1

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

2

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

3

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

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

5

1 2 1

1 2 2 3 1 1 1

n,N 1 n,1

Each    represents a sequence of  N indices that can ta

s

ke the values 1.  V i

s

2 2

| | | | | |

 

| .

.

s

|N

N N

N NNZ V V V V TrV

pbc

1

, , 1 , 1,

1

1, 1, 1,

( )

1 2

(

2 1

, )

is defined on t

( , ,

wo neighbo

..., ) ( , ,

u

..

r

| |

| |

ing rows

. )

.

,

k k k k

N N

E E

NJs s Js s

k

V

V

e

s s s s s

e e

s

largest eigenvalue ( in thermodynamic lim ., , it)N Ni max max

i

Z N

2Free energy per spin log log . B B

max

K T K TF Z

N N

n,N 1 n,1pbc: s s

last factor =first

maxThe problem is: find

First,we need thestructure of as a function of .V K J

6

Definition: direct product of matrices | | | | | | .ij A B kl i A k j B l

Multi-index matrices

11 12 11 12ij,kl

21 22 21 22

, , D ik jl

a a b bA B a b

a a b b

11,11 11,12 11,21 11,22 11 11 11 12 12 11 12 12

12,11 12,12 12,21 12,22 11 21 11 22 12 21 12 22

21,11 21,12 21,21 21,22 21 11 21 12 22 11 22 12

22,11 22,12 22,21 22,22 21 21 21 22

D D D D a b a b a b a b

D D D D a b a b a b a bD A B

D D D D a b a b a b a b

D D D D a b a b a

22 21 22 22b a b

Example:

11 12 12 12

11 21 11 22 12 21 12 22

21 12 22 12

21 21 2

11 11 12 11

2

1 22 22 21 22 2

1 1 2 11

2

1 2

If one restricts toany element of the block matrix acts like ,

but if one restricts toany element of the b

B A

a b a b

a b a b a b a bD

a b a b

a b a

a b

b a b a b

A

a b

a b a b

lock matrix acts likeB

this is a special multi-index matrix.this is a special multi-index matrix.

With many indices multidimensional block structure.

77

Example: consider a chain with N sites. Starting from the Pauli matrices X,Y,Z one defines by direct product:letting 1 denote the 2X2 identity,

2 2 1 1 1 1.

Since

enlarged

1 1 1 1, the enlarged matri

Pauli matr

ce

ices

s

N N xm m

m mY

X X

X XY

mth place

,have the property that [ , ] [ , ] etc.m n m nX Y X Y

11 11 11 22 22 11 22 22 11 22 11 22( )( )TrD a b a b a b a b a a b b TrATrB

,kp ij ki pj kk jjkj

D a b TrD a b TrATrB

Property:

11 11 12 11

21 11

11 12 12 12

11 21 11 22 12 21 12 22

21 12 22 12

2

22

1 21 2

1

1 22 22 2 2

1

1 2 22

a b a b

a b a b a b a bD

a b a b

a b

a b a b

a

a b a b a

b b

b

a

This represents X at site m.

8

Factorization of V 2 1 where:V V V1

2 ' '| | ( , ) all in a row

2 2 matrix, diagonal in row index

k kJs s

k

N N

V e

X

, 1,

1 1| |

2 2 matrix, diagonal in k, s' is in same position as s in next row.

k kJs s

k

N N

V e

X

1 1

2

1 2 2 1 12| | | | | | .k k k k

NJs s Js s

k

V e e V V

Indeed,

, , 1 , 1,1

1 1, 1, 11 2 1 2 ,

( ) ( , )

( , ,..., ) ( , ,..., )

is a multi-index matrix defined on two neighb

|

ouring rows

|

.

| |

k k k k

N N

NJs s Js sE E

k

V V

e e

s s s s

e

s s

Example:N=3

,1 ,2 ,2 ,3 ,3 ,1

,1 1,1 ,2 1,2 ,3 1,3

,1 ,2 ,2 ,3 ,3 ,1

3 3

3 31

2

1

1

is 2 2 direct product

is 2 2 direc

| |

| |

| |

t product

Js s Js s Js s

Js s Js s Js s

Js s Js s Js s

V e e e

V e e e

V e

X

e e e

X

,1 1,1 ,2 1,2 ,3 1,3 3 3 is 2 2 .Js s Js s Js se Xe

1 11 12 13

2 21 22 23

3 31 32 33

( , , )

( , , )

( , , )

is a multiindex, we can write all matrices in terms of 2X2 matrices.

s s s

s s s

s s s

10

Block matrices inside V1

, 1, can take 4 values that can be arranged

Notation:

associated to the pair of classical spins at site k

a

.

sk kKs s

K K

K K

e

e ea

e e

K J

, 1,

1 1| |

2 2 matrix,depends on interaction with next row.

Fix same place k in both rows:

k kJs s

k

N N

V e

X

,1 ,2 ,3 ,{ , , , } , Ns s s s

11

111

2

Here we need the diagonal Z matrices (enlarged Pauli) to write:

m m

m m m mm

K Z ZKZK s

m

s Z

m

V ee e

V2 matrix in terms of Pauli and related matrices

12 ' '| | ( , ) all in a rowk kJs s

k

V e

Pauli matrices X,Y,Z

elarged Pauli matrices 2 2 1 1 1 1 ,N N xm mX X

mth place

,[ , ] [ , ] ,etc.m n m nX Y X Y

Remark: For eack site k we have a matrix, and V is a direct product of all these matrices. It would be fine if we could have a sum, instead! How can we take logarithms?

,1 ,2 ,2 ,3 ,3 ,1

, 1 21 ,2

3 32

1

is 2 2 direct product

where z 1 for spin up and-1

3 3 example: |

for spin down.

| Js s Js s Js s

J JJs s

J J

Jz z

X V e e e

e ee

e e

X

e

12

x

2

θτ

, , , , areenlarged Pauli matrices

and tanh( ) .

The spins are classical, b

Now we show that a=

ut Pauli matrices

are associated toeach site an

2sinh(K)e ,

wh

o

re

y w.

e

h

x y z

K

X Y Z

e

*We shall derive this by a direct method. (Schulz Mattis and Lieb use K for θ.)

, 1,

1

2

1 1

The 2 2 matrix

The 2 2 matrix | | is written as

a direct product 2x2 matrice

is an exponential

.

.

s

k k

m mm

K ZN N

Js sN N

k

K K

K

Z

K

X

X V e

e eof a

e

V e

e

IDEA:all in exponential form!

13

2

To show that

2 ( ) , tanh( ) ,x

K KK

K K

e ea a sinh K e e

e e

note that

0 1, .

1 0

Now set , with , unknowns.

cosh( ) sinh( )By the Taylor expansion: cosh( ) sinh( ) .

sinh( ) cosh( )

K KK K

xK K

X

X

e ea e e X X

e e

a Ce C

e X

cosh( ) sinh( )so we must find C and such that: .

sinh( ) cosh( )

K K

K K

e ea C

e e

2

-K

-2KThesecond

cosh( )(dividing) tanh( )

sinh( )

e;substituting, .

sinh( ) sinh(arcgive

ta e ))s

nh(

KK

K

K

e Ce

e C

eC C

14

Simplifying (e.g. by Mathematica)

42

2 with

1 2 (2 ).si

tanh( ) ,

nh( ( ))

2 (2 )

2sinh(2 ) x

KK K

K

X

K

K

KK

K

eC e e sinh K

arctanh e

a sinh K e

e ea K e QED

e ee

. 1,N N1

1

The 2 ×2 (N times direct product)matrix m m

NKs s

m

V e a a a

2 21becomes ( ) [2 (2 )] [2 (2 )] .

mm m

N N XXV sinh K e sinh K e

15

Canonical transformation

that is, ,x z z z x

2 21 [2sinh(2 )] [2sinh(2 )]

x zm m

m m

M M

V K e K e

1 1

2

z z x xm m m mK K

V e e

(the method by Schulz, Mattis and Lieb, Rev. Mod. Physics (1964)) makes the problem much easier. Start by a p/2 ‘spin’ rotation around y axis ,x z z x

2Evidently, [ , ] 0, ; ( ) 0.m n mm n

11 1

2 22

Remark: ( )

x xm mK

V e

16

Symmetrized form of V

1 2( )N NZ TrV Tr VV We can redefine1 1

2 22 1 2( ) ( )V V V V

which has the same trace as the previous form since TrAB=TrBA but is preferable for its symmetry. In particular it will turn out that this new form is Hermitean.

1 1

2 22 1 2 is an exponentialof a chain hamiltonian with a Pauli matrix

for each site and lends itself to Fermionization. To this end the Pauli

matrices must be rewritten in terms of shift op

( ) ( )

erators.

V V V V

17

†

Introduce fermion creation/annihilation by Jordan-Wigner string

j j j jj m j m

i i

m m m mC e C ep p

Fermionization of the Transfer Matrix VFermionization of the Transfer Matrix V

0 1 0 0 1, 2( ), .

0 0 1 0 2z x

m m m m m m m

m m

elarged Pauli matrices 2 2 1 1 1 1 ,N N xm mX X

mth place

We already introduced those Pauli matrices on every site:

However we must write everything in terms of shift operators.

18

Pauli matrices: write everything by shift operators

Operators defined on different sites obviously commute, and

2 0anticommut

0 1 0 0 1 0 0 0 0 1 0 0, [ , ] 1,

0 0 1 0 0 0 1 0 0 0 0 1

0,

0 0eon-site.

m m m m m m

m m m m m m

m m mans do

Moreover,

1 0 1 0, ] 0

1 1

0 0 0 0[

0 1 0 1 0 0m

m m

zm

m m

19

1 1 1

2

( )( )

2

12( )

221

[2s

12( ), and

2

[2sin inh(2 )]

since ,

.

h(2 )]zm

m

x xm m m m m m

m mm

z xm m m

K

M

m m m

M

K

KV Ke

V e e

e

1 2 Canonically transformed V and V in terms of shift operators

Fermionization of the Transfer Matrix VFermionization of the Transfer Matrix V

†1 12 ( ) 2 ( )

2 2† 2 21 [2sinh(2 )] [2sinh(2 )]

m m m mm m

N N C C

m m m mC C V K e K e

†

Introduce fermion creation/annihilation by Jordan-Wigner string

j j j jj m j m

i i

m m m mC e C ep p

Thus we shall work with a chain of sites with Fermions living there upon.

20

1 1 1 1 1 1( )( ) ( )

2m m m m m m m m m m m mK K

V e e

Next, we must rewrite

in terms of Fermions, using

†

†

,

,

j j j jj m j m

i i

m m m m

m m m m

C e C e

C C

p p

†

,j

j m

jj m

i n

m m

i n

m m

C e

C e

p

p

†1† †

1 1 1For instance, .j j

j m j m m m

i n i ni C C

m m m m m mC e C e C e Cp p

p

Fermionization of the Transfer Matrix VFermionization of the Transfer Matrix V

21

†

†

†1

† †1 1

†1

1

1

1

occupied occ

So, one concludes

upied 0

emty emty em

th

ty .

at .

m m

m m

i C Cm m

i C Cm m m m

m m

m m

m m

m m

m mC e C

C e C C C

C

m m

C

m

p

p

But wait! Annihilation on m brings a – sign!†

1 1m mi C C

m m m mC e Cp

†

†

1 1

1 1

†1 1 1 1

emty 0 emty ,

occupied occupied

a .

but

nd also

m m

m m

i C Cm m m m

i C Cm m m m

m m m m m m m m

m mC e C C C

C e C C C

C C

m m

C C

p

p

††1 1Consider for instance the action of .m mi C C

m m m mC e Cp

†† †1 1Next,consider the action of .m mi C C

m m m mC e Cp

†

†

† †1

† † †

1

†11 1

† †1 1

occupied occu

So,

pied 0

emty emty

one concludes tha

mty

.

e

t

m m

m m

i C Cm m

i C Cm m m m

m m m

m

m

m

m m

m mC e C

C e C C C

C

m m

C

m

p

p

We succeded in converting V to a second-quantization operator with the structure of an exponential. At the exponent there is something similar to a periodic tight-binding chain with nn hoppings + interaction terms. This fictitious system will contain an unspecified number of interacting particles and we must seek the largest eigenvalue.

We shall be able to solve the problem by using the periodicity and the fact that the interactions are of the pairing type.

† † †1 1 1 11 1( ) ( )( )

2 .m m m m m m m m m mm mK C C C C C C C C K C C C CV e e

1 1

2 22 1 2( ) ( )V V V V

† 12 ( )

221 [2sinh(2 )]Recall

m mm

N C C

V K e

1d periodic Fermi system.

Therefore,

23

Fourier, of course/ 4

/ 4

( , )

† ( ) †

( , ) ( , )

† †,0

( , )

† †

(the will be useful later)

.

We prefer to have q>0 and write

1

1

(

iimq i

m qq

Nim q q

m m q qm q q

N Nimq

q m m q qm m q

N

m m q qm

eC e e

N

C C eN

e C CN

C C

pp

p p

p p p p

p p

È

†

(0, )

). Then,q qq p

( , ) (0, )

11

2q qp p p

† †

(0, )

2 ( 1)2

1 [2sinh(2 )] .q q q q

q

N

V K e p

† 12 ( )

221 [2sinh(2 )]

m mm

N C C

V K e

Jean Baptiste Joseph Fourier

24

2† † ( 1) † † † † † †

1 1, ( , ) ( , )

iN

imq i m q iqm m q q m m q q

q q m q

eC C e C C i e

N

p

p p p p

and restricting to positive q† † † †

1(0, )

2 ( )N

m m q qm q

C C sin qp

Similarly1

(0, )

2 ( )N

m m q qm q

C C sin qp

† † † †1 1

( , ) ( , )

,N N

iq iqm m q q m m q q

m q m q

C C e C C ep p p p

† † †1 1

( , )

( ) 2 ( )N

m m m m q qm q

C C C C cos qp p

We need V2 too

† †11( )( )

2m m mmK C C C C

V e

All toghether

/4

( , )

iimq

m qq

eC e

N

p

p p

25

†

1

† † †2

(0, )

22(to get just divide exponent by

exp 2 cos( )

2)

( ) ( )( )q q q q q q q

q

q qq

V K q sin

V

qp

Thus, skipping some complications arising from periodic/antiperiodic boundary conditions and odd/even numbers of particles (since all such distinctions lose importance in the thermodynamic limit) we write:

† †

(0, )

2

(0, )

1 1

2 22 1 2 2 2

(0, )

1† † † †2

2

2 ( 1)

1

[2sinh(2 )] ,

exp cos( )( ) sin( )( )

.q q q q

q

N

qq

q q q q qq

q q q q q q q q q

q

V K V

V V V V V V

V K q q

V e p

p

p

1 1

2 22 1 2( ) ( )V V V V

26

† †

† † † † † † † †(0, )

2 ( 1)cos( )( ) sin( )( ) cos( )( ) sin( )( )q q q q

q q q q q q q q q q q q q q q qq

q

K q q K q qe e

V

e p

Different q are decoupled and the problem reduces to diagonalizing

† †( )q q q q

That is to diagonalizing simultaneously

† †( )q q q q

Since simultanous eigenstates of those operators are eigenstates of Vq.