Ising Problem

Let’s walk through a classic, simple example of the Ising Model from statistical mechanics.

To keep it clean and solvable by hand, we will look at a 1D Ising Model with 3 Spins in a chain with periodic boundary conditions (meaning they form a ring where the last spin connects back to the first).

The Setup

We have 3 spins: $s_1,s_2,s_3$.

Each spin can point either up ($+1$) or down ($-1$).

The Hamiltonian (energy equation) for this system with no external magnetic field is:

• $J$ is the coupling constant.

• If $J>0$, the system is ferromagnetic (spins want to align).

• If $J<0$, it is antiferromagnetic (spins want to alternate). Let's assume $J>0$ for this example.

Step 1: Find All Possible States and Their Energies

Since there are 3 spins and each has 2 choices, there are $2^3=8$ possible configurations. Let's calculate the energy $H$ for each.

Notice the symmetry: We have 2 states where all spins align (lowest energy, $-3J$) and 6 states where one spin is mismatched (higher energy, $+J$).

Step 2: Calculate the Partition Function ($Z$)

The partition function encodes all the statistical properties of the system at a given temperature $T$. It is the sum of the Boltzmann factors $e^{-\beta H}$ over all states, where $\beta =\frac{1}{k_{B}T}$.

Plugging in our 8 states from the table:

Step 3: Find the Average Energy ($⟨E⟩$)

Now that we have $Z$, we can easily find thermodynamic properties like the expected (average) energy of the system using the formula:

Let's take the derivative of $Z$ with respect to $\beta$:

Now, plug this back into the average energy formula:

What does this tell us?

• At Absolute Zero ($T\to 0$, meaning $\beta \to \mathrm{\infty }$): The $e^{3\beta J}$ terms completely dominate. The fraction approaches $1$, so $⟨E⟩\to -3J$. The system freezes into its lowest energy, perfectly ordered ground states.

• At Infinite Temperature ($T\to \mathrm{\infty }$, meaning $\beta \to 0$): The exponentials all become $e^0=1$. The numerator becomes $1-1=0$, so $⟨E⟩\to 0$. Thermal fluctuations completely randomize the spins, destroying any order.