Entanglement in spin-1/2 dimerized Heisenberg systems

Entanglement in spin-1/2 dimerized Heisenberg systems Zhe Sun, XiaoGuang Wang, AnZi Hu, and You-Quan Li Zhejiang Institute of Modern Physics, Department of Physics, Zhejiang University, HangZhou 310027, China (Dated: February 1, 2008) We study entanglement in dimerized Heisenberg systems. In particular, we give exact results of ground-state pairwise entanglement for the four-qubit model by identifying a Z2 symmetry. Although the entanglements cannot identify the critical point of the system, the mean entanglement of nearest-neighbor qubits really does, namely, it reaches a maximum at the critical point. arXiv:quant-ph/0501117v1 21 Jan 2005 PACS numbers: 03.65.Ud, 03.67.-a The study of entanglement properties of many-body condensed matter systems has received much attention [1]-[30]. It covers both the rapid developing area of quantum information and the area of condensed matter. Various quantum models have been considered in the literature and some of them are exactly solvable. Specifically, it was well-known that the anisotropic Heisenberg model can be solved formally by Bethe’s Ansatz method [31, 32] for arbitrary number of qubits N , however, we have to solve a set of transcendental equations. It is interesting to see that entanglement in a system with a few sites displays general features of entanglement with more sites. In the anisotropic Heisenberg model with a large number of qubits, the pairwise entanglement shows a maximum in the isotropic point [33]. This feature was already shown in a small system with four or five qubits [34]. So, the study of small systems is meaningful in the study of entanglement as they can reflect general features of larger or macroscopic systems. Dimerized systems play an important role in condensed matter theory, and entanglement in dimerized systems has been considered [28, 35]. Here, we aim at obtaining some analytical results of entanglement in a small system. Chen et al [35] found that the concurrence [36], quantifying entanglement of a pair of qubits cannot enable us to identify the critical point J2 /J1 = 1. However, we will show that although the single concurrence cannot be used to identify the critical point, the mean concurrence really does. For N ≤ 7, the isotropic Heisenberg Hamiltonian can be analytically solved [37, 38] due to the translational invariant symmetry which can be used to reduce the Hamiltonian matrix to smaller submatrices by a factor of N [39]. However, for our dimerized systems, the translational invariant symmetry breaks. We identify a Z2 symmetry in the dimerized model, and using this symmetry we can further reduce Hamiltonian matrices by a factor of two. We will first consider the simplest representative four-qubit dimerized system, and by using the Z2 symmetry, all eigenvalues can be obtained. Eigenvalue problem. The Heisenberg Hamiltonian for the dimerized chain with even number of qubits N reads H= N/2  X i=1  1 1 J1 (2S2i−1 · S2i + ) + J2 (2S2i · S2i+1 + ) 2 2 N/2 = X i=1 (J1 S2i−1,2i + J2 S2i,2i+1 ). (1) where Si is the spin-half operator for qubit i , Sj,j+1 = 1 σi · ~σi+1 ) is the swap operator between qubit i and 2 (1 + ~ j, ~σi = (σix , σiy , σiz ) is the vector of Pauli matrices, and J1 and J2 are the exchange constants. We have assumed the periodic boundary condition, i.e., N + 1 ≡ 1. In the following discussions, we also assume J1 , J2 ≥ 0 (antiferromagnetic case). For the four-qubit case, the dimerized Hamiltonian simplifies to H = J1 (S12 + S34 ) + J2 (S23 + S41 ). (2) Although we have imposed the periodic boundary condition, the Hamiltonian is no longer translational invariant except the case of J1 = J2 . As we stated above, the translational invariant symmetry can be used to reduce the Hamiltonian matrix to smaller submatrices by a factor of N [39]. However, for our case, we cannot have this reduction. To exactly solve the eigenvalue problem of the Heisenberg model, we first note that [H, Jz ] = 0, the whole 16dimensional Hilbert space can be divided into invariant subspaces spanned by vectors with a fixed number of reversed spins. Then, the largest subspace P is 6-dimensional with 2 reversed spins. Here, Jz = 4i=1 σiz /2. Due to the Z2 symmetry [H, Σx ] = 0, it is sufficient to solve the eigenvalue problems in the subspaces with r reversed spins, where r ∈ {0, 1, 2} and Σx = σx ⊗ σx ⊗ σx ⊗ σx . (3) The subspace with r = 0 only contains one vector |0000i, which is the eigenvector with eigenvalue E0 = 2(J1 + J2 ). (4) For subspace with 2 ≥ r > 0, the smallest submatrix is 4 × 4. We need further reduce submatrices to smaller 2 ones. Although we do not have a translational invariance, from the four-qubit Hamiltonian (2), we can identify another symmetry Z2 symmetry, given by [H, S12,34 ] = [H, S13 S24 ] = 0. (5) The operator exchange the state of qubits 1 and 2 with the state of qubits 3 and 4, namely, S12,34 |m1 , m2 , m3 , m4 i = |m3 , m4 , m1 , m2 i. (6) The eigenvalue of the operator S12,34 on all energy eigenstates is either 1 or -1. The subspace with r = 1 is spanned by four basis vectors {|1000i, |0100i, |0010i, |0001i}. Taking into account the Z2 symmetry, we choose the following basis 1 |ψ1± i = √ (|1000i ± |0010i), 2 1 |φ1± i = √ (|0100i ± |0001i). 2 (7) Obviously, they are all eigenstates of operator S12,34 , and basis {|ψ+ i, |φ+ i} ({|ψ− i, |φ− i}) spans an invariant twodimensional subspace of Hamiltonian H. Let the Hamiltonian act on the basis, then we obtain H|ψ1+ i =(J1 + J2 )(|ψ1+ i + φ1+ i), H|φ1+ i =(J1 + J2 )(|ψ1+ i + φ1+ i), H|ψ1− i =(J1 + J2 )|ψ1− i + (J1 − J2 )|φ1− i, H|φ1− i =(J1 + J2 )|φ1− i + (J1 − J2 )|ψ1− i. (8) From the above equation, the eigenvalues of Hamiltonian H is obtained as E1,1 = 2(J1 + J2 ), E1,2 = 0, E1,3 = 2J1 , E1,4 = 2J2 , (9) where the first subscript denotes the number of reversed spins. For the case of r = 2, due to the existence of the Z2 symmetry, we choose the following basis for the sixdimensional subspace 1 |ψ2± i = √ (|1100i ± |0011i), 2 1 |φ2± i = √ (|1001i ± |0110i), 2 1 |ϕ2± i = √ (|0101i ± |1010i). 2 (10) Then, we can reduce the 6 × 6 Hamiltonian matrix to a block-diagonal form with two 3 × 3 matrices. After the action of the Hamiltonian on the above basis, we obtain H|ψ2− i =2J1 |ψ2− i, H|φ2− i =2J2 |φ2− i, H|ϕ2− i =0, H|ψ2+ i =2J1 |ψ2+ i + 2J2 |ϕ2+ i, H|φ2+ i =2J2 |φ2+ i + 2J1 |ϕ2+ i, H|ϕ2+ i =2J2 |ψ2+ i + 2J1 |φ2+ i. One 3 × 3 block is of the diagonal form, and the eigenvalues read E2,1 = 2J1 , E2,2 = 2J2 , E2,3 = 0. Another block is written as   2J1 0 2J2 2J2 2J1  . H =0 2J2 2J1 0 (12) (13) Although this is a 3 × 3 matrix, we can further reduce it to 2 × 2 matrix since, from the last three equations of Eq. (11), one eigenvalue is found to be E2,4 = 2(J1 + J2 ). With the help is this eigenvalue, the characteristic polynomial of the Hamiltonian matrix can be brought to a quadratic form by dividing it with E − E2,4 . Then, we obtain q E2,5 =2 J12 + J22 − J1 J2 , q EGS =E2,6 = −2 J12 + J22 − J1 J2 . (14) Thus, all eigenvalues are analytically obtained for the four-spin dimerized Heisenberg model. We see that ground state is non-degenerate and the energy is given by E2,6 . Although the eigenstates can be easily obtained, they are not given explicitly here as the knowledge of eigenvalues is sufficient for discussions of entanglement properties as we will see below. Entanglement in the four-qubit model. We first study the ground-state entanglement of qubits 1 and 2. Due to the SU(2) symmetry in our Hamiltonian, the concurrence quantifying the entanglement of two qubits is given by [40] Ci,i+1 = −2hSi · Si+1 i − 1/2 = −hSi,i+1 i, (15) where we have ignored the max function in the usual definition of the concurrence, and thus the negative concurrence implies no entanglement. We see that the entanglement is determined by the expectation value of the SWAP operator. There are two distinctive concurrences C12 and C23 , corresponding to two spins coupled by bond J1 and J2 . The expectation value hS12 i can be calculated via Feynman-Hellman theorem. By applying the theorem to the ground state, we obtain hS12 i + hS34 i = J2 − 2J1 ∂EGS . = p 2 ∂J1 J1 + J22 − J1 J2 (16) Due to the Z2 symmetry [H, S12,34 ], we have hS12 i = hS34 i. Then, from the above equation, the expectation value hS12 i is obtained. Substituting it to Eq. (15) leads to (11) 2J1 − J2 C12 = p . 2 2 J1 + J22 − J1 J2 (17) 3 1 0.8 1 C23 C12 0.8 C 0.4 Concurrence Concurrence C 12 23 0.6 0.6 mean 0.2 0 0.4 C mean 0.2 0 −0.2 −0.2 −0.4 −0.4 −0.6 0 C 0.5 1 1.5 2 J2 2.5 3 3.5 4 −0.6 0 0.5 1 1.5 2 J 2.5 3 3.5 4 2 FIG. 1: The concurrences versus J2 in the four-qubit dimerized model (J1 = 1). FIG. 2: The concurrences versus J2 in the six-qubit dimerized model (J1 = 1). In a similar way, we can obtain the concurrence C23 = C14 as Substituting Eqs. (17) and (18) to the above equation leads to 2J2 − J1 C23 = p . 2 2 J1 + J22 − J1 J2 J1 + J2 Cmean = p 2 , 4 J1 + J22 − J1 J2 (18) Thus, the analytical expressions of two types of concurrence are obtained, and the concurrence C12 can be transformed to C23 by exchanging J1 and J2 . From Eqs. (17) and (18), we read that  when J2 = 0, =1 C12 = 1/2 when J2 = J1 , (19)  (12) ≤0 when J2 ≥ J2th = 2J1 , and  (23) ≤0 when J2 ≤ J2th = J1 /2, C23 = 1/2 when J2 = J1 ,  =1 when J2 = ∞, (20) In the case of J2 = 0, the ground state is just the uncoupled two singlets, and thus C12 = 1. When J2 = J1 , the dimerized Hamiltonian is reduced to the isotropic one, and the concurrence C12 = C23 = 1/2 [6]. When (23) J2 > J2th , the entanglement between qubits 2 and 3 (12) builds up, and when J2 ≥ J2th = 2J1 , the entanglement between qubits 1 and 2 vanishes. It was argued that the pairwise entanglements quantified by the concurrence cannot identify the critical point (J2 = J1 ) of the dimerized system [35], however here we use the mean entanglement of nearest-neighbor qubits as the system we are considering is not a uniform system, namely, it has two distinct exchange constants. There are four pairs of nearest-neighbor qubits, and thus Cmean = C12 + C23 C12 + C23 + C34 + C41 = . 4 2 (21) (22) from which we read Cmean   1/4 when J2 = 0, = 1/2 when J2 = J1 ,  1/4 when J = ∞. 2 (23) In Fig. 1, we plot the concurrences and the mean concurrence versus J2 . We see that the mean entanglement takes its maximum at the critical point J2 =J1 . From Eq. (22), we can also check that the maximal point is at the critical point. In the following, we will numerically show that this fact still holds for dimerized chains with more qubits. Numerical results. Now we consider more general cases of even-number qubits. For even N > 4 , the analytical results of entanglement are hard to obtain, and here, we use exact diagonalization method to numerically calculate the entanglement. For even-number case, the mean entanglement is still given by Eq. (21). In Figs. 2 and 3, we plot the concurrences versus J2 for N = 6 and N = 8, respectively. We observe that the entanglement properties are similar to those in the fourqubit model, namely, they are qualitatively the same. The mean entanglement reaches its maximum at the critical point. From Figs. 1-3, we see that the threshold value (12) (23) J2th decreases, while J2th increases when the number of qubits increases. Finally, we give a relation between the two types of entanglement with the ground-state energy EGS . From Eq. (1), we immediately have the relation between the ground-state energy and the correlators hS12 i and hS23 i, 4 state energy and the two typical concurrences. If we know the entanglement of qubits 1 and 2 and the ground-state energy, we can know the entanglement of qubits 2 and 3 from the relation. 1 C C 23 Concurrence 12 0.5 C mean 0 −0.5 0 0.5 1 1.5 2 J 2.5 3 3.5 4 2 FIG. 3: The concurrences versus J2 in the eight-qubit dimerized model (J1 = 1). EGS /N = 1 (J1 hS12 i + J2 hS23 i). 2 (24) Then, from Eq. (15), we obtain 1 EGS /N = − (J1 C12 + J2 C23 ). 2 Conclusion. We have studied entanglement in spin1/2 dimerized Heisenberg systems. As a representative system, the four-qubit model was studied in detail. By identifying the Z2 symmetry, we have solved the eigenvalue problem completely. From the ground-state energy, the analytical results of the two types of pairwise entanglement were obtained. We have located the threshold value of J2 after which the entanglement between qubits 1 and 2 vanishes and the threshold value before which there exists no entanglement between qubits 2 and 3. In order to identify the critical point of the dimerized system, we have proposed the mean entanglement of nearest-neighbor qubits as an efficient indicator. The mean entanglement displays a maximum at the critical point. We have also considered the case of more qubits, and the numerical results show that the entanglement properties are similar to those in the four-qubit model. 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