I’m trying to compare the energies of the $N$ particle free Fermi gas with the following two wave functions. They are supposed to have the same energy
Let $\psi_l(\vec{r})$ be the single-electron wavefunction that obeys
$$\left(\frac{-h^2 \nabla^2}{2m} + U(\vec{r}) \right) \psi_l(\vec{r}) = E_l \psi_l(\vec{r})$$
where $E_l$ is the energy.
Then, we have two wavefunctions of $N$ electrons, made up with the functions for single electrons, which are supposed to have the same energy. They are
$$\Psi_A = \prod_{l=1}^N \psi_l (\vec{r}_l)$$
$$\Psi_B = \frac{1}{N!} \sum_s (-1)^s \prod_{l=1}^N \psi_{sl}(\vec{r}_l)$$
Where the sum goes over the permutations of $N$ elements, $(-1)^s$ is the parity of the permutation and the sub index $sl$ goes over the elements in each permutation.
I was able to find that the eigenvalue for the $A$ wavefunction is $$\sum_{l=1}^N E_l$$
But I am unable to do it for $\Psi_B$
How is this done?
Thank you for your help!
Each term in the sum contains the same one-particle wavefunctions as $\psi_A$, just with different permutations of the coordinates. Each of these terms is an eigenfunction of the $N$-particle Hamiltonian just like $\psi_A$, with the same eigenvalue. Forming a linear combination of these eigenfunctions again yields an eigenfunction with the same eigenvalue.