Phys 711 Topics in Particles & Fields — Spring 2013 — Lecture 8 — v0.1.2

Lecture notes: Fermi-Hubbard model

Jeffrey Yepez

Department of Physics and AstronomyUniversity of Hawai‘i at ManoaWatanabe Hall, 2505 Correa RoadHonolulu, Hawai‘i 96822E-mail: yepez@hawaii.eduwww.phys.hawaii.edu/∼yepez

(Dated: March 7, 2013)

Contents

I. Crystallographic potential 1A. Bloch waves and quasimomentum 1B. Sine squared potential 2C. Localized Wannier functions 2

II. One-band Hubbard Hamiltonian 3A. Method of solution 3

1. Block diagonalization 32. Site interchange operator 53. Implementation of the S2 operator 6

B. Exact solutions 61. Double-empty symmetry above and below half

filling 72. Check of Nagaoka’s theorem 73. Quantum informational derivation of the ground

state 94. Triangular cluster above half filling 95. Predicted energy eigenstates for the triangular

cluster 9

III. Acknowledgements 11

A. Triangular cluster solutions 111. half filling: Ne = 3 and Sz = 1

211

a. A1 State with S = 12

13

b. A2 States with S = 12, 32

13

c. E States with S = 12

132. Above half filling: Ne = 4 and Sz = 0 13

a. A1 States with S = 0 13b. A2 State with S = 1 13c. E States with S = 0, 1 14

3. Above half filling: Ne = 4 and Sz = 1 14a. A2 State with S = 1 14b. E state with S = 1 14

4. Below half filling: Ne = 2 and Sz = 0 14a. A1 states with S = 0 14b. A2 State with S = 1 15c. E States with S = 0, 1 15

5. Below half filling: Ne = 2 and Sz = 1 15a. A2 State with S = 1 15b. E State with S = 1 15

B. Square cluster solutions 151. Half filling: Ne = 4 and Sz = 0 16

a. A1 States with S = 0, 1 16b. A2 States with S = 0, 1 16c. B1 States with S = 0, 1 17d. B2 States with S = 0, 1, 2 17e. E States with S = 0, 1 17

References 18

I. CRYSTALLOGRAPHIC POTENTIAL

Consider a nonrelativistic particle in a crystallographicperiodic potential with period a. For simplicity, we willtreat this problem in 1+1 dimensions. Let us begin bywriting down the single-particle Hamiltonian

H = − ~2

2m∂xx + Vcrys.(x), (1)

where Vcrys.(x) = Vcrys.(x+ a).

A. Bloch waves and quasimomentum

The eigenstates of (1) may be written in the form

φ(n)k (x) = eikxu

(n)k (x), (2)

where u(n)k (x) = u

(n)k (x+ a) is a periodic function known

as the Bloch wave for the crystal and where k is a real-valued wave number. Since

− ∂xxφ(n)k (x)

(2)= eikx

(−∂xx − 2ik∂x + k2

)u

(n)k (x) (3)

and since

(−i∂x + k)2 = −∂xx − 2ik∂x + k2 (4)

the energy eigenequation

Hφ(n)k (x) = E

(n)k φ

(n)k (x) (5)

reduces to an equation the Bloch waves((−i~∂x + ~k)2

2m+ Vcrys.(x)

)u

(n)k (x) = E

(n)k u

(n)k (x),

(6)where ~k is called the quasimomentum of the crystallo-graphic potential. The wave number range k ∈ (π/2, π/a]defines the primitive cell in reciprocal space1, also known

1 The wave number k-space (the Fourier transform of position x-space) is called reciprocal space in the condensed matter litera-ture.

A Bloch waves and quasimomentum I CRYSTALLOGRAPHIC POTENTIAL

as the first Brillouin zone. The eigenvalue problem (6)has a family of solutions (a discrete energy spectrum) la-belled by the quantum number n for each wave numberk. The Bloch wave solutions u(n)

k (x) represents the bandstructure of the crystal with energy bands E(n)

k .

B. Sine squared potential

Let us consider a scalar quantum particle in a sinesquared potential

Vcrys.(x) = V◦ sin2(kx). (7)

Then (1) may be written as

− ~2

2m∂xxφ

(n)k +V◦

(1− cos 2kx

2

)φ

(n)k = E

(n)k φ

(n)k , (8a)

which after rescaling x 7→ kx ≡ w becomes a differentialequation in w for the function y(w) ≡ φ(n)

k (kx)

− ~2k2

2m∂wwy + V◦

(1− cos 2w

2

)y = E

(n)k y. (8b)

With the atomic recoil energy ER ≡ ~2k2/(2m), theeigenequation may be written as Mathieu’s differentialequation

∂wwy + [a− 2q cos(2w)] y = 0, (8c)

with constant coefficients

a ≡E

(n)k − V◦/2ER

(9a)

q ≡ − V◦4ER

. (9b)

According to Floquet’s theorem, any Mathieu function(i.e. solution to (8c)) can be written in the form

y(w) = eirwf(w), (10)

where f(w) has period 2π and the integer-valued param-eter r is known as the Mathieu exponent. This is themathematician’s version of the physicist’s Block wave so-lution with quasimomentum rk discussed above.

An example Bloch wave solution is shown in Fig. 1.The band structure of the quantum particle is shown inFig. 2 for the lowest three energy eigenstates. Band gapsappear and then widen as the V◦ increases. For very largeV◦ the bands become flat (i.e. the band width of eachenergy band approaches zero and the forbidden energyregions become maximally large).

C. Localized Wannier functions

The advantage of the Bloch wave solutions is that theyrepresent orthogonal energy eigenstates. However, they

0 2 4 6 80.0

0.5

1.0

1.5

2.0

x

ÈΦ2

FIG. 1 A Mathieu function for a = 0.4 and q = −0.563078plotted over the range x ∈ [0, 3π]. Both the sine squaredpotential (red) and the periodic Block wave (blue) are shown.

-1.0 -0.5 0.0 0.5 1.00

2

4

6

8

k a�ΠE

kHnL �E

R

(a)No gaps for q = 0(V◦/ER = 0)

-1.0 -0.5 0.0 0.5 1.00

2

4

6

8

k a�Π

EkHn

L �ER

(b)1st gap appears at q = − 18

(V◦/ER = 1/2)

-1.0 -0.5 0.0 0.5 1.00

2

4

6

8

k a�Π

EkHn

L �ER

(c)2nd gap appears at q = − 12

(V◦/ER = 2)

-1.0 -0.5 0.0 0.5 1.0

0

2

4

6

8

k a�Π

EkHn

L �ER

(d)Gaps widen at q = −2(V◦/ER = 8)

FIG. 2 Band structure for period potential Vcrys.(x) =V◦ sin2(kx) obtained using Mathieu exponent solutions forq = 0, 1/8, 1/2, 2. First (blue), second (red), and third (green)energy bands for the sine squared potential. The energy bandsare shown using the reduced zone scheme.

are delocalized states. If one is interested in localizedorthogonal states2, then these can be constructed as asuperposition of Bloch waves

wn(x− xi) =1

2N

N−1∑ν=−N

e−ikνxiφ(n)kν

(x), (11)

2 A localized state is one where the quantum particle’s wave func-tion is narrow and primarily positioned near the minimum of onewell in the periodic potential. The quantum particle’s averageposition is centered at the well’s minimum.

2

II ONE-BAND HUBBARD HAMILTONIAN

for kν = πν/(Na) with a crystal cellsize a. Such localizedstates are Wannier functions (Wannier, 1937) when wemake the assumption that the Bloch wave is independentof wave number. Hence, writing the energy eigenstateas φ(n)

kν(x) = eikνxu(0)(x), then we express the Wannier

functions as

wn(x− xi) =u(0)(x)

2N

N−1∑ν=−N

eikν(x−xi) (12a)

=u(0)(x)

2N

N−1∑ν=−N

eπiνNa (x−xi). (12b)

We can make use of finite geometric series identity

12N

N−1∑ν=−N

zν =1

2NzN − z−N

z − 1(13)

to rewrite (12b) as

wn(x− xi) =u(0)(x)

2Neπia (x−xi) − e−πia (x−xi)

eπiNa (x−xi) − 1

(14a)

=iu(0)(x)N

sin(π(x− xi)/a)eπiNa (x−xi) − 1

(14b)

= u(0)(x)sin(π(x− xi)/a)π(x− xi)/a

+ · · · (14c)

≈ u(0)(x) sinc(π(x− xi)/a), (14d)

which is indeed a localized wave packet centered at xi.

II. ONE-BAND HUBBARD HAMILTONIAN

Here we briefly review the one-band Hubbard model(Hubbard, 1967) that describes a system of spin one-halffermions on a lattice by the Hamiltonian

H = −t∑〈ij〉σ

(a†iσajσ + a†jσaiσ

)+U

∑i

ni↑ni↓+U∑iσ

niσ,

(15)where a†iσ and aiσ are the Wannier fermionic creation anddestruction operators creating and destroying an electronof spin σ at site i of the lattice.

The physical meaning of each term comprising theHubbard Hamiltonian is straightforward. The first termof (15), the kinetic energy, accounts for electron tunnel-ing from site j to site i and thereby reduces the energy ofthe system by a factor t for each tunneling event. The sitesummation denoted by 〈ij〉 is over all bonds of the lat-tice. In our investigation we restrict ourselves to nearest-neighbor tunneling or hopping. The interaction part ofthe Hubbard Hamiltonian accounts for the Coulomb in-teraction between two electrons residing on the same siteand therefore is a simplification of the full Coulomb in-teraction term

H int =∫d3rd3r′ψ†(r)ψ(r)V (r − r′)ψ†(r′)ψ(r′), (16)

where ψ(r) is the field operator at the space point r.The last two terms of (15) are obtained from (16) if onesubstitutes for the field operator a linear combinationof Wannier states, denoted φiσ(r), as follows, ψ(r) =∑iσ aiσφiσ(r). U denotes the intra-site matrix elements

of the Wannier states so that the significant short-rangedpart of the Coulomb interaction is modeled. Then (16)reduces to

H int = U∑iσσ′

niσniσ′ = U∑i

ni↑ni↓ + U∑iσ

n2iσ. (17)

The last term of (17) is a constant self-energy, UNe,where Ne is the total number of electrons, because n2

iσ =niσ for fermions. Since this term only causes a constantenergy shift of the spectrum it is typically ignored.

Obtaining analytic solutions of the Hubbard model isdifficult. Exact solutions are possible for certain cases,for example when U/t = 0 or when U/t = ∞, the latterbeing the strong-coupling limit where doubly occupiedsites are prohibited. A well known approximation ap-proach involves altering the Hubbard Hamiltonian by aunitary transformation to express it in a t/U expansion(Gros et al., 1987). The lowest order terms of this ex-pansion are kept in the large-U limit, both large on-sitemutual interaction and self-interaction. This is the t-Jmodel.

Here we shall focus on exact solutions. As a firstexercise toward understanding strongly-correlated Fermimatter governed by the Hubbard Hamiltonian, we nowconsider exact solutions of (15) for small clusters.

A. Method of solution

Here we demonstrate our quantum informationmethod by applying it to the one-band Hubbard modelfor small clusters. As examples, we consider the trian-gular and square clusters. We implement operators usedto diagonalize the Hamiltonian with respect to symme-try and total-spin. Joint number operators are employedto represent a swap gate which in turn is used to rep-resent a fermionic interchange operator and the total-spin-squared operator. Finally, we implement the secondquantized form of Wannier creation and annihilation op-erators in terms of qubit creation and annihilation oper-ators in a numbered state representation.

1. Block diagonalization

The first step in solving the Hubbard model for smallclusters, given a lattice of size L and number of electronsNe, is to determine a set of basis states. Since the z-component of the spin commutes with the Hamiltonian,we have chosen to work with a set of basis states in thenumber representation with a given Sz. This is done forconvenience since one can immediately determine Sz forany such state by inspection. Throughout the remainder

3

A Method of solution II ONE-BAND HUBBARD HAMILTONIAN

of this lecture we shall denote an Sz basis by the symbol:{φn}.

As an example of the size of the basis, in the case ofhalf filling, where Ne = L and Sz = 0, for the triangularand square clusters there are 9 and 36 states, respectively.

The triangular cluster possesses the C3v point groupsymmetry and the square C4v.

TABLE I C3 (with ε = ei2π/3) and C3v character tables.

C3 E C3 C23

A 1 1 1

E 1 ε ε∗

E∗ 1 ε∗ ε

C3v E 2C3 3σv

A1 1 1 1

A2 1 1 -1

E 2 -1 0

TABLE II C4 and C4v character tables.

C4 E C4 C2 C34

A 1 1 1 1

B 1 -1 1 -1

E 1 i -1 -i

E∗ 1 -i -1 i

C4v E 2C4 C2 2σv 2σd

A1 1 1 1 1 1

A2 1 1 1 -1 -1

B1 1 -1 1 1 -1

B2 1 -1 1 -1 1

E 2 0 -2 0 0

The second step is to determine matrices representingthe point group operations in the basis {φn}. This isaccomplished by successively employing site-interchangeoperations, as detailed below, to rotate or reflect anyparticular state according to the appropriate group op-eration desired. Let us denote these various operationsby the generic symbol R. The C3v group has six groupoperations and three irreducible representations, and C4v

has eight and five, respectively; see Table I and Table II.Let us denote the ith irreducible representations by ΓRi .Given that we have constructed a reducible matrix repre-sentation, denoted {MR

mn}, of the symmetry group in ourbasis {φn}, we then construct projection operators, de-noted PΓi

mn, for each of the irreducible representations ofthe group (Chen et al., 1988; Freericks and Falicov, 1991;Freericks et al., 1991; Reich and Falicov, 1988). This isis done as follows

PΓimn =

∑R

tr(ΓRi )MRmn, (18)

where tr(ΓRi ) is the character or trace listed in the ithrow and Rth column of the character tables. We canalso determine the number of times the ΓRi irreduciblerepresentation occurs in MR

mn by knowing the character,tr(MR

mn) (Cotton, 1964). In practice the useful quantitybi is determined by

bi =∑R

tr(ΓRi )tr(MRmn), (19)

giving the size of the ith block of the resulting block-diagonalized Hamiltonian matrix of elements.

The third step is to apply PΓimn onto the {φn} basis to

find linear combinations that possess a definite symmetry

ψΓim =

∑n

PΓimnφn. (20)

Throughout this lecture we shall denote a symmetry ba-sis (equivalence class of quantum states with respect toa finite-point group symmetry) by the symbol: {ψΓi

n }.Specifically, for the case of C3v, we obtain four sets ofstates: {ψA1

n }, {ψA2n }, {ψEn }, and {(ψEn )∗}. For the case

of C4v, we obtain six sets of states: {ψA1n }, {ψB1

n }, {ψA2n },

{ψB2n }, {ψEn }, and {(ψEn )∗}. The E-representation al-

ways appears twice3. In the {ψn} basis the Hamiltoniantherefore decouples into four diagonal blocks for the tri-angular cluster and six diagonal blocks for the squarecluster. The group theory has helped in reducing thecomplexity of the problem.

The fourth step is to block diagonalize the Hamiltonianstill further. This is done by calculating the matrix ofelements of the total spin-squared operator, S2, in eachof the {ψΓi

m } subspace and diagonalizing this matrix todetermine linear combinations of the ψΓi

m states whichhave a definite total spin S. That is, we find eigenvectorsof S2 in the {ψΓi

m } subspace and these have eigenvaluesS(S + 1). We denote these eigenvectors of S2 as ϕΓi,S

m .Below, we shall denote a total spin-squared basis by thesymbol ϕΓi,S

m . Since S2 commutes with the Hamiltonianwe thereby partition each symmetry block into smallerblocks, each having a definite total spin eigenvalue S.

The final step is to calculate the matrix elements of theHubbard Hamiltonian in the {ϕΓi,S

m } basis, and analyt-ically find, if possible, the eigenvalues and eigenvectorsof each resulting spin-and-symmetry block and therebycomplete the analytical solution. In the case of the trian-gle, the largest ϕΓi,S

m block is 3× 3 and for the square itis 4× 4. So these two clusters are analytically tractable.

In summary, we denote the various sectors of theHilbert space with quantum state symbols given by

Ne and Sz Γi S2

φNe,Sz ψΓi ϕΓi,S, (21)

for integer Ne ∈ [0, 6] and for Sz = 0,± 12 ,±1,± 3

2 andfor i ∈ {A1, A2, E} of the C3v point group for the L = 3lattice (triangular cluster) and for integer Ne ∈ [0, 8] andfor Sz = 0,± 1

2 ,±1,± 32 ,±2 and i ∈ {A1, A2, B1, B2, E} of

the C4v point group for the L = 4 lattice (square cluster).

3 An important point must be mentioned here. We have foundusing projection operators constructed from the C3 and C4 char-acter tables breaks the size of the E-type blocks in two for thehalf-filled cases.

4

A Method of solution II ONE-BAND HUBBARD HAMILTONIAN

2. Site interchange operator

We can use the qubit creation and annihilation op-erators, a†α and aα for the αth qubit, to represent Wan-nier spin creation and annihilation operators, a†iσ and aiσfor the spin component (say along the z-axis) of valueσ, at the ith lattice point. Two qubits are requiredto encode one quantum spin-position state, both havingfour basis states per point. The isomorphism betweenthe Wannier and qubit operators are mapped as follows(i = 1, . . . , LD):

Wannier operator ↔ qubit operatora†i↑ ↔ a†2i−1 (22a)

ai↑ ↔ a2i−1 (22b)

a†i↓ ↔ a†2i (22c)

ai↓ ↔ a2i. (22d)

We use the following bit encoding and notation for theelectrons at a point:

|e〉 = |00〉, 0 (empty)|p〉 = |01〉, e−↑ (plus or spin-up)|m〉 = |10〉, e−↓ (minus or spin-down)|d〉 = |11〉, e−↑ e

−↓ (double).

(23)

All symmetry operations of rotation or reflection are im-plemented by a particular successive application of afermionic interchange operator, which we denote by χijwhere the interchange of electrons occurs between sites iand j.χij may be implemented using SWAP gates and the fol-

lowing is one derivation of its unitary form in terms ofWannier creation and annihilation operators. We wishto construct χij so as to correctly handle any necessaryphase change due to the fermion anticommutation rela-tions. The simplest implementation of χij should involveonly products of aiσ and a†iσ.

For an site interchange operator, we require thatχij

2 = 1, that the site interchange operator conservethe number of particles, [χij , N ] = 0, that χij is hermi-tian, and that χij |0〉 = |0〉. Let us assume we have aone-particle state φ1e = a†mσ|0〉 where m = i or j. A firstguess at the analytical form of a spin-dependent SWAP

gate between sites i and j would be

χijσ = a†jσaiσ + a†iσajσ. (24)

This acts correctly on φ1e. The problem with (24) is thatits application on to the vacuum violates our requirementthat χijσ|0〉 = |0〉. This is remedied easily enough byslightly modifying our first guess

χijσ = a†jσaiσ + a†iσajσ + 1. (25)

Although (25) repairs the vacuum problem, now its ap-plication onto φ1e would interchange the electron but in-correctly would also give back φ1e, an extra unwanted

electron. We administer the final remedy by includingtwo more terms that have the effect of subtracting offthe unwanted electron

χijσ = a†jσaiσ + a†iσajσ + 1− a†iσaiσ − a†jσajσ. (26)

Although we have constructed (26) by considering onlya one-particle state, a remarkable fact is that χijσ workson any arbitrary state.

Let us investigate the structure of χijσ further. Let usrewrite (26) by factoring the creation operators,

χijσ = 1 + a†jσ(aiσ − ajσ) + a†iσ(ajσ − aiσ), (27)

which leads one to write

χijσ = 1− (a†jσ − a†iσ)(ajσ − aiσ). (28)

Now of course, one is motivated to define the followingjoint creation, annihilation, and number operators

Aijσ =

√12

(ajσ − aiσ

)(29)

A†ijσ =

√12

(a†jσ − a

†iσ

)(30)

NAijσ = A†ijσAijσ (31)

so that (28) may be written as

χijσ = 1− 2NAijσ. (32)

Our joint operators work in a number representationformed of entangled states

A†ijσ|0〉 =

√12

(| jσ〉− | iσ〉

)≡| Aijσ〉. (33)

The joint number operator satisfies (NAijσ)2 = NA

ijσ andconsequently we have

χijσ2 = (1− 2NA

ijσ)2 = 1− 4NAijσ + 4NA

ijσ

2= 1, (34)

satisfying another of our requirements.It can be written in exponential form

χijσ = ezNA (35a)

= 1 + zNAijσ +

z2

2!NAijσ

2+ · · · (35b)

= 1 +(z +

z2

2!+ · · ·

)NAijσ (35c)

= 1 + (ez − 1)NAijσ. (35d)

Comparing (32) with (35d), allows us to choose z = πi,so we have our SWAP gate

χijσ = χijσ = eiπNA , (36)

5

A Method of solution II ONE-BAND HUBBARD HAMILTONIAN

1

2 3 4

1 2

3

FIG. 3 Symmetry axes for the triangular and square clusters.

appearing as a rotation by 180◦.To affect the interchange of electrons betweens the sites

i and j, one defines the site interchange operator as theproduct operator

χij ≡ χij↑χij↓. (37)

For the triangular cluster, we implement the C3v pointoperators as follows:

RC3 = χ12χ23 (38a)

RC23 = χ23χ12 (38b)

Rσ(1)v = χ23 (38c)

Rσ(2)v = χ31 (38d)

Rσ(3)v = χ12, (38e)

which are 120◦ and 240◦ rotations, and reflections aboutsites 1, 2, and 3, respectively. For the square cluster, weimplement the C4v point group operators as follows:

RC4 = χ34χ23χ12 (39a)RC2 = χ23χ12 (39b)

RC34 = χ12χ23χ34 (39c)

Rσ(13)v = χ24 (39d)

Rσ(24)v = χ13 (39e)

Rσ(13)d = χ12χ34 (39f)

Rσ(24)d = χ14χ23, (39g)

which are 90◦, 180◦ and 270◦ rotations, and reflectionsabout two diagonals and the vertical and horizontal, re-spectively.

3. Implementation of the S2 operator

The site-specific total-spin operator is compactly writ-ten in terms of the Wannier creation and annihilationoperators and the Pauli spin matrices as

Si =12a†iασαβaiβ , (40)

where ~ = 1, σ = (σx, σy, σz), and

σx =

(0 11 0

)σy =

(0 −ii 0

)σz =

(1 00 −1

).

The components of (40) are then

Six = 12a†iασ

xαβaiβ =

12

(a†i ↑ai↓ + a†i ↓ai↑

)(41a)

Siy = 12a†iασ

yαβaiβ =

i

2

(−a†i ↑ai↓ + a†i ↓ai↑

)(41b)

Siz = 12a†iασ

zαβaiβ =

12

(a†i ↑ai↑ − a

†i ↓ai↓

). (41c)

Using the mapping (22), the spin operators may be ex-pressed in terms of the qubit creation and annihilationoperators

Six =12

(a†2i−1a2i + a†2ia2i−1

)(42a)

Siy =i

2

(−a†2i−1a2i + a†2ia2i−1

)(42b)

Siz =12

(a†2i−1a2i−1 − a

†2ia2i

). (42c)

Then, the the spin along z of the numbered state is

Sz =L∑i=1

Siz, (43)

where Szϕ = Szϕ. Furthermore, the square of the totalspin operator is readily constructed

S2 =L∑i=1

L∑j=1

(SixSjx + SiySjy + SizSjz

), (44)

where S2ϕ = S(S+1)ϕ. We use the eigenvalues Sz and Sas good quantum numbers to specify the quantum stateof the system.

B. Exact solutions

Here we analyze our findings for the triangular andsquare cluster tabulated in Appendices A and B. We eval-uate these solutions for the expected symmetries of theHubbard Hamiltonian and find that the method leads toconsistent results. We check a theorem proved by Na-gaoka concerning whether or not the ferromagnetic stateabove and below half filling is the ground state. Then,we compare spin-correlations of the ground states of thetriangular and square clusters at half filling. Finally wegive some examples and interpretation of eigenvalue andphotoemission energies as a function of U/t.

6

B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

1. Double-empty symmetry above and below half filling

It is well known that, energetically, the HubbardHamiltonian possesses a symmetry above and below halffilling. This symmetry ensures a one-to-one correspon-dence between states above half filling with those below.The structure of the states, their exact symmetry andtotal-spin, above and below half filling is identical pro-vided that the respective Sz basis, {φn}, are labeled sothat any state above half filling with a particular dou-bly occupied site corresponds to a state below half fillingwith no electrons on that site.

Lets consider the solutions for the triangular clusterand, further, consider the cases with Sz = 0 above andbelow half filling, presented in Sec. A.2 and Sec. A.4. Thefirst point to note is that {ψNe=4,Sz=0

n } can be mappedto {ψNe=2,Sz=0

n } by taking d → e. The structure of thestates are seen to be identical by comparing the eigenketsin Sec. A.2 to those in Sec. A.4. Finally, the energies arealso seen to be isomorphic by taking t→ −t and if, abovehalf filling, we shift the energy down by 3U . The origin ofthis shift is two-fold: (a) the second term of the HubbardHamiltonian (15) causes a shift of U since d→ e; and (b),the third term of the Hamiltonian (self energy) causes ashift of 2U in our case. The same symmetry is seen toexist for the cases with Sz = 1; see Sec. A.3 and A.5.

For the case of the square cluster this symmetry is es-sentially the same, differing only in that the asymmetryin t is not present. The underlying reason for differencebetween the triangle and square arises from the freedomwe have to divide the square into two sublattices suchthat for any lattice point has nearest-neighbors belongingto the co-lattice (Nagaoka, 1966). Then a phase differ-ence of −1 between the wave function of the sublatticesaccounts for a change in the sign of t.

2. Check of Nagaoka’s theorem

Let us consider the triangular cluster below half filling,Ne = 2. The ground state of this system could haveeither Sz = 0 or 1. In the Sz = 0 case, the numberedbasis states are

φ1 = |emp〉 φ2 = |pem〉 φ3 = |mpe〉φ4 = |epm〉 φ5 = |mep〉 φ6 = |pme〉φ7 = |dee〉 φ8 = |ede〉 φ9 = |eed〉

. (45)

In this {φ} basis, the matrix elements of the A1 symme-try operator

ΓA1 =16

(1 +RC3 +RC

23 +Rσ

(1)v +Rσ

(2)v +Rσ

(3)v

)(46)

are

〈φi|ΓA1|φj〉 =16

1 −1 1 −1 1 −1 0 0 0−1 1 −1 1 −1 1 0 0 01 −1 1 −1 1 −1 0 0 0−1 1 −1 1 −1 1 0 0 01 −1 1 −1 1 −1 0 0 0−1 1 −1 1 −1 1 0 0 00 0 0 0 0 0 2 2 20 0 0 0 0 0 2 2 20 0 0 0 0 0 2 2 2

ij

.

(47)From this result, we see that the {ψ} basis is

ψ1 = 1√

6

„|emp〉 − |pem〉+ |mpe〉 − |epm〉+ |mep〉 − |pme〉

«(48)

ψ2 =1√3

(|dee〉+ |ede〉+ |eed〉

), (49)

which are antiferromagnetic states. The lowest energyeigenstate in the Ne = 2 and Sz = 0 sector has A1 sym-metry and total spin S = 0. In the {ψ} basis, the matrixelements of the Hubbard Hamiltonian are

〈ψi|H|ψj〉 =

(−2t 2

√2 t

2√

2 t U

)ij

. (50)

In comparing (50) with (A16) presented in Appendix A,note that we have subtracted 2U (self-energy) from thediagonals of the matrix elements for simplicity. Diago-nalizing (50), we find the exact solution for the lowestenergy eigenstates to be

ΨSz=0,S=0A12

= (51)

α√6

(|emp〉 − |pem〉+ |mpe〉 − |epm〉+ |mep〉 − |pme〉

)+

β√3

(|dee〉+ |ede〉+ |eed〉

),

where the normalized coefficients α and β are

α =µ√

1 + µ2(52a)

β =1√

1 + µ2, (52b)

with

µ = −2t+ U ±√

36t2 + 4tU + U2

4√

2t. (53)

The coefficients α and β (for the positive root in µ) areplotted in Fig. 4b for unity hopping coefficient t = 1.

In the Sz = 1 case, the lowest energy state, the ferro-magnetic state, has E symmetry and total spin S = 1

ΨSz=1,S=1E = (2|epp〉+ |pep〉 − |ppe〉)/

√3

→ 〈HES=1〉 = −t,

(54)

7

B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

5 10 15 20 25 30 35 40

0.2

0.4

0.6

0.8

15 10 15 20 25 30 35 40

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

U

U

!"

# A12, S=0

E, S=1

Energy

(a) (b)

$%12 = " &1 + # &2

5 10 15 20 25 30 35 40

0.2

0.4

0.6

0.8

15 10 15 20 25 30 35 40

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

U

U

!"

# A12, S=0

E, S=1

Energy

(a) (b)

$%12 = " &1 + # &2

FIG. 4 Below half fillng for the triangular cluster with unity hop-ping coefficient. Variation with U of: (a) lowest eigenvalues’ for S =

0 and S = 1; and (b) coefficients of the ΨNe=2,Sz=0A12

= αψ1 + βψ2

eigenstate.

where again we have subtracted 2U (the self-energy). Wesee in Fig. 4a that below half filling the ferromagnetic

state is not the ground state of the system for all valuesof U . Nagaoka has proven, in the limit U → ∞, for anfcc or hcp lattice that the ferromagnetic state with themaximum total spin would not be the ground state of thesystem (Nagaoka, 1966). Our results are in agreementwith this statement, if we make an analogy between ourtriangular cluster and fcc or hcp.

Let us see in more detail why ΨSz=0,S=0A12

is indeed theground state of the system. Let us consider the U =0 limit. Here the Hubbard Hamiltonian has only thekinetic energy term

H◦ = −t∑〈ij〉σ

(a†iσajσ + a†jσaiσ

)(55)

which we can transform to k-space to determine the bandenergy. To convert from Bloch states (in reciprocal space)to Wannier states (in position space) we use the discreteinverse Fourier transform

aiσ =1N

∑k

e−ikRiakσ a†iσ =1N

∑k

eikRia†kσ,

(56a)and to go from position space back to reciprocal spacewe use the discrete Fourier transform

akσ =∑{Ri}

eikRiaiσ a†kσ =∑{Ri}

e−ikRia†iσ, (56b)

where the possible momenta are k = 2nπ/Na and latticevectors are Rn = an, n = 0, 1, .., N − 1 and a is thecellsize. Substituting (56a) into (55) gives

H◦ = − t

N

∑〈ij〉σ

(∑k

eikRia†kσ

∑k′

e−ik′Rjak′σ + h.c.

)(57a)

= − t

N

∑i

∑σ

∑kk′

a†kσak′σeikRi

(e−ik

′(Ri+a) + e−ik′(Ri−a)

), (57b)

where we have went from (57a) to (57b) by replacing thesum over the bonds, 〈ij〉, by a sum only over the latticesites since for nearest-neighbor interactions, Rj = Ri± ain the 〈ij〉 sum. We then have

H◦ =∑kσ

εka†kσakσ, (58)

where εk = −2t cos ka and where we have used

1N

∑i

ei(k−k′)Ri = δkk′ . (59)

Referring to Fig. 5a and Fig. 5b, clearly for Ne = 2 theSz = 0 energy eighstate has lowest energy eigenvalue,−4t, while the Sz = 1 energy eigenstate has energy eigen-value −t. This agrees with Fig. 4a. Let us consider the

(52) at U = 0. Here, expanding about U = 0, we havefor the positive root of µ

α = −√

23− U

9√

6+O(U2) (60a)

β =1√3− U

9√

3+O(U2), (60b)

so substituting (60) into (51) gives

ΨSz=0,S=0A12,U=0 → (61)

1√9

(|dee〉+ |ede〉+ |eed〉)−√19

(|emp〉 − |pem〉+ |mpe〉 − |epm〉+ |mep〉 − |pme〉) .

8

B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

-2t

t

0 2!/3 4!/3

0

"k

k

-2t

t

0 2!/3 4!/3

0 k

-2t

t

0 2!/3 4!/3

0

"k

k

-2t

t

0 2!/3 4!/3

0 k

Sz=0 , Ne=2

Sz=1, Ne=4

Sz=1, Ne=2

Sz=0 , Ne=4 "k

"k

(a) (b)

(c) (d)

FIG. 5 Lowest energy momentum-states above and below halffilling for Sz = 0, 1.

3. Quantum informational derivation of the ground state

Remarkably, and as an additional consistency check,we can arrive at (61) much more simply and directly byusing an a maximally entangled quantum state that islocalized in k-space with zero momentum (k = 0). Thatis, we may construct this maximally entangle state, whichFig. 5a represents, using the momentum-state Wanniercreation operators, (56b), as follows:

a†k=0↑a†k=0↓|0〉 = (a†1↑ + a†2↑ + a†3↑)(a

†1↓ + a†2↓ + a†3↓)|0〉

(62a)= |dee〉+ |ede〉+ |eed〉− |emp〉+ |epm〉 − |mep〉 − |mpe〉+ |pem〉+ |pme〉,

(62b)

where the vacuum state for the L = 3 is |0〉 ≡ |eee〉. Nor-malizing this by

√9, we see that (62) is indeed identical

to the solution (61) obtained by exact diagonalization ofthe Hubbard Hamiltonian for L = 3 when we evaluatethe exact (t, U)-dependent solution at U = 0.

The vacuum state |0〉 may be added in quantum su-perposition with (62) for the following reasons:

1. it too has quantum numbers Sz = 0 and total spinS = 0;

2. it too is an eigenstate of ΓA1, which represents A1symmetry operator4 ;

3. and it contributes zero energy to the superpositionstate (i.e. 〈0|H|0〉 = 0).

4 The vacuum is an eigenstate of all the symmetry operators be-cause they are formed out of unitary swap gates which by con-struction have the vacuum as an eigenstate with unity eigenvalue.

Therefore, the U = 0 ground state may be written as

ΨSz=0,S=0A12,U=0 =

(u+ v a†k=0↑a

†k=0↓

)|0〉, (63)

where u and v are c-numbers that satisfy the normaliza-tion condition

|u|2 + |v|2 = 1. (64)

Therefore, the ground state is superconducting and itcontains one Cooper pair. We will explore this genericproperty of the ground state in the context of large Fermisystems in Lecture 9.

4. Triangular cluster above half filling

Now let us consider the triangular cluster above halffilling, Ne = 4. Like the case just considered, the groundstate of this system could have either Sz = 0 or 1. Re-markably, however, in the above half-filled case the fer-romagnetic and nonferromagnetic states are degeneratewith eigenvalue 5U − 2t, see Sec. A.2 and A.3. Thereason for this degeneracy is depicted in plots (c) and(d) in Fig. 5. The degeneracy arises from the fact thatεk = −2t cos ka has the same value for k = 2π/3 andk = 4π/3, an effect due to the L = 3 cluster.

5. Predicted energy eigenstates for the triangular cluster

Here we give some examples illustrating the insightsone may obtain from energy eigenvalue plots. Let us seewhat physics is contained in an eigenvalue plot such asFig. 6, where we have plotted all the energies of the half-

2 4 6 8 10

-4

-2

2

4

6

8

10

E2

U

Energy

E3

E1 A1

A21

A22

3t

0

-3t

FIG. 6 Eigenvalues for the triangular cluster at half filling.

filled triangular cluster as a function of U . Each curveis labelled according to its symmetry. The A1 and A22

energies are degenerate for all U . In Fig. 7, we have de-picted all the possible Bloch states for Ne = 3. There arefive types of Bloch states we can possibly choose. These,in turn, correspond to the five nondegenerate energies forfinite U plotted in Fig. 6. Consequently, we see why thereare only three nondegenerate energies at U = 0 in Fig. 6:

9

B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

the Bloch states in panels (1), (2), and (3) of Fig. 7 havethe same energy at U = 0. It is possible to assign eachBloch state a given symmetry. Immediately, we assignpanels (1) and (5) of Fig. 7 symmetries E1 and E2, re-spectively, since they are the highest and lowest energystates. Then, we assign panel (2) A21 symmetry sincefrom Fig. 6 the energy of this state is independent of Uand we know

ΨA21 =1√3

(|mpp〉+ |pmp〉+ |ppm〉

). (65a)

Next, we assign panel (3) A1 and A22 symmetry since weknow that these states must have a doubly occupied site

ΨA1 =1√6

[(|dpe〉+ |edp〉+ |ped〉)

+ (|dep〉+ |pde〉+ |epd〉)] (65b)

ΨA22 =1√6

[(|dpe〉+ |edp〉+ |ped〉)

− (|dep〉+ |pde〉+ |epd〉)].

(65c)

This explains why A1 and A22 are degenerate for all U .Finally, we then can explain why the E3 energy is lowerthan the A1 and A22 energy for finite U . This is becausethe E3 state can have a component of Bloch state in panel(4) which is energetically lower for finite U than in panel(3).

The eigenvalues of the triangular cluster above half filling, Ne = 4, and Sz = 0 are the following

λNe=4,Sz=0A11

=11U + 2t+

√(−11U − 2t)2 − 4 (30U2 + 12Ut− 8t2)

2(66a)

λNe=4,Sz=0A12

=11U + 2t−

√(−11U − 2t)2 − 4 (30U2 + 12Ut− 8t2)

2(66b)

λNe=4,Sz=0A2

= 5U − 2t (66c)

λNe=4,Sz=0E1

= 5U + t (66d)

λNe=4,Sz=0E2

=11U − t+

√(−11U + t)2 − 4 (30U2 − 6Ut− 2t2)

2(66e)

λNe=4,Sz=0E3

=11U − t−

√(−11U + t)2 − 4 (30U2 − 6Ut− 2t2)

2(66f)

and the eigenvalues below half filling are

λNe=2,Sz=0A11

=5U − 2t+

√(−5U + 2t)2 − 4 (6U2 − 6Ut− 8t2)

2(67a)

λNe=2,Sz=0A12

=5U − 2t−

√(−5U + 2t)2 − 4 (6U2 − 6Ut− 8t2)

2(67b)

λNe=2,Sz=0A2

= 2U + 2t (67c)

λNe=2,Sz=0E1

= 2U − t (67d)

λNe=2,Sz=0E2

=5U + t+

√(−5U − t)2 − 4 (6U2 + 3Ut− 2t2)

2(67e)

λNe=2,Sz=0E3

=5U + t−

√(−5U − t)2 − 4 (6U2 + 3Ut− 2t2)

2. (67f)

Plots of the eigenvalues above (and below) half filling for Ne = 4 (and 2) and Sz = 0 are plotted in Fig. 8 (with theconstant term of the Hubbard Hamiltonian, 4U , subtracted off) showing the variation of the photoemission energies

10

B Exact solutions A TRIANGULAR CLUSTER SOLUTIONS

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

-2t

t

0 2!/3 4!/3

0 k

"k

"=3t

"=-3t

"=0

"=0

"=0

-2t

t

0 2!/3 4!/3

0 k

"k

#1

$21

#2

$1,$22

(1)

(2)

(3)

(4)

(5)

FIG. 7 Bloch states with definite symmetry for the triangular cluster at half filling.

with U . The ground state energy of the half-filled system has been subtracted off of λNe=4,Sz=0 and λNe=2,Sz=0.Fig. 9 show the photoemission spectrum for different values of U .

III. ACKNOWLEDGEMENTS

I would like to thank Eric Jensen, Bulbul Chakraborty, and Hugh Pendleton for helpful discussions regarding theFermi-Hubbard Hamiltonian.

Appendix A: Triangular cluster solutions

Here we present the exact solutions to the Hubbard Hamiltonian for the triangular cluster in the following cases:half filling, Ne = 3, for Sz = 1

2 ; above half filling, Ne = 4, for Sz = 0, 1; and, below half filling, Ne = 4, for Sz = 0, 1.

1. half filling: Ne = 3 and Sz = 12

For Sz = 12 in the case of half filling where Ne = 3 we have 9 elements in our basis. We choose our {φn} basis as

follows:

φ1 = |mpp〉 φ2 = |pmp〉 φ3 = |ppm〉φ4 = |dpe〉 φ5 = |edp〉 φ6 = |ped〉φ7 = |dep〉 φ8 = |pde〉 φ9 = |epd〉.

(A1)

11

1 half filling: Ne = 3 and Sz = 12 A TRIANGULAR CLUSTER SOLUTIONS

2 4 6 8 10

-10

-5

5

10

15

20A11 A12

A2

E1 E2 E3

A11

A12

A2

E1

E2

E3

U

!"!G

FIG. 8 Photoemission energies versus the tight-binding energy U : λNe=4,Sz=0 − λG in black and λNe=2,Sz=0 − λG in gray.

Out[130]=

U!0

U!1

U!2

U!3

U!4

"10 "5 0 5Energy

0.51.01.52.02.53.0Spectrum

U!5

FIG. 9 Photoemission spectra for the triangular cluster with different tight-binding strengths.

In the case of half filling, the {ψn} basis states generated by the C3v projection operators happen all to be eigenvectorsof S2, so here we do not get any reduction in block size using the total-spin. Therefore, the {ϕn} basis is equivalentto {ψn}.

12

2 Above half filling: Ne = 4 and Sz = 0 A TRIANGULAR CLUSTER SOLUTIONS

a. A1 State with S = 12

ψ1 = (φ4 + φ5 + φ6 + φ7 + φ8 + φ9)/√

6 → 〈HA1S= 1

2〉 = (4U) . (A2)

b. A2 States with S = 12, 3

2

ψ2 = (φ1 + φ2 + φ3)/√

3 → 〈HA2S= 3

2〉 = (3U) (A3)

ψ3 = (φ4 + φ5 + φ6 − φ7 − φ8 − φ9)/√

6 → 〈HA2S= 1

2〉 = (4U) (A4)

c. E States with S = 12

ψ4 = (φ1 + εφ2 + ε∗φ3)/√

3ψ5 = (φ4 + εφ5 + ε∗φ6)/

√3

ψ6 = (φ7 + εφ8 + ε∗φ9)/√

3

→ 〈HES= 1

2〉 =

3U (ε∗ − 1)t (1− ε)t(ε− 1)t 4U (ε∗ − 1)t(1− ε∗)t (ε− 1)t 4U

, (A5)

where ε = e−i2π3 .

2. Above half filling: Ne = 4 and Sz = 0

For Sz = 0, above half filling where Ne = 4, we have 9 elements in our basis. We choose our {φn} basis as follows

φ1 = |dmp〉 φ2 = |pdm〉 φ3 = |mpd〉φ4 = |dpm〉 φ5 = |mdp〉 φ6 = |pmd〉φ7 = |edd〉 φ8 = |ded〉 φ9 = |dde〉

(A6)

The {ϕn} and {ψn} basis are equivalent in the A1 and A2 representations, for n = 1, 2, 3.

a. A1 States with S = 0

ϕ1 = ψ1 = (φ1 − φ2 + φ3 − φ4 + φ5 − φ6)/√

6ϕ2 = ψ2 = (φ7 + φ8 + φ9)/

√3

}→ 〈HA1

S=0〉 =

(5U + 2t 4t√

24t√

26U

). (A7)

b. A2 State with S = 1

ϕ3 = ψ3 = (φ1 − φ2 + φ3 + φ4 − φ5 + φ6)/√

6 → 〈HA2S=1〉 = (5U − 2t) . (A8)

13

4 Below half filling: Ne = 2 and Sz = 0 A TRIANGULAR CLUSTER SOLUTIONS

c. E States with S = 0, 1

ψ4 = (2φ1 + φ2 − φ3)/√

5 (A9a)

ψ5 = (2φ4 + φ5 − φ6)/√

5 (A9b)

ψ6 = (2φ7 − φ8 − φ9)/√

5 (A9c)

ϕ4 = (ψ4 + ψ5)/√

2 → 〈HES=1〉 = (5U + t) (A10)

ϕ5 = (ψ4 − ψ5)/√

2ϕ6 = ψ6

}→ 〈HE

S=0〉 =

(5U − t −2t√

2−2t√

26U

). (A11)

3. Above half filling: Ne = 4 and Sz = 1

For Sz = 1, above half filling where Ne = 4, we have 3 elements in our basis. We choose our {φn} basis as follows

φ1 = |dpp〉 φ2 = |pdp〉 φ3 = |ppd〉. (A12)

The {ϕn} and {ψn} basis are equivalent. There are no states in the A1 representation.

a. A2 State with S = 1

ϕ1 = ψ1 = (φ1 − φ2 + φ3)/√

3 → 〈HA2S=1〉 = (5U − 2t) . (A13)

b. E state with S = 1

ϕ2 = ψ2 = (2φ1 + φ2 − φ3)/√

3 → 〈HES=1〉 = (5U + t) . (A14)

4. Below half filling: Ne = 2 and Sz = 0

For Sz = 0, below half filling where Ne = 2, we have 9 elements in our basis. We choose our {φn} basis as follows

φ1 = |emp〉 φ2 = |pem〉 φ3 = |mpe〉φ4 = |epm〉 φ5 = |mep〉 φ6 = |pme〉φ7 = |dee〉 φ8 = |ede〉 φ9 = |eed〉

(A15)

The {ϕn} and {ψn} basis are equivalent in the A1 and A2 representations, for n = 1, 2, 3.

a. A1 states with S = 0

ϕ1 = ψ1 = (φ1 − φ2 + φ3 − φ4 + φ5 − φ6)/√

6ϕ2 = ψ2 = (φ7 + φ8 + φ9)/

√3

}→ 〈HA1

S=0〉 =

(2U − 2t 2

√2t

2√

2t 3U

). (A16)

14

B SQUARE CLUSTER SOLUTIONS

b. A2 State with S = 1

ϕ3 = ψ3 = (φ1 − φ2 + φ3 + φ4 − φ5 + φ6)/√

6 → 〈HA2S=1〉 = (2U + 2t). (A17)

c. E States with S = 0, 1

ψ4 = (2φ1 + φ2 − φ3)/√

5 (A18a)

ψ5 = (2φ4 + φ5 − φ6)/√

5 (A18b)

ψ6 = (2φ7 − φ8 − φ9)/√

5 (A18c)

ϕ4 = (ψ4 + ψ5)/√

2 → 〈HES=1〉 = (2U − t) (A19)

ϕ5 = (ψ4 − ψ5)/√

2ϕ6 = ψ6

}→ 〈HE

S=0〉 =

(2U + t −2t√

2−2t√

23U

). (A20)

5. Below half filling: Ne = 2 and Sz = 1

For Sz = 1, below half filling where Ne = 2, we have 3 elements in our basis. We choose our {φn} basis as follows

φ1 = |epp〉 φ2 = |pep〉 φ3 = |ppe〉 . (A21)

The {ϕn} and {ψn} basis are equivalent. There are no states in the A1 representation.

a. A2 State with S = 1

ϕ1 = ψ1 = (φ1 − φ2 + φ3)/√

3 → 〈HA2S=1〉 = (2U + 2t) . (A22)

b. E State with S = 1

ϕ2 = ψ2 = (2φ1 + φ2 − φ3)/√

5 → 〈HES=1〉 = (2U − t) . (A23)

Appendix B: Square cluster solutions

In this section we present our findings for the square cluster for the case of half filling, Ne = 4, for Sz = 0.

15

1 Half filling: Ne = 4 and Sz = 0 B SQUARE CLUSTER SOLUTIONS

1. Half filling: Ne = 4 and Sz = 0

For Sz = 0 in the case of half filling where Ne = 4 we have 36 elements in our basis. We choose our {φn} basis asfollows

φ1 = |mpmp〉 φ2 = |pmpm〉φ3 = |ppmm〉 φ4 = |mppm〉 φ5 = |mmpp〉 φ6 = |pmmp〉φ7 = |demp〉 φ8 = |pdem〉 φ9 = |mpde〉 φ10 = |empd〉φ11 = |dpme〉 φ12 = |edpm〉 φ13 = |medp〉 φ14 = |pmed〉φ15 = |depm〉 φ16 = |mdep〉 φ17 = |pmde〉 φ18 = |epmd〉φ19 = |dmpe〉 φ20 = |edmp〉 φ21 = |pedm〉 φ22 = |mped〉φ23 = |dmep〉 φ24 = |pdme〉 φ25 = |epdm〉 φ26 = |mepd〉φ27 = |dpem〉 φ28 = |mdpe〉 φ29 = |emdp〉 φ30 = |pemd〉

φ31 = |dede〉 φ32 = |eded〉φ33 = |ddee〉 φ34 = |deed〉 φ35 = |edde〉 φ36 = |eedd〉.

(B1)

a. A1 States with S = 0, 1

ψ1 = (φ3 − φ4 + φ5 − φ6)/2 (B2a)

ψ2 = (φ7 − φ8 + φ9 + φ10 − φ11 − φ12 + φ13 − φ14)/√

8 (B2b)

ψ3 = (φ15 − φ16 + φ17 + φ18 − φ19 − φ20 + φ21 − φ22)/√

8 (B2c)

ψ4 = (φ23 − φ24 − φ25 + φ26 − φ27 + φ28 + φ29 − φ30)/√

8 (B2d)

ψ5 = (φ31 + φ32)/√

2 (B2e)ψ6 = (φ33 + φ34 + φ35 + φ36)/2 (B2f)

ϕ1 = (ψ2 + ψ3)/√

2 → 〈HA1S=1〉 = (5U) . (B3)

ϕ2 = ψ4

ϕ3 = ψ1

ϕ4 = (ψ2 − ψ3)/√

2ϕ5 = ψ5

ϕ6 = ψ6

→ 〈HA1

S=0〉 =

5U 0 0 0 00 4U 2t 0 00 2U 5U 4t√

22t

0 0 4t√2

6U 0

0 0 2t 0 6U

. (B4)

It is interesting that the eigenstate with eigenvalue 5U does not mix with any of the other states in its subbasis.

b. A2 States with S = 0, 1

ψ7 = (φ1 − φ2)/√

2 (B5a)

ψ8 = (φ7 − φ8 + φ9 + φ10 + φ11 + φ12 − φ13 + φ14)/√

8 (B5b)

ψ9 = (φ15 − φ16 + φ17 + φ18 + φ19 + φ20 − φ21 + φ22)/√

8 (B5c)

ψ10 = (φ23 − φ24 − φ25 + φ26 + φ27 − φ28 − φ29 + φ30)/√

8 (B5d)

ϕ7 = ψ7

ϕ8 = ψ10

ϕ9 = (ψ8 + ψ9)/√

2

→ 〈HA2S=1〉 =

4U 0 4t√

2

0 5U −4t√2

4t√2−4t√

25U

. (B6)

ϕ10 = (ψ8 − ψ9)/√

2 → 〈HA2S=0〉 = (5U). (B7)

16

1 Half filling: Ne = 4 and Sz = 0 B SQUARE CLUSTER SOLUTIONS

c. B1 States with S = 0, 1

ψ11 = (φ7 + φ8 + φ9 − φ10 − φ11 + φ12 + φ13 + φ14)/√

8 (B8a)

ψ12 = (φ15 + φ16 + φ17 − φ18 − φ19 + φ20 + φ21 + φ22)/√

8 (B8b)

ψ13 = (φ23 + φ24 − φ25 − φ26 − φ27 − φ28 + φ29 + φ30)/√

8 (B8c)

ψ14 = (φ31 − φ32)/√

2 (B8d)

ϕ11 = (ψ11 + ψ12)/√

2 → 〈HB1S=1〉 = (5U) . (B9)

ϕ12 = (ψ11 − ψ12)/√

2ϕ13 = ψ13

ϕ14 = ψ14

→ 〈HB1S=0〉 =

5U −4t√

24t√

2−4t√

25U 0

4t√2

0 6U

. (B10)

d. B2 States with S = 0, 1, 2

ψ15 = (φ1 + φ2)/√

2 (B11a)ψ16 = (φ3 + φ4 + φ5 + φ6)/2 (B11b)

ψ17 = (φ7 + φ8 + φ9 − φ10 + φ11 − φ12 − φ13 − φ14)/√

8 (B11c)

ψ18 = (φ15 + φ16 + φ17 − φ18 + φ19 − φ20 − φ21 − φ22)/√

8 (B11d)

ψ19 = (φ23 + φ24 − φ25 − φ26 + φ27 + φ28 − φ29 − φ30)/√

8 (B11e)

ψ20 = (φ33 − φ34 − φ35 + φ36)/√

4 (B11f)

ϕ15 = (ψ15/√

2 + ψ16)/√

3/2 → 〈HB2S=2〉 = (4U) . (B12)

ϕ16 = (ψ17 + ψ18)/√

2ϕ17 = ψ19

}→ 〈HB2

S=1〉 =

(5U 00 5U

). (B13)

ϕ18 = (−2ψ15/√

2 + ψ16)/√

3ϕ19 = (ψ17 − ψ18)/

√2

ϕ20 = ψ20

→ 〈HB2S=0〉 =

4U −6t√3

0−6t√

35U −2t

0 −2t 6U

. (B14)

Although ϕ16 and ϕ17 have the same total-spin, not that they happen to be eigenvectors the Hamiltonian.

e. E States with S = 0, 1

ψ21 = (φ3 − iφ4 − φ5 + iφ6)/2 (B15a)ψ22 = (φ7 − iφ8 − φ9 − iφ10)/2 (B15b)ψ23 = (φ11 + iφ12 + φ13 − iφ14)/2 (B15c)ψ24 = (φ15 − iφ16 − φ17 − iφ18)/2 (B15d)ψ25 = (φ19 + iφ20 + φ21 − iφ22)/2 (B15e)ψ26 = (φ23 − iφ24 + φ25 − iφ26)/2 (B15f)ψ27 = (φ27 − iφ28 + φ29 − iφ30)/2 (B15g)ψ28 = (φ33 − iφ34 + iφ35 − φ36)/2 (B15h)

17

1 Half filling: Ne = 4 and Sz = 0 B SQUARE CLUSTER SOLUTIONS

ϕ21 = ψ21

ϕ22 = (ψ26 + ψ27)/√

2ϕ23 = (ψ23 + ψ25)/

√2

ϕ24 = (ψ22 + ψ24)/√

2

→ 〈HES=1〉 =

4U 0 −2t√

22it√

2

0 5U (−1 + i) t (−1− i) t−2t√

2(−1− i) t 5U 0

−2t√2

(−1 + i) t 0 5U

. (B16)

ϕ25 = (−ψ23 + ψ25)/√

2ϕ26 = (−ψ22 + ψ24)/

√2

ϕ27 = (ψ26 − ψ27)/√

2ϕ28 = ψ28

→ 〈HES=0〉 =

5U 0 (−1 + i) t 2t√

2

0 5U (1 + i) t 2it√2

(−1− i) t (1− i) t 5U 02t√

2−2it√

20 6U

. (B17)

References

Chen, C., A. Reich, and L. M. Falicov, 1988, Phys. Rev. B38(18), 12823.

Cotton, F., 1964, Chemical applications of group theory.Freericks, J. K., and L. M. Falicov, 1991, Phys. Rev. B 44(7),

2895.Freericks, J. K., L. M. Falicov, and D. S. Rokhsar, 1991, Phys.

Rev. B 44(4), 1458.Gros, C., R. Joynt, and T. Rice, 1987, Physical Review B 36,

381.Hubbard, J., 1967, Phys. Lett. 25A, 709.Nagaoka, Y., 1966, Physical Review 147, 382.Reich, A., and L. M. Falicov, 1988, Phys. Rev. B 37(10),

5560.Wannier, G. H., 1937, Phys. Rev. 52, 191, URL http://link.

aps.org/doi/10.1103/PhysRev.52.191.

18