Zero finite-temperature charge stiffness within the half-filled 1D Hubbard model

Even though the one-dimensional (1D) Hubbard model is solvable by the Bethe ansatz, at half-filling its finite-temperature T>0$T>0$ transport properties remain poorly understood. In this paper we combine that solution with symmetry to show that within that prominent T=0$T=0$ 1D insulator the charge stiffness D(T)$D\left(T\right)$ vanishes for T>0$T>0$ and finite values of the on-site repulsion U$U$ 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$T$ and U>0$U>0$ 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$U=0$ and vanishes for U>0$U>0$ is found to result from a general transition from a conductor to an insulator or resistor occurring at U=U_c=0$U=U_c=0$ for all finite temperatures T>0$T>0$. (At T=0$T=0$ such a transition is the quantum metal to Mott-Hubbard-insulator transition.) The interplay of the η$\eta$-spin SU(2)$SU\left(2\right)$ symmetry with the hidden U(1)$U\left(1\right)$ symmetry beyond SO(4)$SO\left(4\right)$ is found to play a central role in the unusual finite-temperature charge transport properties of the 1D half-filled Hubbard model.