1 1

1

1 1

Rescaled length  (rescaling factor )

Rescaled ( , )

( , ) 0

( , )

c

c

c c

b

probability p p p b

p p p p b for b

p p nonconducting islands disappear for b

p p system unchanged for b p p p b

1

Decimation

Consider only paths involving

1 1To proceed  find a simple approximation to ( , )p p p b

1

( , , , ) probability that , , , are connected,

( , , ) probability that α,β,γ are conected but δ is not, and so on.

( , , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , ) ( , )

Let P

P

p P P P P P P P

4 3 2 2 2 41renormal : 4 (1izati )on of (p 2 1 ) 2p p p p p p p p

A

1 B

2

C

D

What is the probability p1 that nodes A and B of new lattice are connected?

2

2 41

2 211

: 2

4 (1 )

Renorm

4 (1 )

at each scaling p changes by a f

aliza

actor ( ).

tion of bond probability p p p

pp p p p p p

p

p

3

2 4 2 41 2 Fixed point equation: 2c c cp p p p p p

0 Trivial Fixed point : repeated scaling no conductance

1 Trivial Fixed point : repeated scaling perfect conductance

1.6 nonsense fixed point

nontrivial solution : 0.618 (Exact result 0.5)

c

c

c

c c

p

p

p

p p

4

Empirical law for various lattices in dimensions 2,3 :

, coordination number= number of nearest neighbors.1

Square lattice: d=2, Z=4, =0.5.

c

c

d

dZ p Z

dp

1

At fixed point: ( ) 1.528 (using 0.618) while 0.5 yields 1.5.

3Exact results known : 0.5 and at fix

Rescaled length  (rescaling factor ),

the correl

ed p

ation length i

oint:

ndex is defined b

)

y

(2

c

c c c c

c c c

b

b

p p p

p p

ln( ) ln( 2)exact 0.8547

3ln( ) ln2 0.

( )2

818 ( ).bb

5

Onset of conductivity (R. Rosman-B. Shapiro Phys. Rev. B16, 5117 (1977))

Conductance of network 0 for

*( ) , at conductance threshold: 1

(generalized Mac Laurin;

this defines the coefficient * and thecriticalexponent

How to evaluate

t).

t ?

c

tc c

p p

p p p p

Principle: conductance is scale invariant at Cp p

A

1 B

2

C

D

1 1 1 1( ,

To get t us

)

e condition

(

on s

,

calin

) * *

g

( (

:

) )t tc cp p p p p p

Path conductances:

conductance of the series of two conductances

1

2conductance of two conductances in parallel

2

6

A

1 B

2

C

D4

1 1

2

2

( ) ( ) (weighted average)

probability of both paths and = ,

1conductance of combination of both paths : ( ) 2

2

probability of path through 1: ,

probability of no path through 2 : 1

probabil

paths

p p path path

path

p

p

p

2 2ity of path through 1 and no path through 2 :

1conductance of path through 1 onl (

2

(1 )

y : )p

p

ath

p

2 11 1For , renormalized ,c c c c cp p p p p p p

2 21 1

24 1 1weighted average .2. 2 1 . .

2 2.( )p pp p p

Evaluation of conductance 1

of renormalized bond AB

11 1

1

1 1

for a small increment of bond conductivity,

( ) ( ) ln( ) ln( ).

1Recall: ln( ) ln( ).

t t

cc

pp p t

p

pp t

p p

7

A

1 B

2

C

D

1 1

1 1

to get t use scale invariance conditi

( , ) ( ,

on

)

*( ) *( )t tc c

p p

p p p p

( ) 1.528 and 0.61Using th 8is, c c cp p

Evaluation of critical exponent t

We get the estimate 1.134.

*( ) . An increase of p near the critical point yields a

more than proportional growth of .

tc

t

p p

2 41

2 211

: 2Recall the renormalization of bond probability

4 (1 ) 4 (1 ).

p p p

pp p p p p p

p

Renormalization group for the 1d Ising model

0 1 1 2 2 3 3 4{ } { }

exp[ ( )], ,is s siB

JZ K s s s s s s s s K

K T

1 log .Hn n B

n

H Js s Z Tre F K T Z

1 2 2 3 3 4 4 5{ }

exp[ ( )]exp[ ( )] ,s

Z K s s s s K s s s s

8

1 2 3 4

Sum over spin 2: 2

1 2 2 3 1 3 1 3exp[ ( )] exp[ ( )] exp[ ( )]s

K s s s s K s s K s s

2 4

1 2 2 3 3 4 4 5

1 3 1 3 3 5 3 5

exp[ ( )]exp[ ( )]

[exp[ ( )] exp[ ( )]][exp[ ( )] exp[ ( )]

s s

K s s s s K s s s s

K s s K s s K s s K s s

Sum over all even sites:

Decimation Eliminating even sites: we must sum over even sites.

9

2 4

1 3 1 3

2

1 3

1 2 2 3 3 4 4 5

( ) ( )1 2 2 3

( )

Result

must be rewritten with appropr

Original: exp[ ( )]exp[ ( )]

decimation: exp[ ( )] .

( iate ( ), ( .) )

s s

K s s K s s

s

b k s s

K s s s s K s s s s

K s s s s e e

A K e A K b K

1 3 3 5( ) ( ) (

(1)

1)2

{ }

( , )

1

( ) ( ) ( ) ( ,

( ) ln cosh(

,2

)2

)

2

Nb k s s b k s s

odd s

NZ K N A K e A K e A K Z K

K b K K

Decimation Eliminating even sites: we must sum over even sites AND set the result back in the original form (as far as we can).

1 3 1 3

1 3 1 3

(

1 3

( ) ( )

1 3

( )

)

(( ) )

2cosh

We want that if and are both 1 or both 1:

2cosh(2 ) We require

If and have opposite signs :

2 We require

(2 ) ( )

2 ( ) .

b k

b

K s s K s s

s s s kK s K

s s

e e K

s s

e

K A K e

Ae K e

2 ( )

( ) 2 cosh(Multiplying :

di c

2 )

osvi h(ding ): 2 b k

A K K

K e

B

JK

K T

( )

( )

2cosh(2 ) ( )

2 ( )

b k

b k

K A K e

A K e

Fixed point: 1ln cosh(2 ) 0

2K K K

no ferromagnetism, no phase transition, no long-range order!

Renormalization group for the 2d Ising model (see Grosso-Pastori,McComb)

0 1 0 2 0 3 0 4 5 1 5 4 5 9 5 10{ }

exp[ ( )]exp[ ( )]s

Z K s s s s s s s s K s s s s s s s s B

JK

K T

0 1 0 2 0 3 0 4 1 2 3 4 1 2 3 4

0

( ) ( ) ( )

2cosh(4 ) all up + all down

2cosh(2 ) three up/down and one down/up (4 ways)

2 two up and two down (6 ways)

K s s s s s s s s K s s s s K s s s s

s

e e e

K

K

Eliminate site 0

New couplings are needed to represent all possibilities:

0

0 1 0 2 0 3 0 4 1 2 3 4 1 2 3 4

1 2 2 3 3 4 4 1 1 1 44 23 2 3

exp[ ( )] exp[ ( )] exp[ ( )]

ex( )exp p[ ( ) ][ ( )( )]

s

c K s s s s

K s s s s s s s s K s s s s K s s s s

A K b K s s s s s s s s s s s s

0

0 1 0 2 0 3 0 4 1 2 3 4 1 2 3 4

1 2 2 3 3 4 4 1 1 3 2 4 1 2 3 4

exp[ ( )] exp[ ( )] exp[ ( )]

( ) exp[ ( )( )]exp[ ( ) ]

2cosh(4 ) ( ) exp[6 ]

2cosh(2 ) /

s

K s s s s s s s s K s s s s K s s s s

A K b K s s s s s s s s s s s s c K s s s s

K all up or all down A K b c

K three up down and two do

/ ( ) exp[ ]

2 ( )exp[ 2 ]

wn up A K c

two up and two down A K b c

11

82( ) 2cosh(2 ) cosh(4 )

1( ) ln cosh(4 )

81 1

( ) ln cosh(4 ) ln cosh(2 )8 2

A K K K

b K K

c K K K

12

Indeed we can fix the uknowns by solving the 16 equations that arise from the possible spin orientations and that we can summarize in three cases:

(1)11

82 2( ) ( ) 2cosh(2 ) cosh(4 )N

HZ A K Tre A K K K

Iterating,

0 1 0 2 0 3 0 4 5 1 5 4 5 9 5 10{ }

exp[ ( )]exp[ ( )]s

Z K s s s s s s s s K s s s s s s s s

(1)

,

, [ , ]

[ ,, ]

2 )( )

,

)( (i ji j

i

i j i j k lii jj

j i j

H b K s s

neareast n

b K s

eighbours next neareast neighbour

s c K s s

s

s s

This is unlike the original H and interations bring interactions of increasing complexity. Approximations needed.

1( ) ln cosh(4 ) iteration lea

Neglect of red

d

and blue

s

terms :

to 08

b K K K K

13

no ferromagnetism, no phase transition, no long-range order!

14

H.J. Maris-L.P.Kadanoff,am.J.Phys.46, 652 (1978):

( 1) ( )

Neglect of red and blue terms but reinforced b(K)

3 3( ) ln cosh(4 ) ln cosh(4 )

8 83

* ln cosh(4 *) 0.507.8

i i

c

b K K iteration leads to K K

The solution of K K is K K

(1)

,

, [ , ]

[ ,, ]

2 )( )

,

)( (i ji j

i

i j i j k lii jj

j i j

H b K s s

neareast n

b K s

eighbours next neareast neighbour

s c K s s

s

s s

0.507 intcK K is the nontrivial fixed po

Prediction: phase transition at 0.507 ( 0.441)c OnsagerK K

Peierls transition in 1d chains

regular 1d chain

Peierls distorted 1d chain

Consider a 1d chain of identical equally spaced atoms at hal filling. Peierls in the thirties has shown that in the Born-Oppenheimer approximation the system is unstable and the ground state shows alternating long and short bond lengths.

The lattice parameter doubles, the BZ becomes the half and a gap opens.This is called a charge density wave (CDW) state, which can be seen in X-Ray diffraction. All the occupied states go down in energy..

aa a

a+u a-u

2

2

The cost in strain energy goes with u ; one finds that for small displacements u

the gain in kinetic energy goes like u ln( ).This is why the regular chain is predicted

to be unstable

u

a

See e.g. Sander van Smaalen, Acta Crystallographica (2005)

Polyacetylene with its anternating singe and double bond is the most common example of the Peierls distortion. Often 1d systems are poor models of 3d crystals because there are qualitative differences.

18

Jordan-Wigner string

Consider a boson field is defined on a chain, with annihilation operators ci † † †[ , ] 0 [ , ] [ , ]i j i j i j ijc c c c c c

How can we use them to define anticomuting fermions? In fact, we can in 1d. The fermions are: †

j j jj i j i

i n i c c

i i if e c e c

with called the Jordan-Wigner string.jj i

i n

18

† †j

j i

i n

i if e c

† †

†

† †

†

The key observation is that s

1 1

1 1

inc

1 1

1 1

e 1, 1

1

1

,

1

a

a a

a

a a

i na a a a a a

i n i na a

i na a a a a a

i n

i even i

i na a a a a

a a a a

odd

a

c e n even c n even n n

e c n eve

c e n odd c n odd n n

e c n odd e n

n e n n odd n

e

n n

e

n

n

† † †

†

0 , that is, anticommutes with .

0 but also .

Instead, and commute with for a b.

a a a

a a a a

b

i n i n ina a a

i n i n i n i na a a a

i na a

c e e c c e

c e e c c e e c

c c e

In 1d one can make Fermions out of Bosons

† † † † 0a b b af f f f So, we can verify that in the following situation,

a b

1 1

1 1

( )† † † † † †

( )† † † †

since we can

chop on sites on left of site a, namely a-1,a-2,... since they occur twice;

one can also write:

(commutatio

j jj a j b a a b

jj a

a b a

i n i ni n n n

a b a b a b

i n

i n n i na b a b

f f e c e c c e c

e

f f e c e c

n on different sites)

1 1

1 1

† † † †

( )† † † †

( )† † † † † † †

† †

On the other hand, ;

deleting those occurring twice, .

So, = [ ] 0

since 0

j jj b j a

a a b

a b a a

a a

i n i n

b a b a

i n n nb a b a

i n n i n i na b b a a a b

i n i na a

f f e c e c

f f e c c

f f f f e c e e c c

c e e c

.†

Thus, is a Fermion.j j j

j i j i

i n i c c

i i if e c e c

There is also a continuous version

' ( ')

( ) ( ).

x

i dx x

f x e c x

20

1 0

0 00 0 0 0 0We can write

0 1 0 1 0 1

j jj m n

ii i

m mn m n mm m n

eC e e

Let us associate Pauli matrices to sites of a chain. Matrices of different sites commute. Then introduce the matricesLet us associate Pauli matrices to sites of a chain. Matrices of different sites commute. Then introduce the matrices

,j j

j m

i

m mC e

†j j

j m

i

m mC e

†Therefore, ( ) ( ).z zm m n m m n

n m n m

C C

È

They anticommute and are Fermion annihilation and creation operatorsThey anticommute and are Fermion annihilation and creation operators

† 2 1 0 0 0[ , ]

0 0[ , ] ( ( )) [ , ] [ , ] 1.

1 0z

m m n m m m mn m

C C

Indeed the Jordan-Wigner factors can be removed when they occur twice, so on site

† .m m m mC C

Jordan-Wigner string with Pauli matrices xyzJordan-Wigner string with Pauli matrices xyz

21

†4 2

†4 1 2 3 4 2 1 2

is also obtained:

for instance let's work out

We may start the Jordan-Wigner stri

Off site,anti

[ , ] .

(

com

)

( )( )(

mutatio

ng from

) ( ) .

n

j=1zm m n

n m

z z z z

C C

C

C C

†4 2 1 2 2 2 23 4 1 1 1 3 4[ , ] ( ) ( ) ( ) ( ) ( ) )( ) ( ()z z zz z zz zC C

†4 2 2 2 3 4[ , ] [ , ] 0 [ , ] 0.sincez z zC C

1

†4 2 2 3 4 2 2 2 3 4

and eliminating the square of we are left wi

[ , ] ( )( ) ( )

t

)

h

(

z

z z z zC C

22

The Onsager solution of the 2d Ising model The Onsager solution of the 2d Ising model

The partition function is:

,Ising

on NxN square lattice

0 otherwise no solution.

i j jadjacent i j

H Js s H s

H

,site notation: is specified as s .i row columns

2Ising

1,1 1,2 ,

.N N

HFN

s s s

Z e e

Notation:

K J

The geometry of a torus is adopted, identifying column N+1 with column 1 and row N+1 with row 1, that is, imposing pbc

n,N 1 n,1 N 1, 1,s s , s =s .m m

23

First step: write Z in terms of the eigenvalues of a Hermitean matrix-the transfer matrix V. This will be chosen with the structure of a one-dimensional

array.

24

Row-by-row description

2

,1 ,2 ,3 ,

The set { , 1, }

assigns one of the 2 configurations

Let denote a configuration of the spins in the row ;

there are 2 configurations

{ , , ,

of the lat

for each ro

tic

w

}

e.

.

,N

N

N

s s s s

N

Ising Ising

1,1 1,2 , 1 2

.N N N

H H

s s s

Z e e

Indeed, in the l.h.s. we are summing on all the configurations taking them site by site, in the r.h.s. we do the same thing row by row.

IsingHowever we have to write in a suitable way for a row by row description.He

25

Ising Ising

1,1 1,2 , 1 2

.N N N

H H

s s s

Z e e

, , 1

1 , 1,1

The interaction energy between the spins within row is

Besides, interaction energy between adjacent rows:

( )

( , )

N

k kk

N

k kk

E J s s

E J s s

For each lattice configuration Z has a contribution

2 2 3 11 1 2 ( ) ( , ) ( ) ( , )( ) ( , ) .N NE E E EE Ee e e

1

,1 ,2 ,3 ,

( ) ( , )1

Define

{ , , , , }

that can

for each row the N-dimension

take 2 values, and the 2

al multiindex

2 Transfer matrix

,|

:

|

N

N N N

E E

s s s s

X

V e

, , 1 , 1,1( ) ( , )1| ., | k k k k

NJs s Js sE E

k

V eHe e enc e

Example:N=3

1 11 12 13

2 21 22 23

3 31 32 33

( , , )

( , , )

( , , )

s s s

s s s

s s s

1 2 3

9The system has 2 512 configurations

Z= He

1, 1, 1 11 ,1 2 2,( )3

3 31 2

1

( )

2

, is | 2 2

since depends on 3 spins, depends on 3 spins.

| k k k kJs s Js sE E

k

V e e e X

For each lattice configuration Z has a contribution

2 2 3 11 1 2 3 3( ) ( , ) ( ) ( , )( ) ( , ) .E E E EE Ee e e

1 2 1

31 2 2 3 1 1 1

32

is written in terms of

| | | | | |

a 8X8 matrix. We can do the same in gene

|

l

|

ra .N

Z V V V V TrV