Abstract

Even though the one-dimensional (1D) Hubbard model is solvable by the Bethe ansatz, at half-filling its finite-temperature T>0 transport properties remain poorly understood. In this paper we combine that solution with symmetry to show that within that prominent T=0 1D insulator the charge stiffness D(T) vanishes for T>0 and finite values of the on-site repulsion U in the thermodynamic limit. This result is exact and clarifies a long-standing open problem. It rules out that at half-filling the model is an ideal conductor in the thermodynamic limit. Whether at finite T and U>0 it is an ideal insulator or a normal resistor remains an open question. That at half-filling the charge stiffness is finite at U=0 and vanishes for U>0 is found to result from a general transition from a conductor to an insulator or resistor occurring at U=Uc=0 for all finite temperatures T>0. (At T=0 such a transition is the quantum metal to Mott–Hubbard-insulator transition.) The interplay of the η-spin SU(2) symmetry with the hidden U(1) symmetry beyond SO(4) is found to play a central role in the unusual finite-temperature charge transport properties of the 1D half-filled Hubbard model.

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