Nagaoka ferromagnetism in large-spin systems -Fermion and Boson systems--
aa r X i v : . [ c ond - m a t . s t r- e l ] S e p , Nagaoka ferromagnetism in large-spin systems-Fermion and Boson systems–
Seiji Miyashita and Masao Ogata
Department of Physics, Graduate School of Science,University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan andCREST, JST, 4-1-8 Honcho Kawaguchi, Saitama, 332-0012, Japan
Hans De Raedt
Department of Applied Physics, Zernike Institute of Advanced Materials,University of Groningen, Nijenborgh 4,NL-9747 AG Groningen, The Netherlands
Abstract
We study magnetic properties of itinerant quantum magnetic particles described by a generalizedHubbard model with large spin (
S > /
2) which may be realized in optical lattices of laser-cooledatom systems. In fermion systems (half-integer spins), an extended form of Nagaoka ferromag-netism may be realized. However, as novel aspects of the large spin cases, we found that thecondition on the lattice connectivity is more stringent than in the case of S = 1 / S > / PACS numbers: 75.10.Jm,75.45.+j,75.75.+a,75.30.Gw,75.40.Mg,75.50.Xx . INTRODUCTION The origin of magnetism is attributed to the quantum mechanical interaction of particleswhich carry spin. In so-called localized spin systems, the Pauli principle plays an essentialrole and the magnetic interaction is expressed as the Heisenberg interaction , the exchangeintegral between atoms being the dominant term. Not only the two-spin interaction but alsomulti-spin interactions may contribute to give rise to exotic magnetic states. In particular,various effects of the multi-spin interaction have been reported in He . For itinerant electronsystems, the origin of the magnetic order, and of ferromagnetic order in particular, hasbeen studied extensively. Itinerant electron systems are often described by tight bindingmodels such as the Hubbard model . Nagaoka pointed out that the ground state of theHubbard model may be a ferromagnetic state (Nagaoka ferromagnetic state) if the numberof electrons is reduced by one from half-filling , where half-filling means that the numberof particles is the same as the number of the lattice sites. The Nagaoka ferromagnetic stateis established if the so-called “connectivity condition” on the lattice is satisfied . In theNagaoka ferromagnetic state, the energy of the system is minimized when the total spinof the system takes the maximum value. On the other hand, the ground state of the halffilled system is a singlet state, that is its total spin is zero. Elsewhere, we have studied howthe magnetic state changes between these two states when an electron is removed from thesystem and demonstrated an adiabatic change between these states .Magnetism has primarily been studied in electron systems in which the spin S = 1 / . In contrast to electron systems, in optical lattices the spin of trapped atoms isnot necessarily S = 1 / S > /
2. The present paper presents the results of such systems.2
I. MODEL
We consider a tight binding model of the form H = − t X
2, we take for U ( n i,M ) the standard form U ( n i,M ) = U n i, − / n i, / , (3)and for larger S > / U ( n i,M ) = 12 U N i ( N i − , (4)where N i = P M n i,M and U is assumed to be positive. Obviously, the interaction U ( n i,M )increases the energy whenever a site is occupied by more than one particle.In order to study magnetic properties of the system, we introduce spin operators for themagnetization S = ( S xi , S yi , S zi ) of the particles, where S zi = S X M = − S M n i,M . (5)Hereafter, we use the operators S + i = S xi + iS yi and S − i = S xi − iS yi which are expressed interms of c † i,M and c i,M : S + i = X M p ( S − M )( S + M + 1) c † i,M +1 c i,M (6)and S − i = X M p ( S + M )( S − M + 1) c † i,M − c i,M , (7)where S is the total spin of each particle, and M is the z -component of the magnetization.To discuss the magnetic properties of the states, we adopt the usual notation | S, M i where3 is the total spin S and M is the magnetization. The action of the operators Eqs. (6) and(7) on the state | S, M i is given by the relations S + i | S, M i = p ( S − M )( S + M + 1) | S, M + 1 i (8)and S − i | S, M i = p ( S + M )( S − M + 1) | S, M − i . (9)In order to explicitly compute matrix elements of the operators, it is convenient to intro-duce orthornormal basis states. In the case of fermion systems, we adopt the form | Ψ i = c † i,M · · · c † j,M ′ | i , (10)where i ≥ j , and M > M ′ if i = j . With this definition, the operations Eqs. (6) and (7) donot change the order of creation operators in Eq. (10). In the case of bosons, more than twoparticles with the same M can occupy the same site and the normalized basis states takethe form | Ψ i = ( c † i,M ) n i,M · · · ( c † j,M ′ ) n j,M ′ | i p n i,M ! · · · p n j,M ′ ! ≡ | n i,M i · · · | n j,M ′ i . (11)For boson systems, the order of i and j and M and M ′ is not relevant.The total spin of the whole system is given by S = ( S x ) + ( S y ) + ( S z ) = S + S − + S − S + S z ) , (12)where S x = P i S xi , S x = P i S xi , and S x = P i S zi , and S ± = P i S xi ± iS yi . We denotethe value of the total spin of the system by S tot , i.e., S tot ( S tot + 1) = h S i . III. CONSERVATION OF NUMBER OF PARTICLES OF DIFFERENT SPINSAND GROUND-STATE DEGENERACY
It should be noted that the Hubbard Hamiltonian conserves number of particles n M = N X i =1 n i,M (13)for each M . To describe the set of n M of states, it is convenient to introduce the notation { n M } = ( n S , n S − , · · · , n − S ) , (14)4here P + SM = − S n M = N . It is important to note that except for S = 1 /
2, the operator S + i S − j + S − i S + j changes the set { n M } . For example, if S = 1 application of S + i S − j + S − i S + j tostates in the set ( n = 0 , n = 2 , n − = 0), created states in the set ( n = 1 , n = 0 , n − = 1).Thus, except for S = 1 /
2, the matrix element h{ n ′ M }| S + i S − j + S − i S + j |{ n M }i can be nonzeroeven if { n M } 6 = { n ′ M } .Thus, even though the Hamiltonian H has the SU(2) symmetry, conserves the total spin,and therefore commutes with S ± , the operation of S + i S − j + S − i S + j on each state with given { n M } must be treated carefully.Let us denote by | G, M i the ground state in the space with the magnetization M . Becausethe Hamiltonian H and S ± commute with each other, S − | G, M i is also a ground statebecause H S − | G, M i = S − H| G, M i = E G S − | G, M i . (15)However, the set of numbers { n M } is not necessarily conserved, that is S − H| G, M, { n M }i = E G S − | G, M, { n M }i = E G X { n ′ M } a { n ′ M } | G, M − , { n ′ M }i , (16)where | G, M, { n M }i denotes one of basis states with fixed { n M } that contribute to theexpansion of | G, M i in terms of basis states. The energy of this state is given by E G = h G, M, { n M }| S + H S − | G, M, { n M }ih G, M, { n M }| S + S − | G, M, { n M }i = P { n ′ M } | a { n ′ M } | h G, M − , { n ′ M }|H| G, M − , { n ′ M }i P { n ′ M } | a { n ′ M } | h G, M − , { n ′ M }| G, M − , { n ′ M }i . (17)Because h G, M − , { n ′ M }|H| G, M − , { n ′ M }ih G, M − , { n ′ M }| G, M − , { n ′ M }i ≥ E G , (18)in order to satisfy the relation (17), the state in each set of numbers { n ′ M } must be theground state, i.e. h G, M − , { n ′ M }|H| G, M − , { n ′ M }ih G, M − , { n ′ M }| G, M − , { n ′ M }i = E G , (19)and thus H| G, M − , { n ′ M }i = E G | G, M − , { n ′ M }i . (20)This shows that the same ground state energy is found for all sets { n ′ M } except if a { n ′ M } = 0.5 a) (b)FIG. 1: Lattices: (a) five sites and six bonds; (b) lattice (a) with two additional bonds. IV. GROUND STATE OF A FERMION SYSTEM WITH S = A. Subspaces due to conservation of particle number of each M Let us now study the dependence of the ground state energy on U . We consider a systemwith four particles with S = 3 / { n M } =( n / , n / , n − / , n − / ) = (4 , , , , , , , , , , , , , , , M , and thus the energy level structure is same for the cases with the same set ofnumbers, e.g. for (3,1,0,0),(3,0,1,0), · · · , (0,0,1,3). For small value of U , the ground stateenergies are all different. As U increases, the ground state energies of (3,1,0,0), (2,2,0,0)and (2,1,1,0) become degenerate with that of (4,0,0,0). This fact indicates that the lowestenergy state in these sets has the total spin S max . However, we find that the lowest energystate in the set (1,1,1,1) is always lower than that of (4,0,0,0). This means that even at large U the ground state has total spin S < S max . Therefore, the Nagaoka-ferromagnetic state,i.e., the state of the maximum total spin, is not realized as the ground state in the presentcase even at large values of U .In general, when there are multiple conserved quantities in a system that do not commutewith each other, each energy state of the system is usually degenerate as shown in Eq. (20) .In the present model, the total magnetization and the set { n M } are conserved. However, theground state in Fig. 2(a) is not degenerate, which gives an exception to the above generalproperty.Figure 3 shows the U dependence of the total spin of the lowest-energy state for the6ets { n M } on the lattice of Fig. 2(a). The total spin of states in the subspace (4,0,0,0)is S max = (3 / × U , the total spin of states in the subspace(3,1,0,0) is less than S max , but it becomes S max when the ground state energy becomesdegenerate to that of the (4,0,0,0) subspace. The ground state energies of (2,2,0,0) and(2,1,1,0) become degenerate to that of (4,0,0,0) at certain values of U . However, the totalspin of the lowest-energy state of these sets does not reach S max . This fact agrees with theearlier argument that if S − | G, M i consists of more than one set of { n M } ’s, neither of thesesets yields an eigenstate of the total spin although each of them have the same ground stateenergy. Therefore, the expectation value of the total spin is not necessarily an integer. B. Connectivity condition for the system with S = Note that the lattice shown in Fig. 1(a) satisfies the so-called connectivity condition forthe case of S = 1 / U . However,for S = 3 /
2, this state is not realized. The reason is that the connectivity condition forfour S = 3 / M = − / , − / , / , / t . Then, we find that the Nagaoka-ferromagneticstate is realized, as is clear from Fig. 2(b). Summarizing, we have shown that S = 3 / C. Structure of the ground state
We found that the ground state in each subspace for a set { n M } is unique but in eachsubspace we have one state with the same ground-state energy. In Table I, we list all the sets { n M } . The structure indicates that there is a state with S tot = 6,4, 2 and 0. In the case of S = 1 /
2, there is a one-to-one correspondence between the magnetization M and { n + , n − } ,as is clear from Table I. Thus, when we create states by applying S − from the all-up state,7 a) E (b) E FIG. 2: Energies of the ground state for several sets of ( n / , n / , n − / , n − / ). (a) A system withfour particles on the lattice shown in Fig. 1(a). The solid line denotes the ground state energy for(4 , , , we have only the state with S tot = 2 for the four spin system. In contrast, in the case of S = 3 /
2, we create different sets by applying S − . As we mentioned in the previous section,the created states are degenerate as ground states. Therefore, the structure displayed in8 S tot FIG. 3: The total spin S tot of the lowest-energy state for the cases shown in Fig.2(a). The legendis the same as in Fig.2. (a) (b)FIG. 4: Nearest-neighbor hopping cannot change configuration (a) into (b) without creating doubleoccupancy. Table I is intrinsic for systems with S = 3 / , / , . . . . V. GROUND STATE OF A BOSON SYSTEM WITH S = 1 Next, we study a system of S = 1 particles, namely a boson system. Figure 5 shows theground state energies for the sets { n M } = ( n , n , n − )= (4 , , , , , , , , , , , , , , ABLE I: Sets of particles of different M ( ≥ { n M } for the cases of S = 3 / /
2. For S = 3 /
2, we list the sets only for positive M . S = 3 / M { n M } (4,0,0,0) (3,1,0,0) (3,0,1,0) (3,0,0,1) (0,4,0,0) (0,3,1,0) (1,0,3,0)(2,2,0,0) (1,3,0,0) (2,1,0,1) (2,0,1,1) (0,3,0,1)(2,1,1,0) (1,2,1,0) (1,2,0,1) (2,0,0,2)(2,0,2,0) (1,1,2,0) (0,2,2,0)(1,1,1,1) S = 1 / M { n M } (4,0) (3,1) (2,2) (1,3) (0,4)TABLE II: Sets of particles of different M ( ≥ { n M } for the case of S = 1. S = 1 M { n M } (4,0,0) (3,1,0) (3,0,1) (2,1,1) (0,4,0)(2,2,0) (1,3,0) (2,0,2)(1,2,1) the same for all the cases regardless of U . The total spin of the different ground states arealso plotted in Fig. 5. Thus, the magnetic properties of the boson system are very differentfrom that of the fermion system. The total spins are less than the maximum value ( S tot =4)except for (4,0,0) and (3,1,0) which reflects the fact that the eigenstate of maximum totalspin contains more than two sets of { n M } , as in the case of the S = 3 / S = 3 /
2. We list the sets in Table II.Here we find again a degenerate ground states with S tot = 4 , S = 1 , , . . . .Moreover, we find that even in the half-filled case, in the boson system the total spintakes the maximum value. For instance, for a system of four atoms on a simple square10attice (corresponding to the half-filled case), the ground state is the same for all sets { n M } (results not shown).This property of boson systems can be understood as follows. Consider the subspace fora fixed set { n M } . All the off-diagonal matrix elements of the Hamiltonian H are − t or 0. Bysubstracting an appropriate multiple of the unit matrix, also the diagonal elements can bemade negative or zero. Thus, all the elements of the shifted matrix e H are either negative orzero. In our model, all the sites are connected by bonds and therefore there exist a number n > e H n are positive. Then, the Perron-Frobenius theoremtells us that there exist a unique eigenstate of e H n with an eigenvalue that is larger thanthe absolute value of all other eigenvalues. This unique eigenstate is therefore the groundstate of H , is totally symmetric with positive coefficients, and contains the state of themaximum total spin. However, as we mentioned above, this ground state has an intrinsicdegeneracy with respect to subspaces that have different { n M } because we always start fromthe state with all spins maximum and let S − create different sets of { n ′ M } . The states thatare generated in this manner have all positive coefficients and are therefore ground statestoo. Cleary, this property does not depend on the connectivity of the lattice or the value of U .In contrast, for fermion systems, it is in general impossible to transform H such that allelements have the same sign but in those cases for which such a transformation exist, whichis precisely the condition of Nagaoka ferromagnetism, we can apply the same arguments asin the boson case to prove that the ground state of H is totally symmetric with positivecoefficients and contains the state of the maximum total spin. VI. SUMMARY AND DISCUSSION
We have studied the magnetic properties of the ground state of itinerant systems with
S >
1. We found that fermion systems ( S = 3 /
2) support an extended form of Nagaokaferromagnetism but that the connectivity condition is more difficult to satisfy because ofthe presence of particles with different magnetization. When the maximum S tot state is theground state, the system has a degenerate ground state in each set { n M } listed in Table I.Thus the ground state manifold consists of not only the state of the maximum S tot but alsoof states with smaller values of S tot . This degenerate structure is intrinsic for the systems11 FIG. 5: Ground state energies E and the corresponding total spin S tot as a function of U for four S = 1 bosons and various ( n , n , n − ) on the lattice shown in Fig. 1(a). The solid line denotesdata for (4,0,0), solid triangles: (3,1,0), open triangles: (3,0,1), open squares: (2,2,0), reversedtriangles: (2,1,1), crosses: (1,2,1), and bullets: (0,4,0). with S > / { n M } exist for a given value of M .For boson systems ( S = 1) we found that there is a ground state in each set of { n M } and that the same type of degenerate ground-state structure appears as the one found for S = 3 /