# Finding Ground State from Hamiltonian in Second Quantization

I am looking at the following mean-field Hamiltonian:

$H=-\sum_{i,j,\sigma}t_{ij}c_{i\sigma}^\dagger c_{j\sigma}-\Delta \sum_i (n_{i\uparrow}-n_{i\downarrow})$

If I didn't have the $\Delta$-dependent term, then the ground state would be

$|GS\rangle=\prod_i c_i^\dagger |0\rangle$

I also understand how to find the energy in a second-quantized Hamiltonian. However, how do I find the ground state wave function of the full Hamiltonian? At first, I was thinking about performing a Slater determinant, but I'm not sure if that would help. I think there's something simple that I'm overlooking--any help would be appreciated.

• Is the number of particles fixed? The expression for $|GS\rangle$ doesn't make sense because the spin is not specified. – higgsss Oct 2 '15 at 22:34
• Yes, the number of particles is fixed. – Mr Frobenius Oct 2 '15 at 23:29
• Regardless of spins, the ground state is not correct, since it is just filling each site with a fermion, so it only makes sense when the number of sites is equal to the number of fermions, otherwise the hopping term $\sum_{ij}t_{ij}c_{i\sigma}^\dagger c_{j\sigma}$ will move the fermions around. The correct ground state, is a Fermi sea in the momentum space. Assuming periodic boundary condition, you can first Fourier transform the Hamiltonian into momentum space so it is diagonalized, then fill the momentum eigenstates up with fermions. – Meng Cheng Oct 3 '15 at 2:10
• You should first try to solve the case where $\Delta=0$. As pointed out, the solution you give in incorrect. – Norbert Schuch Oct 5 '15 at 8:04