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?
Best Answer
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
$$ c_a^{\dagger}c_b^{\dagger}|0\rangle\neq c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2 $$