# Slater-Determinant: When is this appopriate?

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?

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.