Ground-state ferromagnetic transition in strongly repulsive one-dimensional Fermi gases

We prove that as a one-dimensional Fermi gas is brought across the resonance adiabatically from large repulsion to large attraction, the singlet ground state will give way to the maximum spin state, which is the lowest energy state among the states accessible to the system in this process. In the presence of tiny symmetry-breaking fields that destroy spin conservation, the singlet ground state can evolve to the ferromagnetic state or a spin segregated state. We have demonstrated these effects by exact calculations on fermion cluster relevant to current experiments, and have worked out the quantum-mechanical wave function that exhibits phase separation.

Figure 1

Schematic plot of the ground-state energy $E_S(-1/g)$ of a Fermi gas with spin $S$ as a function of $-1/g$. As $-1/g$ is swept across resonance adiabatically, each state with spin $S$ (solid line on the repulsive side $-1/g<0$) continuously evolves into a gaseous excited state in the super-Tonks family, rather than jumping to the true ground state consisting of deep molecules (dashed line far below super-Tonks family). The positive slope of all $E_S$ is due to the Hellmann-Feynman theorem. On the repulsive side of the resonance, $-1/g<0$, the Lieb-Mattis theorem requires that $E_S$ increases with $S$, and the ground state is a singlet, ($N/2>S_2>S_1>0$). The exact solution of the Schrödinger equation shows that $E_S$ must converge at the same point $E^{*}$ at resonance. On the attractive side, $E_S$ increases with decreasing $S$. The super-Tonks family is bounded below by the maximum spin state $S=N/2$. Thus, in a sweep process, the lowest energy state will change abruptly from a singlet on the repulsive side to the maximum spin state on the attractive side.

Figure 2

Density distributions for $\uparrow$ (black square) and $\downarrow$ (red circle) spins in a harmonic trap with frequency ${\omega }_T$ obtained from exact diagonalization. $N_{\uparrow }=N_{\downarrow }=2$. $a_{ho}=\hslash /\sqrt{M{\omega }_T}$. (a) $\gamma =0,G{a}_{ho}/{\omega }_T=0.05$; (b) $\gamma a_{ho}=15,G{a}_{ho}/{\omega }_T=0.05$; and (c) $\gamma a_{ho}=15,G=0$. In (b), the additional yellow (blue) dashed lines show the density of $\uparrow$ ($\downarrow$) spin obtained from performing degenerate perturbation on the six spin states at infinite repulsion (see text). The result matches well with that from exact diagonalization for large but finite repulsion.