Ferromagnetism in the Hubbard Model with a Gapless Nearly-Flat Band

Abstract

We present a version of the Hubbard model with a gapless nearly-flat lowest band which exhibits ferromagnetism in two or more dimensions. The model is defined on a lattice obtained by placing a site on each edge of the hypercubic lattice, and electron hopping is assumed to be only between nearest and next nearest neighbor sites. The lattice, where all the sites are identical, is simple, and the corresponding single-electron band structure, where two cosine-type bands touch without an energy gap, is also simple. We prove that the ground state of the model is unique and ferromagnetic at half-filling of the lower band, if the lower band is nearly flat and the strength of on-site repulsion is larger than a certain value which is independent of the lattice size. This is the first example of ferromagnetism in three dimensional non-singular models with a gapless band structure.

Appendix: Proof of Theorem 1 for \(s=0;\) Flat-Band Ferromagnetism

Throughout this section, we assume that $s=0$ and the other parameters t and U take arbitrary positive values.

As we have seen in Sect. 3, the single-electron ground state energy is zero and $|\mathcal {V}|$-fold degenerate for $s=0,$ and the single-electron ground states are given by $a_{\alpha ,\sigma }^\dagger \varPhi _0$ with $\alpha \in \mathcal {V}.$ Since both $H_\mathrm {hop}$ with $s=0$ and $H_\mathrm {int}$ are positive semidefinite, we find that the energy eigenvalue of H is bounded from below by zero. Taking a state $\varPhi _\mathrm {ferro}$ in (90) as a variational state, we conclude that the ground state energy of H for $N_\mathrm {e}=|\mathcal {V}|$ is exactly zero.

To show the uniqueness of the ground state, we will use the following property which was pointed out by Mielke [10]. Consider Hubbard models with M-fold degenerate single-electron ground state energy. Then, we can construct a set of M single-electron ground states which satisfy the following condition. Let $\varPhi _{1,\sigma }^{(i)}$ with $i=1,\,2,\ldots ,M$ be linearly independent single-electron ground states with spin $\sigma .$ For each i,  there exists site $x_i$ such that $c_{x_i,\sigma }\varPhi _{1,\sigma }^{(i)} \ne 0$ and $c_{x_i,\sigma }\varPhi _{1,\sigma }^{(j)}=0$ for $j\ne i.$

In our model, we can construct single-electron ground states which possess the above property in the following manner. For each $\alpha \in {\mathcal {V}},$ let us define a new fermion operator by

$$\begin{aligned} \bar{a}_{\alpha ,\sigma } = a_{\alpha ,\sigma } -\sum _{n=1}^{L-1}(-1)^{n}a_{\alpha +n\delta _\nu ,\sigma } \end{aligned}$$

(98)

(recall that $\mathcal {V}$ is periodic). Since $\bar{a}_{\alpha ,\sigma }$ is a linear combination of $a_{\beta ,\sigma }$ with $\beta \in \mathcal {V},$ the single-electron states $\bar{a}_{\alpha ,\sigma }^\dagger \varPhi _0$ apparently have energy 0. Furthermore, we have

(99)

These anticommutation relations imply that the single-electron states $\bar{a}_{\alpha ,\sigma }^\dagger \varPhi _0$ with $\alpha \in \mathcal {V}$ have the desired property. We note that (99) also implies the linear independence of these single-electron states.

Let us prove the uniqueness of the ground state by using $\{\bar{a}_{\alpha ,\sigma }^\dagger \}_{\alpha \in {\mathcal {V}}}$ introduced as above. Let $\varPhi _\mathrm {G}$ be a ground state, which must be a zero energy state for both $H_\mathrm {hop}$ and $H_\mathrm {int}.$ Since $\varPhi _\mathrm {G}$ is a zero energy state for $H_\mathrm {hop},$ we expand it as

$$\begin{aligned} \varPhi _\mathrm {G}= \mathop {\sum _{V_\uparrow ,V_\downarrow \subset \mathcal {V}}}_{|V_\uparrow |+|V_\downarrow |=|\mathcal {V}|} \varphi ^\prime \left( V_\uparrow ;\,V_\downarrow \right) \left( \prod _{\alpha \in V_\uparrow } \bar{a}_{\alpha ,\uparrow }^\dagger \right) \left( \prod _{\alpha \in V_\downarrow } \bar{a}_{\alpha ,\downarrow }^\dagger \right) \varPhi _0 \end{aligned}$$

(100)

with complex coefficients $\varphi ^\prime (V_\uparrow ;\,V_\downarrow ).$ To be a zero energy state of the on-site interaction, $\varPhi _\mathrm {G}$ in the form of (100) must further satisfy the condition $c_{x,\downarrow }c_{x,\uparrow }\varPhi _\mathrm {G}=0$ for all x in $\varLambda .$ We examine this condition for sites $x=m(\alpha ,\,\alpha +\delta _\nu )$ with $\alpha \in \mathcal {V}.$ Then, taking account of (99), we find that $\varphi ^\prime (V_\uparrow ;\,V_\downarrow )$ is vanishing if $V_\uparrow \cap V_\downarrow \ne \emptyset .$ A ground state $\varPhi _\mathrm {G}$ is thus rewritten as

$$\begin{aligned} \varPhi _\mathrm {G}= \sum _{\{\sigma \}} \varphi ^\prime (\{\sigma \}) \left( \prod _{\alpha \in \mathcal {V}} \bar{a}_{\alpha ,\sigma _\alpha }^\dagger \right) \varPhi _0 \end{aligned}$$

(101)

with new coefficients $\varphi ^\prime (\{\sigma \}).$ Now we have expressed $\varPhi _\mathrm {G}$ in the same fashion as in (94). Examining repeatedly the condition $c_{x,\downarrow }c_{x,\uparrow }\varPhi _\mathrm {G}=0$ for sites $x=m(\beta ,\,\beta +\delta _l)$ with $l\ne \nu ,$ we obtain the same relation for $\varphi ^\prime (\{\sigma \})$ as that in (96). Therefore, we conclude that the ground state is unique apart from the degeneracy due to the spin rotation symmetry. This completes the proof of Theorem 1 for $s=0.$