One Dimensional Ising Model

1. Introduction

Consider a linear chain of N spins. Each spin can take values:

\[ S_i = \pm 1 \]

Only nearest neighbour interactions are considered.

2. Hamiltonian

The Hamiltonian of the 1D Ising model in an external magnetic field H is:

\[ H = -J \sum_{i=1}^{N} S_i S_{i+1} - \mu H \sum_{i=1}^{N} S_i \]

J = Interaction strength
J > 0 → Ferromagnetic case
μ = Magnetic moment
H = External magnetic field

3. Partition Function

The canonical partition function is:

\[ Z = \sum_{\{S_i\}} e^{-\beta H} \]

where:

\[ \beta = \frac{1}{k_B T} \]

4. Exact Solution (H = 0)

For zero magnetic field, the partition function becomes:

\[ Z = \lambda_+^N + \lambda_-^N \]

where

\[ \lambda_{\pm} = e^{\beta J} \pm e^{-\beta J} \]

5. Free Energy

\[ F = -k_B T \ln Z \]

For large N:

\[ F = -N k_B T \ln \left( 2 \cosh (\beta J) \right) \]

6. Magnetization

In one dimension and at finite temperature:

There is NO spontaneous magnetization at any finite temperature.

Thus, the 1D Ising model does NOT exhibit a phase transition at finite temperature.

7. Important Result

The 1D Ising model shows that short-range interactions in one dimension are not sufficient to produce phase transition at finite temperature.

One Dimensional ising Model

the same direction, yielding a net magnetic moment which is macroscopic in size. The simplest theoretical description of ferromagnetism is called the Ising model. This model was invented by Wilhelm Lenz in 1920: it is named after Ernst Ising, a student of Lenz who chose the model as the subject of his doctoral dissertation in 1925. Mathermatical model that explains Ferromagnetism . It can explain Phase transition between,
a). Ferromagnetism to Antiferromagnetism
b). gas to liquid
c). liquid to gas and many more

Ising Model considers the system as an array of $ N $ fixed lattice sites that form an $ n $-dimensional periodic lattice.
The structure may be square, cubic, hexagonal, etc. Each lattice site is associated with a spin variable represented by $ S_{1}, S_{2}, S_{3}, \ldots $ or by $ S_{i} \ (i = 1, 2, 3, \ldots) $.
The spin variable is a number which is either $ +1 $ or $ -1 $. It is $ +1 $ for up spin and $ -1 $ for down spin.
If the $ i $-th site has an up spin, then $ S_{i} = +1 $; and if the $ i $-th site has a down spin, then $ S_{i} = -1 $.
The given set of spins $ S_{i} $ represents the configuration of the whole system.
Each spin interacts with its nearest spins only.
The energy of the system in a given configuration is given by:

Hamiltonina of the system is :

$$ \begin{aligned} H\left(S_{i}\right)&=-J \sum_{n, n} S_{i} S_{j}-\mu B \sum_{i} S_{i} \\ \Rightarrow H\left(S_{i}\right)& = -J\left(S_{1} S_{2}+S_{2} S_{3}+S_{3} S_{4}+\ldots .\right)-\mu B\left(S_{1}+S_{2}+S_{3}+\ldots\right) \end{aligned} $$

We disregard the KE of atoms at the lattice sites Phase transition is essentially due to interaction energy among nearest neighbour-atoms .\\ To study the property of susceptibility we subject the lattice to an external magnetic field (B) .\\ In this case an additiona PE is associated with the spin $S_{i}$ given by $P . E=-\mu B S_{i}$\\ Where $\mu=g \mu_{B} \sqrt{j(j+1)} \mathrm{g}=$ Lande g factors,$\mu_{B}=$ Bohr Magneton, $j=$ Total angular momentum. The first term on the right-hand side of Equation above shows that the overall energy is lowered when neighbouring atomic spins are aligned. This effect is mostly due to the Pauli exclusion principle. Electrons cannot occupy the same quantum state, so two electrons on neighbouring atoms which have parallel spins (i.e., occupy the same orbital state) cannot come close together in space. No such restriction applies if the electrons have anti-parallel spins. Different spatial separations imply different electrostatic interaction energies, and the exchange energy, , measures this difference. Note that since the exchange energy is electrostatic in origin, it can be quite large. This is far larger than the energy associated with the direct magnetic interaction between neighbouring atomic spins.However, the exchange effect is very short-range; hence, the restriction to nearest neighbour interaction is quite realistic.\\ J is exhchange interaction energy. It occurs between identical \href{http://particles.It}{particles.It} is a quantum Mechanical effect. It is due to wavefunctions of indistinguishable particles subeject to exchange symmetry.

If $\mathbf{J}>\mathbf{0}$, there is Ferromagnetism

If $\mathbf{J}<\mathbf{0}$, there will be Antiferromagnetism

First term is interaction energy among spins . The second term is due to the interaction of the ith state spin with applied Magnetic field.\\ $Z=\sum_{S_{1}} \sum_{S_{2}} \sum_{S_{3}} \ldots \ldots . . \sum_{S_{n}} e^{-\beta H\left(S_{i}\right)}$\\ ( When Magnetic field is not subjected )\\ To simplify the partition function let us simplify for a one dimensional ISING chain with spins $S_{1}, S_{2}$ and $S_{3}$. Then The number of spin interactions taking into nearest spins is $2^{3}=8$. Which is:

$$ \begin{aligned} & S_{1} S_{2}+S_{2} S_{3}+S_{3} S_{1} \\ & =(1+1+1)+(1+1+1) \\ & +(1-1-1)+(-1-1+1)+(-1+1-1) \\ & +(1-1-1)+(-1-1+1)+(-1+1-1) \end{aligned} $$

There will be 8 microstates $. \mathrm{g}=2$ (degeneracy) with $\mathrm{E}=\mathrm{H}=3 \mathrm{~J}, \mathrm{~g}=6$ for $\mathrm{E}=\mathrm{H}=-\mathrm{J}$

$$ \begin{array}{ccc} S_{1} & S_{2} & S_{3} \\ +1 & +1 & +1 \\ -1 & -1 & -1 \\ +1 & +1 & -1 \\ +1 & -1 & +1 \\ -1 & +1 & +1 \\ -1 & -1 & +1 \\ -1 & +1 & -1 \\ +1 & -1 & -1 \end{array} $$

The ground state is expected to be one in which all spins are at the same state, i.e., . Let us calculate the energy for these configurations for a chain of N spins and N spins embedded on a line

\begin{align*} & Z=\sum_{S_{1}} \sum_{S_{2}} \ldots \ldots . \sum_{S_{N}} e^{-\beta H} \\ & =\sum_{S_{1}} \sum_{S_{2}} \ldots \ldots . \sum_{S_{N}} e^{\beta J \sum_{1}^{N} s_{i} S_{i+1}} \\ & =\sum_{S_{1}} \sum_{S_{2}} \ldots \ldots \sum_{S_{N}} \prod_{1}^{N}\left(\cosh \beta J+S_{i} S_{i+1} \sinh \beta J\right) \\ & =\sum_{S_{1}} \sum_{S_{2}} \ldots \ldots . \sum_{S_{N}}\left[\prod_{1}^{N-2}\left(\cosh \beta J+S_{i} S_{i+1} \sinh \beta J\right)\right] \\ & \times \\ &\left[\left(\cosh \beta J+S_{N-1} S_{N} \sinh \beta J\right)\left(\cosh \beta J+S_{N} S_{1} \sinh \beta J\right)\right]\\ &=\sum_{S_{1}} \sum_{S_{2}} \cdots \ldots . \sum_{S_{N}}\left[\prod_{1}^{N-2}\left(\cosh \beta J+S_{i} S_{i+1} \sinh \beta J\right)\right]\\ &\times\\ &\left.\left\{\left(\cosh \beta J+S_{N-1} \sinh \beta J\right)\left(\cosh \beta J+S_{1} \sinh \beta J\right)\right\}\left(\cosh \beta J-S_{N-1} \sinh \beta J\right)\left(\cosh \beta J-S_{1} \sinh \beta J\right)\right] \\ &=2 \sum_{S_{1}} \sum_{S_{2}} \ldots \ldots . \sum_{S_{N}}\left[\prod_{1}^{N-2}\left(\cosh \beta J+S_{i} S_{i+1} \sinh \beta J\right)\right]\left(\cosh ^{2} \beta J+S_{N-1} S_{N} \sinh ^{2} \beta J\right)\\ &Z=2^{N}\left\lfloor(\cosh \beta J)^{N}+(\sinh \beta J)^{N}\right\rfloor\\ & for N \gg 1, \quad(\cosh \beta J)^{N} \gg(\sinh \beta J)^{N}\\ &\Rightarrow Z=2^{N}(\cosh \beta J)^{N}=(2 \cosh \beta J)^{N}\\ \end{align*}

Thermodynamic Properties According to ISING MODEL

HelmHoltz free energy is :

\begin{align*} F&=-k T \ln Z\\ &=-k T\left(\ln \left(2 \cosh \frac{J}{k T}\right)^{N}\right)\\ &=-N k T \ln \left(2 \cosh \frac{J}{k T}\right) \end{align*}

The entropy $S$ is

\begin{align*} S &=-\frac{\partial F}{\partial T}\\ &=N k \frac{\partial}{\partial T}\left(T \ln \left(2 \cosh \frac{J}{k T}\right)\right)\\ &=N k \ln \left(2 \cosh \frac{J}{k T}\right)-\frac{N J}{T} \tanh \left(\frac{J}{k T}\right) \end{align*}

The Total energy of the system is :

\begin{align*} U&=F+S T\\ &=-N J \tanh \left(\frac{J}{k T}\right) \end{align*}

The Specific heat at constant Volume is :

\begin{align*} C_{V}&=\frac{\partial U}{\partial T}\\ &=\frac{N J^{2}}{k T^{2}} \sec h^{2}\left(\frac{J}{k T}\right) \end{align*}

As specific heat varies smoothly with T, there is no transition temperature. Hence One-dimensional Ising Model can not explain the Ferromagnetism behaviour of a Metal.

Net Magnetisation

$$ \begin{aligned} \bar{M}&=-\left(\frac{\partial F}{\partial \beta}\right)_{T}\\ &=-\frac{\partial}{\partial \beta}(-k T \ln (2 \cosh \beta J))\\ &=\frac{\partial}{\partial \beta}(k T \ln (2 \cosh \beta J)) \\ &=\frac{\partial}{\partial \beta}\left(\frac{1}{\beta} \ln (2 \cosh \beta J)\right)\\ &=-\frac{1}{\beta^{2}} \ln (2 \cosh \beta J)+\frac{1}{2 \cosh \beta J}(2 \sinh \beta J) \frac{J}{\beta} \\ &=-\frac{1}{\beta^{2}} \ln (2 \cosh \beta J)+\frac{J}{\beta} \tanh \beta J \\ & \text { If } \mathbf{M}=\mathbf{0} \text {, there will be spontaneous magnetisation. } \end{aligned} $$

Fermi-Dirac Distribution and Chemical Potential

The Fermi-Dirac distribution is given by:

$ f(\varepsilon) = \frac{1}{e^{(\varepsilon - \mu)/k_B T} + 1} $

where:

$ \mu $: Chemical potential
$ k_B $: Boltzmann's constant
$ T $: Temperature

Total Number of Fermions

The total number of particles is given by:

$ N = \int_0^\infty g(\varepsilon) f(\varepsilon) \, d\varepsilon $

For a 3D free electron gas, the density of states is:

$ g(\varepsilon) = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\varepsilon} $

So:

$ N = \int_0^\infty \frac{g(\varepsilon)}{e^{(\varepsilon - \mu)/k_B T} + 1} \, d\varepsilon $

At T = 0 K (Zero Temperature)

$ f(\varepsilon) = \begin{cases} 1, & \varepsilon < \mu \\ 0, & \varepsilon > \mu \end{cases} $

So the number of particles becomes:

$ N = \int_0^{\mu} g(\varepsilon) \, d\varepsilon $

Substitute $ g(\varepsilon) $:

$ N = \int_0^{\mu} \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\varepsilon} \, d\varepsilon $

After integration:

$ N = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \cdot \frac{2}{3} \mu^{3/2} $

Solving for $ \mu $:

$ \mu = \left( \frac{3N}{V} \cdot \frac{\pi^2 \hbar^3}{(2m)^{3/2}} \right)^{2/3} $

This value at $ T = 0 $ is called the Fermi energy:

$ \mu(T = 0) = E_F $

At Low Temperatures

Using Sommerfeld expansion, for low $ T $:

$ \mu(T) \approx E_F \left[ 1 - \frac{\pi^2}{12} \left( \frac{k_B T}{E_F} \right)^2 \right] $

Fermi Function at $ T = 0 $: Step Function