当前位置：文档库 > The ground state of the Kondo model with large spin

a r X i v :c o n d -m a t /9703145v 1 [c o n d -m a t .s t r -e l ] 14 M a r 1997

The ground state of the Kondo model with large spin

Shun-Qing Shen

Max Planck Institute for Physics of Complex Systems,Bayreuther Strasse 40,Haus 16,D-01187Dresden,Germany

(Received:December 26,1996)

In this paper,we prove that the ground state of the Kondo model with large spin is nondegenerate,apart from a SU(2)spin degeneracy in the case of half ?lling.The ground state spin is found for the system,and the energy level orderings are discussed.Finally,the existence of ferrimagnetism in some cases is proved.

PACS numbers:75.10.Lp 75.30.Mb,75.50.Gg

The Kondo models,or single and lattice impurity mod-els,are one of the most challenging subjects in strongly correlated systems 1–3.Recent developments of re?ection positivity technique in the spin space make it possible to establish some rigorous results for the half-?lled strongly correlated electron systems 4–9.Theory of re?ection pos-itivity in the spin space for the single-and multi-channel Kondo models with spin 1/2was developed recently and a series of rigorous results on the ground state prop-erties were proved 8,9.However some materials are de-scribed by the Kondo models with large spin,such as (La 1?x X x )MnO 3with X =Ba,Ca,Sr etc.and the lo-calised spin s =3/210.Due to more degrees of freedom in the case of large spin than in the case of spin 1/2,usu-ally it is very hard to extract rigorous results for those systems.As a generalisation of the theory for the Kondo model with spin 1/2,we will investigate the Kondo model with large spin and provide some rigorous results on the ground state of the Kondo model in the case of half ?ll-ing.

Let us ?rst write down the Hamiltonian we will inves-tigate:H =

ij ∈∧,σ

t ij c ?i,σc j,σ+

i ∈∧

U i (n i,↑?

1

2

)+

i ∈∧

J i S i ·S ci + ij ∈∧d

K ij S i ·S j ,

(1)

where c ?i,σand c i,σare the creation and annihilation op-erators for the conduction electron (c-electron)at site

i with spin σ(=↑,↓)and n i,σ=c ?i,σc i,σ.

S i is the localised spin operator with spin s i at site i .S ci = σ,σ′c ?

i,σσσ,σ′c i,σ′/2,and σα(α=x,y,z )are the Pauli matrices.The model is de?ned on a bipartite lattice ∧with the site numbers N A and N B of the two sublattices A and B .t ij =t ji is possibly non-zero only when i and j belong to two di?erent sublattices.The lattice ∧is con-nected by the hopping terms {t ij },which implies that for any two sites k and l on ∧,we can always ?nd a sequence {(k,i 1),(i 1,i 2),···,(i n ,l )}such that t ki 1t i 1i 2···t i n l =0.∧d is the distribution of localised spins on ∧.If we re-gard the same site for conduction electron and localised spin as two independent sites in a generalised lattice,the site i for conduction electron and the site i for localised

spin belong to the same sublattice when J i <0,and to two di?erent sublattice when J i >0.In the case the generalised lattice can be still regarded as a generalised bipartite lattice.Assume K ij J i J j ≤0if i and j belong to the same sublattice,and K ij J i J j ≥0if i and j belong to two di?erent sublattices.When J i and J j are constant or have the same sign,K ij ≤0if i and j belong to the same sublattice and K ij ≥0if i and j belong to two dif-ferent sublattices as in the antiferromagnetic Heisenberg model on a bipartite lattice.This condition guarantees that the last term in Eq.(1)does not introduce any frus-tration for the generalised bipartite lattice.Physically,it is completely absent of frustration in this case.The model is reduced to the Hubbard model when ∧d =0,i.e.,there is no magnetic impurity or localised spin in the systems.The structure function of the bipartite lattice ∧is de?ned as ?(i )=1if i ∈A and ?1if i ∈B .The main results are summarised as follows:

Theorem :Assume the model in Eq.(1)with t ij and K ij is de?ned on a connected bipartite lattice ∧with sublattice sites N A and N B and with the distribution of localised spins ∧d .All U i >0and J i =0and the number of conduction electrons is N e =N ∧=N A +N B .The signs of K ij and J i satisfy the previously stated con-dition.Denote the lowest energy state in the subspace

decomposed by the z-component of the total spin S z

tot by |Ψ .

i).The state |Ψ is non-degenerate in each subspace of S z tot .The ground state of Eq.(1)is unique apart from a

(2S 0

tot +1)-fold spin degeneracy.

ii).The total spin S tot in the lowest energy state is

S tot = S 0

tot ,if |S z tot |≤S 0

tot ;

|S z

tot |,otherwise (2)where

S 0tot

=

1|J i |

?(i ) .

(3)

iii).The spin-spin correlation functions obey,

Ψ|S +ci ·S ?

cj |Ψ =?(i )?(j )C ij ; Ψ|S +i ·S ?j |Ψ =?(i )?(j )F ij ;

(4)

Ψ|S +ci ·S ?

j

|Ψ =??(i )?(j )J ⊥

where C ij,F ij and G ij≥0if U i≥0,and>0if all U i>0.

iv).When all U i=0,(at least one of)the ground state(s) (if degenerate)has the total spin as that in Eq.(2). Before we present the proof,several remarks or corol-laries are made:

1).In the case of all J i>0or<0and s i=s,suppose that N Ad spins are on the sublattice A,and N Bd spins on the sublattice B,the total spin is

S0tot= 1|J|s(N Ad?N Bd) .(5)

In the Kondo lattice case,N A=N Ad and N B=N Bd, and the total spin is

S0tot= s?1|J| |N A?N B|.(6)

When all s i=1/2,we recover the result for the case of spin1/28,9.The ground state is a singlet only when N A=N B,or s=1/2and J>0.When N A=N B and (N A?N B)/(N A+N B)=0when the system becomes su?ciently large,we obtain a state with ferromagnetic long-range order.

2).Theorem(iii)indicates that strong antiferromag-netic correlations exist between the conduction electrons or localised spins.A direct corollary is that the antifer-romagnetic correlation is always stronger than the ferro-magnetic correlation.For example on a cubic lattice,

S+

Q

·S?Q =max{ S+q·S??q },(7)

where S+=1

N∧ i∈∧S+i e i q· r i and Q=(π,π,···). If the state possesses ferromagnetic long-range order,it must also possess antiferromagnetic long-range order.In other words,the ferromagnetism in the case of N A=N B is,strictly speaking,the ferrimagnetism.

3).The energy level orderings can be obtained from Theorems(i-ii)and the SU(2)symmetry with a varia-tional principle:denote E(S z tot)the lowest energy state with S z tot.As|Ψ(S z tot) is the lowest energy state with S z tot,we can construct an eigenstate( i∈∧S?ci+ i∈∧d S?i)|Ψ(S z tot) (suppose S z tot≥1)due to the spin SU(2)symmetry.This state has its z-component of the total spin S z tot?1and the eigenvalue E(S z tot).Meanwhile it has the same total spin as in|Ψ(S z tot) .In the varia-tional principle the lowest energy state in the subspace of S z tot?1should not be higher than E(S z tot).If the low-est energy in the subspace S z tot?1is equal to the lowest energy in the subspace S z tot,the lowest energy state or one of the states if degenerate with S z tot?1must has the same total spin in the state with S z tot.Since the lowest energy states in each subspace are non-degenerate when all U i>0and we assume S z tot≥0,we have

E(S z tot)=E(?S z tot);(8)

E(S z tot)

E(S z tot)=E(S0tot)if0≤S z tot≤S0tot.(10)For S z tot≤0,the energy level ordering can be obtained from Eqs.(8-10).This is similar to that of the Heisen-berg model12.It is worth of mentioning that in the case of U i≥0we have to use“lower or equal”instead of “lower than”if we could not determine whether the low-est energy state in each subspace is non-degenerate or not

4).In the one-dimensional chain of the Kondo lattice with all J i>0or<0,Theorems(ii)and(iii)on the to-tal spin and spin-spin correlation functions are the same as those in a two-chain spin-ladder system.This coin-cides with the analysis on the resemblance of these two systems by White and A?eck13.

5).When the system has no localised spin or magnetic impurity,N Ad=N Bd=0and the model in Eq.(1)is reduced to a Hubbard model.Theorem(i-iv)hold for the Hubbard model.In the case,one can regard the spin formula in Eq.(2)as a generalisation of Lieb’s theorem for the Hubbard model4.The conditions of even number of the lattice sites and S z tot=0in Lieb’s theorem are re-moved.As a corollary,the energy level ordering for the Hubbard model in the case of half?lling is the same as shown in Eqs.(8-10).

The main steps to prove this theorem are1).to ex-press the spin operator in a multi-fermion representation;

2).to de?ne the generalised bipartite lattice after we consider the localised spins;3)to introduce a complete and orthonormal set of basis for the system;4)to prove the positive de?niteness or semide?niteness of the ground state on the chosen basis14;and5)to prove(iii)and(iv) by utilising the positive de?niteness of the ground state and the known theorems on a positive de?nite state. Except for the re?ection positivity approach in the spin space as in the case of spin1/28,9,the key trick used in this paper is to express the localised spin op-erator S i as the summation of2s i spin1/2operators in a multi-fermion representation(d-fermion)and these 2s i spins are coupled http://www.wendangku.net/doc/6e12a67a31b765ce05081403.htmlpletely fer-romagnetic coupling could be realized by introducing a strong ferromagnetic coupling limit between these spins in the Hamiltonian.The spin operator is written as

S i=

2s i

αi=1d?iαi,σσσ,σ′

2

)(n i,↓?

1

2

d iα

i

,σ′

)·(c?i,σ

σσ,σ′

2

d iα

i

,σ′

)·(d?jα

j

,σ

σσ,σ′

2

d iα,σ′)2?s i(s i+1)],(12)

where all λi are

positive 15

.We shall take all λi →∞at the last step (as a matter of fact our theorem is true for any ?nite λi ),and this guarantees that any deviation from

Ψ| 2s

i αi =1

d ?iαi ,σ

σσ,σ′

|J i |

?(i )d ?iαi ↑

.

(15)

Under this transformation,we have

T c i ↑T ?=(?1)N t ?(i )c ?i ↑;(16)T c i ↓T ?=(?1)N t c i ↓;

(17)

T d iαi ↑T ?=(?1)N t +1

J i

2

(N t ?2N 0)T |φ↑α ?|φ↓

β .

(21)

Any lowest energy state of H ′with S z

tot =

1

2

(n i,σ?1

2)

+

ij ∈∧d ,αi ,αj

K ij

2

)(n jαj ,σ?

1

2

(n iαi ,σ?

1

2

)|φσβ ,(24)

and

1

(V i)αβ= φσα|(n iσ?

|J i|2s i?(i).We get Theorem(ii)and(iv)combin-

ing the theorem above and the positive de?niteness of

the lowest energy state.Theorem(iii)is also obtained as

we have decomposed the spin S i into2s i1/2-spins and

the whole system is still on a generalised bipartite lattice.

For example,

Ψ|S+i·S?j|Ψ = αi,αj Ψ|d?iαi,↑d iαi,↓d?iαi,↓d iαi,↑|Ψ

=?(i)?(j)F ij(40)

where

F ij= αi,αj T r(W?V diαi jαj W(V diαi jαj)?).(41)

F ij>0when the state is positive de?nite,and≥0when

the state is positive semide?nite.

In summary,we have provided several theorems on the

ground state properties of the Kondo model at half-?lling

and in the case with large spin.The uniqueness and the

total spin in the ground state are found.Furthermore

we have also investigated the spin-spin correlation in the

system.The co-existence of both antiferromagnetic and

ferromagnetic long-range correlations is proved,which is

similar to the case of spin1/2.Hopefully,these exact re-

sults are useful for understanding this highly correlated

electron-impurity lattice.

I would like to thank Prof.P.Fulde for conversations.I

also wish to thank Dr.D.F.Wang for fruitful discussions

and for his encouragement.This work was supported by

the Alexander von Humboldt foundation of Germany.

1For a review,please see P.Fulde,J.Keller,and G.Zwick-nagl,In Solid State Physics,Vol.41,ed.by H.Ehrenre-ich and D.Turnbull,1(Academic,San Diego1988),P.

A.Lee,T.M.Rice,J.W.Serene,L.J.Sham and J.W. Willkins,Comments Condens.Matt Phys.12,99(1986), and P.Coleman,Phys.Rev.B28,5255(1983).References therein.

2P.Coleman,E.Mirada and A.Tsvelik,Phys.Rev.B49, 7252(1989).

3I.A?eck and A.Ludwig,Nucl.Phys.B360,641(1991), B352,849(1991).

4E.H.Lieb,Phys.Rev.Lett.62,1201;1927(E)(1989).

5K.Ueda,H.Tsunetsugu,and M.Sigrist,Phys.Rev.Lett. 68,1030(1992).

6S.Q.Shen,Z.M.Qiu,and G.S.Tian,Phys.Rev.Lett. 72,1280(1994).

7T.Yanagisawa and Y.Shimoi,Phys.Rev.Lett.74,4939 (1995).

8S.Q.Shen,Phys.Rev.B53,14252(1996).

9S.Q.Shen,Phys.Rev.B54,4397(1996).

10J.Inoue and S.Maekawa,Phys.Rev.Lett.74,3407(1995). 11F.C.Pu and S.Q.Shen,Phys.Rev.B50,16086(1994). 12E.H.Lieb and D.C.Mattis,J.Math.Phys.3,749(1962). 13S.R.White and I.A?eck,Phys.Rev.B54,9862(1996). 14Usually we say a square matrix or an operator is positive de?nite or semide?nite when all its eigenvalues are positive or non-negative.In theory of re?ection positivity,the state is expressed as a square matrix on a chosen basis.When we say a state is positive de?nite or semide?nite,it means that the matrix standing for the state is positive de?nite or semide?nite.

15We can also regard the last term in Eq.(12)as Lagrange multipliersλi.In this case,they are variable,not constant. The resulting variational conditions are the same as those in Eq.(13).As s i(s i+1)is the maximum of the eigenval-ues of S2i,the resulting conditions are equivalent to those in Eq.(14).

- 二年级下册数学【课件】认识有余数的除法课件ppt
- 犯罪化理论根据的抉择与范围界限
- 高中英语随笔作文
- 腾讯设计中心校园招聘
- 产品造型设计材料与工艺――第五章玻璃及其加工技术PPT课件
- 【精品】重庆涪陵生猪标准化规模养殖场40小区41建设项目可研报告
- 子网掩码练习题(含答案)
- 学习型党组织创建活动总结
- 《马克思主义政治经济学》学习笔记-第七章
- 龙门吊安全检查要点
- 标杆企业如何进行设计管理
- 课堂刮刮乐课件ppt
- 东财网院 201109 考试模拟题 含答案 【全面预算管理】
- 幂指函数的极限与导数问题_朱美玉
- 免费2013幼儿园小班数学课件PPT_认识形状
- 二年级下册数学【课件】数据收集整理课件ppt(1)
- 安徽省往年公务员录用考试行测考试练习题：其他常识(672)
- 2020压力与情绪管理考试继续教育90分
- 第11章 企业发展能力分析习题
- 农田机井工程施工组织设计
- 小学数学经验总结论文
- 硅表面清洗对热氧化13nm SiO_2可靠性的影响
- 初三数学圆知识点集合
- 劳务派遣招投标书
- 北师大版二年级上册语文1～2单元测试卷
- 哈尔滨海关门户网站连接你我
- 消防安全管理人考试题库(含答案)