The two-dimensional Fermi–Hubbard model is important when attempting to understand strongly correlated electronic systems.^[1–5] In the model, the electrons can hop from one site to its nearest neighbors and there will be an interaction between two electrons when they are on the same site. The competition between the electron hopping and the on-site electron interaction makes it hard to find exact eigenstates of the model.^[6–8] In 1990, Yang and Zhang showed that many new exact eigenstates of the model could be constructed from the known ferromagnetic eigenstates based on the SO₄ symmetry of the model.^[9] There are also other methods to construct exact eigenstates of the model.^[10] Finding out more exact eigenstates of this model not only is an intellectual challenge, but also can increase our understanding of the model especially when the SO₄ symmetry is considered.^[9,11] In this paper, we will present some new methods to construct more exact eigenstates of the model by connecting the problem with the Slater determinant. The eigenstates we constructed are independent of the on-site electron interaction, i.e., they are eigenstates for any value of the on-site electron interaction strength, which were ever believed to have been all figured out by Yang and Zhang and no new such eigenstates could be constructed.^[9]

The two-dimensional Fermi–Hubbard model we considered is defined on a periodic two-dimensional L × L lattice with the Hamiltonian 1 2 3 where the lattice sites are described by r = (x,y) with x and y being integers 0, 1, …, L−1. The operators a_r and b_r are the annihilation operators in coordinate space for the spin-up and spin-down electrons, respectively, and they obey the usual commutation relations for Fermi operators. The notation ⟨r,r′⟩ indicates that the sum is over the nearest neighbors and the parameter U describes the on-site electron interaction strength.

When the on-site electron interaction strength U is zero, it is easy to find exact eigenstates of the model Hamiltonian H. By introducing the following annihilation operators in momentum space: 4 where k = (k_x,k_y), k_x and k_y are limited to the range (−π,π] and are 2π/L multiplied by some integers, the operator T can be rewritten in the diagonal form 5 where 6 However, it is usually hard to find exact eigenstates of H for nonzero U.

In the following, we will give some methods to construct exact eigenstates of the Hamiltonian H for nonzero U. To reach this goal, we consider the following quantum state with j spin-up electrons and (n−j) spin-down electrons: 7 8 where |0⟩ is the vacuum state and the function f is required to satisfy two conditions: (i) when r_α = r_β for any α ≠ β, there is f = 0; (ii) the state |Ψ⟩ is an exact eigenstate of the operator T. The first condition guarantees that there will not be two electrons in the same lattice site in coordinate space so that we can neglect the operator V in the Hamiltonian H. The second condition ensures that the state |Ψ⟩ is an exact eigenstate of the Hamiltonian H. To satisfy the first condition, we assume that the function f is in the following form: 9 where k_i (i = 1, …, n) are different vectors in momentum space. Since the function f is proportional to a Slater determinant, the first condition will be satisfied due to the characteristics of the determinant. The problem is now reduced to find a group of k_i and a function g(r₁, r₂, …., r_n) that make the function f satisfy the second condition.

2.2. The case g(r₁,r₂, …, r_n) ≠ 1

To give some new eigenstates of H, in this case we assume 19 where q_α (α = 1, …,n) are vectors in momentum space and they are required not to be all the same, otherwise the function f in the following Eq. (20) will have the same form as Eq. (10). Substituting Eq. (19) into Eq. (9), we obtain 20 which leads to 21 with 22 according to Eq. (7). It can be seen that for each permutation p, the state |Ψ_p⟩ is an eigenstate of the operator T, and the corresponding eigenvalue is 23 Usually the eigenvalue E_p is dependent on p, and the state |Ψ⟩ in Eq. (21) as a sum of |Ψ_p⟩ is not an eigenstate of the operator T. To make the state |Ψ⟩ in Eq. (21) an eigenstate of the operator T (or H), we can choose proper vectors k_α and q_α (α = 1, …, n) in momentum space such that the eigenvalue E_p is not a function of p. This can be done at least in the following two situations.

(i) Assume k_α = (k_α, 0) and q_α = (0, q_α) for α = 1,2, …, n, which are illustrated in Fig. 1. Now there is 24 according to Eq. (23), which leads to 25 according to Eq. (23). Since the eigenvalue E_p does not depend on the permutation p, the state |Ψ⟩ in Eq. (21) is an eigenstate of the Hamiltonian H under the above assumptions. In other words, the state 26 with different k_i is an eigenstate of the Hamiltonian H, and the corresponding eigenvalue is 27

Fig. 1.

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	Fig. 1. (color online) Lattice sites in momentum space. The black dots represent qα = (0, qα) and the white dots represent kα = (kα, 0), where L = 6 is chosen as an example.

Since k_α = (k_α, 0) for α = 1,2, …, n are assumed to be different from each other, the number of electrons n of the eigenstate |Ψ⟩ in Eq. (26) cannot be larger than L as we are considering an L × L lattice. When the eigenstate |Ψ⟩ in Eq. (26) has L electrons, the corresponding eigenvalue will be 28 where L is assumed to be an even number. As q_α = (0, q_α) for α = 1,2, …, L cannot all be the same, the minimal E_p(L) will be 29 which can be achieved when q₁ = 2π/L and q_α = 0 for α = 2,3, …, L. The average energy E_p(L)_min/L for each electron will be −2t for large L, which is also true when L is an odd number.

(ii) Assume k_α = (k_α, k_α+ λ) and q_α = (0,0) or (λ, −λ) for some nonzero λ (α = 1,2, … n). Now there is 30 for any permutation p according to Eq. (6), which leads to 31 according to Eq. (23). Note that the eigenvalue E_p does not depend on the permutation p again, therefore the state |Ψ⟩ in Eq. (21) is an eigenstate of the Hamiltonian H under the above assumptions.

In the above two situations, we have given two methods to construct the eigenstate of the Hamiltonian H by choosing proper vectors k_α and q_α (α = 1, …, n) in momentum space. These eigenstates do not depend on the electron interaction strength U. We need to show that our eigenstates have never been constructed, because Yang and Zhang have claimed that they have constructed all U-independent eigenstates of H.^[9] The eigenstates without double occupation in coordinate space constructed by Yang and Zhang are described by |Ψ_y⟩ in Eq. (12), which has a total spin s = n/2 for any j (the state has n electrons).^[9] However our eigenstates with n electrons such as |Ψ⟩ in Eq. (26) do not have a total spin, so that they are different from the eigenstates constructed by Yang and Zhang. To demonstrate it, we assume n = 2 and j = 1 for our eigenstate |Ψ⟩ in Eq. (26), which can now be written as 32 When k₁ ≠ k₂ and q₁ ≠ q₂, the vectors (k₁, q₁), (k₂, q₂), (k₂, q₁), and (k₁, q₂) are four different vectors in momentum space, so that the state |Ψ⟩ in Eq. (32) cannot have a total spin, i.e., it is the sum of a state with the total spin s = 0 and a state with the total spin s = 1.

We have considered the construction of exact eigenstates of the two-dimensional Fermi–Hubbard model. One may wonder whether our methods are also applicable for constructing exact eigenstates of the two-dimensional Bose–Hubbard model, where the operators obey the Bose commutation relations and many particles can be on the same site. Since our constructions prevent double occupation on the same site and do not use the Fermi commutation relations, the eigenstates we constructed for the Fermi–Hubbard model will also be eigenstates for the Bose–Hubbard model when the Bose commutation relations do not make the states zero. For example, the state |Ψ⟩ in Eq. (26) will also be an exact eigenstate of the two-dimensional Bose–Hubbard model, where a_k and b_k describe two kinds of Boson particles, under the condition that k_α = (k_α, 0) for α = 1,2, …, n are different from each other and q_α = (q_α, 0) for α = 1,2, …, n are also different from each other. This is because each term behind the sum of |Ψ⟩ in Eq. (26) is orthogonal to the other terms under the above condition, so that |Ψ⟩ is not zero.

The number of electrons in our constructed eigenstates cannot reach or be close to the Hall filling, i.e., L² electrons in the L × L lattice, which is believed to have a rich physics.^[12,13] However, Yang has shown that applying the η^† pair operator on a known eigenstate can obtain a new eigenstate with two more electrons.^[14] Therefore, we can apply the η^† pair operator many times on our constructed eigenstates to obtain new eigenstates with the number of electrons reaching or being close to the Hall filling. The eigenstates obtained in this way are not the ground state, but their corresponding eigenvalues can give some upper bound for the ground state energy and the knowing of these excited states can reduce the size of the space to numerically search the ground state, as the ground state is orthogonal to all excited states.

In summary, we have given some methods to construct exact eigenstates of the two-dimensional Fermi–Hubbard model. The basic idea is to prevent two electrons from being on the same site through the Slater determinant, therefore the eigenstates constructed are independent of the on-site interaction strength U. The finding that there are many U-independent eigenstates itself is of interest, and the connection with the Slater determinant can give some inspiration for understanding the two-dimensional Fermi–Hubbard model.