# Bipartite lattices

```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
Let  ( x, t )  Fermi field (x,t) = scalar field, coupled through the mass,
and the Lagrangian
L=L B  LF
with:
2
1  2  2 1 2
L B  2 [( )  ( )     1 ]
2g
t
x
2
L F   (i     m( x, t )) 
2
1  2  2 1 2
 2  2
L B  2 [( )  ( )     1 ]  nonlinear field equation ( 2 - 2 )=( x, t )  ( x, t )3 .
2g
t
x
2
t
x
1
 2  2
The nonlinear field equation ( 2 - 2 )=( x, t )  ( x, t ) 3 solved by
t
x
 ( x, t )  1 . This is called vacuum sector, with 2 vacua : semiclassically,
 ( x, t )  1  L F   (i     m)   ( x, t )  1  L F   (i     m) 
Additional static (time-independent) solutions: soliton sector
 2
x
The nonlinear field equation - 2 =( x, t )  ( x, t )3 solved by ( x)  Tanh[ ].
x
2
x
]
x
d ( x)
1
1
d 2( x)
2
Indeed,  ( x)  Tanh[ ] 

,

2
x
dx
dx
2
2 Cosh[ x ]2
Cosh[ ]2
2
2
x
Tanh[ ]
x
2
 ( x)  Tanh[ ]   ( x)   ( x)3 =
x
2
Cosh[ ]2
2
Tanh[
2
kink
x
 2  2
( x)  Tanh[ ] is the
static solution of ( 2 - 2 )=( x, t )  ( x, t )3
antikink
t
x
2
1.0
0.5
4
f x_
2
Tanh
2
4
x
2
0.5
1.0
3
semiclassical solution
Vacuum sector:
for   1
Dirac’s theory
LF   (i     m ( x, t )) ( x, t )   (i     m) Dirac's Lagrangian
 uk 
 ( x, t )    with uk , uk two component spinors, E k = k 2  m 2
 uk 
H D   p   m,
 0
usually  k  
 k
k 
I 0 
, since they anticommute.
 and   

0
 0 I 
field in 1+1 dimensional theory
In 1+1 d spinless case two components are enough;
we may take    2 ,  = 1. Then,
(-i x   m)uk ( x)  Ek uk ( x)
(-i x   m)uk ( x)   Ek uk ( x)
4
Electron-Positron field in Dirac’s theory
 (r , t )  
k

s 1,2
[cs (k )us (k )ei ( k .r  ( k )t )  bs† (k )v s (k )e i ( k .r  ( k )t ) ]
cs annihilates electron with spin s, bs† creates positron with spin s,
 (k)= (c k ) 2  m 2c 4 .
Field in 1+1 dimensional theory
 ( x, t )   [bk uk e iEk t  d k†uk e  iEk t ]
k
Dirac's vacuum:
bk v  0
d k v  0.
Charge conjugation in Dirac’s theory
 1 
 
ieA
2 


If  
solves      
 3 
c

 
 4 

mc

(
x
,
t
)

 ( x, t )  0


 4* 
 * 
ieA


then  C   3*  = 2 * solves       
 2 
c

 * 
 1 
 0 i k 
k  

0 
 i k
 C
mc C

(
x
,
t
)

 ( x, t )  0


Charge conjugation in 1+1 dimensional theory
 uk 
 ( x, t )   
 uk 
 3uk  uk sends particles to antiparticles; in addition, [ 3 , H D ]  0.
 3 (-i x   m)uk ( x)  Ek 3uk ( x)  (-i x   m) 3uk ( x)
  3  charge conjugation operator.
Charge operator in Dirac’s theory
Standard charge density in Dirac's theory:
1
2
 ( x, t )= [  † ( x, t ), ( x, t )]_
1
ˆ
Indeed te charge is Q   dx ( x, t )   ([bk† , bk ]  [d k , d k† ])
2 k
1
  {bk†bk  (1  bk†bk )  d k† d k  (1  d k† d k )} 
2 k
  {bk†bk  d k† d k }.
k
number of particles - number of antiparticles
7
Soliton sector
Positive energy
continuum
Positive energy
continuum
1.0
0.5
4
2
2
0.5
1.0
negative
energy
continuum
4
localized
state
negative
energy
continuum
Solution in 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 &frac12; and - &frac12; .
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, &frac12; 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
Analogy
field
The Dirac
equationwith
is in one
spatialtheory
dimensionmodel
and involves two
components, corresponding to two Dirac points:
( p  g  )  E ,  = 3  = 2
d 2 ( x)
Phonon field:
 V '( ( x))
2
dx
[ 1 , H ]  0   1 sends eigenfunction from E to -E.
The zero mode has eigenvalue
1
.
2
14
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
H  t


 x , y L
cx† c y  U  nx nx
x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)
15
Hubbard Model with nearest-neighbor hopping on a lattice L
H  t


 x , y L
c c y  U  nx nx
†
x
xL
John Hubbard
London 1931-San Jose
1980
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
Bipartite lattices
Chain (d=1), square (d=2) , cubic (d=3) lattices
H  th  ci†c j
i , j 
 i , j  n.n. th  0
Chain:
1 ikn
e
N
 k  2th cos  k 
 (n) 
common eigenfuctions of
translation T
and
H
17
17
Bipartite lattice
Square lattice
H  th  ci†c j
i , j 
 ( n) 
1
N
2
e
i ( k x nx  k y n y )
 i , j  n.n. th  0
common eigenfuctions of
 k  2th [cos  k x   cos  k y ]
translation T
and
H
Bipartite lattice
Cubic lattice
 ( n) 
1
N
2
e
i ( k x nx  k y n y  k z n z )
common eigenfuctions of
 k  2th [cos  k x   cos  k y   cos  k z ]
translation T
and
H
H  t


 x , y L
cx† c y  U  nx nx
xL
Spin in Hubbard Model
The total spin is conserved. Recalling the general rule for writing operators in
second quantization,
Vˆ  

 cx  
†
dx

(
x
)
V

(
x
),
with

(
x
)

  , one gets




 c 
1
Sz  
2 xL
S 

xL
c
†
x
c
†
x
cx†   1 0   cx  1

c   
 0 1  x  2 xL
cx†   0 1   cx 
†

  c    cx  cx  ,
 0 0   x  xL
n
x
 nx  
S    cx† cx
xL
S S   S S 
and one finds S  S 
2
2
2
z
20
.
Trasformazione a U negativo
H (U )  t 


 x , y L
cx† c y  U  nx nx  H  H (U )  UN 
xL
Possiamo fare una trasformazione canonica sui soli stati di spin su introducendo le buche
d x  cx†
col che il termine cinetico di spin su diventa
T  t

x ,y L

d x d y  t

x ,y L

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 c x
 x   1
21
Con questo, quando scambiamo creazione e distruzione, solo il termine in U cambia segno:
H t
 [d
x , yL
†
x
d y   cx† c y  ]  U  d x† d xcx†cx  UN  ,
xL
N    nx 
xL
N   nx e ' conservato, quindi e’ una costante.
xL
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:
Sz 
1

2 x
n
 nx    S z 
x
1

2 x
  ( x ) 2 d x  d † x   nx   
1  n  n
1
†
1

d
d

n

  x x x 
2 x
2
S    cx† cx  S     ( x)d x cx .
x
x
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.
22
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  ....   N
single particle levels. If 0  U   Ne  1
(that is U&lt; band width)  Emin ( Smax  1)  Emin (Smax )
Pieri, Daul, Baeriswyl, Dzierzawa, and Fazekas theorem:
no ferromagnetism in Hubbard model at very low
electron density.
23
Teorema di Lieb-Mattis
Phys. Rev. 125, 164 (1962)
Consideriamo il modello di Hubbard repulsivo in 1d con obc (open boundary conditions)
H 


 x, y 
t x cx† c y   U x nx nx
x
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,
Emin ( S )  Emin ( S  1).
Piu’ basso lo spin piu’ bassa e’ l’energia. Questo e’ ovvio per U=0
quando i livelli si riempiono secondo l’aufbau
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.
24
Strong coupling half filled Hubbard Model, d=2 or 3
H  t


 x, y 
cx† c y  U  nx nx , U
x
t
Half filling: number of sites=
number of electrons
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:
In first order in t one gets nothing (H takes to doubly occupied states orthogonal to g.s.)
configuration
is forbidden for infinite U
25
H  t


 x, y 
cx† c y  U  nx nx , U
t
x
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:
H '   H ba
1
H a b , H ba  t (ca† cb   ca† cb  ), H a b  t (cb† ca   cb† ca  ),
HU
HU  Unb  nb 
t2 †
H '   (ca  cb   ca† cb  )(cb† ca   cb† ca  )
U
t2 †
  ca  cb  cb† ca   ca† cb  cb† ca   ca† cb  cb† ca   ca† cb  cb† ca  
U
t2
  (1  nb  )na   ca† ca  cb† cb   ca† ca  cb† cb   (1  nb  )na  
U
a
b
This is impossible for parallel spins. Scond-order corrections are always negative. So
antiparallel configurations are lower.
26
Important case: Three-band Hubbard Model of CuO
The Cuprates are known to be
antiferromagnets at half filling. By hole
doping (and sometimes also by electron
doping) they become high temperature
superconductors.
In the CuO structure with half filled Cu band this mechanism favors antiferromagnetism.
The AF configuration allows the second-order
spin exchange. This further lowers the energy.
27
Let us consider the contributions inside the second-order perturbation
t2
H '   (1  nb )na   ca†ca  cb† cb  ca† ca cb†cb  (1  nb )na  
U
Non-interaction terms linear in n just renormalize all site energies. We ignore them.
t2
H '    na  nb  na  nb  Sb Sa  Sa Sb 
U
Sz 
1

2 x
 nx   n x  
with S x†   cx† cx
x
1
S a Sb  S az Sbz  ( S a Sb  Sb S a )
2
Besides aba processes one must include bab and all other sites on same
footing. Neglecting parallel-spin interactions , part of the physics corresponds to an
Hamiltonian of the form
H Heisenberg  J  Sm  Sn
mn
sometimes written as:
1
H Heisenberg  J  (Sm  Sn  )
4
mn
28
1
H Heisenberg  J  (Sm  Sn  )
4
mn
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.
I just mention a related model that P.W. Anderson likes for Cuprate
superconductors: the t-J model, appropriate for strong coupling:
1
H  t  ci† c j  J  (Si S j  ni n j )
4
ij
ij 
29
Toy Model:
2 electrons on 3 sites
2
t'
(Hal Tasaki , Cond-mat/9512169)
1
t'
3
t
for t=t’/2 the ground
state has S=1 if U is
large (ferro) and S=0
otherwise
For t=t’&gt;0 lo stato
fondamentale ha S=1
per ogni U (ferro).
30
Perron-Frobenius theorem for real symmetric matrices
Given a matrix M  {mij }, mij  m ji
mij  0
with M  a tight  binding model that represents
a connected tight-binding cluster:
1) the lowest eigenvalue is nondegenerate
2)the corresponding eigenvector can be taken with all strictly
positive components
Proof.
Consider the eigenvalue equation Mu= u.

mij u j   ui .
j
If all u j  0 ,

mij u j  0 strictly, then ui cannot vanish.
j
So, all u j  0  all u j  0.
Hence either the components ui have both signs or they must be all strictly positive (or
all strictly negative).
31
Let  mij v j  0 vi
0  minimum eigenvalue
0  (v, Mv)
j
Pick a vector u having for components the absolute values of those of v,
ui=|vi| . We use a variational argument:
(u , Mu )  (v, Mv) because mij  0, but cannot go below
0  minimum eigenvalue  (v, Mv)   mij v i v j . Therefore,
ij
(u , Mu )  (v, Mv)   mij ui u j   mij vi v j .
ij
ij
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 .
32
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).
Informal Proof
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, )   (1, 2 , 3 , 4 , 5 )  c2,†  c3,†  c4,†  c5,†  vac
2
4
3
4
5
We need the action of the hopping Hamiltonian on this:
3
1
5
2
for instance the matrix element:
 (5,  1 ,  2 ,  3 ,  4 ) H  (1,  2 ,  3 ,  4 ,  5 )
  (5,  1 ,  2 ,  3 ,  4 ) t15 (c1† c5  c1† c5 )  (1,  2 ,  3 ,  4 ,  5 )
 t15 vac c4, 4 c3, 3 c2, 2 c1,1 (c1† c5  c1† c5 )c2,†  2 c3,†  3 c4,†  4 c5,†  5 vac
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Evaluating:
4
3
1
5
t15 vac c4, 4 c3, 3 c2, 2 c1,1 (c1† c5  c1†c5 )c2,†  2 c3,†  3 c4,†  4 c5,†  5 vac .
2
c1,1 annihilates the fermion created by c1† or by c1† and yields 1.
If c1,1  c1, one gets:
t15 vac c4, 4 c3, 3 c2, 2 c5 c2,†  2 c3,†  3 c4,†  4 c5,†  5 vac ,
If c1,1  c1, one gets:
t15 vac c4, 4 c3, 3 c2, 2 c5 c2,†  2 c3,†  3 c4†, 4 c5†, 5 vac .
However c5 ...c5,†  5 or c5 ...c5,†  5 yields -1 because of the 3 operators in between. Thus,
 (5,  1 ,  2 ,  3 ,  4 ) H  (1,  2 ,  3 ,  4 ,  5 ) =- t15 vac c4, c3, c2, c2,†  c3,†  c4,†  vac   ( 1 ,  5 )t15
4
3
2
2
3
4
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
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.
Informal Proof
All the matrix elements like
 (5,  1 ,  2 ,  3 ,  4 ) H  (1,  2 ,  3 ,  4 ,  5 )
=- t15 vac c4, 4 c3, 3 c2, 2 c2,†  2 c3,†  3 c4,†  4 vac   ( 1 ,  5 )t15
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.
Remark
At half filling there is no ferromagnetism. This theorem is surprising and
cannot be understood in terms of mean field!
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