The Hubbard Model for Dummies Introduction Some math

2nd quantization The partition function

Limiting cases t = 0 U = 0 Mean field theory

Matrix notation and the t-J model Extensions of the Hubbard model

Bruce Patterson13.6.07

The Hubbard Model: Introduction J. Hubbard, Proc. Roy. Soc. London, A266, 238 (1963).

Simplest model of interacting particles in a lattice: extension of the tight-binding model to include short-range el-el interactions.

One-band Hubbard Hamiltonian (one orbital per site):

Half-filling: on average one electron per site.

Electrons may hop from site to site, but for large U/t, we get a Mott insulator.

Note: One of the first correlated electron calculations: Heitler+London treatmentof H2-molecule (1927).

H = t c j

ij ci +U ni

i ni impliesadjacent sites

hopping term interaction term

Second Quantization Boson annihilation and creation operators for the harmonic oscillator:

From , we verify that and that the oscillator Hamiltonian is:

From the ground state , we build up the excited states:

a m2h

x + i 12mh

p

a m2h

x i 12mh

p

x, p[ ] = ih

a,a[ ] =1

Hosc =12m

p2 + m2

2x 2

= h aa + 12

= h n +

12

n=aa = number operator.

0

a n = n +1 n +1 Hosc n = h n +12

n

Electron states for the (N=8) Hubbard model in terms of fermionic operators:

c1 0 = ,,,,,,,

c5 0 = ,,,,,,,

c5 c1

0 = ,,,,,,,

c4 c4

0 = ,,,,,,,

c1 0 = 0

c1 c1

0 = 0

ci ,c j '{ } ci c j ' + c j 'ci = 0ci ,c j '

{ } = 0ci ,c j '

{ } = ij '

{} = anticommutator

c1 c1,,,,,,, n1,,,,,,, =1,,,,,,,

c5 c5,,,,,,, n5,,,,,,, = 0,,,,,,,

Partition Function Grand Canonical Ensemble: particles can be exchanged with a resevoir.

Thermal expectation value of operator A:

Consider a single state with 0 or 1 electrons with energy :

Average occupation:

partition function : Z Tr e Hn( )[ ]

1kT

"chemical potential"

H = cc = n states : 0 , 1{ }

Z = 0 e Hn( ) 0 + 1e Hn( ) 1 =1+ e ( )

A = 1ZTr Ae Hn( )[ ]

n = 1Z

0 ne Hn( ) 0 + 1 ne Hn( ) 1[ ] = 1Z 0 + e ( )[ ] = e

( )

1+ e ( )

=1

e ( ) +1= Fermi - Dirac distribution

Hubbard Model with t=0 No hopping implies independent sites. We thus consider a single site and

calculate its average occupation.

possible states = 0 , , ,{ }

H =Unn

Z =

e Hn( ) =1+ e + e + eU +2

n = 1Z

n + n( )e Hn( ) =1Z0 + e + e + 2eU +2[ ]

=2 e + eU +2( )1+ 2e + eU +2

Note: Since , at half-filling, the energy to add another particle to the

site jumps by U. This is the Hubbard gap.

0,6

0,8

1

1,2

1,4

1,6

1,8

2

0 0,5 1 1,5 2 2,5

T=0.05

T=0.1

T=0.5

U=1

"half filling"

"Hubbard gap" (=U)

=En

Hubbard Model with U=0 Non-interacting limit: we go into reciprocal space:

where k takes the values kn=2n/N. (1-d, periodic boundary conditions)

ck =

1N

eik ll cl

H = t c j cl

jl = tN e

ikjeik' lck ck '

jl

k,k '

=tN

ck ck' e

ikjeik' j+1( ) + eikje ik' j1( )[ ] = t ck ck ' eik' + eik'[ ] 1N ei(kk' ) j

j

k,k',

j

k,k',

= t ck ck' 2cosk'[ ]k,k'

k,k ', = knk

k k = 2t cosk

For U=0, the energy-eigenvalues of H are:

The bandwidth is equal to 4t.

k = 2t cosk kn =2nN

n = 0,1,2,...N 1

-3

-2

-1

0

1

2

3

-5 0 5

k

[t]

k [2/N]

4t

The Mott Transition: U>>t: diagonalize H in real space - at half-filling we obtain a static

(insulating) system.

t>>U: diagonalize H in reciprocal space - obtain itinerant electron band,width 4t.

At some intermediate value of U/t, there will be a metal-to-insulaltortransition: the Mott transition.

Mean Field Theory We wish to treat the full Hubbard model (in 1-d):

Problem: the second term is proportional to c4.

Mean field approximation: each electron feels an average interaction from all others. This ignores possibly important electron correlations.

Write:

and assume is small, and analogously for the down spins. Expand:

H = t c j c +U nlnl

l

jl

nl = n + nl n[ ] n + l

nlnl n n + l n + l n nl n + nl n n n

In reciprocal space:

H is now diagonal:

We define spin-dependent Fermi wavevectors:

After integrating, we obtain for the total energy:

cl =

1N

e iklk ck

H = 2t cosk nk + nk( ) +U nk n + nk n( )[ ] UN nk n

= knk + knk[ ] UN nk n k 2t cosk +U n

=k N

dk

0

kF

kF = nN

Etot =2Nt

sin n + sin n[ ] +UN n n

For half-filling: write Etot as a function of

n + n =1

Etot n( )4Nt

= sin n +U4t

n 1 n( )

-0,3

-0,2

-0,1

0

0,1

0,2

0,3

0 0,2 0,4 0,6 0,8 1

E /4

Nt

U/4t = 33.25

3.53.754

4.25

4.5

4.75

5 small U a paramagnetlarge U a ferromagnetThis is completely wrong!The 1-dimensional Hubbard model hasbeen exactly solved by Lieb and Wu, PRL20, 1445 (1998). At infinitesimal U/t anantiferromagnetic state forms.

n

Hubbard Model in Matrix NotationToy Model

(simplified H2 molecule) Two orthonormal orbitals separated by some distance:

Add a single electron, which moves with an amplitude -t. Tight-binding" Hamiltonian:

Eigenvectors:

Energy eigenvalues: Note: + is the ground state.

H =0 tt 0

=12

1 2( ) =1211

= mt

1 =10

, 2 =

01

1 = , , 2 = ,schematically:

We now consider two electrons:

Hamiltonian

ionic states

covalent states

, = c2 c1

0

, = c2 c1

0

, = c1 c1

0

, = c2 c2

0

H =

0 0 t t0 0 +t +tt +t U 0t +t 0 U

,,,,

Note: +t due to fermion transposition

Diagonalize H:

Energy levels:

=U2

U 2 +16t 2

2 =

, , 2t

, + [ ]

2 + 2 / 2t 2( )

cov = 0 cov =12

, + ,( )

ion =U ion =12

, ,( )

-2

0

2

4

6

8

0 1 2 3 4 5 6 7 8

/t

U/t

+

ion

-

cov

Consider the limit of large U:

- is now largely covalent, with a small ionic admixture.

With downfolding, we will project out the interesting, low-energy, covalent-like part of Hilbert space.

cov = 0

4t 2

U

A note on matrix inversion by partitioning:

Ref: Ch. 2.7, Numerical Recipes, 2nd Ed. (1992)

A =A0 A1A2 A3

We divide a matrix A into theblocks A0-A3.

A1 =B0 B1B2 B3

The inverse can also be written inblock form:

B0 = A0 A1 A31 A2( )1

B1= B0 A1 A31

B2 = A31 A2 B0B3 = A31 B2 A1 A31

We begin downfolding by partitioning H into blocks:

The Greens function can also be partitioned:

and the (energy-dependent) covalent part G00 can be written:

This looks like the Greens function for the effective Hamiltonian:

H =

0 0 t t0 0 +t +tt +t U 0t +t 0 U

H00 T01T10 H11

H00 = covalent HamiltonianH11= ionic HamiltonianTij=covalent-ionic transitions

G ( ) H( )1 = H00 T01T10 H11

1

G00 G01G10 G11

Heff ( ) = H00 + T01 H11( )1T10 Heff 0( ) 0 is a typical covalent energy.

G00 ( ) = H00 + T01 H11( )1T10[ ]( )

1

We evaluate the effective Hamiltonian between the covalent states:

which diagonalizes to:

These states correspond to singlet and triplet states of the H2 molecule.

The other triplet states, , also have zero energy. (There is no

Coulomb energy, and the electrons cannot hop, due to the Pauli principle.)

Heff 0( ) = H00 + T01 0 H11( )1T10 =

0 00 0

+

t tt t

0 U 00 0 U

1t tt t

Heff 0( ) = 2t 2

U1 11 1

,,

s = 4t 2

Us

12, ,( )

t = 0 t =12, + ,( )

Note that we choose 0=0.

, , ,

Recall that s includesa small ionic admixture.

Kinetic Exchange

Anti-parallel alignment of the spins is favored, since it allows theelectrons to hop to the neighboring site.

We can write Heff in terms of the fermion operators:

Consider the properties of the Pauli spin matrices:

Heff = 2t 2

Uc1 c1c2

c2 c1 c1c2

c2 c1 c1c2

c2 + c1 c1c2

c2( )

+ 10

01

r S = 1

2 x

) x + y) y + z

) z ( )

x =0 11 0

y =

0 ii 0

z =

1 00 1

x = cc + c

c( ) = m y = i c

c cc( ) = i m

z = cc c

c( ) = n n( ) =

We can therefore rewrite Heff :

The Hubbard Hamiltonian, for large U/t, has led us to an

antiferromagnetic exchange coupling J=4t2/U.

This is the basis of the t-J model.

Heff =4t 2

Ur S 1

r S 2

n1n24

Heisenberg Hamiltonian

ni ni + ni

We have solved the 4x4 Hubbard Hamiltonian by exact diagonalization. Forlarger systems, matrix expressions for H grow quickly in size. For N sites:

For 30 sites and 15 spins of each orientation, this dimension is2.4 x 1016.

This is why we use 2nd quantization.

No exact solution the Hubble model in >1 dimension is known.

dimension of Hilbert space = Nn

Nn

Nn

N!n! N n( )!

21

21

= 4

(mutually inconsistent) approximate solutions of the 2-DHubbard Model

Results of 6 groups,compiled by M.P. March,Condensed MatterPhysics, Wiley, 2000.

n 1 = 0 half - filling

Extensions of the Hubbard Model Three-Band Hubbard Model To describe the 2-d CuO2 layers in superconducting cuprates, this modelincludes:

Cu(3dx2-y2) states (Ud)O(2px,y) states (Up)their nn interaction (Upd)

For large Upd, this Hamiltonian reduces to the t-J model.

2-d hole-pairing:

AF-coupling causes holes to form pairs.

unpaired: 8 bonds broken paired: 7 bonds broken

Charge-Spin separation:

In 1-d, Fermi liquid theory breaks down, due to the destruction ofthe Fermi surface (Fermi points) by an arbitrarily small perturbation.In this case, separate gapless charge (holon) and spin (spinon)excitations arise. This state is a Luttinger liquid.

Creation in photoemissionof a holon and a spinon.

The Hubbard model with an infinite number of dimensions Each site of the Hubbard model can then be treated as an isolated

impurity interacting with an electron resevoir. This problem can beattacked with the dynamical mean-field theory.

Ref: G. Kotliar and D. Vollhardt, Phys. Today March, 53 (2004).

At intermediate U/W (=U/t), we now see asharp DOS peak at the Fermi level (Kondoresonance).

The Hubbard-Holstein model (with electron-phonon interaction)

Ref: W. Koller, et al, cond-mat/0312367v2 (2004).

HHH = t c j ci +U ni

i

ij ni g bi + bi( )ni

i +0 bi

i bi

bi = creation operator of the

local phonon mode at site i

g = electron - phonon coupling constant

In the limit of large 0, theH-H model maps onto theHubbard model withHeff=U-2g2/ 0.

Introductory References to the Hubbard Model

E. Koch, Electronic Structure of Matter: Electron Correlations, inProbing the Nanoworld, lecture manuscripts of the 38th SpringSchool, Jlich (2006).

R.T. Scalettar, Elementary Introduction to the Hubbard Model,lecture notes, UC Davis,http://leopard.physics.ucdavis.edu/rts/p210/hubbard7.pdf

M.P. March, Condensed Matter Physics, Wiley, 2000.

Some examples of 2-electron calculations:

c1 c2, = c1

c2 c2 c1

0 = c1 c2

c2 c1 0 = +c1

c2 c1

c2 0 = 0

c1 c2 , = c1

c2c2 c2

0 = c1 c2

0 c1 c2

c2 c2 0 = c2

c1 0 = ,

c2 c1, = c2

c1c2 c1

0 = c2 c2

c1c1 0 = c2

c2 0 + c2

c2 c1

c1 0 = c2 c2

0 = ,