Experience the CMOS Annealing Machine

# What is Ising model?

### Overview of the hamiltonian of Ising model

The Ising model is the simplest model in the field of "statistical mechanics". It is a model that expresses what kind of behavior as a whole (macro) when an enormous number of microelements interact with each other and when a force is given to each microelement. The micro element in the Ising model is called spin and takes two values of ±1. The Ising model is defined on an undirected graph $G=(V,E)$. Here $V$ is a set of vertices, $E$ is a set of edges. The spin variable exists on the vertex and it is expressed as $s_i$. That means $s_i=±1$.

#### Graph $G=(V,E)$

In order to use an annealing machine, it is necessary to express cost function of combinatorial optimization problem and constraint condition by Hamiltonian (energy function) of Ising model.
The energy function is given as follows.

#### $$E (\lbrace s _i \rbrace) = \sum _ { i \in V } h _ { i } s _ { i } + \sum _ { ( i , j ) \in E } J _ { i j } s _ { i } s _ { j }$$

Here, $h_i$ is a magnetic field that applied to the site $i \in V$, $J_{ij}$ is a interaction that applied to the site $i \in V$ and $j \in V ( ( i , j ) \in E )$. In terms of physics, the minimum state of energy $E({s_i})$ is called a stable state.

### A specific example of the Hamiltonian of Ising model (in case of one spin)

Let's consider a case where there is only one spin.
At this time, the energy function is given as

#### $$E \left( s _ { 1 } \right) = h _ { 1 } s _ { 1 }$$

The energy value according to the value of $s_1$ is as follows.

Spin state Energy value
$s_1=+1$ $h_1$
$s_1=-1$ $-h_1$

When $h_1>0$, $s_1=-1$ is stable. Conversely, when $h_1<0$, $s_1=+1$ is stable.

### A specific example of the Hamiltonian of Ising model (in case of two spins - 1)

Let's consider a case where there are only two spins. Here we assume only the interaction is working between the two spins.

At this time, the energy function is given as

#### $$E \left( s _ { 1 } , s _ { 2 } \right) = J _ { 12 } s _ { 1 } s _ { 2 }$$

The energy value according to the value of $s_1$ and $s_2$ is as follows.

Spin state Energy value
$s_1=+1$ $s_2=+1$ $J_12$
$s_1=+1$ $s_2=-1$ $-J_12$
$s_1=-1$ $s_2=+1$ $-J_12$
$s_1=-1$ $s_2=-1$ $J_12$

When $J_12>0$, $s_1≠s_2$ is stable. Conversely, when $J_12<0$, $s_1=s_2$is stable.

### A specific example of the Hamiltonian of Ising model (in case of two spins - 2)

Let's consider a case where there are only two spins. Here we assume not only the interaction is working between the two spins but also a magnetic field is applied to each spin.

At this time, the energy function is given as

#### $$E \left( s _ { 1 } , s _ { 2 } \right) = J _ { 12 } s _ { 1 } s _ { 2 } + h _ { 1 } s _ { 1 } + h _ { 2 } s _ { 2 }$$

The energy value according to the value of $s_1$ and $s_2$ is as follows.

Spin state Energy value
$s_1=+1$ $s_2=+1$ $J_12+h_1+h_2$
$s_1=+1$ $s_2=-1$ $-J_12+h_1-h_2$
$s_1=-1$ $s_2=+1$ $-J_12-h_1+h_2$
$s_1=-1$ $s_2=-1$ $J_12-h_1-h_2$

The stable state will change according to the value of $J_12, h_1, h_2$. For example, when $J_12=1, h_1=-2, h_2=3$, $s_1=+1, s_2=-1$ is stable.

### A specific example of the Hamiltonian of Ising model (in case of four spins)

Finally, let's consider the case where there are only four spins. See the case like the figure.

Here we assume the interaction is working between $s_1,s_2$, $s_2,s_3$, and $s_3,s_4$ and a magnetic field is applied to only $s_1$ and $s_4$.

At this time, the energy function is

#### $$E ({s_i}) = J _ { 12 } s _ { 1 } s _ { 2 } + J _ { 23 } s _ { 2 } s _ { 3 } + J _ { 34 } s _ { 3 } s _ { 4 } + h _ { 1 } s _ { 1 } + h _ { 4 } s _ { 4 }$$

The energy value according to the value of $s_1,s_2,s_3,s_4$ is as follows.

Spin state Energy value
$s_1=+1$ $s_2=+1$ $s_3=+1$ $s_4=+1$ $J_12+J_23+J_34+h_1+h_4$
$s_1=+1$ $s_2=+1$ $s_3=+1$ $s_4=-1$ $J_12+J_23-J_34+h_1-h_4$
$s_1=+1$ $s_2=+1$ $s_3=-1$ $s_4=+1$ $J_12-J_23-J_34+h_1+h_4$
$s_1=+1$ $s_2=+1$ $s_3=-1$ $s_4=-1$ $J_12+J_23+J_34+h_1-h_4$
$s_1=+1$ $s_2=-1$ $s_3=+1$ $s_4=+1$ $-J_12-J_23+J_34+h_1+h_4$
$s_1=+1$ $s_2=-1$ $s_3=+1$ $s_4=-1$ $-J_12-J_23-J_34+h_1-h_4$
$s_1=+1$ $s_2=-1$ $s_3=-1$ $s_4=+1$ $-J_12+J_23-J_34+h_1+h_4$
$s_1=+1$ $s_2=-1$ $s_3=-1$ $s_4=-1$ $-J_12+J_23+J_34+h_1-h_4$
$s_1=-1$ $s_2=+1$ $s_3=+1$ $s_4=+1$ $-J_12+J_23+J_34-h_1+h_4$
$s_1=-1$ $s_2=+1$ $s_3=+1$ $s_4=-1$ $-J_12+J_23-J_34-h_1-h_4$
$s_1=-1$ $s_2=+1$ $s_3=-1$ $s_4=+1$ $-J_12-J_23-J_34-h_1+h_4$
$s_1=-1$ $s_2=+1$ $s_3=-1$ $s_4=-1$ $-J_12-J_23+J_34-h_1-h_4$
$s_1=-1$ $s_2=-1$ $s_3=+1$ $s_4=+1$ $J_12-J_23+J_34-h_1+h_4$
$s_1=-1$ $s_2=-1$ $s_3=+1$ $s_4=-1$ $J_12-J_23-J_34-h_1-h_4$
$s_1=-1$ $s_2=-1$ $s_3=-1$ $s_4=+1$ $J_12+J_23-J_34-h_1+h_4$
$s_1=-1$ $s_2=-1$ $s_3=-1$ $s_4=-1$ $J_12+J_23+J_34-h_1-h_4$

The stable state will change according to the value of $J_12, J_23, J_34, h_1, h_4$. For example, when $J_12=1, J_23=3, J_34=-1, h_1=2, h_4=3$, $s_1=-1, s_2=+1, s_3=-1, s_4=-1$ is stable.