Transcript The Potts and Ising Models of Statistical Mechanics

The Potts and Ising Models of
Statistical Mechanics
The Potts Model
The model uses a lattice, a regular,
repeating graph, where each vertex
is assigned a spin. This was
designed to model behavior in
ferromagnets; the spins interact
with their nearest neighbors align in
a low energy state, but entropy
causes misalignment. The edges
represent pairs of particles which
interact.
The q-state Potts Model

The spins of the model can be represented as
vectors, colors, or angles depending on the
application.
q=2
q=3
Unit vectors point in the
q orthogonal directions
n  2 n / q, n  0,1, . . . , q  1.
q = 2 is the special
case Ising Model
q=4
Phase Transitions


The Potts Model is used to analyze behavior in
magnets, liquids and gases, neural networks, and
even social behavior.
Phase transitions are failures of analycity in the
infinite volume limit. For ferromagnetics, this is
the point where a metal gains or loses
magnetism.
Magnetism
Magnetism
N 
Temperature
Temperature
The Hamiltonian

In mechanics, the Hamiltonian
corresponds to the energy of a system.
The lowest energy states exist where all
spins are in the same direction.
h( )   J   ( i , j )
{i , j}E ( G )
Where ω is a state of graph G, σi is the spin at vertex i, δ
is the Kronecker delta function, and J is the interaction
energy
This measures the number of pairings between vertices
with the same spins, weighted by the interaction energy.
Ising model Hamiltonian
For the Ising model, the Kronecker delta may be defined
as the dot product of adjacent spins where each spin is
valued as the vector <1,0> or <0,1>.
0
1
0
1
0
1
0
h( )   J
0
1
1
1
0
1
0
1
0
0
0
1
1
1
1
{i , j}E ( G )
 i  j
Where J is uniform throughout the graph.
1
1

An edge between two like spins has
a value of 1, while an edge between
unlike spins has a value of 0.
For this state, h(ω) = -14J
The Partition Function

The probability of a particular state:
1
exp(
h ( ))
kT
Pr(i  T  
1
 exp( h( ))
all 
kT
k=1.38x10-23 joules/Kelvin, the Boltzmann
constant

The denominator is the same for any
state, and is called the partition function.
Z   exp( 
all 
1
h( ))
kT
Z   exp(
all  {i , j }
J
 i  j )
kT
4-cycle example
Recall the
partition
function:
 exp(
all 
1
h( ))
kT
H  4 J
H  2 J
H  2 J
H  2 J
H  2 J
H  2 J
H  2 J
H 0
H 0
H  2 J
H  2 J
H  2 J
H  2 J
H  2 J
H  2 J
H  4 J
In this example, the partition
function is:
4J
2J
2exp( )  12exp( )  2
kT
kT
The probability of an
all blue spin state:
exp(4
2 exp(
J
)
kT
4J
2J
)  12 exp(
)2
kT
kT
Why use another approach?

The partition function is difficult to
calculate for graphs of useful size, as the
number of possible states increases
rapidly with the number of vertices.
|i | 2
N
|i | is the number of states
2 is number of possible spins,
and N is the number of vertices

An effective approach for estimating the
partition function for large lattices is
through a transfer matrix. This estimation
is exact for infinite lattices.
Ising Model Transfer Matrix

Define a matrix P such that
1
 i P j T  exp( J ( i   j ))
kT


For like spins:
1 
 0
J
1,0 P    0,1 P    exp( )
kT
0 
 1
Unlike spins:
0
 1
1,0 P    0,1 P    exp(0)  1
1
 0
This is a symmetric 2x2 transfer matrix, describing all
possible combinations of spins between two vertices
J


exp(
)
1


kT


J

1
exp( ) 

kT 
The exact partition function using a
transfer matrix
Using the transfer matrix for a
cycle, the partition function is:
1
exp(

h( ))    P  P ... P  P

kT

 Which can be simplified to become:
Z    1 P N 1T
As    I the identity matrix

T
1
all
T
2
2
3
N 1
T
N
N
T
1
all
T
n
n
all i
=Tr (P N )
=  i N
i
=1N  2 N
Since σ1 = σ1 in all possible states
Where λ1 and λ2 are the eigenvalues of
the transfer matrix:
1  exp(
J
) 1
kT
2  exp(
J
) 1
kT
Lets check if this works:

We calculated the partition function
for a 4-cycle earlier:
2 exp(
4J
2J
)  12 exp(
)2
kT
kT
Using the transfer matrix approach,
Z  14  2 4

J
J
4
4
Z  (exp( )  1)  (exp( )  1)
kT
kT
4J
2J
=2exp( )  12exp( )  2
kT
kT
Phase Transitions

As the partition function is positive, the
eigenvalue with the largest absolute value must
also be positive. And, as
Z   i N  1N
N 
i


Failure of analycity occurs where two or more
eigenvalues have the greatest absolute value.
This is proven not occur in the one dimensional
Ising model, where   exp( J )  1 and   exp( J )  1
1

kT
2
kT
but does occur in greater dimensions, or in Potts
Models with more spin possibilities.
Thank You