Lecture Notes: Fermi-Hubbard Model

Phys 711 Topics in Particles & Fields — Spring 2013 — Lecture 8 — v0.1.2 Lecture notes: Fermi-Hubbard model

Jeﬀrey Yepez Department of Physics and Astronomy University of Hawai‘i at Manoa Watanabe Hall, 2505 Correa Road Honolulu, Hawai‘i 96822 E-mail: [email protected] www.phys.hawaii.edu/∼yepez

(Dated: March 7, 2013)

Contents I. CRYSTALLOGRAPHIC POTENTIAL

I. Crystallographic potential 1 Consider a nonrelativistic particle in a crystallographic A. Bloch waves and quasimomentum 1 periodic potential with period a. For simplicity, we will B. Sine squared potential 2 C. Localized Wannier functions 2 treat this problem in 1+1 dimensions. Let us begin by writing down the single-particle Hamiltonian II. One-band Hubbard Hamiltonian 3 2 A. Method of solution 3 H = − ~ ∂ + V (x), (1) 1. Block diagonalization 3 2m xx crys. 2. Site interchange operator 5 2 3. Implementation of the S operator 6 where Vcrys.(x) = Vcrys.(x + a). B. Exact solutions 6 1. Double-empty symmetry above and below half filling 7 A. Bloch waves and quasimomentum 2. Check of Nagaoka’s theorem 7 3. Quantum informational derivation of the ground state 9 The eigenstates of (1) may be written in the form 4. Triangular cluster above half filling 9 5. Predicted energy eigenstates for the triangular (n) ikx (n) φk (x) = e uk (x), (2) cluster 9 (n) (n) III. Acknowledgements 11 where uk (x) = uk (x + a) is a periodic function known as the Bloch wave for the crystal and where k is a real- A. Triangular cluster solutions 11 valued wave number. Since 1 1. half filling: Ne = 3 and Sz = 2 11 a. A State with S = 1 13 (n) (2) ikx 2 (n) 1 2 − ∂xxφ (x) = e −∂xx − 2ik∂x + k u (x) (3) 1 3 k k b. A2 States with S = 2 , 2 13 1 c. E States with S = 2 13 and since 2. Above half filling: Ne = 4 and Sz = 0 13 2 2 a. A1 States with S = 0 13 (−i∂x + k) = −∂xx − 2ik∂x + k (4) b. A2 State with S = 1 13 c. E States with S = 0, 1 14 the energy eigenequation 3. Above half filling: Ne = 4 and Sz = 1 14 (n) (n) (n) a. A2 State with S = 1 14 Hφ (x) = E φ (x) (5) b. E state with S = 1 14 k k k 4. Below half filling: Ne = 2 and Sz = 0 14 reduces to an equation the Bloch waves a. A1 states with S = 0 14 b. A2 State with S = 1 15 2 ! c. E States with S = 0, 1 15 (−i~∂x + ~k) (n) (n) (n) + Vcrys.(x) uk (x) = Ek uk (x), 5. Below half filling: Ne = 2 and Sz = 1 15 2m a. A2 State with S = 1 15 (6) b. E State with S = 1 15 where ~k is called the quasimomentum of the crystallo- B. Square cluster solutions 15 graphic potential. The wave number range k ∈ (π/2, π/a] 1. Half filling: Ne = 4 and Sz = 0 16 defines the primitive cell in reciprocal space1, also known a. A1 States with S = 0, 1 16 b. A2 States with S = 0, 1 16 c. B1 States with S = 0, 1 17 d. B2 States with S = 0, 1, 2 17 e. E States with S = 0, 1 17 1 The wave number k-space (the Fourier transform of position x- space) is called reciprocal space in the condensed matter litera- References 18 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- 2.0 belled by the quantum number n for each wave number (n) 1.5 k. The Bloch wave solutions uk (x) represents the band (n) 2 Φ structure of the crystal with energy bands Ek . È 1.0

0.5 B. Sine squared potential 0.0 Let us consider a scalar quantum particle in a sine 0 2 4 6 8 squared potential x

2 Vcrys.(x) = V◦ sin (kx). (7) FIG. 1 A Mathieu function for a = 0.4 and q = −0.563078 plotted over the range x ∈ [0, 3π]. Both the sine squared Then (1) may be written as potential (red) and the periodic Block wave (blue) are shown.

2 1 − cos 2kx − ~ ∂ φ(n) +V φ(n) = E(n)φ(n), (8a) 2m xx k ◦ 2 k k k 8 8 which after rescaling x 7→ kx ≡ w becomes a diﬀerential 6 6 R R E E

(n) L L

n 4 n 4 H H equation in w for the function y(w) ≡ φk (kx) k k E E 2 2 2k2 1 − cos 2w − ~ ∂ y + V y = E(n)y. (8b) 0 0 2m ww ◦ 2 k -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0 k aΠ k aΠ 2 2 (a)No gaps for q = 0 (b)1st gap appears at q = − 1 With the atomic recoil energy ER ≡ ~ k /(2m), the 8 eigenequation may be written as Mathieu’s diﬀerential (V◦/ER = 0) (V◦/ER = 1/2) equation 8 8 6 ∂ y + [a − 2q cos(2w)] y = 0, (8c) 6 R ww R E E L

L 4 n n H k

H 4 k E with constant coeﬃcients E 2 2 0 E(n) − V /2 0 a ≡ k ◦ (9a) -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0 ER k aΠ k aΠ (c)2nd gap appears at q = − 1 (d)Gaps widen at q = −2 V◦ 2 q ≡ − . (9b) (V◦/ER = 2) (V◦/ER = 8) 4ER

According to Floquet’s theorem, any Mathieu function FIG. 2 Band structure for period potential Vcrys.(x) = 2 (i.e. solution to (8c)) can be written in the form V◦ sin (kx) obtained using Mathieu exponent solutions for q = 0, 1/8, 1/2, 2. First (blue), second (red), and third (green) y(w) = eirwf(w), (10) energy bands for the sine squared potential. The energy bands are shown using the reduced zone scheme. where f(w) has period 2π and the integer-valued param- eter r is known as the Mathieu exponent. This is the mathematician’s version of the physicist’s Block wave solution with quasimomentum rk discussed above. are delocalized states. If one is interested in localized An example Bloch wave solution is shown in Fig. 1. orthogonal states2, then these can be constructed as a The band structure of the quantum particle is shown in superposition of Bloch waves Fig. 2 for the lowest three energy eigenstates. Band gaps appear and then widen as the V◦ increases. For very large 1 N−1 V◦ the bands become ﬂat (i.e. the band width of each X −ikν xi (n) wn(x − xi) = e φ (x), (11) energy band approaches zero and the forbidden energy 2N kν regions become maximally large). ν=−N

C. Localized Wannier functions 2 A localized state is one where the quantum particle’s wave function is narrow and primarily positioned near the minimum of one The advantage of the Bloch wave solutions is that they well in the periodic potential. The quantum particle’s average represent orthogonal energy eigenstates. However, they position is centered at the well’s minimum.

2 II ONE-BAND HUBBARD HAMILTONIAN

ˆ for kν = πν/(Na) with a crystal cellsize a. Such localized where ψ(r) is the field operator at the space point r. states are Wannier functions (Wannier, 1937) when we The last two terms of (15) are obtained from (16) if one make the assumption that the Bloch wave is independent substitutes for the field operator a linear combination of wave number. Hence, writing the energy eigenstate of Wannier states, denoted φiσ(r), as follows, ψ(r) = (n) ik x (0) P as φ (x) = e ν u (x), then we express the Wannier iσ aiσφiσ(r). U denotes the intra-site matrix elements kν functions as of the Wannier states so that the significant short-ranged N−1 part of the Coulomb interaction is modeled. Then (16) u(0)(x) X w (x − x ) = eikν (x−xi) (12a) reduces to n i 2N ν=−N int X X X 2 H = U niσniσ0 = U ni↑ni↓ + U n . (17) N−1 iσ (0) 0 u (x) X πiν (x−x ) iσσ i iσ = e Na i . (12b) 2N ν=−N The last term of (17) is a constant self-energy, UNe, 2 We can make use of finite geometric series identity where Ne is the total number of electrons, because niσ = niσ for fermions. Since this term only causes a constant N−1 1 X 1 zN − z−N energy shift of the spectrum it is typically ignored. zν = (13) 2N 2N z − 1 Obtaining analytic solutions of the Hubbard model is ν=−N difficult. Exact solutions are possible for certain cases, to rewrite (12b) as for example when U/t = 0 or when U/t = ∞, the latter being the strong-coupling limit where doubly occupied (0) πi (x−x ) − πi (x−x ) u (x) e a i − e a i w (x − x ) = sites are prohibited. A well known approximation ap- n i πi (x−x ) 2N e Na i − 1 proach involves altering the Hubbard Hamiltonian by a (14a) unitary transformation to express it in a t/U expansion (Gros et al., 1987). The lowest order terms of this ex- iu(0)(x) sin(π(x − x )/a) = i (14b) pansion are kept in the large-U limit, both large on-site πi (x−x ) N e Na i − 1 mutual interaction and self-interaction. This is the t-J sin(π(x − x )/a) = u(0)(x) i + ··· (14c) model. π(x − xi)/a Here we shall focus on exact solutions. As a first ≈ u(0)(x) sinc(π(x − x )/a), (14d) exercise toward understanding strongly-correlated Fermi i matter governed by the Hubbard Hamiltonian, we now which is indeed a localized wave packet centered at xi. consider exact solutions of (15) for small clusters.

II. ONE-BAND HUBBARD HAMILTONIAN A. Method of solution

Here we briefly review the one-band Hubbard model Here we demonstrate our quantum information (Hubbard, 1967) that describes a system of spin one-half method by applying it to the one-band Hubbard model fermions on a lattice by the Hamiltonian for small clusters. As examples, we consider the trian- X X X gular and square clusters. We implement operators used H = −t a† a + a† a +U n n +U n , iσ jσ jσ iσ i↑ i↓ iσ to diagonalize the Hamiltonian with respect to symme- hijiσ i iσ try and total-spin. Joint number operators are employed (15) † to represent a swap gate which in turn is used to rep- where aiσ and aiσ are the Wannier fermionic creation and resent a fermionic interchange operator and the total- destruction operators creating and destroying an electron spin-squared operator. Finally, we implement the second of spin σ at site i of the lattice. quantized form of Wannier creation and annihilation op- The physical meaning of each term comprising the erators in terms of qubit creation and annihilation oper- Hubbard Hamiltonian is straightforward. The first term ators in a numbered state representation. of (15), the kinetic energy, accounts for electron tunneling from site j to site i and thereby reduces the energy of the system by a factor t for each tunneling event. The site 1. Block diagonalization summation denoted by hiji is over all bonds of the lattice. In our investigation we restrict ourselves to nearest- The first step in solving the Hubbard model for small neighbor tunneling or hopping. The interaction part of clusters, given a lattice of size L and number of electrons the Hubbard Hamiltonian accounts for the Coulomb in- Ne, is to determine a set of basis states. Since the z- teraction between two electrons residing on the same site component of the spin commutes with the Hamiltonian, and therefore is a simplification of the full Coulomb in- we have chosen to work with a set of basis states in the teraction term number representation with a given Sz. This is done for Z int 3 3 0 ˆ† ˆ 0 ˆ† 0 ˆ 0 convenience since one can immediately determine Sz for H = d rd r ψ (r)ψ(r)V (r − r )ψ (r )ψ(r ), (16) any 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: giving the size of the ith block of the resulting block- {φn}. diagonalized Hamiltonian matrix of elements. Γi As an example of the size of the basis, in the case of The third step is to apply Pmn onto the {φn} basis to half filling, where Ne = L and Sz = 0, for the triangular find linear combinations that possess a definite symmetry and square clusters there are 9 and 36 states, respectively.

The triangular cluster possesses the C3v point group Γi X Γi ψm = Pmnφn. (20) symmetry and the square C4v. n

Throughout this lecture we shall denote a symmetry ba- TABLE I C (with = ei2π/3) and C character tables. 3 3v sis (equivalence class of quantum states with respect to 2 C3 EC3 C C3v E 2C3 3σv Γi 3 a ﬁnite-point group symmetry) by the symbol: {ψn }. A 1 1 1 A1 1 1 1 Speciﬁcally, for the case of C3v, we obtain four sets of ∗ A1 A2 E E ∗ E 1 A2 1 1 -1 states: {ψn }, {ψn }, {ψn }, and {(ψn ) }. For the case ∗ ∗ A1 B1 A2 E 1 E 2 -1 0 of C4v, we obtain six sets of states: {ψn }, {ψn }, {ψn }, B2 E E ∗ {ψn }, {ψn }, and {(ψn ) }. The E-representation al- 3 ways appears twice . In the {ψn} basis the Hamiltonian therefore decouples into four diagonal blocks for the triangular cluster and six diagonal blocks for the square

TABLE II C4 and C4v character tables. cluster. The group theory has helped in reducing the complexity of the problem. 3 C4v E 2C4 C2 2σv 2σd C4 EC4 C2 C4 A1 1 1 1 1 1 The fourth step is to block diagonalize the Hamiltonian A 1 1 1 1 still further. This is done by calculating the matrix of A2 1 1 1 -1 -1 B 1 -1 1 -1 elements of the total spin-squared operator, S2, in each B 1 -1 1 1 -1 1 Γi E 1 i -1 -i of the {ψm } subspace and diagonalizing this matrix to ∗ B2 1 -1 1 -1 1 Γi E 1 -i -1 i determine linear combinations of the ψm states which E 2 0 -2 0 0 have a definite total spin S. That is, we find eigenvectors 2 Γi of S in the {ψm } subspace and these have eigenvalues 2 Γi,S S(S + 1). We denote these eigenvectors of S as ϕm . The second step is to determine matrices representing Below, we shall denote a total spin-squared basis by the Γi,S 2 the point group operations in the basis {φn}. This is symbol ϕm . Since S commutes with the Hamiltonian accomplished by successively employing site-interchange we thereby partition each symmetry block into smaller operations, as detailed below, to rotate or reflect any blocks, each having a definite total spin eigenvalue S. particular state according to the appropriate group op- The final step is to calculate the matrix elements of the eration desired. Let us denote these various operations Γi,S Hubbard Hamiltonian in the {ϕm } basis, and analyt- by the generic symbol R. The C3v group has six group ically find, if possible, the eigenvalues and eigenvectors operations and three irreducible representations, and C4v of each resulting spin-and-symmetry block and thereby has eight and five, respectively; see Table I and Table II. complete the analytical solution. In the case of the trian- R Let us denote the ith irreducible representations by Γ . Γi,S i gle, the largest ϕm block is 3 × 3 and for the square it Given that we have constructed a reducible matrix repre- is 4 × 4. So these two clusters are analytically tractable. sentation, denoted {M R }, of the symmetry group in our mn In summary, we denote the various sectors of the basis {φn}, we then construct projection operators, de- Γi Hilbert space with quantum state symbols given by noted Pmn, for each of the irreducible representations of the group (Chen et al., 1988; Freericks and Falicov, 1991; N and S Γ S2 Freericks et al., 1991; Reich and Falicov, 1988). This is e z i , (21) is done as follows φNe,Sz ψΓi ϕΓi,S

Γi X R R Pmn = tr(Γi )Mmn, (18) 1 3 for integer Ne ∈ [0, 6] and for Sz = 0, ± , ±1, ± and R 2 2 for i ∈ {A1,A2,E} of the C3v point group for the L = 3 lattice (triangular cluster) and for integer N ∈ [0, 8] and where tr(ΓR) is the character or trace listed in the ith e i for S = 0, ± 1 , ±1, ± 3 , ±2 and i ∈ {A ,A ,B ,B ,E} of row and Rth column of the character tables. We can z 2 2 1 2 1 2 R the C4v point group for the L = 4 lattice (square cluster). also determine the number of times the Γi irreducible R representation occurs in Mmn by knowing the character, R tr(Mmn) (Cotton, 1964). In practice the useful quantity b is determined by i 3 An important point must be mentioned here. We have found using projection operators constructed from the C3 and C4 char- X R R bi = tr(Γi )tr(Mmn), (19) acter tables breaks the size of the E-type blocks in two for the R half-ﬁlled cases.

4 A Method of solution II ONE-BAND HUBBARD HAMILTONIAN

2. Site interchange operator electron. We administer the final remedy by including two more terms that have the effect of subtracting off We can use the qubit creation and annihilation op- the unwanted electron † erators, aα and aα for the αth qubit, to represent Wan- † † † † † χijσ = a aiσ + a ajσ + 1 − a aiσ − a ajσ. (26) nier spin creation and annihilation operators, aiσ and aiσ jσ iσ iσ jσ for the spin component (say along the z-axis) of value σ, at the ith lattice point. Two qubits are required Although we have constructed (26) by considering only to encode one quantum spin-position state, both having a one-particle state, a remarkable fact is that χijσ works four basis states per point. The isomorphism between on any arbitrary state. the Wannier and qubit operators are mapped as follows Let us investigate the structure of χijσ further. Let us (i = 1,...,LD): rewrite (26) by factoring the creation operators, Wannier operator ↔ qubit operator † † χijσ = 1 + ajσ(aiσ − ajσ) + aiσ(ajσ − aiσ), (27) a† ↔ a† (22a) i↑ 2i−1 which leads one to write ai↑ ↔ a2i−1 (22b) † † † † χijσ = 1 − (a − a )(ajσ − aiσ). (28) ai↓ ↔ a2i (22c) jσ iσ ai↓ ↔ a2i. (22d) Now of course, one is motivated to define the following We use the following bit encoding and notation for the joint creation, annihilation, and number operators electrons at a point: ! r1 A = a − a (29) |ei = |00i, 0 (empty) ijσ 2 jσ iσ − |pi = |01i, e↑ (plus or spin-up) r ! − (23) † 1 † † |mi = |10i, e↓ (minus or spin-down) Aijσ = ajσ − aiσ (30) − − 2 |di = |11i, e↑ e↓ (double). A † Nijσ = A Aijσ (31) All symmetry operations of rotation or reflection are im- ijσ plemented by a particular successive application of a so that (28) may be written as fermionic interchange operator, which we denote by χij where the interchange of electrons occurs between sites i A χijσ = 1 − 2Nijσ. (32) and j. χij may be implemented using SWAP gates and the fol- Our joint operators work in a number representation lowing is one derivation of its unitary form in terms of formed of entangled states Wannier creation and annihilation operators. We wish r ! to construct χij so as to correctly handle any necessary 1 A† |0i = | jσi− | iσi ≡| A i. (33) phase change due to the fermion anticommutation rela- ijσ 2 ijσ tions. The simplest implementation of χij should involve † only products of aiσ and aiσ. The joint number operator satisfies (N A )2 = N A and For an site interchange operator, we require that ijσ ijσ 2 consequently we have χij = 1, that the site interchange operator conserve the number of particles, [χij,N] = 0, that χij is hermi- 2 A 2 A A 2 χijσ = (1 − 2Nijσ) = 1 − 4Nijσ + 4Nijσ = 1, (34) tian, and that χij|0i = |0i. Let us assume we have a † one-particle state φ1e = amσ|0i where m = i or j. A first satisfying another of our requirements. guess at the analytical form of a spin-dependent SWAP It can be written in exponential form gate between sites i and j would be

zNA † † χijσ = e (35a) χijσ = a aiσ + a ajσ. (24) jσ iσ 2 A z A 2 = 1 + zNijσ + Nijσ + ··· (35b) This acts correctly on φ1e. The problem with (24) is that 2! its application on to the vacuum violates our requirement z2 = 1 + z + + ··· N A (35c) that χijσ|0i = |0i. This is remedied easily enough by 2! ijσ slightly modifying our ﬁrst guess z A = 1 + (e − 1) Nijσ. (35d) † † χijσ = ajσaiσ + aiσajσ + 1. (25) Comparing (32) with (35d), allows us to choose z = πi, Although (25) repairs the vacuum problem, now its ap- so we have our SWAP gate plication onto φ1e would interchange the electron but in- iπNA correctly would also give back φ1e, an extra unwanted χijσ = χijσ = e , (36)

5 A Method of solution II ONE-BAND HUBBARD HAMILTONIAN

where ~ = 1, σ = (σx, σy, σz), and 1 1 2 ! ! ! 0 1 0 −i 1 0 σx = σy = σz = . 1 0 i 0 0 −1

The components of (40) are then

2 3 3 4 1 S = 1 a† σx a = a† a + a† a (41a) ix 2 iα αβ iβ 2 i ↑ i↓ i ↓ i↑ i S = 1 a† σy a = −a† a + a† a (41b) iy 2 iα αβ iβ 2 i ↑ i↓ i ↓ i↑ FIG. 3 Symmetry axes for the triangular and square clusters. 1 S = 1 a† σz a = a† a − a† a . (41c) iz 2 iα αβ iβ 2 i ↑ i↑ i ↓ i↓ appearing as a rotation by 180◦. To aﬀect the interchange of electrons betweens the sites Using the mapping (22), the spin operators may be ex- i and j, one deﬁnes the site interchange operator as the pressed in terms of the qubit creation and annihilation product operator operators

χ ≡ χ χ . (37) 1 ij ij↑ ij↓ S = a† a + a† a (42a) ix 2 2i−1 2i 2i 2i−1 For the triangular cluster, we implement the C3v point i operators as follows: S = −a† a + a† a (42b) iy 2 2i−1 2i 2i 2i−1 C3 R = χ12χ23 (38a) 1 † † 2 C Siz = a2i−1a2i−1 − a2ia2i . (42c) R 3 = χ23χ12 (38b) 2 σ(1) R v = χ (38c) 23 Then, the the spin along zˆ of the numbered state is σ(2) R v = χ31 (38d) L σ(3) R v = χ12, (38e) X Sz = Siz, (43) which are 120◦ and 240◦ rotations, and reﬂections about i=1 sites 1, 2, and 3, respectively. For the square cluster, we where S ϕ = S ϕ. Furthermore, the square of the total implement the C point group operators as follows: z z 4v spin operator is readily constructed

C4 R = χ34χ23χ12 (39a) L L C2 R = χ23χ12 (39b) 2 X X S = SixSjx + SiySjy + SizSjz , (44) C3 R 4 = χ12χ23χ34 (39c) i=1 j=1 σ(13) R v = χ24 (39d) 2 where S ϕ = S(S+1)ϕ. We use the eigenvalues Sz and S σ(24) R v = χ13 (39e) as good quantum numbers to specify the quantum state (13) of the system. σ R d = χ12χ34 (39f) σ(24) R d = χ14χ23, (39g) B. Exact solutions which are 90◦, 180◦ and 270◦ rotations, and reﬂections about two diagonals and the vertical and horizontal, re- Here we analyze our ﬁndings for the triangular and spectively. square cluster tabulated in Appendices A and B. We evaluate these solutions for the expected symmetries of the

2 Hubbard Hamiltonian and find that the method leads to 3. Implementation of the S operator consistent results. We check a theorem proved by Na- gaoka concerning whether or not the ferromagnetic state The site-specific total-spin operator is compactly writ- above and below half filling is the ground state. Then, ten in terms of the Wannier creation and annihilation we compare spin-correlations of the ground states of the operators and the Pauli spin matrices as triangular and square clusters at half filling. Finally we 1 give some examples and interpretation of eigenvalue and S = a† σ a , (40) i 2 iα αβ iβ photoemission 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 are  1 −1 1 −1 1 −1 0 0 0  It is well known that, energetically, the Hubbard  −1 1 −1 1 −1 1 0 0 0  Hamiltonian possesses a symmetry above and below half     filling. This symmetry ensures a one-to-one correspon-  1 −1 1 −1 1 −1 0 0 0    dence between states above half filling with those below.  −1 1 −1 1 −1 1 0 0 0  1   The structure of the states, their exact symmetry and hφi|ΓA1|φji =  1 −1 1 −1 1 −1 0 0 0  . 6   total-spin, above and below half filling is identical pro-  −1 1 −1 1 −1 1 0 0 0    vided that the respective Sz basis, {φn}, are labeled so  0 0 0 0 0 0 2 2 2  that any state above half filling with a particular dou-    0 0 0 0 0 0 2 2 2  bly occupied site corresponds to a state below half filling   with no electrons on that site. 0 0 0 0 0 0 2 2 2 ij Lets consider the solutions for the triangular cluster (47) From this result, we see that the {ψ} basis is and, further, consider the cases with Sz = 0 above and below half filling, presented in Sec. A.2 and Sec. A.4. The 1 „ « Ne=4,Sz =0 ψ1 = √ |empi − |pemi + |mpei − |epmi + |mepi − |pmei first point to note is that {ψn } can be mapped 6 Ne=2,Sz =0 to {ψn } by taking d → e. The structure of the (48) states are seen to be identical by comparing the eigenkets 1 in Sec. A.2 to those in Sec. A.4. Finally, the energies are ψ2 = √ |deei + |edei + |eedi , (49) also seen to be isomorphic by taking t → −t and if, above 3 half filling, we shift the energy down by 3U. The origin of which are antiferromagnetic states. The lowest energy this shift is two-fold: (a) the second term of the Hubbard eigenstate in the Ne = 2 and Sz = 0 sector has A1 sym- Hamiltonian (15) causes a shift of U since d → e; and (b), metry and total spin S = 0. In the {ψ} basis, the matrix the third term of the Hamiltonian (self energy) causes a elements of the Hubbard Hamiltonian are shift of 2U in our case. The same symmetry is seen to √ ! exist for the cases with Sz = 1; see Sec. A.3 and A.5. −2t 2 2 t hψi|H|ψji = √ . (50) 2 2 t U For the case of the square cluster this symmetry is es- ij sentially the same, differing only in that the asymmetry in t is not present. The underlying reason for difference In comparing (50) with (A16) presented in Appendix A, between the triangle and square arises from the freedom note that we have subtracted 2U (self-energy) from the we have to divide the square into two sublattices such diagonals of the matrix elements for simplicity. Diago- that for any lattice point has nearest-neighbors belonging nalizing (50), we find the exact solution for the lowest to the co-lattice (Nagaoka, 1966). Then a phase differ- energy eigenstates to be ence of −1 between the wave function of the sublattices ΨSz =0,S=0 = (51) accounts for a change in the sign of t. A12 α √ |empi − |pemi + |mpei − |epmi + |mepi − |pmei 6 β + √ |deei + |edei + |eedi , 2. Check of Nagaoka’s theorem 3 where the normalized coefficients α and β are Let us consider the triangular cluster below half filling, µ N = 2. The ground state of this system could have α = (52a) e p1 + µ2 either Sz = 0 or 1. In the Sz = 0 case, the numbered 1 basis states are β = , (52b) p1 + µ2 φ1 = |empi φ2 = |pemi φ3 = |mpei with φ = |epmi φ = |mepi φ = |pmei. (45) √ 4 5 6 2t + U ± 36t2 + 4tU + U 2 φ = |deei φ = |edei φ = |eedi µ = − √ . (53) 7 8 9 4 2t The coefficients α and β (for the positive root in µ) are In this {φ} basis, the matrix elements of the A1 symme- plotted in Fig. 4b for unity hopping coefficient t = 1. try operator In the Sz = 1 case, the lowest energy state, the ferromagnetic state, has E symmetry and total spin S = 1 1 2 (1) (2) (3) √ C3 C σ σ σ S =1,S=1 ΓA1 = 1 + R + R 3 + R v + R v + R v Ψ z = (2|eppi + |pepi − |ppei)/ 3 6 E (54) E (46) → hHS=1i = −t,

7 B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

1 5 10 15 20 25 30 35 40 U state is not the ground state of the system for all values -0.5 !" of U. Nagaoka has proven, in the limit U → ∞, for an E, S=1 0.8 -1 fcc or hcp lattice that the ferromagnetic state with the -1.5 0.6 maximum$%12 = total" &1 spin+ # & would2 not be the ground state of the -2 system (Nagaoka, 1966). Our results are in agreement -2.5 0.4 # A12, S=0 with this statement, if we make an analogy between our -3 0.2 triangular cluster and fcc or hcp. -3.5 Let us see in more detail why ΨSz =0,S=0 is indeed the -4 A12 5 ground10 15 state20 25 of30 the35 system.40 U Let us consider the U = Energy 0 limit. Here the Hubbard Hamiltonian has only the (a) kinetic (b) energy term U 1 5 10 15 20 25 30 35 40 -0.5 !" X † † E, S=1 0.8 H◦ = −t aiσajσ + ajσaiσ (55) -1 hijiσ -1.5 0.6 $%12 = " &1 + # &2 -2 which we can transform to k-space to determine the band -2.5 0.4 # A12, S=0 energy. To convert from Bloch states (in reciprocal space) -3 0.2 to Wannier states (in position space) we use the discrete -3.5 inverse Fourier transform -4 5 10 15 20 25 30 35 40 U Energy 1 X 1 X a = e−ikRi a a† = eikRi a† , (a) (b) iσ N kσ iσ N kσ k k (56a) FIG. 4 Below half fillng for the triangular cluster with unity hop- and to go from position space back to reciprocal space ping coefficient. Variation with U of: (a) lowest eigenvalues’ for S = we use the discrete Fourier transform Ne=2,Sz =0 0 and S = 1; and (b) coefficients of the ΨA = αψ1 + βψ2 12 X ikRi † X −ikRi † eigenstate. akσ = e aiσ akσ = e aiσ, (56b)

{Ri} {Ri} where the possible momenta are k = 2nπ/Na and lattice where again we have subtracted 2U (the self-energy). We vectors are Rn = an, n = 0, 1, .., N − 1 and a is the see in Fig. 4a that below half ﬁlling the ferromagnetic cellsize. Substituting (56a) into (55) gives

! t 0 X X ikRi † X −ik Rj H = − e a e a 0 + h.c. (57a) ◦ N kσ k σ hijiσ k k0

t 0 0 X X X † ikRi −ik (Ri+a) −ik (Ri−a) = − a a 0 e e + e , (57b) N kσ k σ i σ kk0

where we have went from (57a) to (57b) by replacing the (52) at U = 0. Here, expanding about U = 0, we have sum over the bonds, hiji, by a sum only over the lattice for the positive root of µ sites since for nearest-neighbor interactions, Rj = Ri ± a r2 U in the hiji sum. We then have α = − − √ + O(U 2) (60a) 3 9 6 X † H◦ = kakσakσ, (58) 1 U β = √ − √ + O(U 2), (60b) kσ 3 9 3 where = −2t cos ka and where we have used k so substituting (60) into (51) gives 1 0 X i(k−k )Ri e = δkk0 . (59) ΨSz =0,S=0 → (61) N A12,U=0 i 1 √ (|deei + |edei + |eedi) − Referring to Fig. 5a and Fig. 5b, clearly for Ne = 2 the 9 Sz = 0 energy eighstate has lowest energy eigenvalue, r1 −4t, while the Sz = 1 energy eigenstate has energy eigen- (|empi − |pemi + |mpei − |epmi + |mepi − |pmei) . value −t. This agrees with Fig. 4a. Let us consider the 9

8 B Exact solutions II ONE-BAND HUBBARD HAMILTONIAN

S =0 , N =2 S =1, N =2 Therefore, the U = 0 ground state may be written as "k z e "k z e t t Sz =0,S=0 † † ΨA ,U=0 = u + v a a |0i, (63) 0 k 0 k 12 k=0↑ k=0↓ where u and v are c-numbers that satisfy the normaliza- -2t -2t 0 2!/3 4!/3 0 2!/3 4!/3 tion condition (a) (b) |u|2 + |v|2 = 1. (64) Sz=0 , Ne=4 Sz=1, Ne=4 "k "k Therefore, the ground state is superconducting and it t t contains one Cooper pair. We will explore this generic 0 k 0 k property of the ground state in the context of large Fermi systems in Lecture 9. -2t -2t 0 2!/3 4!/3 0 2!/3 4!/3 (c) (d) 4. Triangular cluster above half filling FIG. 5 Lowest energy momentum-states above and below half filling for Sz = 0, 1. Now let us consider the triangular cluster above half filling, Ne = 4. Like the case just considered, the ground state of this system could have either Sz = 0 or 1. Re- markably, however, in the above half-filled case the fer- 3. Quantum informational derivation of the ground state romagnetic and nonferromagnetic states are degenerate with eigenvalue 5U − 2t, see Sec. A.2 and A.3. The Remarkably, and as an additional consistency check, reason for this degeneracy is depicted in plots (c) and we can arrive at (61) much more simply and directly by (d) in Fig. 5. The degeneracy arises from the fact that using an a maximally entangled quantum state that is k = −2t cos ka has the same value for k = 2π/3 and localized in k-space with zero momentum (k = 0). That k = 4π/3, an effect due to the L = 3 cluster. is, we may construct this maximally entangle state, which Fig. 5a represents, using the momentum-state Wannier creation operators, (56b), as follows: 5. Predicted energy eigenstates for the triangular cluster

(62b) Energy 10 where the vacuum state for the L = 3 is |0i ≡ |eeei. Nor- √ 8 E malizing this by 9, we see that (62) is indeed identical 1 6 A1 to the solution (61) obtained by exact diagonalization of A22 4 the Hubbard Hamiltonian for L = 3 when we evaluate 3t E the exact (t, U)-dependent solution at U = 0. 2 3 A 0 21 The vacuum state |0i may be added in quantum su- 2 4 6 8 10 U perposition with (62) for the following reasons: -2 E -3t 2 -4 1. it too has quantum numbers Sz = 0 and total spin S = 0;

2. it too is an eigenstate of ΓA1, which represents A1 FIG. 6 Eigenvalues for the triangular cluster at half filling. symmetry operator4 ; 3. and it contributes zero energy to the superposition filled triangular cluster as a function of U. Each curve state (i.e. h0|H|0i = 0). is labelled according to its symmetry. The A1 and A22 energies are degenerate for all U. In Fig. 7, we have depicted all the possible Bloch states for Ne = 3. There are five types of Bloch states we can possibly choose. These, 4 The vacuum is an eigenstate of all the symmetry operators be- in turn, correspond to the five nondegenerate energies for cause they are formed out of unitary swap gates which by con- finite U plotted in Fig. 6. Consequently, we see why there struction have the vacuum as an eigenstate with unity eigenvalue. are 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 have This explains why A1 and A22 are degenerate for all U. the same energy at U = 0. It is possible to assign each Finally, we then can explain why the E3 energy is lower Bloch state a given symmetry. Immediately, we assign than the A1 and A22 energy for ﬁnite U. This is because panels (1) and (5) of Fig. 7 symmetries E1 and E2, re- the E3 state can have a component of Bloch state in panel spectively, since they are the highest and lowest energy (4) which is energetically lower for ﬁnite U than in panel states. Then, we assign panel (2) A21 symmetry since (3). from Fig. 6 the energy of this state is independent of U and we know 1 ΨA = √ |mppi + |pmpi + |ppmi . (65a) 21 3

The eigenvalues of the triangular cluster above half ﬁlling, Ne = 4, and Sz = 0 are the following q 11U + 2t + (−11U − 2t)2 − 4 (30U 2 + 12Ut − 8t2) λNe=4,Sz =0 = (66a) A11 2 q 11U + 2t − (−11U − 2t)2 − 4 (30U 2 + 12Ut − 8t2) λNe=4,Sz =0 = (66b) A12 2 λNe=4,Sz =0 = 5U − 2t (66c) A2 λNe=4,Sz =0 = 5U + t (66d) E1 q 11U − t + (−11U + t)2 − 4 (30U 2 − 6Ut − 2t2) λNe=4,Sz =0 = (66e) E2 2 q 11U − t − (−11U + t)2 − 4 (30U 2 − 6Ut − 2t2) λNe=4,Sz =0 = (66f) E3 2 and the eigenvalues below half ﬁlling are q 5U − 2t + (−5U + 2t)2 − 4 (6U 2 − 6Ut − 8t2) λNe=2,Sz =0 = (67a) A11 2 q 5U − 2t − (−5U + 2t)2 − 4 (6U 2 − 6Ut − 8t2) λNe=2,Sz =0 = (67b) A12 2 λNe=2,Sz =0 = 2U + 2t (67c) A2 λNe=2,Sz =0 = 2U − t (67d) E1 q 5U + t + (−5U − t)2 − 4 (6U 2 + 3Ut − 2t2) λNe=2,Sz =0 = (67e) E2 2 q 5U + t − (−5U − t)2 − 4 (6U 2 + 3Ut − 2t2) λNe=2,Sz =0 = . (67f) E3 2

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

10 B Exact solutions A TRIANGULAR CLUSTER SOLUTIONS

"k "k (1) t t "=3t 0 k 0 k #1

-2t -2t 0 2!/3 4!/3 0 2!/3 4!/3

"k (2) t "=0 $21 0 k

-2t 0 2!/3 4!/3

"k "k (3) t t "=0 0 k 0 k $1,$22

-2t -2t 0 2!/3 4!/3 0 2!/3 4!/3

"k "k (4) t t "=0 0 k 0 k

-2t -2t 0 2!/3 4!/3 0 2!/3 4!/3

"k "k (5) t t "=-3t 0 k 0 k #2

-2t -2t 0 2!/3 4!/3 0 2!/3 4!/3

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 the Fermi-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: 1 half filling, Ne = 3, for Sz = 2 ; above half filling, Ne = 4, for Sz = 0, 1; and, below half filling, Ne = 4, for Sz = 0, 1.

1 1. half ﬁlling: Ne = 3 and Sz = 2

1 For Sz = 2 in the case of half ﬁlling where Ne = 3 we have 9 elements in our basis. We choose our {φn} basis as follows:

φ1 = |mppi φ2 = |pmpi φ3 = |ppmi

φ4 = |dpei φ5 = |edpi φ6 = |pedi (A1)

φ7 = |depi φ8 = |pdei φ9 = |epdi.

11 1 1 half ﬁlling: Ne = 3 and Sz = 2 A TRIANGULAR CLUSTER SOLUTIONS

!"! A E G 11 2 E1 A12 20 E3 A2

A11 E 2 U 2 4 6 8 10

-5

E3 -10 A 12 E1

N =4,S =0 N =2,S =0 FIG. 8 Photoemission energies versus the tight-binding energy U: λ e z − λG in black and λ e z − λG in gray.

U!0

U!1

U!2

Out[130]=

U!3

U!4 Spectrum 3.0 2.5 2.0 U!5 1.5 1.0 0.5 Energy "10 "5 0 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 eigenvectors 2 of S , so here we do not get any reduction in block size using the total-spin. Therefore, the {ϕn} basis is equivalent to {ψn}.

12 2 Above half ﬁlling: Ne = 4 and Sz = 0 A TRIANGULAR CLUSTER SOLUTIONS

1 a. A1 State with S = 2

√ A1 ψ1 = (φ4 + φ5 + φ6 + φ7 + φ8 + φ9)/ 6 → hH 1 i = (4U). (A2) S= 2

1 3 b. A2 States with S = 2 , 2

√ A2 ψ2 = (φ1 + φ2 + φ3)/ 3 → hH 3 i = (3U) (A3) S= 2

√ A2 ψ3 = (φ4 + φ5 + φ6 − φ7 − φ8 − φ9)/ 6 → hH 1 i = (4U) (A4) S= 2

1 c. E States with S = 2

√    ψ = (φ + φ + ∗φ )/ 3 3U (∗ − 1)t (1 − )t 4 1 2 3 √  ∗ E ∗ ψ5 = (φ4 + φ5 + φ6)/ 3 → hH 1 i =  ( − 1)t 4U ( − 1)t , (A5) √ S= 2   ∗  ∗ ψ6 = (φ7 + φ8 + φ9)/ 3  (1 − )t ( − 1)t 4U

−i 2π where = e 3 .

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 = |dmpi φ2 = |pdmi φ3 = |mpdi

φ4 = |dpmi φ5 = |mdpi φ6 = |pmdi (A6)

φ7 = |eddi φ8 = |dedi φ9 = |ddei

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

a. A1 States with S = 0

√ ) 4t ! ϕ = ψ = (φ − φ + φ − φ + φ − φ )/ 6 5U + 2t √ 1 1 1 2 3 √ 4 5 6 → hHA1 i = 2 . (A7) ϕ = ψ = (φ + φ + φ )/ 3 S=0 √4t 6U 2 2 7 8 9 2

b. A2 State with S = 1

√ A2 ϕ3 = ψ3 = (φ1 − φ2 + φ3 + φ4 − φ5 + φ6)/ 6 → hHS=1i = (5U − 2t). (A8)

13 4 Below half ﬁlling: Ne = 2 and Sz = 0 A TRIANGULAR CLUSTER SOLUTIONS c. E States with S = 0, 1

√ ψ = (2φ + φ − φ )/ 5 (A9a) 4 1 2 3 √ ψ = (2φ + φ − φ )/ 5 (A9b) 5 4 5 6 √ ψ6 = (2φ7 − φ8 − φ9)/ 5 (A9c)

√ E ϕ4 = (ψ4 + ψ5)/ 2 → hHS=1i = (5U + t) (A10)

√ ) ! 5U − t −√2t ϕ5 = (ψ4 − ψ5)/ 2 E 2 → hHS=0i = . (A11) ϕ = ψ −√2t 6U 6 6 2

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 = |dppi φ2 = |pdpi φ3 = |ppdi. (A12)

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

a. A2 State with S = 1

√ A2 ϕ1 = ψ1 = (φ1 − φ2 + φ3)/ 3 → hHS=1i = (5U − 2t). (A13) b. E state with S = 1

√ E ϕ2 = ψ2 = (2φ1 + φ2 − φ3)/ 3 → hHS=1i = (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 = |empi φ2 = |pemi φ3 = |mpei

φ4 = |epmi φ5 = |mepi φ6 = |pmei (A15)

φ7 = |deei φ8 = |edei φ9 = |eedi

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)/ 6 A1 2U − 2t 2 2t √ → hHS=0i = √ . (A16) ϕ2 = ψ2 = (φ7 + φ8 + φ9)/ 3 2 2t 3U

14 B SQUARE CLUSTER SOLUTIONS

b. A2 State with S = 1

√ A2 ϕ3 = ψ3 = (φ1 − φ2 + φ3 + φ4 − φ5 + φ6)/ 6 → hHS=1i = (2U + 2t). (A17)

c. E States with S = 0, 1

√ ψ = (2φ + φ − φ )/ 5 (A18a) 4 1 2 3 √ ψ = (2φ + φ − φ )/ 5 (A18b) 5 4 5 6 √ ψ6 = (2φ7 − φ8 − φ9)/ 5 (A18c)

√ E ϕ4 = (ψ4 + ψ5)/ 2 → hHS=1i = (2U − t) (A19)

√ ) ! 2U + t −√2t ϕ5 = (ψ4 − ψ5)/ 2 E 2 → hHS=0i = . (A20) ϕ = ψ −√2t 3U 6 6 2

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 = |eppi φ2 = |pepi φ3 = |ppei . (A21)

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

a. A2 State with S = 1

√ A2 ϕ1 = ψ1 = (φ1 − φ2 + φ3)/ 3 → hHS=1i = (2U + 2t). (A22)

b. E State with S = 1

√ E ϕ2 = ψ2 = (2φ1 + φ2 − φ3)/ 5 → hHS=1i = (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 as follows

φ1 = |mpmpi φ2 = |pmpmi

φ3 = |ppmmi φ4 = |mppmi φ5 = |mmppi φ6 = |pmmpi

φ7 = |dempi φ8 = |pdemi φ9 = |mpdei φ10 = |empdi

φ11 = |dpmei φ12 = |edpmi φ13 = |medpi φ14 = |pmedi

φ23 = |dmepi φ24 = |pdmei φ25 = |epdmi φ26 = |mepdi

φ27 = |dpemi φ28 = |mdpei φ29 = |emdpi φ30 = |pemdi

φ31 = |dedei φ32 = |ededi

φ33 = |ddeei φ34 = |deedi φ35 = |eddei φ36 = |eeddi.

a. A1 States with S = 0, 1

ψ = (φ − φ + φ − φ )/2 (B2a) 1 3 4 5 6 √ ψ = (φ − φ + φ + φ − φ − φ + φ − φ )/ 8 (B2b) 2 7 8 9 10 11 12 13 14 √ ψ = (φ − φ + φ + φ − φ − φ + φ − φ )/ 8 (B2c) 3 15 16 17 18 19 20 21 22 √ ψ = (φ − φ − φ + φ − φ + φ + φ − φ )/ 8 (B2d) 4 23 24 √ 25 26 27 28 29 30 ψ5 = (φ31 + φ32)/ 2 (B2e)

ψ6 = (φ33 + φ34 + φ35 + φ36)/2 (B2f) √ A1 ϕ1 = (ψ2 + ψ3)/ 2 → hHS=1i = (5U). (B3)    5U 0 0 0 0 ϕ2 = ψ4    0 4U 2t 0 0  ϕ3 = ψ1    √  A  4t  ϕ = (ψ − ψ )/ 2 → hH 1 i =  0 2U 5U √ 2t  . (B4) 4 2 3 S=0  2  ϕ = ψ   0 0 √4t 6U 0  5 5   2   ϕ6 = ψ6 0 0 2t 0 6U 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

√ ψ = (φ − φ )/ 2 (B5a) 7 1 2 √ ψ = (φ − φ + φ + φ + φ + φ − φ + φ )/ 8 (B5b) 8 7 8 9 10 11 12 13 14 √ ψ = (φ − φ + φ + φ + φ + φ − φ + φ )/ 8 (B5c) 9 15 16 17 18 19 20 21 22 √ ψ10 = (φ23 − φ24 − φ25 + φ26 + φ27 − φ28 − φ29 + φ30)/ 8 (B5d)    4U 0 √4t ϕ7 = ψ7  2  A  −4t  ϕ = ψ → hH 2 i = 0 5U √ . (B6) 8 10 √ S=1  2    4t −4t  ϕ9 = (ψ8 + ψ9)/ 2  √ √ 5U 2 2 √ A2 ϕ10 = (ψ8 − ψ9)/ 2 → hHS=0i = (5U). (B7)

16 1 Half ﬁlling: Ne = 4 and Sz = 0 B SQUARE CLUSTER SOLUTIONS

c. B1 States with S = 0, 1

√ ψ = (φ + φ + φ − φ − φ + φ + φ + φ )/ 8 (B8a) 11 7 8 9 10 11 12 13 14 √ ψ = (φ + φ + φ − φ − φ + φ + φ + φ )/ 8 (B8b) 12 15 16 17 18 19 20 21 22 √ ψ = (φ + φ − φ − φ − φ − φ + φ + φ )/ 8 (B8c) 13 23 24 √ 25 26 27 28 29 30 ψ14 = (φ31 − φ32)/ 2 (B8d) √ B1 ϕ11 = (ψ11 + ψ12)/ 2 → hHS=1i = (5U). (B9)

√   −4t 4t  ϕ = (ψ − ψ )/ 2 5U √ √ 12 11 12  2 2 → hHB1 i =  −√4t 5U 0  . (B10) ϕ13 = ψ13 S=0  2    4t  ϕ14 = ψ14  √ 0 6U 2

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

√ ψ15 = (φ1 + φ2)/ 2 (B11a) ψ = (φ + φ + φ + φ )/2 (B11b) 16 3 4 5 6 √ ψ = (φ + φ + φ − φ + φ − φ − φ − φ )/ 8 (B11c) 17 7 8 9 10 11 12 13 14 √ ψ = (φ + φ + φ − φ + φ − φ − φ − φ )/ 8 (B11d) 18 15 16 17 18 19 20 21 22 √ ψ = (φ + φ − φ − φ + φ + φ − φ − φ )/ 8 (B11e) 19 23 24 25 26 √ 27 28 29 30 ψ20 = (φ33 − φ34 − φ35 + φ36)/ 4 (B11f) √ p B2 ϕ15 = (ψ15/ 2 + ψ16)/ 3/2 → hHS=2i = (4U). (B12)

√ ) ! ϕ16 = (ψ17 + ψ18)/ 2 B2 5U 0 → hHS=1i = . (B13) ϕ17 = ψ19 0 5U

√ √   4U −√6t 0  ϕ18 = (−2ψ15/ 2 + ψ16)/ 3  3 √  B ϕ = (ψ − ψ )/ 2 → hH 2 i =  −√6t 5U −2t . (B14) 19 17 18 S=0  3   ϕ20 = ψ20  0 −2t 6U

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

   4U 0 −√2t √2it ϕ21 = ψ21  2 2 √    ϕ22 = (ψ26 + ψ27)/ 2   0 5U (−1 + i) t (−1 − i) t √ → hHE i =   . (B16) ϕ = (ψ + ψ )/ 2 S=1  −√2t (−1 − i) t 5U 0  23 23 25 √   2   −2t ϕ24 = (ψ22 + ψ24)/ 2  √ (−1 + i) t 0 5U 2

√   2t  ϕ = (−ψ + ψ )/ 2 5U 0 (−1 + i) t √ 25 23 25 √  2   0 5U (1 + i) t √2it  ϕ26 = (−ψ22 + ψ24)/ 2 E  2  √ → hHS=0i =   . (B17) ϕ27 = (ψ26 − ψ27)/ 2 (−1 − i) t (1 − i) t 5U 0      2t −2it ϕ28 = ψ28  √ √ 0 6U 2 2

References

Chen, C., A. Reich, and L. M. Falicov, 1988, Phys. Rev. B 38(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.