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