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Low Temperature Physics

On the ground state of the Hubbard model with strong repulsion

Ju. V. Mikhailova

Published Jun 1, 1999• 0 citations
Publisher

Abstract

The exact result is obtained for a square (cubic) Hubbard lattice with an infinitely large repulsion energy U: the ferromagnetic state with the maximum spin is not the ground state of the system if the number of holes is equal to two.
DOI
10.1063/1.593759

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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.
The ground state of the two-dimensional Hubbard model on a square lattice is studied in the large-U limit in a half-filled band with a dynamical hole. For a 4\ifmmode\times\else\texttimes\fi{}4 lattice with hopping t=1 and coupling JJc{J}_{c}Jc​\ensuremath{\approxeq}0.075, the ground state has zero momentum (k) and spin 15/2 (ferromagnetic background). For J\ensuremath{\ge}Jc{J}_{c}Jc​ the ground state is degenerate with nonzero k and spin 1/2, in agreement with recent variational calculations. The nonzero k of the ground state comes from a nontrivial phase in the overlap of the spin states before and after a hole move. For small lattices we show that the effective Hamiltonian of the hole is that of a particle moving in a magnetic field. We also find that many of the features of the weak-coupling region carry over continuously to the strong-coupling region.