## Ising model

Boltzman distribution: $\pi (\text{x}) = \frac{1}{Z}e^{-U(\text{x})/kT}$

Partition function $Z=Z(T) = \int_D e^{-U(\text{x})/kT}d\text{x}$ (also called normalizing constant)

Potential energy $U(\text{x}) =-J\sum_{\sigma \thicksim \sigma^{'}} x_\sigma x_{\sigma^{'}} + \sum_\sigma h_\sigma x_\sigma$

The symbol $\sigma \thicksim \sigma^{'}$ means that they are neighboring pair, $J$ is the interaction strength, $h_\sigma$ is the external magnetic field.

In zero field we can also define the potential energy as

$U(\text{x})=\sum_{\sigma \thicksim \sigma^{'}} 1_{x_\sigma \ne x_{\sigma^{'}}}$

We define the expectation of $U(\text{x})$ with respect to $\pi$ as internal energy:

$= E_\pi \{U(\text{x})\} = \int_D U(\text{x})\pi(\text{x})d\text{x}$

The free energy of the system is $F = -kT\text{log}Z$

The specific heat of the system is $C = \frac{\partial }{\partial T}$

The mean magnetization per spin is $ = E_\pi \{\frac{1}{N^2} |\sum_{\sigma \in S}x_\sigma|\}$

1.

Let $\beta = 1/kT$, we have

$\partial \text{log}Z /\partial \beta = \frac{1}{Z}\partial Z/\partial \beta =$ $\frac{1}{Z}\partial[\int_D$ $e^{-\beta U(\text{x})}d\text{x}]/ \partial\beta$

$=\frac{1}{Z}\int_D \{\partial[e^{-\beta U(\text{x}) }] / \partial \beta\}d{\text{x}} =\frac{1}{Z}\int_D-U(\text{x}) e^{-\beta U(\text{x}) } d{\text{x}}$

$=\frac{1}{Z}\int_D \frac{\partial e^{-\beta U(\text{x}) }} {\partial \beta} d{\text{x}}$

$= -$

Thus $\frac{\partial \text{log}Z}{\partial \beta} = -$

2. $C = \frac{\partial }{\partial T} = \frac{1}{kT^2}Var_\pi\{U(\text{x})\}$