Analytic solutions to 1D Ising model#

H=ijJsisiBisi
βH=ijKsisihisi
K=JkBT,h=BkBT

Case-1: h0 and J=0#

Z=[s]ehsi=[s]i=1Nehsi=iNZi=ZiN
Zi=siehsi=eh+eh=2cosh(h)
m=si=logZih=eheheh+eh=tanh(h)
  • Free energy

F=β1logZ=β1Nlog[2cosh(βB)]
  • Energy. Obtained by ensemble average of Ising Hamiltonian. Which comes down as simply the average of single spin multipled by number of spins

U=NBs=NBtanh(βB)
  • Entropy

S=NkB(βBtanh(βB)+log[2cosh(βB)])
  • Heat capacity

c=1NUT=kBβ2B2sech2(βB)

We see that heat capacity goes to zero at both T=0 and T=

import ipywidgets as widgets
import matplotlib.pyplot as plt
import numpy as np
import scipy as sp
@widgets.interact(kbT=(0.1,10))
def mag1(kbT=1):
    
    x = np.linspace(-10,10,1000)
    plt.plot(x, np.tanh(x/kbT), lw=3, label=f'$k_B T={kbT}$')
    plt.xlabel('h, magnetic field',fontsize=16)
    plt.ylabel('m, magnetization',fontsize=16)
    plt.legend(fontsize=16)
1.00

Case-2: h=0 and J0#

H=Ji=1N1sisi+1
  • We use free boundary conditions

  • Let τ1=s1 and τj=sj1sj for j2

H=Ji=2Nτi
  • There will be an overall factor of 2 in front of partition function becasue of summation over τ1 $Z=τj=1NeKτj=2j=2NτjeKτj=2j=2NZj$

Zj=eK+eK=2cosh(K)=2cosh(βJ)
Z=2ZjN1=2(2cosh(K))N1(2cosh(K))N
F=β1logZ=kBTNlog[2cosh(βJ)]

Case-3: h and J0 and Transfer Matrix technique#

H=Ji=1N1sisi+1B2(sj+sj+1)
  • We write hamiltonain in this symmetric form consisting of sums of (sj,sj+1) terms for presenting partiion function as product of terms. $Z=[s]jeKsjsj+1+12h(sj+sJ+1)=[s]jT(sj,sj+1)$

  • Transfer matrix has been introduced:

T(sj,sj+1)=eKsjsj+1+12h(sj+sJ+1)
T(sj,sj+1)=(eK+heKeKeKh)

While compared to previous examples partition function did not factor out into single particle contributions we nevertheless have factored the partition function as product of 2 by 2 matrices!

Note the close connection of matrix technique applied to partion functions with mathematical formalism of quantum mechanics!#

In quantum mecchanical notation Tj,j+1=sj|T|sj+1 can be seen as an operator that propagates or transfers state from spin j+1 to spin state j. $Z=[s]eβH=s1,s2,...sNs1|T|s2s2|T|s3...sN1|T|sN$

Z=s1=±1s1|TN|s1=Tr(TN)

Trace of matrix is invariant to unitary trasnformation U1TU=D which we can use to diagonalize the matrix T which then allows to us to write the N product in terms of two diagonal elements:

Z=λ1N+λ2N

Problem is reduced to diagonalizing the transfer matrix

det(eK+hλeKeKeKhλ)=0
λ1,2=eK(cosh(h)±cosh2(h)2e2Ksinh(2K))
  • Thus we have arrived at an exact solution for the one dimension Ising model with external field:

Z=λ1N+λ2N=λ1N(1+(λ2λ1)N)λ1N

No phase transition at finite T>0 is posisble for 1D ising model as free energy remains analytic for T>0

F=kBTNlogλ1