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
This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys