Analytic solutions to 1D Ising model#

\[H = -\sum_{\langle ij \rangle} J s_i s_i -B\sum_i s_i\]
\[\beta H = -\sum_{\langle ij \rangle} K s_i s_i - h \sum_i s_i\]
\[K = \frac{J}{k_B T}\,\, , \,\, h = \frac{B}{k_B T} \]

Case-1: \(h\neq 0\) and \(J=0\)#

\[Z = \sum_{[s]} e^{h \sum s_i } = \sum_{[s]} \prod^{N}_{i=1} e^{h s_i } = \prod^{N}_i Z_i = Z^N_i\]
\[Z_i = \sum_{s_i} e^{h s_i} = e^{h}+e^{-h} = 2cosh(h)\]
\[ m = \langle s_i \rangle = \frac{\partial log Z_i}{\partial h} = \frac{e^{h}-e^{-h}}{e^{h}+e^{-h}} =tanh(h)\]

  • Free energy

\[F = -\beta^{-1} log Z = -\beta^{-1} N log \Big[ 2cosh(\beta B)\Big]\]

  • Energy. Obtained by ensemble average of Ising Hamiltonian. Which comes down as simply the average of single spin multipled by number of spins

\[U = -N B\langle s \rangle = -N B \cdot tanh(\beta B)\]

  • Entropy

\[S = N k_B\Big(-\beta B \cdot tanh(\beta B) + log \Big[ 2cosh(\beta B)\Big] \Big)\]

  • Heat capacity

\[c = \frac{1}{N}\frac{\partial U}{\partial T} = k_B \beta^2 B^2 sech^2 (\beta B)\]

We see that heat capacity goes to zero at both \(T=0\) and \(T=\infty\)

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)

Case-2: \(h= 0\) and \(J\neq 0\)#

\[H = -J \sum^{N-1}_{i=1} s_i s_{i+1}\]

  • We use free boundary conditions

  • Let \(\tau_1 =s_1\) and \(\tau_j =s_{j-1} s_j\) for \(j\geq 2\)

\[H = -J \sum^{N}_{i=2} \tau_i\]

  • There will be an overall factor of 2 in front of partition function becasue of summation over \(\tau_1\) $\(Z = \sum_{\tau} \prod^{N}_{j=1} e^{K \tau_j} = 2 \prod^{N}_{j=2} \sum_{\tau_j} e^{K \tau_j} =2 \prod^{N}_{j=2} Z_j\)$

\[Z_j = e^K +e^{-K} = 2cosh(K) =2cosh(\beta J)\]
\[Z =2Z^{N-1}_j = 2 (2 cosh(K))^{N-1} \approx (2 cosh(K))^N\]
\[F = -\beta^{-1}logZ = -k_B T N log [2cosh(\beta J)]\]

Case-3: \(h \neq\) and \(J\neq 0\) and Transfer Matrix technique#

\[H = -J \sum^{N-1}_{i=1} s_i s_{i+1} - \frac{B}{2} (s_j +s_{j+1})\]

  • We write hamiltonain in this symmetric form consisting of sums of \((s_j, s_{j+1})\) terms for presenting partiion function as product of terms. $\(Z = \sum_{[s]}\prod_j e^{K s_j s_{j+1}+\frac{1}{2}h(s_j+s_{J+1})} = \sum_{[s]}\prod_j T(s_j, s_{j+1})\)$

  • Transfer matrix has been introduced:

\[T(s_j, s_{j+1}) = e^{K s_j s_{j+1}+\frac{1}{2}h(s_j+s_{J+1})}\]
\[\begin{split}T(s_j, s_{j+1}) = \begin{pmatrix} e^{K+h} & e^{-K} \\ e^{-K} & e^{K-h} \end{pmatrix} \end{split}\]

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 \(T_{j,j+1} = \langle s_j | T | s_{j+1}\rangle\) can be seen as an operator that propagates or transfers state from spin \(j+1\) to spin state \(j\). $\(Z = \sum_{[s]}e^{-\beta H} = \sum_{s_1,s_2,...s_N} \langle s_1 | T | s_{2}\rangle \langle s_2 | T | s_{3}\rangle ...\langle s_{N-1} | T | s_{N}\rangle\)$

\[Z = \sum_{s_1 = \pm 1} \langle s_1 | T^N | s_{1}\rangle = Tr(T^N)\]

Trace of matrix is invariant to unitary trasnformation \(U^{-1} T U =D\) 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:

\[Z = \lambda^N_1 +\lambda^N_2\]

Problem is reduced to diagonalizing the transfer matrix

\[\begin{split}det \begin{pmatrix} e^{K+h}-\lambda & e^{-K} \\ e^{-K} & e^{K-h}-\lambda \end{pmatrix} =0 \end{split}\]
\[\lambda_{1,2} = e^K \big ( cosh(h) \pm \sqrt{cosh^2(h)-2e^{-2K}sinh(2K)}\Big)\]

  • Thus we have arrived at an exact solution for the one dimension Ising model with external field:

\[Z = \lambda^N_1 +\lambda^N_2 = \lambda^N_1 \Big(1+\big (\frac{\lambda_2}{\lambda_1}\big)^N \Big)\rightarrow \lambda^N_1\]

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

\[F = -k_B T N log \lambda_1 \]