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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.