# Nagaoka ferromagnetism in a four-site lattice

I'm reading Altland & Simons' Condensed Matter Field Theory, and when presenting the t-J model, $$H = -t \sum_{<mn>}P_s a^{\dagger}_{m\sigma} a_{n\sigma} P_s + J \sum_{<mn>} S_m \cdot S_n$$ they say that Nagaoka's theorem asserts that, in the $U=\infty\ (J=0)$ limit, when we remove one electron from the half-filled Hubbard model, the ground state becomes ferromagnetic. I'm having a hard time trying to show this for a four-site square lattice with three electrons.

1) The way I understand it, we can consider separately the cases $S^z_{total} = 3/2$ and $S^z_{total} = 1/2$, because this Hamiltonian does not allow the total z-spin to change. Is this correct?

2) For the $S^z_{total} = 3/2$ case, we can take the basis $\{a^{\dagger}_{1\uparrow}a^{\dagger}_{2\uparrow}a^{\dagger}_{3\uparrow}|0\rangle, a^{\dagger}_{2\uparrow}a^{\dagger}_{3\uparrow}a^{\dagger}_{4\uparrow}|0\rangle, a^{\dagger}_{1\uparrow}a^{\dagger}_{3\uparrow}a^{\dagger}_{4\uparrow}|0\rangle, a^{\dagger}_{1\uparrow}a^{\dagger}_{2\uparrow}a^{\dagger}_{4\uparrow}|0\rangle\}$, and the ground state (with energy $-2t$) is

$$|\psi_{-2t}\rangle = \frac{1}{\sqrt{4}} \left(a^{\dagger}_{1\uparrow}a^{\dagger}_{2\uparrow}a^{\dagger}_{3\uparrow} + a^{\dagger}_{2\uparrow}a^{\dagger}_{3\uparrow}a^{\dagger}_{4\uparrow} + a^{\dagger}_{1\uparrow}a^{\dagger}_{3\uparrow}a^{\dagger}_{4\uparrow} + a^{\dagger}_{1\uparrow}a^{\dagger}_{2\uparrow}a^{\dagger}_{4\uparrow}\right)|0\rangle$$

and it's total spin is indeed $3/2$. However, if we take any other eigenstate, like $|\psi_{0}\rangle = \frac{1}{\sqrt{2}}\left( -a^{\dagger}_{1\uparrow}a^{\dagger}_{2\uparrow}a^{\dagger}_{3\uparrow} + a^{\dagger}_{1\uparrow}a^{\dagger}_{3\uparrow}a^{\dagger}_{4\uparrow} \right)|0\rangle$, isn't its spin also $3/2$? So I don't see how it is surprising that the ground state is ferromagnetic... but I probably got something wrong.

3) For $S^z_{total} = 1/2$ I have no idea how to proceed without having to diagonalize an enormous 12x12 matrix. The book gives a hint to arrange the basis in the order in which they are generated by application of the Hamiltonian, but I don't know how this helps. I understand that there is also a mirror symmetry in the lattice.

I looked at Patrik Fazekas' "Lecture notes on Electron Correlation and Magnetism" solution to this problem, but he seems to conclude that the $S^z_{total}=1/2$ case has not a ferromagnetic ground state, so I'm a little confused.

• Could you give all operators appearing in the Hamiltonian in terms of the creation/annihilation operators ? You could then show that $S_z$ commutes with $H$. – Adam Feb 28 '17 at 17:28
• Ok, thanks! $P_s$ is $(1-n_{i,-\sigma})$ and $S^z=\sum_{k}\frac{1}{2} (n_{k,\uparrow} - n_{k,\downarrow})$, hence $H,S^z$ commute.This tells us that $S_z$ and $H$ share eigenstates, hence we can look for $H$ eigenstates for the cases $3/2$ and $1/2$ separately. Can you help me with my other questions? – Patrick Feb 28 '17 at 18:38
• I don't really understand your notation for the eigenstates. You don't apply the operators on a ket, and there are no spin index either... – Adam Mar 1 '17 at 7:04
• I'm sorry, it should be clear now. – Patrick Mar 1 '17 at 12:05
• Since there are only 3 electrons, and if I understand the problem correctly, only one electron allowed per site, then the total spin is either 3/2 or 1/2 (or the opposite, which are easily found by symmetry). A state 3/2 is ferromagnetic (all spins are aligned), while 1/2 is not (since one spin is down, and the other two up). – Adam Mar 1 '17 at 17:10