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The Onsager solution of the 2d Ising model (continued). Summary of first part. Notation:. Fourier, of course. By fermionization V becomes an infinite interacting 1d periodic Fermi system . Jean Baptiste Joseph Fourier. - PowerPoint PPT Presentation
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The Onsager solution of the 2d Ising model (continued)
,Ising on NxN square latticei j
adjacent i j
H Js s
2Ising
1,1 1,2 ,
.N N
HFN
s s s
Z e e
Notation:K J
, , 1 , 1,1, 1, 11 2 1 2 ,
, ,
|
largest eigenvalue ( in thermodynamic limit)of transfer matrix
( , ,..., ) ( , ,
is defined on two neighbourin
|
g rows and
..., )
+
k k k k
Nmax max
NJs s Js s
N Nk
s s s
Z N
V e es s s
1.
Summary of first part
1
Fourier, of course
a second-quantization operator with the structure of an exponential. This fictitious system will contain an unspecified number of interacting particles and we must seek the largest eigenvalue.
By fermionization V becomes an infinite interacting 1d periodic Fermi system.
†† † † †1 11 1
12 ( )( )( ) ( )( )22 2 2
2
[2sinh(2 )]
Recall: K= J, tanh( ) .
m mm m m m m mm mm
N K KC CC C C C C C C C
K
V K e e e
e
/4/4
( , )
† † †
(0, )
(the will be useful later)
We prefer to have q>0 and write ( ). Then,
iimq i
m qq
N
m m q q q qm q
eC e eN
C C
Jean Baptiste Joseph Fourier
2
3
† † † †
† †† † † †
(0, )
cos( )( ) sin( )( )
2 ( 1)cos( )( ) sin( )( )
*
q q q q q q q q
q q q qq q q q q q q qq
K q qq
K q q
e
e
V
e
different q are decoupled and the problem reduces to diagonalizing
† †( ) pair creation and annihilationq q q q
† †( ) total fermion numberq q q q
We shall find simultaneous eigenvectors of
Simultanous eigenstates of those operators are eigenstates of Vq.
2where tanh( ) .Ke
In states labelled by q one finds either 0 or 1 particles, and one can create in both q and –q but only once. The dimension of the Hilbert space is just 4.
End of summary of first part
3
4
Diagonalization of Vq
What is the Hilbert space? 4 basis vectors per q † † †0 0 0vacuu ,m, .q q qq q q
† † † †
† † 2 cos
†0
eigenare already one-Fermion vectors :
0 0
( )q q q q q q
K q
q
q q q
q q q qq q q q qV e
† †
† † † † † † † †(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 qqK q q K q qqV ee e
The average over the pair state is obtained by setting occupation numbers =1.† †
(0, )
2 ( 1)2
1 0 0
2
1 2
| | | | .
00
q q q qq
q q qq
q
q V e e
eV
e
† †
(0, )
† †
(0, )
2 ( 1)
1
2 ( 1)2
0 1 0 0 0
Start with . It depends on number operators only.
The vacuum average is immedi | | |at | .e:
q q q qq
q q q qq
q
q
V e
V e e
In states labelled by q one finds either 0 or 1 particles, and one can create in both q and –q but only once.
0
We must seek more eigenvectors so we must construct the matrix of V
in pair space ( ,
two
). q
5
1
† † † †2
0
2
Next, we need the matrix
e
in pair
xp cos( )( ) sin( )
space ( , )
( ) .q q q q q
q
q
q
q q q
of
V K q q
† † 2 00 0 0
In pair space ( , )) ( 1 ., q q q qqq z
† †
†
† † †
† † †
0
†
0
†
creates the pair and k
s
ills it.
,
the pair operin( )( ) involves ,
where
sin( )( ) sin( )( ) ,
0 1 0 0
at
, .
ors
0 0
and
1 0 q q x
q q q q
q q
q q q q q
qq qq
q
q q q q
q q
q q
q
b b
q q b
b b
b
b b b b
1
† †22 exp cos( )( 1) sin( )( ) .q z q q q qV K q q
2
12
2
0
12 in pair space ( , ) is:Therefore, the matrix of
exp cos( )( 1) sin( ) .
qqq
q z x
V
V K q q
6
1
cos( )22
cos( )
exp cos( )( 1) sin( ) exp c
One can evaluate the exp
os( ) sin( )
, wherecos( ) sin( )
cos( ) sin( )
on
sin( ) cos( )
ential.
K qq z x z x
K q K M
z x
V K q q e K q q
e eq q
M q qq q
�;
1 0 cos( ) sin( ) cos( ) sin( ) 1 0M is a root of unity: since .
0 1 sin( ) cos( ) sin( ) cos( ) 0 1
One can do the exponential by resumming the Taylor series.
q q q qM
q q q q
osh(K) M inh(K)! !
m mKM
m even m odd
K Ke M c sm m
� � �
6
7
1cos( )2
2
cosh( ) sinh( )cos( ) sinh( )sin( )we get .
sinh( )sin( ) cosh( ) sinh( )cos( )K q
q
K K q K qV e
K q K K q
2 cos( )
2
2
We must perform the matrix product; let
cosh( ) sinh( )cos( ) sinh( )sin( )sinh( )sin( ) cosh( ) sinh( )cos( )
cosh( ) sinh( )cos( ) sinh( )sin( )0s0
q qK qq
q q
q q
q q
A CV e
C B
A CC B
K K q K qK q K K q
K K q K qee
inh( )sin( ) cosh( ) sinh( )cos( )K q K K q
1cos( )2
2
cos( ) sin( )Inserting into the result Cosh(K)+M Sinh(K), ,
sin( ) cos( )K q KM KM
q
q qV e e e M
q q
� � �
21 12 2
2 1 2 1 2using previous result
0Finally, compute: .
0 q q q q q
eV V V V V
e
8
2 cos( ) q qK qq
q q
A CV e
C B
By hand this product is tedious.Fullsimplify on Mathematica yelds:
22 2 2
22 2 2
cosh( ) sinh( )cos( ) (sinh( )sin( ))
(sinh( )sin( )) cosh( ) sinh( )cos( )
2sinh( )sin( )(cosh(2 )cosh( ) sinh(2 )sinh( )cos( ).
q
q
q
A e K K q e K q
B e K q e K K q
C K q K K q
0
2
2 cos( )
Recall: K= J, tanh( ) .
is a known function of .
Getting the eigenvalues of a 2X2 matrix is
in
t
pa
rivial, but we need to find the simplest expressi
ir space
o
,
n
( ), q
K
q qK qq
q qq
eA C
V e qC B
.
We show that det 1 and use T .rq q q qq q
q q q q
A C A CA B
C B C B
9
From
22 2 2
22 2 2
cosh( ) sinh( )cos( ) (sinh( )sin( ))
(sinh( )sin( )) cosh( ) sinh( )cos( )
2sinh( )sin( )(cosh(2 )cosh( ) sinh(2 )sinh( )cos( )
q
q
q
A e K K q e K q
B e K q e K K q
C K q K K q
2
2 2
, , expanding the and collecting,Using theabove expressions for
2((cosh( ) sinh( ) )cosh(2 ) 2sinh(2 )cosh( )sinh( ) cos( );
q q
q qq q
q q
A B
A CTr A B K K K K q
C B
2 product of eigenvalues 1.
it is expedient to write the eigenvalues in the form , 0
The
1 det 1,
2cosh .)n (
q
q q
q qq q q
q q
q qq q
q
qq q
A CA
e
e e
B CC B
A Cr A BC B
T
one can work out, recalling that V is hermitean and eigenvalues must be real:
9
10
2 2
Simplify using the identities:
2cosh( )sinh( ) sinh(2 ), cosh( ) sinh( ) cosh(2 )q qA B
K K K K K K
cosh( ) cos
is know
h(2 )
n! It is the p
cosh(2 ) sinh(2 )sinh(2 )cos( )
ositive root ofq
q K K q
2[cosh(2 )cosh(2 ) sinh(2 )sinh(2 )cos( )].
This must be 2cosh( )
q
q q
q
qA CTr K K q
C B
2 22((cosh( ) sinh( ) )cosh(2 ) 2sinh(2 )cosh( )sinh( ) cos( );q q
q q
A CTr K K K K q
C B
2 cos( )0
2 cos( )
Summarizing, in pair space ( , ), .
This has eigenvalues , 0.q
q qK qqq q
q q
K qq
A CV e
C B
e e
2 cosare already one-Fermion vectorRecall: s :e e ;ig n K qqq q qV e
2 cos( ) 2 cos( )
0 the m
that has eigenvalues
atrix of V in pair space (
, 0
i :
.
,
) s
qq qK q
q
K q
qqq
q
A CeV e e
C B
The 4 eigenvalues of Vq
2
(0, )
2
2 cos( )
2
largest eigenvalue of
The transfer matrix is [2sinh(2 )] and so we want
[2sinh(2 )] .
The largest eigenvalue of largest exponent choose
[2sinh(2 )]
q
N
N
max qq
q
K qq
N
max
V K V
K V
V
K
ee
2 co 2
0
s( ) [2sinh(2 )] e 2 cxp os( ].)[q
NK
q
qqe K K qe
( , )
q
2
122
(- , )
[2sinh(2 )] co
Extend to q (- , ) and divide by 2 exponent:
exp [ ].
We
s( )2
cos can clean up since ( ) 0. Therefore,
[2sinh(2 )] .q
q
Nq
max
N
max
q
K K q
q
K e
ò
11
( , )
122[2sinh(2 )] .
N
max K e
ò
12
422 , In the thermodynamic lim [2sinh(2 )] .it qNN dq
maxdq K eN
ò
The free energy per spin is1 1[ ln(2sinh(2 ) ].2 4 qF KT K dq
ò
2
qTheeigenvalues
cosh( ) cosh(2 )cosh(2 ) sinh(2 )sinh
ε are k
(2 )cos( )
where tanh( )
nown. Reca l
.
l
K
q K K q
e
The solution is a bit implicit, but complete !
13
2critical temperature .ln(1 2)B C
Jk T
Temperature dependence of energy per spin Temperature dependence
of specific heat
Actually, one can obtain (see Huang page 386) the slightly more explicit expression
2 220
1 1 2sinh(2 )ln[2cosh(2 )] ln{ (1 1 sin( ) } where .2 2 cosh(2 )
KF K dK
1 1[ ln(2sinh(2 ) ].2 4 qF KT K dq
ò
14
2critical temperature .ln(1 2)B C
Jk T
Temperature dependence of spontaneous magnetization (Yang’s calculation, not Onsager’s, where H=0),see Huang page 391
Phase transition
18
4
1[1 ](sinh(2 ))
mJ
14
Monte Carlo simulations
The exact solution offers a unique benchmark for computer simulations by a Metropolis algorithm. Lattices of up and down spins are produced by random number generation using an algorithm based on importance sampling. Given a lattice one spin is turned and the new energy evaluated. The new lattice is accepted if the energy is not too high. Then averages are computed.
Average:
r
r
Er
rE
r
A eA
e
22
22
in this way one estimates (fluctuation-Dissipation theorem),
Magnetic susceptibility , etc.
vC E ET
M M
15
(from Lectures by Lisa Larrimore)16
17
Temperature dependence of specific heat
18
19
20
(Weiss theory)
(Weiss theory)
20
21
d ferromagnetic Ising ModelThe Ising model including H is soluble if we assume N>>1 nearest neighbours (many dimensions or very long-range interaction): we must divide interaction by N to have a finite answer for large N. So we take:
,
Hamilto nian 02 n n
mm
n nNJJ s s H s
2,
,
2 2
{ } { }
2
,sum over all confi
Since the total spin is :
gurati s.
,
onm n
n
i
nm n
i
m
H s
n nn m n
J JsN
s
HSN
s
Ss
S Ss s s
Z e e
We calculate Z exactly via the Hubbard-Stratonovich transformation
0
Non-interacting case J=0
( ) ( ) [2cosh( )]
log(2cosh( )).
kk
i
k
k
H sHs H H N
s sk k
Z H e e e e H
F KTN H
22
22
22Why exp[ ] ?2
02
,J Sx Na N J N Je dx x JSxdxe a
a
21 exp[ ]2 2N J N Jdx x
2By a shift of coordinate 1= exp[ ( ) ].2 2N J N J Sdx x
N
2 2 222 2 2Multiply both sides by : exp[ ( ) ].
2 2
J J JS S SN N N N J N J Se e e dx x
N
2
2 2 2
2 22
22
2At the exponent: ( ) 22 2 2 2
exp[ ], qed2 2 2
2 2
J SN
J N J SJ N J S N J N J SS x x xN N N
N J N J N Jx JSx e dx
SN N
x JSx
The Hubbard-Stratonovich transformation
2
1 , with a=2
axa N Jdxe
23
2 2
( )( )
0
2
(
( ) 2
)
0
( )( , ) ( ) [2cosh( ( ))] .
( , )2 2
kk
i
i
k
k
H Jx sH Jx s H Jx H N
s sk k
N J N Jx H x x
Jx
J S
s
Z x H e e e e H Jx
N J N JZ dxe e dxe Z x H
22
{ } }
2
{
becomes exp[ .]2 2
i is s
J S HSN HS N JZ e Z eN Je dx x JSx
2
)
2 ( )
0(
Next, exchange with .2
weighted average of ( , ) (noninteracting spins in field H+Jx);i i
i
N JH Jx S
H Jx
x
s s
s
S
N Jdx Z dxe e
Z Z x H e
2
2
20
}
22
{ } {
Noninteracting case: . .
Substi
( )
etute xp[ ]2 2
i i
HSHS
s
J
s
SN
J SN
e
N J N Je dx x JSx
H eZ e Z
The Hubbard-Stratonovich transformation
24
0 tanh( ) tanh( )N Jx NJ H Jx NJx NJ H Jx
2
20 0
the optimum x is tanh( ),effective magn
( , ) ( , ).2
etic fieldN J xN JZ dxe Z x H Z x H
x H J xJ x
Evaluation of Z by the steepest descent method
( , , ...) ( , , ...) ( , , ,...)const. , such that 0.f x a b f x a b
x x
df x a bdxe e xdx
20ln( ( , ))
2
2
N J x Z x HN JZ dxe
the minimum condition isWith
20 [ ln(cosh( ))] { tanh( )}2
d N J x N H Jx NJ x H Jxdx
2
20 0( , ), ( , ) [2cosh( )] .
2
N J x NN JZ dxe Z x H Z x H H Jx
25
PIERRE-ERNEST WEISS born March 25, 1865, Mulhouse, France. died Oct. 24, 1940, Lyon, France.
Recall the Weiss mean field theory (1907)
effH H M
1 , 2 tanh2
BB
B
HNJ g MV K T
( )tanh BB
B
H MNMV K T
The behavior of an Ising model on a fully connected graph may be completely understood by mean field theory,because each site has a very large number of neighbors. Each spin interacts with all the others, Only the average number of + spins and − spins is important, since the fluctuations about this mean will be very small. Mean field = exact in infinite d
Exchange field
Mermin-Wagner theorem in Statistical Mechanics, = (Coleman theorem (1973) in QFT)In d=1 and d=2 continuous symmetries cannot be spontaneously broken at finite T in systems with sufficiently short ranged interactions. This does not apply to discrete symmetries (2d Ising model)
Thin films do show phase transitions experimentally, so perhaps they are not really 2d.
X-Y model and Kosterlitz-Thouless topological transitionVortices canot occur in 1d. Since KT deal with vortices we digress a little about fluids. For a 2d fluid moving with velocity u(x,y,t) in 2d the vorticity is
. ( ) ( )yxzS S
C
uuk dl u dS rot u dS
y x
k>0
In addition topological transitions (no broken symmetry) can occur in 2d.
26
In polar coordinates with basis vectorsˆ( , ), =z ( , )x y y x
r r r r
( , ) (0, )2rku u ur
consider a circular vortex
If the density is , the kinetic energy of the vortex is finite if we introduce two cutoff lengths: size R of the fluid, and intermolecular distance r0
0
2 22
20
1 1 2 ln( ).2 2 (2 ) 4
R
r
k k RE dSu drrr r
In a similar way one can compute the energy of two vortices of opposite vorticity at a distance d and find that they attract each other. One finds that the interaction energy grows with the distance as
2
0
( ) log( )2 2k dU d
r
Vortex and antivortex attract each other and annihilate.
27
,
cos( ), J>0j jj
H J
X-Y model and the KT transition Classical spins confined in (x,y) plane in a 2d lattice; the one at site j makes an angle j with the x axis. The Hamiltonian is:
In the ground states all spins are aligned in some direction, and every cos =1
Excitations are vortices and monopoles. They are topological, that is, they are like holes in the system.I take figures from Mahan’s Nutshell book
28
29
2
Entropy of vortex (or monopole) =K log(number of sites)=K log( )
Energy of vortex-antivortex (or monopole-antimonopole) interaction
estimated by Kosterlitz-Thouless =2 J log( ).
Same story for vortic
B BRa
Ra
es and monopoles. Both are topological excitations.
JTransition temperature T=BK
Above, the system creates free vorticesBelow, excitations consist of vortex-antivortex pairs.No symmetry is broken in the transition.
30
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