Why two-particle wavefunctions are separable and their corresponding particles are indistiguishable at the same time?

If the wavefunction $\psi(r_1,r_2)$ doesn't represents an entangled state, it is separable: $$\psi(r_1,r_2)=\psi_a(r_1)\psi_b(r_2)$$ In this treatment, we ignore the interaction between two particles so that the initial wavefunction can be written as a product. However, since they are indistinguishable particles, their wavefunction must overlap more or less, otherwise we need infinite potential wells.

My question is: what is the relationship between the interaction of particles and the overlap of their wavefunctions? I know they are not the same thing but I feel confused about their relation. And how both of them affect the indistinguishablility of particles?

• "In this treatment, we ignore the interaction between two particles so that the initial wavefunction can be written as a product. However, since they are indistinguishable particles, their wavefunction must overlap more or less, otherwise we need infinite potential wells." [citation needed] We don't ignore the interation, non-entangled states are perfectly possible as results of interactions, and "overlap of wavefunctions" has intrinsically nothing to do with this. – ACuriousMind Dec 21 '14 at 14:44

The way you wrote it, they are distinguishable (unless $a=b$ of course). For the particles to be indistinguishable their wavefunction must be of the form $$\psi(r_1,r_2) = \frac{1}{\sqrt{2}} [ \psi_a(r_1)\psi_b(r_2) \pm \psi_a(r_2)\psi_b(r_1) ]$$ where the sign depends on the fermionic/bosonic nature of the particles.