardless, now I have a state written in two different but supposedly equivalent ways, with the same partition of the Hilbert space, yet it is unentangled in one way and entangled in the other.
Help?
 A: Let me remind you that the Fock space of multiple fermions is defined to be the antisymmetric (fermionic) subspace of the full Fock space
$$
\Gamma_a=\bigoplus_{n=0}^{\infty}H^{\wedge n},
$$
where $\wedge$ stands for the antisymmetric tensor product
$$
v_1\wedge\ldots\wedge v_n=\frac{1}{n!}\sum_{p\in P_n}\sigma_p v_{p(1)}\wedge\ldots\wedge v_{p(n)}.
$$
Here $\sigma_p$ is the sign of the permutation $p$ in the group of permutations $P_n$.
Thus the confusion here comes from the fact that $c_a^{\dagger}c_b^{\dagger}|0\rangle\neq|ab\rangle$ as you seem to state. 
Recall that the creation and annihilation operators are defined within the occupation number representation, i.e. $c_a^{\dagger}c_b^{\dagger}|0\rangle=|11\rangle$, where the first number denotes the number of fermions in state $a$ while the second denotes the number of fermions in state $b$. On the other hand, states written in the occupation number representation are defined to be properly antisymmetrized (for fermions) many-body basis states, as forced upon us by the particle indistinguishability. Therefore they are defined within the fermionic Fock space. Any textbook will show so, take a look at the first chapter of Many-Body Quantum Theory in Condensed Matter Physics: An Introduction by Bruus and Flensberg for example. For two fermions described via a single particle basis $\{|a\rangle,|b\rangle\}$ one possible choice is:
$$
|11\rangle=\frac{1}{\sqrt{2}}\left(|ab\rangle-|ba\rangle\right).
$$
Therefore
$$
c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right)
$$
The familiar anticommutativity of these operators is now obvious from this from
$$
c_b^{\dagger}c_a^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2-c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2\right)=-c_a^{\dagger}c_b^{\dagger}|0\rangle
$$
In fact one of the great advantages of creation and annihilation operators is that they include the antisymmetry (for fermions) of the wave function implicitly. 
Dotting this with $\langle x_1x_2|$ we obtain:
$$
\langle x_1x_2|c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(\phi_a(x_1)\phi_b(x_2)-\phi_b(x_1)\phi_a(x_2)\right).
$$
Thus there is no inconsistency, both representations show that the particles are entangled.
On the other hand dotting $\langle x_1x_2|$ with $c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2$ would simply produce
$$
\psi(x_1,x_2)=\phi_a(x_1)\phi_b(x_2)
$$
Therefore there is no inconsistency here also, however, as I said, the important thing to remember is that
$$
c_a^{\dagger}c_b^{\dagger}|0\rangle\neq c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2
$$
A: I'm posting a modified version of my comment as an answer, as more people will see it this way.
I think the confusion hinges crucially on what kind of partitioning you're doing. The $\nu=1$ QH state is pure under orbital partitioning, but not under "particle partitioning". Maybe arXiv:0905.4204 will help. IIRC, they work out a simple example about this detail, in the 2nd section.
@nervxxx, Your 2-particle state might be pure under orbital partitioning, but it is entangled under particle partitioning. Due to the antisymmetrization, it looks like a singlet Bell state. 
So the bottomline is that entanglement is completely dependent on how you choose to partition your system. The subtlety is not widely appreciated. For a nuanced discussion, see this article http://rspa.royalsocietypublishing.org/content/463/2085/2277.full
A: We have to be careful with the bra-ket formalism and its meaning. Unlike $|x_1\rangle$, I am not sure that the notation $|x_1 x_2\rangle$ where $x_1$ and $x_2$ are positional coordinates makes any sense. In literature [1] the notation $|ab\rangle$ designates the Slater determinant or Hartree-Fock state, i.e.:
$$|ab\rangle=c_a^{\dagger}c_b^{\dagger}|0\rangle=\phi_a(x_1) \phi_b(x_2)-\phi_a(x_2) \phi_b(x_1)$$
My feeling is that your confusion is related to mixing of the occupation numbers formalism and real space representation.
[1] Szabo, Ostlund, "Modern quantum chemistry: introduction to advanced electronic structure theory"
