Download Renormalization Group Seminar Exact solution to the Ising model

Renormalization Group Seminar
Exact solution to the Ising model
Alois Dirnaichner
12.05.2009
Exact Solution to the Ising-Model
Outline:
Starting point: Hamiltonian derived in previous lecture
Transformation to free fermion-style Hamiltonian
Diagonalization
Partition sum, free energy and specific heat
Starting point
H =∑  3 n− ∑  1 n 1  n1
n
n
describing one-dimensional quantum chain in a
transverse magnetic field with periodic boundary
conditions
Remarks:
Unitary transformation 1 − 3 ;  3   1
 1 n=1×...× 1×...×1
Transformation I
1
±n ≡  1  n±i  2 n
2

−
satisfying{ n ,  n }=1
[ n , −m ]=0 ; n≠m
⇒ H =∑ 1−2 −n n − ∑ n −n n1 −n1 
n
n
and
then
n−1
n−1
c n ≡expi  ∑   
l=1

l
−
l
−
n
and
c ≡ exp−i  ∑ l −l 
†
n

n
l=1
†
{c n , cm }=nm
to arrive at Fermion operators with
Motivation:
{n , expi  n −n  }=0
n
n−1
exp ∑
l=1
± 2
and   =0
Transformation II
⇒ H =∑ 2cn c n − ∑  c n−c n c n1 c n1 
†
n
†
†
(Free fermion
Hamiltonian)
n
in momentum space:a =
k
†
N
1
2
2N
ikn
e c n ; k=0,±
, ... ,±
∑
2N1
2N1
 2N1 n=−N
†
†
†
⇒ H = ∑ 21−cos k a k a k a−k a−k 2i  ∑ sin k a k a−k a k a −k 
k 0
after symmetrization.
k 0
Diagonalization
construct explicitly the matrix in Fock space
calculate the eigenvalues:
a=−2i  sin k ; b=1− cos k
0/1 =2b
2/3  k =2 b±2  12 −2 cos k
(twice)
Partition sum and free energy
The partition sum yields Z =tr e

− H
=∑ 

 exp− N ∫ dq 3 q
N
 TD limit
3 k =1− cos k −  1 −2  cos k 
2
Thus we get for the free energy density

F
1
1
˙
=−
ln Z = f =− ∫ dk  1−2 cos k 2 const.
0
N
N
0
Specific heat
We analyze the free energy integral for small values of k.
To do so, we cut the integral in two parts.



0
0

k2
cos k ≈1−
2
∫  ∫ ∫
When we integrate this and identify
t≡∣1−∣~
T −T c
Tc
we get

2
−∂
f =−∫ dk  t 21−t  k ~−t ln∣t∣ ⇒ c= 2 f ~−ln∣t∣
∂t
0
2
2
2
Free energy, entropy
Specific heat