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?
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