aa r X i v : . [ c ond - m a t . s t r- e l ] F e b PAIR FORMATION IN A t-J MODEL
COURSE PH-614M.Sc. Project Report IIAYAN KHAN Reg. No. 03212103M.Sc IVth. SemesterProject Instructor :Dr. SAURABH BASU Assistant ProfessorDepartment of PhysicsIIT Guwahati [email protected] [email protected] ERTIFICATE
It is certified that the work contained in the project titled
Pair Formation in a t-JModel has been carried out by Ayan Khan, under my supervision.
Date:
20. 04. 2005
Dr. Saurabh Basu
Place:
Guwahati
Assistant ProfessorDepartment of PhysicsIIT Guwahati CKNOWLEDGMENT
I am deeply indebted to my project supervisor, Dr. Saurabh Basu, whose able guidence,thoughtful instruction invaluable criticism were instrumental in the progress of mywork.To my parents, I owe a special debt of gratitude for there blessings and support.Above all I like to thank all my friends, my juniors and senior P.hD. scholars fortheir constant encouragement and invaluable suggetions.
Date:
AYAN KHAN ontents Second Quantization: . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2
Theoretical Models in Many Particle Systems: . . . . . . . . . . 52.2.1 Tight Binding Model: . . . . . . . . . . . . . . . . . . . . . . . . 52.2.2 Heisenberg Model: . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.3 Hubbard Model: . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.4 t-J-U Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3
Construction of Equation of Motion (EOM) for Two ParticleSystem: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
One Dimensional Chains: . . . . . . . . . . . . . . . . . . . . . . . . 83.2
Two Dimensional Square Lattice: . . . . . . . . . . . . . . . . . . 113.3
Two Leg Ladder: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.4
Three Dimensional Cubic Lattice: . . . . . . . . . . . . . . . . . . 17
BSTRACT
We have investigated the formation of bound state of two electrons in different kindof lattices using a t-J-U model. In the model hopping parameter t tries to delocalizethe electrons where as pairing of electrons comes via Heisenberg exchange integral J and hence it becomes necessary to calculate the threshold value of J , viz. J c , for whichformation of bound states between two electron system is possible. The analysis isrepeated for one dimensional chains, two dimensional square lattices, two leg laddersand three dimensional simple cubic lattices. Further we calculated the bound stateenergies for J > J c . Also we have tried to shed some light to the symmetry operationof the lattices to understand the characteristic of two electron pairing.1 INTRODUCTION:
The discovery of superconductivity in 1911 by Kammerlingh Onnes had given a new di-mension in Condensed Matter Physics research. From the very beginning the scientificand commercial potential of superconductors had been well understood by the commu-nity. So as the days progressed different exciting features of superconductivity startedto come to light and in 1987 with the discovery of high temperature superconductorsthe field of interest is further broadened.Normal superconductors, which are usually good metals, are quite well understoodby BCS theory where electron electron interaction is mainly controlled by phonons.But in high temperature superconductors the interaction picture among the electronsare still not clear.Here our motivation is to study strongly correlated systems because it is well agreedthat the origin of high temperature superconductivity is purely from electronic inter-action, as for example here we are interested to study two dimensional square latticewhich is analogous to the
CuO planes in a high temperature superconductor.There are several models to study many particle systems and we are here usingt-J model for studying electron pairing in metals with t denoting kinetic energy andJ denoting Heisenberg exchange integral.We are starting with the assumption thatthe parent compounds are quite well represented by the antiferromagnetic Heisenbergmodel with localized electrons of spin 1 / J c , such that when spin exchange integral J exceeds J c , the hole rich phase has no electrons and for J slightly less than J c the hole richphase is a low density superfluid of electron pairs.Here we are investigating the critical value of J for which pairing of electron ispossible in different kind of lattices. On the due course in the earlier semester we hadconcentrated on one dimensional chains. One dimensional analysis always carries asignificant importance for its relative simplicity, also if one dimensional lattice featuresare well understood then it becomes relatively easy to understand the higher dimen-sional lattices. Then we had taken one step ahead to two dimensional square lattice.As we have stated the features of the two dimensional lattice has become very im-portant after the discovery of high temperature superconductors in ceramic materials.The structural feature of ceramic materials is CuO plane which are the main sourcefor high temperature superconductivity. So it is important to understand the electroncorrelation in two dimensional lattices. Further we had extended our study on twoleg ladder lattice. Since for the one-dimensional chain system quite a few things areknown exactly. One approach to tackle the superconducting cuprates is to investigatethe quasi one-dimensional lattices known as ‘spin ladder’ structures, which are stripsof square lattice with a finite width and infinite length. An example of spin ladderlattice system is Sr n − Cu n +1 O n For these above mentioned lattices we had verified2he critical value of J in square lattice which is [2]. Also we had investigated thecritical value for one dimensional chain like lattice and two leg ladder where again weland up with the same result as J = 2 t .In this semester we were curious to look at the critical value of J in a simple cubiclattice. To understand the electronic interaction of the heavy-fermionic superconduc-tors such as CeCu Si , one needs to deal with different three dimensional crystalstructures. So it is a natural to look at a three dimensional structure and try to findout the critical value of J for which pair formation among two electrons is possible. Sowe calculated J c for a simple cubic lattice and this constitutes a new and central resultin this project work. To understand the many body theory in condensed matter physics the essential tech-nique is the method of second quantization. Soon after the foundation of quantumtheory, the formalism of creation and annihilation operator (second quantization) wasintroduced. The physics of creation and annihilation operators can be explained in abetter way from relativistic quantum field theory. So second quantization is is nothingbut a alternative formulation of quantum mechanics. the creation and annihilationoperators are nothing but a tool that permits different process such as creation andannihilation of operators. Such process can not be discussed in the framework ofSchr¨ o dinger equation[3].We know that quantum mechanical wave function which represents a collection of elec-trons is antisymmetric with respect to the operation which exchanges the space andspin coordinates of any two electrons. Thus, if ψ (1 , , , ..., N ) is an N electron wavefunction, and if P ij is the operator which exchanges the coordinates if electron i andelectron j, then P ij Ψ = − ΨJ. C. Slater has introduced a method to represent such many electron wave functions.We need to begin with an orthonormal set of one-electron functions: φ , φ , .... , where R dτ φ ⋆i (1) φ j (1) = δ ij [3] , where∆ = 1 √ N ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ ( r ) ... ... φ ( r N ) ... ... ... ...... ... ... ...φ N ( r ) ... ... φ N ( r N ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≡ | φ φ φ ... | Such a function is called ”Slater determinant.” Since interchange of any two columnsof a determinant changes its sign, ∆ is antisymetric with respect to the exchange op-erator P ij .We can also represent antisymmetric many electron wave function in a different man-ner. Let us now define a set of electron creation operators, b † , b † , ..., , correspondingto the one electron spin orbitals, φ , φ , , ..., . When the creation operator b † j acts on3n N-electron state, it produces an (N+1) electron state by creating an electron inthe spin-orbital φ j . We used to denote no electron state or ”vacuum state” as | > .Similarly we define annihilation operator as b j .The commutation relations of the operators as follows: b j | > = 0, < | b † j = 0 | b j | N > = | N − > , b † j | N > = 0 b † i b j + b j b † i = δ ij b † i b † j + b † j b † i = 0 b i b j + b j b i = 0Now we can also represent Slater determinant with creation and annihilationoperators, √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ ( r ) φ ( r ) φ ( r ) φ ( r ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≡ | φ φ | = b † i b † j | > Also b † i b † j | > = − b † j b † i | > Now we can write kinetic energy operator as b T = X i,j < i | T | j > b † i b j Similarly, the potential energy operator can be written as, b V = 12 X i,j,k,l < ij | V ( r , r ) | kl > b † i b † j b l b k Now if we define field operator asΨ( r ) = X i b n i φ ni ( r )Ψ † ( r ) = X i b ⋆n i φ † ni ( r )then < i | b T | j > = Z d rφ n i † ( r ) T ( r ) φ ni ( r ) < ij | b V | kl > = Z Z d r d r φ † n i ( r ) φ † n j ( r ) V | r − r | φ n i ( r ) φ n i ( r )If we consider Bloch wave function i.e φ ( r ) = e i k · r √ V u kn ( r ), where u kn ( r ) signifies particleis in periodic potential the Hamiltonian will be, c H = X k,σ ǫ k b † kσ b kσ + 12 X q,σ,σ ′ V q b † k − q,σ b k ′ + q,σ b k ′ σ ′ b kσ this is second quantized Hamiltonian in many electron system.As for example if we consider ǫ k = − t P σ e i k · δ , where t is hopping strength and δ isnearest neighbour where a electron can hop. If our system is a two dimensional square4attice with lattice parameter a then, δ = ± b xa, ± b ya ,So our energy dispersion relation will be then ǫ k = − t ( e ± ikx + e ± iky ) = − t (cos k x + cos k y ) In tight binding model the Hamiltonian describes the kinetic energy (hopping) of elec-trons for nearest neighbour pairs. c H = − t X ,σ ( c † iσ c jσ + c † jσ c iσ )In this approximation we consider the wave function of the electrons are sharply lo-calized neglecting any overlap between them and they are confined in the lattice sitesby an infinite potential barrier[4]. From here on we like to fix the notation of creationand annihilation operator as c i , c † j respectively. The simplest model in quantum many body theory is isotropic spin half Heisenbergchain. The Heisenberg Hamiltonian is given by: c H = J X ( S i S j −
14 )where S j is a local spin variable at j th. state. For antiferromagnets J >
0. The sumis over the distinct nearest neighbours. For spin 1/2 particle the spins are representedby Pauli’s spin matrices.
The Hubbard model describes the strongly correlated electron systems. The model inmore than one dimension has not been solved. In describing the
CuO planes in hightemperature superconductivity Hubbard model is a good starting point[5]. The basicingredients of Hubbard model are: • The kinetic energy (electron hopping) delocalizes the electron in Bloch state,leading to metallic behavior.[6] • The electron electron interaction (approximated by onsite Coulomb interaction)wants to localize the electron on to sites.[6]5he Hubbard model contains only one orbital per site and is defined as (consideringnearest and next nearest neighbour interactions)[5]: H = − t X < i , j >,σ ( c † i σ c j σ + c † j σ c i σ ) − t ′ X < i , i ′ >,σ ( c † i σ c i ′ σ + c † i ′ σ c i σ ) + U X i n i ↑ n i ↓ If we only take into account the nearest neighbour interaction it reduces to: H = − t X < i , j >,σ ( c † i σ c j σ + c † j σ c i σ ) + U X i n i ↑ n i ↓ where c † iσ are creation operators and n iσ = c † iσ c iσ are occupation number operator. Bymeans of on site Coulomb U the singlet band of the Hubbard model is split into a lower(LHB) and an upper Hubbard band (UHB). But the validity of the three band modelto the single band model is still controversial [5]. It has been questioned whether thestrong coupling version of the Hamiltonian, i.e the t-J model, is appropriate to describecorrectly the low energy physics of the original three band model. In the large U limitand at half filling (one electron per site) the Hubbard ladder is equivalent to the spin1/2 Heisenberg ladder [7]. c H = J X S i · S j When the Hubbard ladder is doped with holes away from half filling, its strong couplingdescription is modified from the Heisenberg model to the t-J model with the constrainof no doubly occupied sites[7]. H = − t X < i , j >,σ ( c † i σ c j σ + c † j σ c i σ ) + J X < i , j > ( S i · S j −
14 n i n j ) The most important local interactions in a doped antiferromagnet are well representedby the large U Hubbard model, the t-J model, and their various relatives. To beconcrete we will focus on the t-J-U model. The t-J-U Hamiltonian is written as: c H = − t X X σ ( c † iσ c jσ + c † jσ c iσ ) + J X ( S i · S j − n i .n j U X i n i ↑ n i ↓ It is a close variant of the familiar t-J model where the sites of the lattice is strictlyprohibited against double occupancy, i.e the doubly occupied sites are projected out.The ”no double occupancy” restriction is suitably achieved by using ”constrained”fermionic operators. The same physics can be achieved by using a t-J-U model givenby the limit U → ∞ we go back to simple t-J model. The exchange integral J arisesthrough virtual processes where in the intermediate state has a doubly occupied site,producing an antiferromagnetic coupling. Dopping is assumed to remove electronsthereby producing a ”hole” or missing spin which is mobile because neighbouringelectrons can hop into its place with amplitude t. So among the other models tostudy the correlation effects in the high temperature superconductors this model issimplest one and gives a exact critical value of J ( J c ) such that J ≥ J c two electronscan form a two particle bound state. 6 .3 Construction of Equation of Motion (EOM) for Two Par-ticle System: The system consisting of two electrons the wave function can be written as | Ψ i = X i ,i Φ( i , i ) c † i ↑ c † i ↓ | > (1)and the model Hamiltonian isˆ H = − t X X σ ( c † iσ c jσ + c † jσ c iσ ) + J X ( S i .S j − n i .n j U X i n i ↑ n i ↓ (2)where | i denotes the vaccum state. For a two body problem the ground state isasinglet i.e Φ( i , i ) = Φ( i i ) and we know that H | Ψ i = E | Ψ i so the equation ofmotion can be written as[ − t X X σ ( c † iσ c jσ + c † jσ c iσ )+ J X ( S i .S j − n i .n j U X i n i ↑ n i ↓ ] X i ,i Φ( i , i ) c † i ↑ c † i ↓ | > = E X i ,i Φ( i , i )(3)ˆ t | Ψ i = − t X X σ ( c † iσ c jσ + c † jσ c iσ ) X i ,i Φ( i , i ) c † i ↑ c † i ↓ | i = − t X X i ,i Φ( i , i )( c † i ↑ c j ↑ c † i ↑ c † i ↓ | i + c † j ↑ c i ↑ c † i ↑ c † i ↓ | i + c † i ↓ c j ↓ c † i ↑ c † i ↓ | i + c † j ↓ c i ↓ c † i ↑ c † i ↓ | i )= X j Φ( j, i ) c † i ↑ c † i ↓ | i + 0 − X j Φ( i , j ) c † i ↓ c † i ↑ | i + 0ˆ t | Ψ i = − t [ X j Φ( j, i ) c † i ↑ c i ↓ | i − Φ( i , j ) c † i ↓ c † i ↑ | i ] (4)ˆ U | Ψ i = U X X i ,i Φ( i , i ) c † i ↑ c i ↑ c † i ↓ c i ↓ c † i ↑ c † i ↓ | i = U X i ,i Φ( i , i ) δ i i c † i ↑ c † i ↓ | i (5)ˆ J | Ψ i = J X X i ,i Φ( i , i )( S i · S j − n i .n j c † i ↑ c † i ↓ | i = J X i ,i Φ( i , i ) c † i ↑ c † i ↓ | i (6) E Φ( i , i ) = X j [ t i j Φ( j, i ) + t i j Φ( i , j )] + [ U δ i ,i − J i ,i ]Φ( i , i ) (7)Fourier transform of the equation yields E Φ( k , k ) = [ t ( k )+ t ( k )]Φ( k , k )+ UN X k Φ( k + k , k − k ) − N X k J (k)Φ( k − k , k + k )(8)7here, Φ( k , k ) = 1 N X i ,i Φ( i , i ) e − i k · r i1 − i k · r i2 t ( k ) = 1 N X i,j t ij e − k · ( r i − r j ) = − t (cos k x + cos k y ) J ( k ) = 2 J (cos k x + cos k y )taking the lattice constant 1. Since the system is translationally invariant, the totalmomentum can be used to specify its eigenstates. let us define Q = k + k , q = ( k − k ), and Φ( k , k ) = Φ Q ( q )then we obtainΦ Q ( q ) = UN P k Φ Q ( k ) − N P k J ( q − k )Φ Q ( k ) E − t ( Q + q ) − t ( Q − q ) (9)This is the starting point of our analysis. t tU U Fig 1For singlet pairing we can take Q=0 so from eq. (9) can be decoupled so that we canwrite, C = U C I − J I x C x (10) C x = U C I x − J I xx C x (11)where C = 1 N X q Φ ( q ) C x = 1 N X k cos k x Φ ( k ) I = 1 N X q E + 4 t cos q x I x = 1 N X q cos q x E + 4 t cos q x xx = 1 N X q cos q x E + 4 t cos q x eq. (11) and (12) can be written in matrix form as follows: U I − − J I x U I x − J I xx − ! C C x ! = 0 (12)In eq. (12) unique solution of C and C x will exit if and only if the determinant of thecoefficient is zero. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U I − − J I x U I x − J I xx − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 (13)Now solving eq. (13) for J − J = 1 − U I I x U + I xx (1 − U I ) (14)Using the lattice symmetry we can write I = −
14 1 √ α − I x = 14 α + √ α − √ α − I xx = − α ( α + √ α − √ α − I x = 14 t − E t I (15) I xx = − E t I x (16)substituting these values with the limit as I → ∞ and U → ∞ in eq. (14) we get J c = If we think in terms of the energy band in one dimension it is 8t for our system, so thebound state energy E of the two particle system can be obtained via equation (14),(15), (16) is we can write 8 tJ = 1 tI − Et (17)Taking t = 1 we can write the energy equation as E = 1 I − t J (18)9 | E _b s | /t J/tBound State Energy of One Dimensional Lattices
Fig 210 .2 Two Dimensional Square Lattice:
U Ut tt t U tt t tU UtUt
Fig 3In a similar fashion as we have done in the previous section we can write the followingequations: C = U I C − J I x C x − J I y C y (19) C x = U C I x − J C x I xx − J C y I xy (20) C y = U C I y − J C x I xy − J C y I yy (21)where C = 1 N X q Φ ( q ) C x = 1 N X k cos k x Φ ( k ) C y = 1 N X k cos k y Φ ( k ) I = 1 N X q E + 4 t (cos q x + cos q y ) I x = 1 N X q cos q x E + 4 t (cos q x + cos q y )11 xx = 1 N X q cos q x E + 4 t cos q x I y = 1 N X q cos q y E + 4 t (cos q x + cos q y ) I xy = 1 N X q cos q x cos q y E + 4 t (cos q x + cos q y ) I yy = 1 N X q cos q y E + 4 t (cos q x + cos q y )eq. (20),(21),(22) can be written in a matrix form as follows: U I − − J I x − J I y U I x − J I xx − − J I xy U I y − J I xy − J I yy − C C x C y = 0 (22)In eq. (23) unique solution of C , C x and C y will exit if and only if the determinant ofthe coefficient matrix is zero. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U I − − J I x − J I y U I x − J I xx − − J I xy U I y − J I xy − J I yy − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 (23)For an isotropic square lattice symmetry permits us to write I y = I x , I yy = I xx and C y = C x hence eq. (23) becomes U I − − J I x − J I x U I x − J I xx − − J I xy U I y − J I xy − J I xx − C C x C x = 0 (24)and our modified determinant will be, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U I − − J I x − J I x U I x − J I xx − − J I xy U I x − J I xy − J I xx − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 (25)Now our motivation is to take U → ∞ limit to project out the possibility of doubleoccupancy. The value of I , I x , I xx , I xy after integration are as follows: I = 12 K ( − α ) πα (26) I x = 12 (1 − α ) K ( − α ) πα −
12 ( α − α , − α ) π (27) I xx = 14 ( α − α + 2) K ( − α ) πα −
12 ( α − α , − α ) π + 14 αE ( − α ) π (28)12 xy = −
12 (1 − α ) K ( − α ) π −
12 ( α − α + 1) K ( − α ) πα − αE ( − α ) π (29)These expression can be written in a simple form as: I xx + I xy = − E t I x (30) I x = 18 t − E t I (31) I = 1 E π K ( 8 tE ) (32)So the lattice integrals are turned out in terms of complete elliptic integral of first kind( K ( k )), second kind ( E ( k )), third kind (Π( ν, k )) respectively. Π( ν, k ) can be evaluatedthrough K ( k ), E ( k ), F ( q, φ ), E ( q, φ ), where F ( q, φ ) and E ( q, φ ) are the incompleteElliptic integrals of first and second kind respectively[8]. The nature of K ( k ) and E ( k )is shown as follows: Fig 4So from the physical nature of the elliptic integral of the first kind we can conclude thatit is of logarithmic diverging nature so to tackle this problem let us consider α = − − δ and expand the elliptic integrals w.r.t δ then take δ → lnδ term should be equal to zero.So − J π + 18 π − Jπ + J π = 0The solution of the quadratic equation for J= 2t and .13 ound state energy of electron in two dimension
From eq. (24) we can write − J = 1 − U I U I x + ( I xx + I xy )(1 − U I ) (33)From eq. (31), (32), (33), and (34) the bound state energy E of the two electron systemcan be obtained via 16 tJ = π Et K ( tE ) − Et (34)Taking t = 1 we rewrite the equation as: E = 16 J ( π K ( α ) −
1) (35) E b s /t Plot 2
Bound State Energy vs. J
Fig 514 .3 Two Leg Ladder: t t UU U Ut ttU U
Fig 6So a ladder like lattice involves a one dimensional wave vector integral ( viz over q x )rather that two dimensional integral (over q x and q y ). More precisely the lattice inte-grals appearing in the calculation of bound states are expressed as, X q = 12 X q y =0 , π π Z π − π dq x The various lattice integrals are as follows: I = 14 Z π − π dxα + 1 + cos q x + 14 Z π − π dxα − q x (36) I x = 14 Z π − π cos q x dxα + 1 + cos q x + 14 Z π − π cos q x dxα − q x (37) I xx = 14 Z π − π cos q x dxα + 1 + cos q x + 14 Z π − π cos q x dxα − q x (38)where α = E t from eq. (13) with t = 1 and U → ∞ we can write, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) I − J I x I x − J I xx − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 (39)expanding the determinant we obtain2 J I x − J I I xx − I = 0 (40)We have substituted the integral values in eq. (42) and also α is replaced by = − δ − δ we got a term of √ δ . This term will divergeas soon as we will take the limit as δ →
0. So collecting the coefficients of this termand equating them to zero the critical value of J is obtained which is . Bound state energy of Two leg ladder
Directly from the integrals one can find out the bound state energy for pairing ofelectrons with proper substitution. From the plot also we can verify that the minimumenergy required for formation of bound state among two electrons in a ladder likelattice is , considering t=1. 15 | E _b s | /t J/tBound State Energy of Two Dimensional Ladder Lattices
Fig 716 .4 Three Dimensional Cubic Lattice:
UU UU Utt t tt tt t t
Fig 8From eq.(9) considering that the lattice is isotropic (Q=0), we can writeΦ ( q ) = UN P k Φ ( k ) − N P k J ( q − k )Φ ( k ) E − t ( q ) (41)Φ ( q ) = UN P k Φ ( k ) − N P k J ( q − k )Φ ( k ) E + 4 t (cos q x + cos q y + cos q z ) (42)This equation can be written as follows: C = U C I − J C x I x − J C y I y − J C z I z (43) C x = U C I x − J C x I xx − J C y I xy − J C z I xz (44) C y = U C I y − J C x I yx − J C y I yy − J C z I yz (45) C z = U C I z − J C x I zx − J C y I zy − J C z I zz (46)where C = 1 N X q Φ ( q ) C x = 1 N X k cos k x Φ ( k )17 y = 1 N X k cos k y Φ ( k ) C z = 1 N X k cos k z Φ ( k ) I = 1 N X q E + 4 t (cos q x + cos q y + cos q z ) I x = 1 N X q cos q x E + 4 t (cos q x + cos q y + cos q z ) I xx = 1 N X q cos q x E + 4 t (cos q x + cos q y + cos q z ) I y = 1 N X q cos q y E + 4 t (cos q x + cos q y + cos q z ) I xy = 1 N X q cos q x cos q y E + 4 t (cos q x + cos q y + cos q z ) I yy = 1 N X q cos q x E + 4 t (cos q x + cos q y + cos q z ) I z = 1 N X q cos q z E + 4 t (cos q x + cos q y + cos q z ) I zz = 1 N X q cos q z E + 4 t (cos q x + cos q y + cos q z ) I zx = 1 N X q cos q z cos q x E + 4 t (cos q x + cos q y + cos q z ) I yz = 1 N X q cos q y cos q z E + 4 t (cos q x + cos q y + cos q z )eq. (44), (45), (46), (47) can be written in a matrix form as follows: U I − − J I x − J I y − J I z U I x − J I xx − − J I xy − J I zx U I y − J I xy − J I yy − − J I yz U I z − J I zx − J I yz − J I zz − C C x C y C z = 0 (47) J c /t can be found out by vanishing of the determinant by choosing an energy E to beslightly between noninteracting 2 electron band. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U I − − J I x − J I y − J I z U I x − J I xx − − J I xy − J I zx U I y − J I xy − J I yy − − J I yz U I z − J I zx − J I yz − J I zz − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 018ince we are employing the constrain that there can not be any doubly occupied sitesso considering U → ∞ , our vanishing determinant be as follows: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) I − J I x − J I y − J I z I x − J I xx − − J I xy − J I zx I y − J I xy − J I yy − − J I yz I z − J I zx − J I yz − J I zz − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 (48)Now because of lattice symmetry we can write I x = I y = I z I xx = I yy = I zz I xy = I yz = I zx Also these lattice integrals can be written as: I x = 13 ( 14 t − E t I ) (49) I xx = − E t I x − I xy (50)So we need to find out I and I xx or I xy . Solution of Lattice Integrals
Previously mentioned lattice integrals are commonly known as lattice Green’s function.The most general representation of lattice green’s function is,[9] G ( s, l, m, n ) = 1 π Z π Z π Z π cos lx cos my cos nzdxdydzs − (cos x + cos y + cos z ) G ( s ) = G ( s, , ,
0) = 1 π Z π Z π Z π dxdydzs − (cos x + cos y + cos z )This integral defines a single valued analytic function G(s) in the complex s-plane cutalong the real axis from -3 to +3. in most physical applications one usually requiresthe limiting behavior of the Green function G(s) as s approaches the real axis[10].Another way we can write as[11]13 G ( s ) = 13 1 π Z π Z π Z π dxdydzs − (cos x + cos y + cos z ) G ( s ) = 1 π Z π Z π Z π dxdydzs − (cos x + cos y + cos z ) (51)The series representation of the Green’s function is[11]: G j ( s ) = 1 s ∞ X n =0 p ( j ) n s n (52)19here j=1, 2, 3, 1 ≤ | s | < ∞ and p ( j ) n = 1 π Z π Z π Z π [ λ j ( θ , θ , θ )] n dθ dθ dθ Now for simple cubic lattice λ ( θ , θ , θ ) = (cos θ + cos θ + cos θ ) The recurrencerelation for simple cubic lattice coefficient p n is36( n + 1) p (1)2 n +2 − n + 1)(10 n + 10 n + 3) p (1)2 n + n (4 n − p (1)2 n − = 0 (53)from eq. (52) and (53) G ( s ) can be written as a solution of the linear third orderdifferential equation( s − s − d G ds + 6 s (9 s − d G ds + 3(21 s − dG ds + 9 sG = 0 (54)So G ( s ) can be written as, G ( s ) = 1 − ξ s (1 − ξ ) (1 + 3 ξ ) [ 2 π κ ( k )] (55)where, k = vuut ξ (1 − ξ ) (1 + 3 ξ ) ξ = ξ ( s ) = s − s ! − / − s − s ! / and κ ( k ) is the complete elliptic integral of the first kind.Also G (1) = 3(18 + 12 √ − √ − √ π κ ((2 − √ √ − √ (56)So from this knowledge with proper adjustment of coefficients we can write: I = 112 t G ( s ) (57)Here we like to mention that elliptic integrals diverges for modulus of 1, the physicalsignificance of the divergence in our work is the electrons are on the energy band there-fore in our earlier analysis we had taken care of that divergence keeping our calculationjust below the band, but here when we substitute our parameters as electrons are onthe energy band we found the modulus is not 1 therefore there is no any sharp diver-gence of elliptic integral rather there is a local maxima, and we are returning with anumerical value of the integral on the band. Its a striking feature of the elliptic integralfor simple cubic lattice. So from this knowledge we can evaluate I . But one of theintegral among I xx or I xy are still to be evaluated. To evaluate our integrals we hadtaken help from the recurrence relation of the Green function. With the knowledge ofthe recurrence relation for fcc lattice[12] and triangular lattice[13], we have constructed20he recurrence relation for simple cubic lattice. For nearest neighbour (6 nn for S.C)interaction only the recurrence is as follows: G ( l +1 , m, n )+ G ( l − , m, n )+ G ( l, m +1 , n )+ G ( l, m − , n )+ G ( l, m, n +1)+ G ( l, m, n − δ l δ m δ n − sG ( l, m, n ) (58)where, G ( l + 1 , m, n ) = 1 π Z π Z π Z π cos ( l + 1) x cos my cos nzs + (cos x + cos y + cos z ) dxdydz and so on, Let l=1, m=0, n=0 then G (2 , ,
0) + G (0 , ,
0) + G (1 , ,
0) + G (1 , − ,
0) + G (1 , ,
1) + G (1 , , −
1) = − sG (1 , , G (1 , ,
0) = G (1 , − ,
0) = G (1 , ,
1) = G (1 , , −
1) Hence, G (2 , ,
0) + G (0 , ,
0) + 4 G (1 , ,
0) + 2 sG (1 , ,
0) = 0Adjusting the coefficients for our calculation, I xx = − E t I x − I xy (59)where, I xx = t ( G (2 , ,
0) + G (0 , , I = t G (0 , , I x = t G (1 , ,
0) Solving eq.(50) and (59) I xy is determined. I xy = − E t I x − I (60)Let us now summaries the lattice integral values: I = 112 t G ( s ) , s = E tI x = 13 ( 14 t − E t I ) I xy = − E t I x − I (61) I xx = − E t I x − I xy Substituting eq. (60) in eq. (48) and solving for J, the values are obtained as , , . In our entire work we were involved in finding out the critical values of J ( J c ) for whichbounstate formation of two electron system is possible, and our results obtained areas follows: 21attice Type J c ValuesOne Dimensional Chains 2tTwo Dimensional Square Lattice 2t, 7.32tTwo Leg Ladder 2tThree Dimensional Cubic Lattice 2t, 2t, 7.88tAlso we have plotted J c /t Vs. | E bs | /t for one dimensional chain, two dimensionalsquare lattice and two leg ladder.For one dimensional chain like system has been already solved for J=2t[14, 15].For square lattice the critical value of J=2t can seen in papers of Lin [1, 2]. In ourwork we are proposing that for two leg ladder lattice the critical value of J is 2t, andit is quite desirable because two leg ladder is a simplified form of n leg ladder systemwhich is nothing but a 2 dimensional system.We are also proposing for the first time that for three dimensional lattice the criticalvalue of J is 4.Let us now try to understand the significance of the multiple value of two dimensionaland three dimensional lattice for the basis of group theory. Square Lattice
We know that the symmetries of a square lattice is represented by C V group. Thecharacter table for C V group is as follows:Characters E C C C m x m y σ u σ v χ (1) χ (2) χ (3) χ (4) χ (5) (1) (2) xy Γ (3) x − y Γ (4) xy ( x − y )Γ (5) ( x, y )The possible pairing symmetries correspond to these irreducible representation and tothe basis functions. In case of square crystals the singlet orders are called s , d x − y , d xy and g . The corresponding basis functions are 1, x − y , xy , xy ( x y ). It is reasonableto classify order parameters as ”s-wave like” and ”d-wave-like”. So in our analysisJ=2t corresponds to s-wave and J=7.32t corresponds to d-wave[7]. Simple Cubic Lattice
It is known that the simple cubic lattice corresponds to O h group. The character tablefor O h group is as follows[17]: 22haracters E 8 C C C C (= C ) i S S σ h σ d A g A g E g T g T g A u A u E u T u T g ( A g ) 1Γ ( A g ) ( k x − k y )( k y − k z )( k z − k x )Γ ( E g ) 2 k nz − k nx − k ny , k nx − k ny n = 2; 4Γ ( T g ) k x k y ( k nx − k ny ) , k y k z ( k ny − k nz ) , k z k x ( k nz − k nx ) n = 2; 4; 6Γ ( T g ) k x k y k nz , k y k z k nx , k z k x k ny n = 0; 2; 4So in a cubic lattice, Γ ( A g ) represents s-wave pairing, Γ ( E g ) (for n=2) and Γ ( T g )(for n=0) represents d-wave pairing[7]. Also Γ ( E g ) is doubly degenerate. So fromthe evaluated value of of J, we can conclude that J=2t value corresponds to Γ ( E g )representation and J=7.88t corresponds to Γ ( A g ) representation. The analytical result of J for simple cubic lattice is a new result. The result can becross examined by bilayer consideration of the three dimensional cubic lattice. Thebilayer consideration is analogous with the two leg ladder consideration. Physicallyit can be vilualised that, n number of layers togather is forming a three dimensionalstructure provided the separation between the layers is sufficiently small. So startingthe calculation for bilayer one can extend the calculation to n number of layer whichcan give the similar result as of the three dimension. A natural extrapolation can beinvestigating superconductivity in these lattices and studying the phase diagram.
References [1] V.J.Emery, S.A.Kivelson and H.Q.Lin, Phys. Rev. Lett. , 475 (1990)232] H.Q.Lin, Phys. Rev. B , 4674 (1991)[3] Creation and Annihilation Operators by J.Avery, McGRAW-HILL InternationalCompany c (cid:13) (cid:13) , 219-351 (1994)[6] Lecture Notes on Electron Correlation and Magnetism by P.Fazekas, World Sci-entific c (cid:13) (cid:13) Springer-Verlag Berlin Heidelberg 2004[8] An Atlas of Functions by Jerome Spanier and Keith B. Oldham, HemispherePublishing Corporation, 1987[9] T. Morita and T. Horiguchi, Journal of Physical Society of Japan, Vol. , 957-964(1971)[10] . S. Joyce, Phil. Trans. R. Soc. (Lond.)A , 705-708 (1974)[13] T. Horiguchi, J. Math. Phys., Vol. , 1411-1419 (1972)[14] P. Schlottmann., Phys. Rev. B, Vol. , 5177, (1987).[15] Exactly Solvable Models of Strongly Correlated Electrons edited by Vladimir E.Korepin, Fabian H. L. E β ler, World Scitific Publishing Co. Pte. Ltd. c (cid:13) (cid:13) (cid:13)(cid:13)