For square (cubic) Hubbard lattice with infinite repulsion energy U exact result has been obtained: the ferromagnetic state with maximum total spin is not the ground state of system, if the hole number is equal to two.
All the energy eigenvalues of the Hubbard model with infinite repulsion on 2×2×2 cubic lattice for different electron numbers N e are calculated numerically. For N e =4, 7 the ground state is completely ferromagnetic; namely, total spin S is N e /2. For N e =5, the total spin of the ground state is 3/2. For N e =2, 6, it is zero. For N e =1, 3, it is 1/2. Thermodynamic quantities such as entropy, susceptibility and specific heat are calculated with the method of canonical ensemble. There are two peaks in the specific heat curve for N e =4, 5, 6, 7, but only one peak for N e =1, 2, 3.
The Hubbard Hamiltonian with infinite on-site repulsion and negative transfer is diagonalized numerically on 3×3 frustrated lattices, such as the square lattice with next-nearest-neighbor transfer and the triangular lattice. The number of electrons is chosen to be less by just one electron than that in the exactly half-filled band. When the strength of the frustration is sufficiently large, a singlet state appears as the ground state. Then, the total spin of the lowest excited state is unity. There is an excitation gap for the square lattice but not for the triangular lattice.
We consider the question of ferromagnetism in the one-band Hubbard model on the square lattice with infinite on-site repulsion using exact diagonalization techniques. Our results show a nonmonotonic behavior as a function of the number of holes Nh away from half filling. We find that the spin per electron is generally finite but less than its maximum value and, for a fixed hole number, increases with increasing system size. We interpret this as indicating that the inifinite-U Hubbard model is ferromagnetic in the thermodynamic limit for a finite fraction of holes, but that there are nonferromagnetic states which are very close in energy.