## What is meant by Ising model?

The Ising model is one of the easiest and most famous theoretical models for ferromagnetism in statistical mechanics. When a group of atomic spins aligns such that their magnetic moments are pointed in the same direction and produce a strong permanent magnetic moment in macroscopic size, the phenomenon is called ferromagnetism. The relative permeability of these magnets is greater than unity and increasing magnetization with the applied external field. Even in the absence of an external magnetic field, these substances retain spontaneous magnetization. An example of ferromagnets is iron.

## Introduction to Ising model

The Ising model was first proposed by Wilhelm Lenz. This model was published in Ising‘s doctoral thesis, but he gives credit to his guide Lenz for inventing the model. The one dimension Ising model was solved by him in 1924. Ising was very disappointed that the model did not exhibit ferromagnetism in one dimension and has no spontaneous magnetization.

In 1941, Kramers and Wannier gave a matrix formulation of the problem. In 1944 Lars Onsager gave a complete solution to the two-dimensional square lattice problem in zero external field.

## Derivation

The Ising model has separate variables that imply magnetic moments of atomic spins that can be in one or two states (+1 or -1). The spins are arranged in the lattice sites, allowing each spin to interact with its neighbors. The Ising model helps to identify the phase transition in a simple way.

Let us consider the lattice of N sites with a spin S on each lattice sites. Each spin S has two value +1 for spin up and -1 for spin down. There are ${2}^{N}$ possible configurations of the system. A configuration is represented by the direction of the spins on all the sites, $\left\{{S}_{i}\right\}$, where S is the spin on ${i}^{th}$lattice.

The interacting energy of the configuration is given by the hamiltonian function. It is expressed as,

${E}_{I}\left\{{S}_{i}\right\}=-\sum _{<i,j>}^{}{J}_{ij}{S}_{i}{S}_{j}-\sum _{i=1}^{N}{B}_{i}{S}_{i}$ (1)

Where, the subscript I denotes Ising model.

B is the magnetic field.

$<i,j>$ is the nearest neighbor pair of spins. $<i,j>$is same as $<j,i>$

${J}_{ij}$ is the exchange constant.

If ${J}_{\mathrm{ij}}>0$, the spins are aligned parallel to one another, that kind of interaction is ferromagnetic.

If ${J}_{\mathrm{ij}}<0,$ the spins are aligned antiparallel to one another, and they are called anti ferromagnetic.

If ${J}_{\mathrm{ij}}=0$, then means the system is non- ineracting.

For better understanding ${J}_{ij}=J>0$, assume The last term in the equation represents the coupling of the spins to an external magnetic field, B. Assume that the spins are arranged along in the Z-axis, as it is the magnetic field. The spins lower their energy by aligning parallel to the magnetic field. Suppose if ${B}_{i}$has a random value, then the model is called the random Ising model. Consider a uniform magnetic field, so that ${B}_{I}=B0.$

Hence, the interaction energy becomes,

${E}_{I}\left\{{S}_{i}\right\}=-\underset{<i,j>}{\overset{}{J\sum}}{S}_{i}{S}_{j}-B\sum _{i=1}^{N}{S}_{i}$ (2)

The partition function (Z) is given by,

$Z=\sum _{{s}_{1}=-1}^{+1}\sum _{{s}_{2}=-1}^{+1}....\sum _{{s}_{N}=-1}^{+1}{e}^{-\beta {E}_{I}\left\{{S}_{i}\right\}}$ (3)

## One-dimensional Ising model

Consider a one-dimensional Ising model where N spins are arranged on a chain. By applying the periodic boundary conditions so the spins are on a ring. Each spin on the ring interacts with the neighbors on both sides and with the external magnetic field B.

The energy is written as,

${E}_{I}\left\{{S}_{i}\right\}=-\underset{<i,j>}{\overset{N}{J\sum}}{S}_{i}{S}_{i+1}-B\sum _{i=1}^{N}{S}_{i}$ (4)

The periodic boundary condition of the system is represented by,

${S}_{N+1}={S}_{1}$ (5)

The partition function is Kramer and Wannier explain that the partition function can be expressed in matrices:

$Z=\sum _{{s}_{1}=-1}^{+1}\sum _{{s}_{2}=-1}^{+1}....\sum _{{s}_{N}=-1}^{+1}exp\left[\beta \sum _{i=1}^{N}(J{S}_{i}{S}_{i+1}+\frac{1}{2}B({S}_{i}+{S}_{i+1})\right]$ (6)

The above product is a $2\times 2$matrices. Let us denote the matrix as A and the elements are defined as,

$<S\left|A\right|S\text{'}>=exp\left\{\beta \left[JSS\text{'}+\frac{1}{2}B(S+S\text{'})\right]\right\}$ (7)

Where S and S’ independently takes the value +1 or -1.

The list of all the matrices element are:

$<+1\left|A\right|+1>={e}^{\left[\beta (J+B)\right]}\phantom{\rule{0ex}{0ex}}<-1\left|A\right|-1>={e}^{\left[\beta (J-B)\right]}\phantom{\rule{0ex}{0ex}}<+1\left|A\right|-1>=<-1\left|A\right|+1>={e}^{\left[-\beta J\right]}$

An explicit representation of A is,

$A=\left({e}^{\left[\beta (J+B)\right]}{e}^{\left[-\beta J\right]}\phantom{\rule{0ex}{0ex}}{e}^{\left[-\beta J\right]}{e}^{\left[\beta (J-B)\right]}\right)$ (8)

From the above definitions, the partition function is expressed as,

$Z=\sum _{{s}_{1}=-1}^{+1}\sum _{{s}_{2}=-1}^{+1}....\sum _{{s}_{N}=-1}^{+1}<{S}_{1}\left|A\right|{S}_{2}><{S}_{2}\left|A\right|{S}_{3}>...<{S}_{N}\left|A\right|{S}_{1}>\phantom{\rule{0ex}{0ex}}Z=\sum _{{s}_{1}=-1}^{+1}<{S}_{1}\left|{A}^{N}\right|{S}_{1}>\phantom{\rule{0ex}{0ex}}Z=Tr{A}^{N}\phantom{\rule{0ex}{0ex}}Z={\lambda}_{+}^{N}+{\lambda}_{-}^{N}$

Where ${\lambda}_{+}and{\lambda}_{-}$ are the eigen values of A. From the above equation, we concluded that Z is the trace of the Nth matrix is a consequence of the periodic boundary conditions of equation (5)

The eigen value equation is,

$\u2206\left({e}^{\left[\beta (J+B)\right]}-\lambda {e}^{\left[-\beta J\right]}\phantom{\rule{0ex}{0ex}}{e}^{\left[-\beta J\right]}{e}^{\left[\beta (J-B)\right]}-\lambda \right)={\lambda}^{2}-2\lambda {e}^{\beta J}\mathrm{cos}h\left(\beta B\right)+2\mathrm{sin}h\left(2\beta J\right)=0$

By solving the above equation in quadratric method, we get,

${\lambda}_{\pm}={e}^{\beta J}\left[\mathrm{cos}h\left(\beta B\right)\pm \sqrt{{\mathrm{cos}}^{2}h\left(\beta B\right)-2{e}^{-2\beta J}\mathrm{sin}h\left(2\beta J\right)}\right]$

If B=0,

${\lambda}_{+}=2\mathrm{cos}\left(\beta J\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{-}=2\mathrm{sin}\left(\beta J\right)$

If $B\ne 0$, then $\frac{{\lambda}_{-}}{{\lambda}_{+}}\le 1$. It is same for J=0. The thermodynamic limit$\left(N\to \infty \right)$, only ${\lambda}_{+}$ is valid. The helmholtz free energy per spin is given by,

$-\frac{F}{N{k}_{B}T}=\underset{N\to \infty}{\mathrm{lim}}\frac{1}{N}\mathrm{ln}Z\phantom{\rule{0ex}{0ex}}=\underset{N\to \infty}{\mathrm{lim}}\frac{1}{N}\mathrm{ln}\left\{{\lambda}_{+}^{N}{\left[1+\left(\frac{{\lambda}_{-}}{{\lambda}_{+}}\right)\right]}^{N}\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{ln}\lambda +\underset{N\to \infty}{\mathrm{lim}}\frac{1}{N}\mathrm{ln}{\left[1+\left(\frac{{\lambda}_{-}}{{\lambda}_{+}}\right)\right]}^{N}\phantom{\rule{0ex}{0ex}}=\mathrm{ln}\lambda $

Therefore,

$-\frac{F}{N}=\frac{{k}_{B}T}{n}lnZ=-{k}_{B}T\mathrm{ln}{\lambda}_{+}\phantom{\rule{0ex}{0ex}}=-J--{k}_{B}T\mathrm{ln}\left[\mathrm{cos}h\left(\beta B\right)\pm \sqrt{{\mathrm{cos}}^{2}h\left(\beta B\right)-2{e}^{-2\beta J}\mathrm{sin}h\left(2\beta J\right)}\right]$

The magnetization per spin is,

$m=\frac{M}{N}\phantom{\rule{0ex}{0ex}}m=\frac{1}{\beta N}\frac{\partial \mathrm{ln}Z}{\partial B}\phantom{\rule{0ex}{0ex}}m=-\frac{1}{N}\frac{\partial F}{\partial B}\phantom{\rule{0ex}{0ex}}m=\frac{\mathrm{sin}h\left(\beta B\right)}{\left[\mathrm{cos}h\left(\beta B\right)-\sqrt{{\mathrm{cos}}^{2}h\left(\beta B\right)-2{e}^{-2\beta J}\mathrm{sin}h\left(2\beta J\right)}\right]}$

At zero field (B = 0), the magnetization is zero for all temperatures. This shows that there is no spontaneous magnetization and it does not exihibits ferromagnetism. Also, it has no phase transistion. Only the two dimensional square lattice shows phase transisiton.

## Monte-Carlo simulation

Monte- Carlo simulation is one of the popular simulation methods widely used in physics. For a spin system, we would want to obtain average values of thermodynamic quantities such as magnetization, energy, etc.

Metropolis algorithm is a classic Monte Carlo methods. Using transition probability, we generate configuration from the previous state which relies on the energy difference $\u2206E$ between the initial and final states. For Ising model, the probability obeys master equation,

$\frac{\partial {P}_{n}\left(t\right)}{\partial t}=\sum _{n\ne m}\left[-{P}_{n}\left(t\right){W}_{n\to m}+{P}_{m}\left(t\right){W}_{m\to n}\right]$

Here, ${P}_{n}\left(t\right)\to $probability of the system and,

${W}_{n\to m}\to $is the transition rate from state n to state m.

In equilibrium $\frac{\partial {P}_{n}\left(t\right)}{\partial t}=0,$ and the two terms on the right-hand side must be equal. The resulting value is known as detailed balance.

${P}_{n}\left(t\right){W}_{n\to m}={P}_{m}\left(t\right){W}_{m\to n}$

The above probability is known as Boltzmann probability.

${P}_{n}\left(t\right)=\frac{{e}^{-{E}_{n}/{k}_{B}T}}{Z}$

By generating a Markov chain of states, i.e. generate each new state directly from the preceding state. If we create the ${n}^{th}$ state from the ${m}^{th}$ state, the relative probability is given by the ratio,

$\frac{{P}_{n}\left(t\right)}{{P}_{m}\left(t\right)}=\frac{{e}^{-{E}_{n}/{k}_{B}T}}{{e}^{-{E}_{m}/{k}_{B}T}}\phantom{\rule{0ex}{0ex}}=\frac{{W}_{n\to m}}{{W}_{m\to n}}\phantom{\rule{0ex}{0ex}}={e}^{-\left({E}_{n}-{E}_{m}\right)/{k}_{B}T}\phantom{\rule{0ex}{0ex}}={e}^{-\u2206E/{k}_{B}T}\phantom{\rule{0ex}{0ex}}$

Where $\u2206E$ is the difference between ${n}^{th}$ and ${m}^{th}$ state.

Metropolis form was the first option of a transition rate used in statistical physics.

${W}_{n\to m}=\left\{\begin{array}{ll}{e}^{\raisebox{1ex}{$-\u2206E$}\!\left/ \!\raisebox{-1ex}{${k}_{B}T$}\right.}& \u2206E0\\ 1& \u2206E0\end{array}\right.\phantom{\rule{0ex}{0ex}}$

The basic method on how to implement metropolis algorithm in the spin system is,

1. Select an initial spin state.

2. Then select a lattice site i.

3. Find the change in energy, which shows that the spin at the state I is overturned or not.

4. If $\u2206E<0$, flip the spin. If $\u2206E>0$, then

a) Create a random number r such that flip the spin $0r1.$

b) if $rexp(-\u2206E/{k}_{B}T)$, flip the spin.

5. Repeat the process.

For the easy calculation of determinant in a spin model, it is written as

$H=-\sum _{i>j}^{}{J}_{ij}{S}_{i}\xb7{S}_{j}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}\sum _{i,j}^{}{J}_{ij}{S}_{i}\xb7{S}_{j}\phantom{\rule{0ex}{0ex}}=-\sum _{i}^{}{S}_{i}(\sum _{j}^{}\frac{1}{2}{J}_{ij}{S}_{j})\phantom{\rule{0ex}{0ex}}H=-\sum _{i}^{}{S}_{i}{h}_{i}$

where $h$ is the local magnetic field due to other spins.

## Context and Applications

The Ising model is an important topic in solid-state physics for the postgraduate student, especially for Masters in physics and Bachelor in Technology (material science).

## Practice Problems

**Question 1: **Who invented the Ising model?

- Wilhelm Lenz
- Onsager
- Kramer
- Ising

**Answer: **The correct option is a.

**Explanation: **The Ising model was first designed by Wilhelm Lenz. Later, Ising solve the model and gave the solution for the one-dimensional model of ferromagnets. And the model is named as Ising model.

**Question 2:** Which of the following exhibits spontaneous magnetization?

- Paramganetic material
- Ferromagnetic material
- Diamagentic material
- None of these

**Answer: **The correct option is b.

**Explanation:** The ferromagnetic material has spontaneous magnetization even in the absence of the external field.

**Question 3:** Of the following model, which dimension does not have phase transition?

- One-dimension
- Two-dimension
- Three-dimension
- All the above

**Answer: **The correct option is a.

**Explanation:** From the following options, the one-dimensional rising model has no phase transition. Because the one-dimensional model does not exhibit ferromagnetism and the tendency to align the spins always loses out. Hence it does not have a phase transition.

**Question 4:** If the interaction energy ${J}_{ij}<0,$ what kind of magnetism it shows?

- Ferromagnetic
- Antiferro magnetic
- Paramagnetic
- Diamagnetic

**Answer: **The correct option is a.

**Explanation:** If the interaction energy,${J}_{ij}<0,$ it means, the spins are aligned in the direction anti parallel to one another. This phenomenon is known as anti- ferromagnetic.

**Question 5**: Which of the following material is a ferromagnetic substance?

- Silver
- Gold
- Iron
- Aluminium

**Answer:** The correct option is c.

**Explanation:** From the following materials, iron is a ferromagnetic substance. Most of the ferromagnetic substances are metals. They have strong magnetic properties because of their magnetic domains. They retain their magnetic field even in the absence of the magnetic field.

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