Slater-Determinant: When is this appopriate?

Question

Imagine we have a N-particle Hamiltonian without any interaction between the electron particles

$$ H = \sum_{i=1}^{N} \frac{p_i^2}{2m} + V(r_i)$$

then the solution to this equation $H\Psi = E \Psi$ will be a product wavefunction $\Psi = \psi_1..\psi_N.$

Now for this Hamiltonian it is true that all permutation operators $P$ commute with the Hamiltonian.

This means that there is a system of eigenfunctions to the Hamiltonian and the permutation operators and the eigenfunctions can be found by antisymmetrizing the single particle wavefunctions (Slater determinant). For this Hamiltonian we can apparently find a system of eigenfunctions that is completely antisymmetric.

Now assume that we have a Hamiltonian that does NOT commute with all permutation operators, because it acts differently on the electrons. For instance

$$H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + V_1(r_1) + V_2(r_2),$$ where $V_1$ is very different from $V_2.$

What does second quantization tell us in this case, when the permutation operator does not commute with the Hamiltonian, are the "right" eigenstates of the Hamiltonian still antisymmetrized states?

Is there any physical difference in a symmetric in $N$-particles Hamiltonian vs a non-symmetric one?

Answer 1

Now assume that we have a Hamiltonian that does NOT commute with all permutation operators, because it acts differently on the electrons. For instance H=K+V_1(r1)+V_2(r2), where V_1 is very different from V_2.

In the honest-to-god physical world that we live in you can not have $V_1$ being very different from $V_2$ because all electrons are identical, i.e., indistinguishable.

The anit-symmetrization postulate of quantum mechanics was derived from observations based on the honest-to-god physical world that we live in--a world in which all electron are indistinguishable. So this question is absurd. [Note to mods: I do not mean "absurd" in a rude way]