# Fermion 1D Hubbard Model ground state in the U = 0 limit

I am trying to determine the ground state of the 1D fermionic Hubbard model at half-filling of $2L$ sites with $L$ electrons with spin-$\uparrow$ and $L$ electrons with spin-$\downarrow$ in the $U=0$ limit. To do this, I first solve for the band structure of the free-fermion Hamiltonian to get $E(k) = -2 t \cos(k a)$, with the usual notations, where $k = 2\pi n/L$ and $-L \leq n \leq L$, $n$ in $\mathbb{Z}$. Then, I fill the momentum levels starting with the one with the lowest energy, until $L$ levels are filled with $2$ electrons each, one with spin-$\uparrow$ and one with spin-$\downarrow$. To convert it into the position basis on the $2L$ sites, since I know that the state can be written as a product of spin-$\uparrow$ and spin-$\downarrow$, I write down the Slater determinant separately for the spin-$\uparrow$ and spin-$\downarrow$, to take care of the antisymmetry. However, the state obtained is not antisymmetric under the exchange of the spin-$\uparrow$ and spin-$\downarrow$. What is wrong in this approach?