Fractional charge in 1d(see e.g. R.Rajaraman, cond-mat/0103366)

Fractional Charge in Field Theory (in 1 and 3 d) was introduced by Jackiw-Rebbi, PRD (1976)

Consider case in 1+1 d , i.e. one space dimension + time

22 2 22

Let ( , ) Fermi field (x,t) = scalar field, coupled through the mass,and the Lagrangian

1 1

L [( ) ( )

L=L with

1 ]2

:

2B

B F

g t x

x t

L

L ( ( , ))F i m x t

2 222 2 2 3

2 2 2

1 1L [( ) ( ) 1 ] nonlinear field equation ( - )= ( , ) ( , ) .2 2B x t x tg t x t x

1

2

22 2

3

2

[ ]( ) 1 1 ( ) 2Indeed, ( ) [ ] ,

2 2 [ ] [ ]2 2

[ ]2( ) [ ] ( ) ( ) =

2 [ ]2

xTanhx d x d xx Tanh x xdx dxCosh Cosh

xTanhxx Tanh x x xCosh

2

2 23

2 2The nonlinear field equation ( - )= ( , ) ( , ) solved by

( , ) 1 . This is called vacuum sector, with 2 vacua : semiclassically,( , ) 1 ( , ) 1 L ( ) L ( )F F

x t x tt x

x tx t i m i mx t

Additional static (time-independent) solutions: soliton sector

23

2The nonlinear field equation - = ( , ) ( , ) solved by ( ) [ ].2xx t x t x Tanh

x

4 2 2 4

1.0

0.5

0.5

1.0

fx_ Tanh x2

3

2 23

2 2

kink( ) [ ] is the static solution of ( - )= ( , ) ( , )

antikink2xx Tanh x t x t

t x

Vacuum sector:classical solution for 1semi

4

2 1

In 1+1 d spinless case two components are enough; we may take , = . Then, (-i ) ( ) ( )(-i ) ( ) ( )

x k k k

x k k k

m u x E u xm u x E u x

2 2

( ( , )) ( , ) ( ) Dirac's Lagrangian

( , ) with , two component spinors, E =

,

0 0usually and

0 0

F

kk k k

k

D

kk

k

L i m x t x t i m

ux t u u k m

u

H p m

II

, since they anticommute.

Dirac’s theory

field in 1+1 dimensional theory

Electron-Positron field in Dirac’s theory

( . ( ) ) † ( . ( ) )

1,2

†

2 2 4

( , ) [ ( ) ( ) ( )v ( ) ]

annihilates electron with spin s, creates positron with spin s,

(k)= ( ) .

i k r k t i k r k ts s s s

k s

s s

r t c k u k e b k k e

c b

c k m c

Field in 1+1 dimensional theory

†( , ) [ ]

Dirac's vacuum: v 0 d v 0.

k kiE t iE tk k k k

k

k k

x t b u e d u e

b

Charge conjugation in Dirac’s theory

1

2

3

4

If solves ( , ) ( , ) 0ieA mcx t x tc

*4

**3

2*2

*1

then = solves ( , ) ( , ) 0

00

C C C

kk

k

ieA mcx t x tc

ii

3 3

3 3 3

3

( , )

sends particles to antiparticles; in addition, [ , ] 0. (-i ) ( ) ( ) (-i ) ( ) charge conjugation operator.

k

k

k k D

x k k k x k

ux t

u

u u Hm u x E u x m u x

†_

† †

† † † †

† †

Standard charge density in Dirac's theory:1 ( , )= [ ( , ), ( , )]2

1ˆIndeed te charge is ( , ) ([ , ] [ , ])2

1 { (1 ) (1 )}2

{ }.

number

k k k kk

k k k k k k k kk

k k k kk

x t x t x t

Q dx x t b b d d

b b b b d d d d

b b d d

of particles - number of antiparticles 7

Charge conjugation in 1+1 dimensional theory

Charge operator in Dirac’s theory

4 2 2 4

1.0

0.5

0.5

1.0

Positive energy continuum

Positive energy continuum

negative energy continuum

localized state

negative energy continuum

Soliton sector

Solving Dirac’s equation one finds a single 0 mode (solution with E=0). There are two Vacuum states with the zero mode filled or unfilled.These ground states:differ by charge e and are connected by the charge conjugation operator C since H anticommutes with C, therefore the conjugate of a g.s. must be a g.s. with opposite charge.

the two ground states must have charge ½ and - ½ .

Solution in Soliton Sector

Fractional Charge in Polyacetilene, Su,Schrieffer and Heeger prl 1979

Unstable vacuum

Stable vacuum B

Stable vacuum A

Soliton

10

Compare stable vacuum A and vacuum+2 solitons: perturbation is local

however solitons can be delocalized:

11

however solitons can be delocalized:

How many bonds change overall? 7

6

It is a matter of 1 electron (per spin) per bond, that is, ½ electron per soliton. 12

Relation to field theory model

The Dirac equation arises by linearizing the energy dispersion near the “Dirac points” which are intersections of the energy dispersion with the Fermi level.

13

The Dirac equation is in one spatial dimension and involves twocomponents, corresponding to two Dirac points:

3 2

2

2

( ) , = =

( )Phonon field: '( ( ))

p g E

d x V xdx

1 1[ , ] 0 sends eigenfunction from E to -E.1The zero mode has eigenvalue .2

H

14

Analogy with field theory model

15

Magnetism from Coulomb interactions

Magnetism si caused by electrostatic interactions independent of the spin, and is essentially an effect of correlation. But how much do we know the way in which this happens? What causes the antiferromagnetic order in CuO2 and NiO? Much progress has been done by the Hubbard model and related models.

The CuO2 antiferromagnetic

order

Hubbard Model with nearest-neighbor hopping on a lattice L

†

,x y x x

x y x

H t c c U n n

L L

Many interesting resulys are known for bipartite lattices.

Bipartite lattice (first neighbors of black sites are red, first neighbors of red sites are black)

16

John HubbardLondon 1931-San Jose 1980

Hubbard Model with nearest-neighbor hopping on a lattice L

†

,x y x x

x y x

H t c c U n n

L L

Many interesting resulys are known for bipartite lattices.

Bipartite lattice (first neighbors of black sites are red, first neighbors of red sites are black)

1717

Bipartite latticesChain (d=1), square (d=2) , cubic (d=3) lattices

†

,, . . 0h i j h

i jH t c c i j nn t

Chain:

1( ) common eigenfuctions of translation an

2 o

d

c sk

i

h

knn e T HN

t k

Square lattice

( )

2

1( ) common eigenfuctions of translation and

2 [cos cos ]

x x y y

k

i k

y

n

h

k

x

nn e TN

t k k

H

Bipartite lattice

†

,, . . 0h i j h

i jH t c c i j nn t

Cubic lattice

( )

2

2 [cos cos co

1(

s

)

]

x x y y z zi k

k h x y z

n k n k nn e common eigenfuctions of translation T and H

t k k kN

Bipartite lattice

20

† † 1 01 1

0 12 2xx x

z x xx xx

cc cS n n

c

L L

† †† †0 1

, 0 0

xx xx x x x

x x xx

cc cS c c S c c

c

L L L

The total spin is conserved. Recalling the general rule for writing operators in second quantization,

Spin in Hubbard Model

†

,x y x x

x y x

H t c c U n n

L L

†ˆ ( ) ( ), with ( ) , one getsxcV dx x V x xc

2 2and one finds S2z

S S S SS

.

Possiamo fare una trasformazione canonica sui soli stati di spin su introducendo le buche

col che il termine cinetico di spin su diventa

T tx ,yL d x d y

tx ,yL d y

d x ;

A questo punto in un reticolo bipartito possiamo ripristinare il segno cambiando segno a t; questa e’una gauge, perche’ equivale a cambiare di segno gli spinorbitali di un sottoreticolo,

d x x cx

x 1 21

†x xd c

Trasformazione a U negativo

†

,

( ) ( )x y x xx y x

H U t c c U n n H H U UN

L L

Con questo, quando scambiamo creazione e distruzione, solo il termine in U cambia segno:

Abbiamo mappato il problema repulsivo e-e in uno attrattivo e-h. Pero’ gli operatori di spin originali una volta espressi in termini e-h cambiano forma:

Questi operatori e-e non hanno piu' nel problema attrattivo il significato di spin (si chiamano infatti pseudospin). Essi seguitano ad essere conservati, insieme a quelli e-h di spin.

2 † † 11 1 1( ) 12 2 2 2z zx x x x x x x x

x x x

n nS n n S x d d n d d n

† ( ) .x x x xx x

S c c S x d c

22

† † † †

,

[ ] , x y x y x x x x xx y x x

H t d d c c U d d c c UN N n L L L

e ' conservato, quindi e’ una costante.xx

N n L

23

Theorems on Ferromagnetism in Hubbard Model I state without proof some theorems

No ferromagnetism at small U. The minimum energy increases with spin.Theorem:

1 2 1

min min

single particle levels..... 0

(that is ) ( 1U ) ( )< ba

If

nd

dth

wieN N

max max

U

E S E S

Pieri, Daul, Baeriswyl, Dzierzawa, and Fazekas theorem: no ferromagnetism in Hubbard model at very low electron density.

24

Teorema di Lieb-Mattis Phys. Rev. 125, 164 (1962)

Consideriamo il modello di Hubbard repulsivo in 1d con obc (open boundary conditions)

Gli hoppings (a primi vicini) e gli U possono dipendere dal sito. Sia Emin(S) l’energia dello stato fondamentale con spin S. Allora, con qualsiasi filling,

min min( ) ( 1).E S E S

Piu’ basso lo spin piu’ bassa e’ l’energia. Questo e’ ovvio per U=0quando i livelli si riempiono secondo l’aufbau

†

,x x y x x x

x y x

H t c c U n n

ma resta vero con U. Quindi non c’e’ ferromagnetismo. L’avevamo visto con il modello di Ising. Niente transizioni di fase in 1d.Rientra nei teoremi di Lieb che dimostreremo piu’ avanti.

25

Strong coupling half filled Hubbard Model, d=2 or 3

At order 0 in hopping t, there is an electron per site, and huge degeneracy (each spin can be +/-). For 2 sites one has 4 ground states:

†

,

,x y x xx y x

H t c c U n n U t

Half filling: number of sites= number of electrons

is forbidden for infinite U

In first order in t one gets nothing (H takes to doubly occupied states orthogonal to g.s.)

configuration

26

In second order there are processes where an electron hops from site a with spin + to neighbouring site b (provided that the ground state spin there is -) producing a doubly occupied virtual state (energy U) and back. One can describe that by an effective Hamiltonian:

† † † †1' , ( ), ( ),b a a b b a a b a b a b b a b aU

U b b

H H H H t c c c c H t c c c cH

H Un n

2† † † †

2† † † † † † † †

2† † † †

' ( )( )

(1 ) (1 )

a b a b b a b a

a b b a a b b a a b b a a b b a

b a a a b b a a b b b a

tH c c c c c c c cU

t c c c c c c c c c c c c c c c cUt n n c c c c c c c c n nU

†

,

,x y x xx y x

H t c c U n n U t

a b

This is impossible for parallel spins. Scond-order corrections are always negative. So antiparallel configurations are lower.

27

In the CuO structure with half filled Cu band this mechanism favors antiferromagnetism.

Important case: Three-band Hubbard Model of CuO

The AF configuration allows the second-order spin exchange. This further lowers the energy.

The Cuprates are known to be antiferromagnets at half filling. By hole doping (and sometimes also by electron doping) they become high temperature superconductors.

28

1 ( )2a b az bz a b b aS S S S S S S S

1

2z x xx

S n n

† †with x x xx

S c c

Non-interaction terms linear in n just renormalize all site energies. We ignore them.

2

' a b a b b a a btH n n n n S S S SU

Besides aba processes one must include bab and all other sites on same footing. Neglecting parallel-spin interactions , part of the physics corresponds to an Hamiltonian of the form

Heisenberg m nmn

H J S S

sometimes written as:1( )4Heisenberg m n

mn

H J S S

2† † † †

Let us consider the contributions inside the second-order perturbation

' (1 ) (1 )b a a a b b a a b b b atH n n c c c c c c c c n nU

29

I just mention a related model that P.W. Anderson likes for Cuprate superconductors: the t-J model, appropriate for strong coupling:

† 1( )4i j i j i j

ij ij

H t c c J S S n n

1( )4Heisenberg m n

mn

H J S S

This model was considered on a bipartite AB lattice in a famous paper by Lieb and Mattis (J. Mathematical Physics 3, 749 (1962)). They were able to show that in the ground state the spin of the elementary cell is 2S=|B|-|A| where |B| and |A| are the numbers of sites in the two lattices. So one can indeed have magnetism.We use this result later.

30

for t=t’/2 the ground state has S=1 if U is large (ferro) and S=0 otherwise

For t=t’>0 lo stato fondamentale ha S=1 per ogni U (ferro).

1

2

3t

't 't

Toy Model: 2 electrons on 3 sites

(Hal Tasaki , Cond-mat/9512169)

Perron-Frobenius theorem for real symmetric matrices

Proof. Consider the eigenvalue equation Mu= u. .ij j i

j

m u u

Hence either the components ui have both signs or they must be all strictly positive (or all strictly negative).

Given a matrix { }, 0

with a tight binding model that represents a connected tight-binding cluster:1) the lowest eigenvalue is nondegenerate2)the corresponding eigenvector can be taken wit

ij ij ji ijM m m m m

M

h all strictly positive components

31

If all 0 , 0 strictly, then cannot vanish.

So, all 0 all 0.

j ij j ij

j j

u m u u

u u

One cannot change sign to some components without changing the summations (and thus violating the above conclusions) except the case of disconnected clusters that we have excluded by hypothesis. Then v=u (or v=-u, which is the same solution)

Then u=v has all strictly positive components.

Two ground eigenvectors cannot exist, because they cannot be orthogonal.Then u is the only ground eigenvector .

0

( , ) (v, v) because 0, but cannot go below

minimum eigenvalue (v, v) v v . Therefore,

( , ) (v, v) v v .

ij

ij i jij

ij i j ij i jij ij

u Mu M m

M m

u Mu M m u u m

� �

�

� �

Pick a vector u having for components the absolute values of those of v,ui=|vi| . We use a variational argument:

32

0 0 0Let v v minimum eigenvalue (v, v)ij j ij

m M �

33

Nagaoka’s saturated Ferromagnetism

Y. Nagaoka’s Theorem (weak version)Consider a Hubbard Model with infinite U on any lattice L in d=2 and in d=3 with (possibly long range ) non negative hopping t. Let Ne=| L |-1 , that is, the electron number is the number of sites minus 1. Then among the ground states there are some with maximal spin (Ne/2).

In the infinite U limit the Hilbert space consists of the configurations with no double occupation. An orthogonal basis for such an Hilbert space can be specified by assigning the position x of the hole in the lattice and the spin configuration on all the other sites. Just for notational simplicity I exemplify by this 5-atom cluster. The basis vectors with the hole in 1 are:

1 23

4

5

2 3 4 5

† † † †2 3 4 5 2, 3, 4, 5,(1, ) (1, , , , ) vacc c c c

We need the action of the hopping Hamiltonian on this:for instance the matrix element:

4 3 2 1 2 3 4 5

1 2 3 4 2 3 4 5

† †1 2 3 4 15 1 5 1 5 2 3 4 5

† † † † † †15 4, 3, 2, 1, 1 5 1 5 2, 3, 4, 5,

(5, , , , ) (1, , , , )

(5, , , , ) ( ) (1, , , , )

vac ( ) vac

H

t c c c c

t c c c c c c c c c c c c

Informal Proof

1 23

4

5

4 3 2 2 3 4 51

1

1,† † † † † †

15 4, 3, 2, 1 5 1 5 2, 3, 4, 5,

† †, 1 11

Evaluating:

vac ( ) vac

annihilates the fermion created by

.

or by and yields 1.

t c c c c c c c c c c c

cc c

c

We must choose the spin configuration that gives maximum hopping and therefore maximum band width and lowest minimum energy. It is clear that by taking all spins parallel the delta is always satisfied and if we look for the ground state among parallel spin configurations we do find a lower ground state. A more formal proof reaches this conclusion by the Schwartz inequality, but this is the essential reason.Details are in Hal Tasaki cond-mat/9712219

5 5

4 3 2 2 3 4

† †5 5, 5 5,

† † †1 2 3 4 2 3 4 5 15 4, 3, 2, 2, 3, 4, 1 5 15

However ... or ... yields -1 because of the 3 operators in between. Thus,

(5, , , , ) (1, , , , ) =- vac vac ( , )

c c c c

H t c c c c c c t

4 3 2 2 3 4

4 3 2 2

1

5

1

4

5

3

† † †15 4, 3, 2, 2, 3, 4,

† † †15 4, 3, 2

1, 1

†5,

†5

5

,, 2, 3,

,

1,

45

1

,

,

If one gets:

vac vac ,

If one gets:

vac vac .

t c c c c c

c

c

t c

c c

c

c

c c

c ccc c c

35

Nagaoka

Y. Nagaoka’s Theorem (strong version)Consider a Hubbard Model with infinite U on any lattice L in d=2 and in d=3 with (possibly long range ) non negative hopping t. Let Ne=| L |-1 , that is, the electron number is the number of sites minus 1. Assume that the connectivity condition is satisfied. Then the ground states has maximal spin (Ne/2) and has no other degeneracy.

RemarkAt half filling there is no ferromagnetism. This theorem is surprising and cannot be understood in terms of mean field!

Informal Proof

All the matrix elements like

4 3 2 2 3 4

1 2 3 4 2 3 4 5

† † †15 4, 3, 2, 2, 3, 4, 1 5 15

(5, , , , ) (1, , , , )

=- vac vac ( , )

H

t c c c c c c t

are negative if both states. The Perron-Frobenius theorem implies that the ground state in the space with a fixed total Sz is unique.

The connectivity condition which is required by Frobenius holds for all common lattices.The theorem showed for the first time that dtrong correlation can lead to ferromagnetism.