Nonperturbative results for few electrons in the two-dimensional Hubbard model

Based on a generalization of the equation which solves exactly the two-electron problem, we have found a method which allows the nonperturbative, asymptotically exact solution for the ground state of any finite number n of electrons in the low-density limit (n/N→0) of the N-site two-dimensional Hubbard model. The three-electron case is checked against the exact numerical diagonalization and the total energy is shown by finite-size scaling to agree well, to leading order of an expansion in powers of 1/lnN, with the asymptotic solution. We then indicate how to solve for an arbitrary electron number, including both closed and open shells. The case of four electrons is explicitly presented as an application. More generally, the closed-shell d=2 ground state is a nondegenerate singlet with zero total momentum. Similar to d=1, all observables differ from their U=0 value by a correction which is U independent. Unlike d=1 and similar to d=3, however, the ground state is a linear combination of the unperturbed U=0 Fermi sea with the set of two-particle excitations. Moreover, the momentum distribution n(k) can be evaluated explicitly, and shows a finite Fermi jump Z=1-4 ln2/${\ln }^2$(n/N). This result appears to argue in favor of a Fermi-liquid-like behavior for d=2 in the zero-density limit n/N→0. The open-shell ground states display an interesting degeneracy between a nondegenerate singlet and high-spin states.