Slater determinant and exact eigenstates of the two-dimensional Fermi–Hubbard model
Ren Jun-Hang1, 2, Ye Ming-Yong1, 2, †, Lin Xiu-Min1, 2
Fujian Provincial Key Laboratory of Quantum Manipulation and New Energy Materials, College of Physics and Energy, Fujian Normal University, Fuzhou 350117, China
Fujian Provincial Collaborative Innovation Center for Optoelectronic Semiconductors and Efficient Devices, Xiamen 361005, China

 

† Corresponding author. E-mail: myye@fjnu.edu.cn

roject supported by the National Natural Science Foundation of China (Grant No. 11674059) and Natural Science Foundation of Fujian Province, China (Grant Nos. 2016J01008 and 2016J01009).

Abstract

We consider the construction of exact eigenstates of the two-dimensional Fermi–Hubbard model defined on an L × L lattice with a periodic condition. Based on the characteristics of Slater determinants, several methods are introduced to construct exact eigenstates of the model. The eigenstates constructed are independent of the on-site electron interaction and some of them can also represent exact eigenstates of the two-dimensional Bose–Hubbard model.

1. Introduction

The two-dimensional Fermi–Hubbard model is important when attempting to understand strongly correlated electronic systems.[15] 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.[68] 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 where the lattice sites are described by r = (x,y) with x and y being integers 0, 1, …, L−1. The operators ar and br 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: where k = (kx,ky), kx and ky 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 where However, it is usually hard to find exact eigenstates of H for nonzero U.

2. Slater determinant and exact eigenstates

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 (nj) spin-down electrons: where |0⟩ is the vacuum state and the function f is required to satisfy two conditions: (i) when rα = rβ for any αβ, 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: where ki (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 ki and a function g(r1, r2, …., rn) that make the function f satisfy the second condition.

2.1. The case g(r1,r2, …, rn) = 1

The simplest case may be g(r1,r2, …, rn) = 1. In this case, from Eq. (9) there is where Sn denotes the set of all permutations of the set {1,2,3, …,n}, p(i) denotes the value in the i-th position after the permutation p, and σ(p) is the inverse number of the permutation p. Substituting Eq. (10) into the state |Ψ⟩ in Eq. (7), we can obtain Now we find that |Ψ⟩ is obviously an eigenstate of the operator T with the eigenvalue , because each term behind the sum of Eq. (11) has the same kinetic energy E(Ψ) independent of the permutation p. As no two electrons of the state |Ψ⟩ can be on the same lattice site in coordinate space, |Ψ⟩ as an eigenstate of T is also an eigenstate of the Hamiltonian H.

An enormous number of eigenstates of H can be constructed through Eq. (11), as long as we choose a group of different ki (i = 1, …, n) in momentum space. However, it is a pity that |Ψ⟩ in Eq. (11) is the same as the following seemingly different state |Ψy⟩ with the definition where the sum is over all vector k in momentum space. The state |Ψy⟩ is obtained from the ferromagnetic state |Ψ0⟩ by flipping the spin of j electrons and it is already known to be an eigenstate of the Hamiltonian H.[9] To show the relation between the state |Ψ⟩ in Eq. (11) and |Ψy⟩, we note that there are for each permutation pSn. Therefore, from Eq. (11), we have which shows that |Ψ⟩ is the same as |Ψy⟩ in physics. In the above, we have used the equation where ki (i = 1, …, n) are different vectors in momentum space. Although our state |Ψ⟩ in Eq. (11) is not a new eigenstate, it shows an interesting relation between the Slater determinant and the eigenstates of the Fermi–Hubbard model Hamiltonian H.

2.2. The case g(r1,r2, …, rn) ≠ 1

To give some new eigenstates of H, in this case we assume where qα (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 which leads to with according to Eq. (7). It can be seen that for each permutation p, the state |Ψp⟩ is an eigenstate of the operator T, and the corresponding eigenvalue is Usually the eigenvalue Ep is dependent on p, and the state |Ψ⟩ in Eq. (21) as a sum of |Ψp⟩ 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 kα and qα (α = 1, …, n) in momentum space such that the eigenvalue Ep is not a function of p. This can be done at least in the following two situations.

(i) Assume kα = (kα, 0) and qα = (0, qα) for α = 1,2, …, n, which are illustrated in Fig. 1. Now there is according to Eq. (23), which leads to according to Eq. (23). Since the eigenvalue Ep 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 with different ki is an eigenstate of the Hamiltonian H, and the corresponding eigenvalue is

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 kα = (kα, 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 where L is assumed to be an even number. As qα = (0, qα) for α = 1,2, …, L cannot all be the same, the minimal Ep(L) will be which can be achieved when q1 = 2π/L and qα = 0 for α = 2,3, …, L. The average energy Ep(L)min/L for each electron will be −2t for large L, which is also true when L is an odd number.

(ii) Assume kα = (kα, kα+ λ) and qα = (0,0) or (λ, −λ) for some nonzero λ (α = 1,2, … n). Now there is for any permutation p according to Eq. (6), which leads to according to Eq. (23). Note that the eigenvalue Ep 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 kα and qα (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 |Ψy⟩ 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 When k1k2 and q1q2, 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.

3. Discussion and summary

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 ak and bk describe two kinds of Boson particles, under the condition that kα = (kα, 0) for α = 1,2, …, n are different from each other and qα = (qα, 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.

Reference
[1] Hubbard J 1963 Proc. R. Soc. 276 238
[2] Hayes B 2009 Am. Sci. 97 438
[3] Yun S J Dong T K Zhu S N 2017 Chin. Phys. Lett. 34 080201
[4] Cheuk L W Nichols M A Lawrence K R Okan M Zhang H Khatami E Trivedi N Paiva T Rigol M Zwierlein M W 2016 Science 353 1260
[5] Wang Y L Huang L Du L Dai X 2016 Chin. Phys. 25 037103
[6] Lieb E H Wu F Y 2003 Physica 321 1
[7] Essler F H L Frahm H Göhmann F Klümper A Korepin V E 2005 The One-dimensional Hubbard Model Cambridge Cambridge University Press 5
[8] Li Y Y Cao J P Yang W L Shi K J Wang Y P 2014 Nucl. Phys. 879 98
[9] Yang C N Zhang S C 1990 Mod. Phys. Lett. 4 759
[10] Ye M Y Lin X M 2018 Phys. Status Solidi 255 1700321
[11] Su G Ge M L 1992 Commun. Theor. Phys. 17 1
[12] Shen S Q Qiu Z M Tian G S 1994 Phys. Rev. Lett. 72 1280
[13] Chen Y H Tao H S Yao D X Liu W M 2012 Phys. Rev. Lett. 108 246402
[14] Yang C N 1989 Phys. Rev. Lett. 63 2144