Introduction to Hubbard Model and Exact Diagonalization
aa r X i v : . [ c ond - m a t . s t r- e l ] J u l Introduction to Hubbard Model and Exact Diagonalization
S. Akbar Jafari
Department of Physics, Isfahan University of Technology, Isfahan 84156, Iran
Hubbard model is an important model in theory of strongly correlated electron systems. In thiscontribution we introduce this model along with numerically exact method of diagonalization of themodel.
PACS numbers:
I. INTRODUCTION
Hubbard model and its variants constitute an im-portant research topic in theoretical condensed matterphysics, particularly in the context of strongly corre-lated electron systems. Most of the many-body tech-niques commonly used in condensed matter physics canbe learnt in this context. Also there are some theoreticaltools and concepts which apply to this model only.There are already some monographs , which can beused by experts, along with some text books whichcan be consulted for further details of the various meth-ods. In this contribution our aim is to provide a smoothintroduction to the model and exact diagonalization tech-nique used in dealing with Hubbard model. This work isbased on the set of lectures given by the author in the firstIUT school on strongly correlated electron systems.Analytical methods of solving the Hubbard model areall approximate, except in 1D, where the so called Betheansatz provides an exact solution . On the other handthere are exact numerical techniques, which are however,either computer time expensive or memory expensive.Therefore one is limited to rather small cluster sizes.A popular method to solve Hubbard model and manyother models in condensed matter physics is exact diag-onalization (ED) of the models for small clusters whichwe will study at length in this set of lectures. We en-courage the reader to implement the method presentedhere in a simple fortran program. In numerically exactdiagonalization method one gets the ’exact’ results at ahigh price, namely limitation to very small cluster sizes(about 18 sites for Hubbard model at half filling), whichis essentially due to limited amount of computer RAMone can typically have. If one accepts some error bars innumerical results (which can however be systematicallyimproved), then a family of the so called Monte Carlo methods are methods of choice . These methods are es-sentially exact. The accuracy of the results depends onhow much computer time one would likes to spend. Inthis sense, these family of methods are time expensive,while ED is memory expensive.One of the important methods to deal with almost anymodel is Mean Field Theory (MFT). MFT ignores quan-tum fluctuations; hence becoming less accurate in lowerspatial dimensions. Despite this, mean field treatmentreveals the wealth of various condensed matter phasescan emerge from a simple Hubbard model . In MFT one ignores both spatial and temporal fluctuations. It is pos-sible to retain the temporal fluctuations by performinga full fledged quantum dynamics of the problem. Thisis the subject of the so called ’dynamical’ mean fieldtheory (DMFT) .There are also a class of approximate analytic methodsknown as auxiliary particle or slave particle meth-ods, which are invented to deal with the large U limitof the Hubbard model . This type of techniques are re-lated to the so called Gutzwiller projection which aredevised to obtain approximate ground state of the Hub-bard model at half filling. Generalization of this methodto deal with excited states is also there in the market .In the limit of large U , the charge fluctuations in theHubbard model are frozen and only the spin of electronscan fluctuate. Thereby reducing the physics of the Hub-bard model to spin physics described by the so called t-JModel . II. HUBBARD MODELA. Non-interacting electrons
The Hamiltonian of a system of non-interactingfermions on a lattice of L sites labeled by i, j , etc canbe represented in second quantization by H = X ij t ij c † i c j where c † j ( c j ) creates (annihilates) a fermion in a single-particle orbital φ j localized at site j . In condensedmatter applications one can assume φ j ’s to be Wan-nier wave functions (Fourier transform of Bloch orbitals).Fermionic operators satisfy the anti-commutation rela-tions, { c † i , c j } = δ ij , others = 0 . (1)The coefficients t ij characterize the single-particle matrixelements t ij = h φ i | (cid:18) − ~ ∇ m + ˆ v (cid:19) | φ j i = Z dxφ i ( x ) (cid:18) − ~ ∇ m + v ( x ) (cid:19) φ j ( x ) (2)For many practical purposes it suffices to assume that t ij is none-zero, only when i, j are nearest neighbors in whichcase it is usually denoted by − t , so that the Hamiltonianwritten in manifestly hermitian format becomes H = − t X h i,j i c † i c j + c † j c i (3)Assuming the periodic boundary conditions (PBC), theHamiltonian of the system will be invariant under trans-lation. The irreducible representations of the translationgroup (due to abelian structure of the group), are one-dimensional (i.e. numbers of type e iθ ). Hence the one-particle (or non-interacting=free) Hamiltonian (3) can bediagonalized by a Fourier transformation ( θ ↔ kj ) c † k = 1 √ L X j e ikj c † j , (4)Then the Hamiltonian in | φ k i basis becomes H = X k ε k c † k c k , ε k = − t cos k, (5)where ε k determines a cosine dispersion and represents aband structure with L allowed k values in the first Bril-louin zone. Of course a simple cosine may not be a goodapproximation of the realistic band structure of solids(spaghetti plots). To mimic the realistic band structures,one can add further neighbors’ hoppings which generatehigher harmonics of the simple cosine band. - π - π/2 π/2 π k-2-1012 ε / t ε=−2 t cos(k)
1D tight-binding band dispersion
FIG. 1: A tight-binding band dispersion in 1D with electron-like Fermi surface.
If the number N of the electrons is equal to number L of the sites, then each allowed k state can be occupiedby two ↑ and ↓ spins. Hence the ground state of H isconstructed by filling the lower half ( ε k <
0) of the banddispersion of Fig. 1 which is denoted by circles on the fig-ure. Since half of the band is filled, the N = L situationis called half-filling . This state is known as a Fermi sea state, usually denoted by | FS i . Second quantized repre-sentation of this state is: | FS i = Y k
2. This excited with a hole left behindin state k and an electron created in state k + q with, say ↑ spin is called a ’particle-hole’ excitation: H | ψ n i = H c † k + q ↑ c k ↑ | FS i = ε ph k,q | ψ n i (9)which is again an eigen state of H with energy ε ph k ( q ) = ε k + q − ε k . This excitation carries a center of mass mo-mentum q . The above discussion can be straightfor-wardly generalized to higher dimensions.Evidently the half-filled state | FS i characterizes ametal, as the energy of the particle-hole excitation can bemade arbitrarily small. The above band picture alwaysgives a metal for an odd (in this case one) number ofelectrons per unit cell. However, as we will argue in thefollowing, there might be situations in which this simpleprediction of the band theory fails drastically. B. Electron-electron interaction
When we have only H (Eq. 3) in the Hamiltonian,then the minimization of energy is achieved by filling the k states independently with electrons of opposite spins.Such k space picture in real space translates to equalprobabilities p = 1 / (cid:13) (cid:13)↑ (cid:13)↓ (cid:13)↑↓ (10)The most general form of interaction in second quanti-zation representation which can be added to H is of theform V = 12 X µναβ V µνβα c † µ c † ν c α c β (11)where α = { α, i } is a collective name for the site index i , and spin index σ the two-particle matrix elements isgiven by V µνβα = Z dxdx ′ ψ ∗ µ ( x ) ψ ∗ ν ( x ′ ) V ( | x − x ′ | ) ψ β ( x ) ψ α ( x ′ )(12)usually in metals with appreciable density of states D ( ǫ F )at the Fermi level, the Coulomb potential V ( | x − x ′ | = r )is screened and obtains the form V ( r ) = e − rk TF r , (13)where k − is the so called screening length. For the d electron systems where the overlap between the atomicwave functions is small (smaller t ), one has narrower bandwhich is synonymous to larger DOS at the Fermi levelthe screening length is usually on the scale of the Bohrradius a B . Therefore the most important term amongall possible µνβα matrix elements is when all the indicescorrespond to the same site j . In such case, the Pauliprinciple forces µ = α = ↑ and β = ν = ↓ . Denoting thecorresponding matrix element with − U one gets for thescreened interaction V = − U X j c † j ↑ c † j ↓ c j ↑ c j ↓ = U X j n j ↑ n j ↓ . (14)If we add this term to H , we obtain the celebrated Hubbard model H = − t X h i,j i σ (cid:16) c † iσ c jσ + c † jσ c iσ (cid:17) + U X j n j ↑ n j ↓ . (15)For the Hamiltonian in U → ∞ limit the doubly occu-pied configuration of a single site in Eq. (10) is going tocost a large energy U n j ↑ n j ↓ = U (1)(1) for each doublyoccupied site. Therefore presence of such term violatesthe equi-probability of four possible states of a single-site,thereby inducing some kind of correlation by minimizingthe double occupancy. Therefore at half-filling, and forlarge U the ground state charge distribution adjust itselfto avoid doubly occupied sites as much as possible; i.e.each site is occupied by a single electron, and that more-over, large value of U makes charge fluctuations around n jσ = 1 configuration very expensive. Therefore chargefluctuations are frozen and one has an insulator knownas Mott insulator .For finite values of U , the two terms of the Hamil-tonian (15) compete with each other. The kinetic en-ergy term (corresponding to U = 0) tends to delocalizeelectrons by putting individual electrons in Bloch states. This limit known as band limit always describes a metal.The U term on the other hand increases the cost of chargefluctuation, leading to an insulator in the opposite limit U → ∞ . Therefore there must be a critical value U c ofthe order of band width W = 2 zt (where z is the coordi-nation number), beyond which one has an insulator. This(first oder at zero temperature) phase transition is knownas Mott metal-insulator transition (MIT). The only tech-nique which can handle this model at arbitrary values of U and for arbitrary filling is ED, which will be describedin next section. III. EXACT DIAGONALIZATION
The easiest way to describe the essence of this methodis by the example of a two-site Hubbard model. Thistoy model consists in two sites labeled 0 ,
1. In this casethe Hubbard model written explicitly (in units in which t = 1) reads: H = H t + H U = − (cid:16) c † ↑ c ↑ + c † ↑ c ↑ + c † ↓ c ↓ + c † ↓ c ↓ (cid:17) + U ( n ↑ n ↓ + n ↑ n ↓ ) (16)where sites are labeled as follows: (cid:3) |{z} (cid:3) |{z} where the site index increases from right to left. A. Organizing the Hilbert space
To organize the Fock space for this Hamiltonian, onefirst notes that the number operator N = P jσ n jσ com-mutes with the Hamiltonian. Therefore one can consideronly Hilbert space corresponding to a fixed value of N .Let us consider N = 2 for this toy model which corre-sponds to half-filling condition. The next question ariseswith regard to the total spin of the electrons: whetherthey are ↑↑ , ↓↓ or ↑↓ ? First two cases represent tripletstate, while the last one corresponds to S = 0 (more pre-cisely ( ↑↓ − ↓↑ ) is a singlet). Formally it can be checkedthat the total S z = 1 / P j ( n j ↑ − n j ↓ ) also commuteswith the Hamiltonian and hence a conserved quantity.Therefore there would be no matrix element of the Hamil-tonian connecting sections of the Hamiltonian with dif-ferent values of S z . The structure of the Hamiltonianwill be block diagonal where each block corresponds to afixed value of S z . To see this block-diagonal structure, weconfine ourselves to N = 2 with both triplet and singletspins.In sector with quantum numbers N = 2 and S z = 0,Hilbert space is six dimensional with six possible basisstates | φ J i with J = 1 . . . J algebraic picture binary I ↓ I ↑ I c † ↑ c † ↑ |i ↑ ↑ | i ↓ | i ↑ c † ↓ c † ↑ |i (cid:13) ↑↓ | i ↓ | i ↑ c † ↓ c † ↑ |i ↑ ↓ | i ↓ | i ↑ c † ↓ c † ↑ |i ↓ ↑ | i ↓ | i ↑ c † ↓ c † ↑ |i ↑↓ (cid:13) | i ↓ | i ↑ c † ↓ c † ↓ |i ↓ ↓ | i ↓ | i ↑ • The second column labeled ’beginner’ is a way abeginner would define and work with a set of 6 basisstates | φ i to | φ i . • The third column labeled ’expert’ is a way an ex-pert works with these basis states. • Columns number 4 to 7 are related to the way acomputer organizes and works with the basis states.Explanation of each columns is as follows:1. In the first column there is an integer J = 1 . . . |i . As an examplelook at the first basis | φ i : Physically, it describestwo ↑ electrons in sites number 0 ,
1. Therefore wehave two choices: | φ a i = c † ↑ c † ↑ |i , or | φ b i = c † ↑ c † ↑ |i . These two choices by Fermionic anti commutationrelations are negative of each other: | φ a i = −| φ b i .Then the question is: which one is correct way ofrepresenting | φ i ? The answer is that, both of themare fine, the same way one can describe ordinary 3dimensional vector space with basis ˆ e = ˆ x, ˆ e =ˆ y, ˆ e = ˆ z , or say, ˆ e = − ˆ x, ˆ e = ˆ y, ˆ e = ˆ z . Theimportant point is to stick to one convention duringthe entire matrix and vector manipulations. Forexample, to construct the above table we choosethe following convention: (i) ↑ spin operators sit tothe right of ↓ spins . (ii) The order of site indicesincreases from right to left .3. Third column is a pictorial representation of thebasis vector in real space. We will learn in thefollowing how to work with this intuitive represen-tation of the basis states.4. The fourth column represents the states as directproduct of spin- ↓ with spin- ↑ state states, wherethe occupations in up and down spin sectors are represented by sequence of 0 , ↑ and ↓ spins do not admix, this separation makes a goodsense. To see this, consider the effect of a term like c † i ↑ c j ↑ on a typical state | φ i = [ c † i ↓ . . . c † i N ↓ ] [ c † j ↑ . . . c † j M ↑ ] |i = [ c † i ↓ . . . c † i N ↓ ] |i ↓ [ c † j ↑ . . . c † j M ↑ ] |i ↑ =: | φ ↓ i| φ ↑ i , which can be written as c † i ↑ c j ↑ | φ ↓ i| φ ↑ i = c † i ↑ c j ↑ [ c † i ↓ . . . c † i N ↓ ][ c † j ↑ . . . c † j M ↑ ] |i = [ c † i ↓ . . . c † i N ↓ ] ( − N ( − N c † i ↑ c j ↑ [ c † j ↑ . . . c † j M ↑ ] |i = | φ ↓ i c † i ↑ c j ↑ | φ ↑ i . (18)Here two ( − N factors arise from moving each ofthe c †↑ operators through a length N sequence of c †↓ operators. Therefore, when operating in spin- ↑ sec-tor, we need not worry about spin- ↓ configurationand vice versa .5. The last three column indicate the way a computerunderstands and stores these basis state. There isonly one integer I stored on computer which fullyrepresents the occupation pattern of the ↑ and ↓ electrons when transformed to binary representa-tion of fourth column in Eq. (17). Given I , one canobviously find out I ↑ and I ↓ through the relation I = 2 L I ↓ + I ↑ (19)and vice versa. First and last columns of (17) doactually tabulates an array I = T ( J ) which for anygiven J , returns the corresponding I . The value of I fully specifies the state. One therefore only toextract the bits of I and find a way to work withbits of integer I .There remains only a final note on how we have labeledthe states. We have labeled the configuration representedby I = 9 as fourth ( J = 4) basis vector, etc. We couldhave labeled them in any order, so that I = 9 wouldhave corresponded to, e.g. first ( J = 1) basis vector.The above convention which was suggested by Lin andcoworkers , has the advantage that the table T can besearched given a value for I in a fast way in order tofind out corresponding J value. The essential idea ofthis convention is the following : For any sector you areinterested in, just choose the labels J in such a way thatwhen the above table T is constructed, the I values areascending function of J . B. Acting with operators on the basis states
If the action of an operator on a complete basis setis known, then the operator is completely specified.Eq. (16) has two type of terms: H U and H t . When H U acts on a basis states, it gives non zero contribution foreach site j in which both n j ↑ and n j ↓ are 1. Thereforethe effect of H U on any basis state gives the same statemultiplied by the number of doubly occupied sites × U .Hence the effect of H U on | φ i i , i = 1 , , , | φ i and | φ i is just U . The H U part is diagonal inoccupation number representation: H U . = U (20)Now let us concentrate on the effect of H t on our basisstates. Consider, e.g. the sate | φ i . H t in Eq. (16) has4 terms. But only 2 of them give non zero contributionwhen they act on | φ i : The one which allows an ↑ spin tohop from site 0 to 1, i.e. c † ↑ c ↑ , and another one whichallows the ↓ spin at site 0 to jump to site 1, i.e. c † ↓ c ↓ . H t | φ i = − t (cid:16) c † ↑ c ↑ + c † ↓ c ↓ (cid:17) c † ↓ c † ↑ |i = − t c † ↑ → z}|{ c ↑ c † ↓ c † ↑ + c † ↓ c ↓ c † ↓ ←← z}|{ c † ↑ |i = − t (cid:16) − c † ↑ c † ↓ c ↑ c † ↑ |i + c † ↓ c † ↑ c ↓ c † ↓ |i (cid:17) = − t (cid:16) − c † ↑ c † ↓ |i + c † ↓ c † ↑ |i (cid:17) where in the third line we have used the Fermionicanti-commutation relations with associated minus signsneeded for each exchange of Fermionic operators with dif-ferent indices. Also in the last step we have used the factthat c jσ c † jσ |i = (1 − n jσ ) |i = |i , as the vacuum state |i contains no particles. Rearranging the Fermionic opera-tors to comply with our convention we get H t | φ i = − t (+ | φ i + | φ i ) . (21)This way of working with commutations and algebra isnot a convenient for putting on computers. As we alreadyshowed in Eq. (18), the up and down spins hop separately.Therefore, when an spin ↑ electron hops from site say j to site i , one only needs to count a ( −
1) factor for each ↑ spin electron over which it passes. Since when c j ↑ hasto moved through a chain of c † j ′ ↑ operators with j ′ = j ,until it reaches c † j ↑ , there they form a (1 − n j ↑ ) operatorwhich commutes with all other remaining c † j ′′ ↑ operatorsto reach the vacuum |i where it produces |i itself.With this argument in mind, one can most conve-niently work with the pictorial representation of states:The effect of H t on | φ i is to either move an ↑ spin to site 1, or to move a ↓ spin to site 1. In the former caseone gets ( − | φ i where the exponent 0 is because the ↑ spin at site 0 passes through no other ↑ spin when it hopsto site 1. Similarly the later case gives | φ i . Accordingto the Hamiltonian (16), each of these processes happenswith an amplitude − t and thus one can pictorially seethat H t | φ i = − t (+ | φ i + | φ i ) . With this in mind it is almost trivial to check that H t | φ i = 0 H t | φ i = − t ( | φ i + | φ i ) H t | φ i = − t ( | φ i + | φ i ) H t | φ i = − t ( | φ i + | φ i ) H t | φ i = − t ( | φ i + | φ i ) H t | φ i = 0 (22)Therefore in this basis the hopping term has the followingmatrix representation: H t . = − t (23)Therefore the matrix representation of the entire Hamil-tonian in this basis becomes H . = U − t − t − t − t − t − t
00 0 − t − t U
00 0 0 0 0 0 (24)Clearly the above block diagonal structure has a parallelwith the following form for the S z matrix: S z . = +1 0 0 0 0 − (25)In other words state | φ i has a quantum number S z =+1, while | φ i has S z = −
1. The set of states | φ i . . . | φ i belong to S z = 0 sector. If we had confined ourselves to S z = 0 sector, we would have obtained a four dimen-sional Hilbert space in which the Hamiltonian could berepresented by the 4 × ↑ and ↓ electrons.But there is only one integer I (7th column) stored oncomputer the binary representation of which preciselycorresponds to column 4. Extracting I ↑ and I ↓ from agiven I in fortran is as simple as I_up=mod(I,2**L); I_dn=I/2**L
Most of the standard high level programming languagessuch as c++ and fortran 90 have appropriate intrinsicfunctions for bitwise operations on integer which are ableto access and examine or change individual bits of a giveninteger p (can be I ↑ or I ↓ ). Therefore an integer I com-pletely specifies the occupation pattern in each spin sec-tor. For example the Intel Fortran compiler has followingcommands for bitwise manipulation of an integer pIBSET(p,b) c † b IBCLR(p,b) c b BTEST(p,b) n b (26)For more details please consult the Intel Fortran languagemanual . Also more implementation notes can be foundin Lin et.al. . C. Diagonalization of the matrix: Spacesymmetries
So far we have made use of the symmetries of the
Hamiltonian itself to reduce the dimension of the ma-trix to be diagonalized. For example conservation of S z (equivalent to [ S z , H ] = 0) reduces the 6 × × × S z and H is quite general, andindependent of the geometry of the lattice in use.Now we would like to make use of the spatial symme-tries in order to diagonalize the 4 × S z = 0sector manually. For a two site problem composed ofsites 0 ,
1, there is a mirror reflection operator M whichhas the following action on real lattice: M (0) = 1 , M (1) = 0 (27)Corresponding to operation M which is a member of spa-tial symmetry group of the underlying lattice, there is anoperator M in Hilbert space which acts on the state vec-tors. For example consider the effect of M on state | φ i which reads M c † ↓ c † ↑ | i = c † M (0) ↓ c † M (1) ↑ | i = c † ↓ c † ↑ | i , (28)or M| φ i = | φ i . Similarly, M| φ i = | φ i , etc. which is summarized as follows M| φ i = −| φ iM| φ i = | φ iM| φ i = | φ iM| φ i = | φ iM| φ i = | φ iM| φ i = −| φ i The first and last lines of the above equation indicate thatstates | φ j i with j = 1 , M (parity op-erator) with eigen values ±
1. Line numbers 3,4 however,indicate that states | φ i and | φ i do not have definite par-ity. Instead a new combination | φ ± i = √ ( | φ i ± | φ i ),has a definite parity: M| φ ± i = ±| φ ± i . Similarly from | φ i and | φ i , symmetric and antisymmetric combina-tions will have definite parity.The eigenvalues ± of parity operator M partition theHilbert space into two pieces which belong either to +1eigen value: M ( | φ i + | φ i ) = + ( | φ i + | φ i ) M ( | φ i + | φ i ) = + ( | φ i + | φ i ) , (29)or to − M| φ i = −| φ iM| φ i = −| φ iM ( | φ i − | φ i ) = − ( | φ i − | φ i ) M ( | φ i − | φ i ) = − ( | φ i − | φ i ) (30)If we knew this when finding out the effect of H t inEq. (22), we would have acted on the following states: | ψ i = | φ i| ψ i = ( | φ i + | φ i ) / √ | ψ i = ( | φ i + | φ i ) / √ | ψ i = ( | φ i − | φ i ) / √ | ψ i = ( | φ i − | φ i ) / √ | ψ i = | φ i (31)Eq. (31) defines the so called, symmetry adopted basis inwhich the action of, say, H t is enormously simplified: H t | ψ i = 0 H t | ψ i = − t | ψ i H t | ψ i = − t | ψ i H t | ψ i = 0 H t | ψ i = 0 H t | ψ i = 0 (32)Similarly the only none zero matrix elements of H U aregiven by: H U | ψ i = U | ψ i , H U | ψ i = U | ψ i (33)How do we generate a symmetry adopted basis? Thereis a very powerful theorem in group representation theorywhich in case of one dimensional representations is easyto implement and reads : ψ ( p ) = X R Γ p ( R ) ∗ Rφ (34)In this equation φ is an arbitrary state to begin with. R isan element of the symmetry group which in this case andbe either I or M . Γ p is the irreducible representationof the group which for group {I , M} are numbers ± φ = | φ i i , with i = 1 , . . . , ~k values whendealing with problems of translational invariance.In symmetry adopted basis total Hamiltonian in S z = ± {| ψ i , | ψ i} sub-space) remains diagonal witheigen values equal to 0, as it was. In S z = 0 sector it willbecome: H . = U − t − t U (35)In this sector as well, there are one eigen value equal to 0which corresponding to eigen state | ψ i . Note that | ψ i isthe S z = 0 component of a triplet. Therefore the tripletstate lies at zero energy: E t = 0 , for triplet state ( S z = +1 , , −
1) (36)The fact that {| φ i , | φ i , | φ i} form a triplet is consistentwith having odd spatial parity in Eq. (30).The eigen value corresponding to | ψ i is U which isalways positive and above the triplet E t = 0.To find out the remaining two eigen values in the {| ψ i , | ψ i} sector, we note that first of all, this sectorhas even spatial parity, and hence is spin singlet. Hamil-tonian is U/ I + U/ σ z − tσ x , where σ x , σ z are Pauli matrices, and I is the unit 2 × E ± s = U/ ± p ( U/ + 4 t (37)Therefore the ground state is a singlet with energy E − s ,while the first excited state is a triplet with E t = 0.Second excited state is | ψ i , with energy U , and finalexcited state has energy E + s . IV. STRONG CORRELATIONS AND SPINPHYSICS
As we already saw in previous section, the ground stateof the two site Hubbard model is a singlet with energy E − s = U/ − p U / t , The ground state wave func-tion | E − s i = 4 | ψ i + (cid:16) U + p U + 16 (cid:17) | ψ i (38)in large U limit is dominated by | ψ i ∼ | φ i + | φ i inwhich there is no doubly occupied configuration, andhence charge fluctuations are suppressed.Since the first excited state is at E t = 0. The splittingbetween these two states for large U ≫ t is: − J = E s − E t = U/ − p U / t ≈ − t U (39)Therefore the singlet state is slightly below ( − t /U ) thetriplet state. This indicates that in large U limit, the low-energy physics of Hubbard model is given by spin fluc-tuations which are anti ferromagnetic (singlet has lowerenergy). This observation in a two site Hubbard model isindeed very general and it can be shown using a unitarytransformation that the Hubbard model at large U limitcan be mapped into the so called t-J model, where thereare AF spin fluctuations along with hoppings restrictedto subspace with no double occupancy . Hubbard J. Proc. Roy. Soc. London,
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Quantum Theory of Moleculesand Solids , Vol. 2. McGraw Hill, 1965. H.Q. Lin, and J.E. Gubernatis, Comp. Phys. (1993) 400. See the Intel Fortran language manual: ftp://download.intel.com/support/performancetools /fortran/linux/v8/for_lang.pdf Isfahan University of Technology, June 200716