\documentclass[../ising_model.tex]{subfiles}

\begin{document}
\appendix
\section{Partition function}\label{sec:partition_function}
Using the values estimated for the $2 \times 2$ case, found in \ref{tab:lattice_config}, 
we find the partition function
\begin{align*}
	Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} + 2 \cdot e^{-\beta (8J)} \\
    & \quad + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\
    &= 2e^{8 \beta J} + 2e^{-8 \beta J} + 12.
\end{align*}
We rewrite the expression using the identity $\cosh(8 \beta J) = 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})$, 
and get
\begin{align*}
    z &= 4 \cosh (8 \beta J) + 12 \ .
\end{align*}

\section{Expectation values}\label{sec:expectation_values}
For a linear function of a stochastic random variable $X$, with a known probability 
distribution, the expected value of $x$ is given by
\begin{align*}
    \mathbb{E}(aX + b) &= a \cdot \mathbb{E}(X) + b & \text{\cite[p. 131]{springer:2012:modernstat}}
\end{align*}
In our case the discrete random variable is the spin configuration, and we want 
to find the expected value of the function $E(\mathbf{s})$. Continuing, we will 
use the notation $\langle E \rangle$ for the expectation value of a given variable, 
in this case $E$. 

Both energy per spin and magnetization per spin are functions of $\mathbf{s}$. 
In addition, the number of spins is given as a constant for each lattice. We can 
use the expression for $\langle E \rangle$ and $\langle M \rangle$ to find the 
expectation values per spin. For energy per spin 
\begin{align*}
    \langle \epsilon \rangle &= \sum_{i=1}^{N} \epsilon(s_{i}) p(s_{i} \ | \ T) \\
    &= \sum_{i=1}^{N} \frac{E(\mathbf{s})}{N} p(s_{i} \ | \ T) \\
    &= \frac{1}{N} \sum_{i=1}^{N} E(\mathbf{s}) p(s_{i} \ | \ T)
\end{align*}
The same applies to magnetization per spin
\begin{align*}
    \langle |m| \rangle = \frac{1}{N} \sum_{i=1}^{N} |M(s_{i})| p(s_{i} \ | \ T) \ .
\end{align*}
Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which give zero we get
\begin{align*}
    \langle E \rangle &= (-8J) \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J) \cdot \frac{1}{Z} e^{-8 \beta J} + (-8J) \cdot \frac{1}{Z} e^{8 \beta J} \\
    &= \frac{16J}{Z} \big(e^{-8 \beta J} - e^{8 \beta J}) \\
    &= -\frac{32J \sinh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
    &= -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
\end{align*}
and
\begin{align*}
    \langle |M| \rangle &= 4 \cdot \frac{1}{Z} \cdot e^{8 \beta J} + 4 \cdot 2 \cdot \frac{1}{Z} \cdot e^{0} \\
    & \quad + 4 \cdot | -2| \cdot \frac{1}{Z} \cdot e^{0} + | -4| \cdot e^{8 \beta J} \\
    &= \frac{8 e^{8 \beta J} + 16}{Z} \\
    &= \frac{4 (2e^{8 \beta J} + 4)}{4(\cosh(8 \beta J) + 3)} \\
    &= \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ .
\end{align*}
The squared function 
\begin{align*}
    \langle E^{2} \rangle &= (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J)^{2} \cdot \frac{1}{Z} e^{-8 \beta J} + (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} \\
    &= \frac{128J^{2}}{Z} \big(e^{8 \beta J} + e^{-8 \beta J} \big) \\
    &= \frac{128J^{2} \cosh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
    &= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
\end{align*}
and
\begin{align*}
    \langle M^{2} \rangle &= 4^{2} \cdot \frac{1}{Z} \cdot e^{8 \beta J} + 4 \cdot 2^{2} \cdot \frac{1}{Z} \cdot e^{0} \\
    & \quad + 4 \cdot (-2)^{2} \cdot \frac{1}{Z} \cdot e^{0} + (-4)^{2}\cdot e^{8 \beta J} \\
    &= \frac{32e^{8 \beta J} + 32}{Z} \\
    &= \frac{4 (8e^{8 \beta J} + 8)}{4(\cosh(8 \beta J) + 3)} \\
    &= \frac{8e^{8 \beta J} + 8}{\cosh(8 \beta J) + 3} \ .
\end{align*}
The squared expectation value is given by 
\begin{align*}
    \langle E \rangle^{2} &= \bigg(-\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \bigg)^{2} \\
    &= \frac{64J^{2} \sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \ ,
\end{align*}
and
\begin{align*}
    \langle |M| \rangle^{2} &= \Big( \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \Big)^{2} \\
    &= \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \ .
\end{align*}

\section{Heat capacity and magnetic susceptibility}\label{sec:heat_susceptibility}
To find the heat capacity in Eq. \ref{eq:heat_capacity}, we normalize to heat 
capacity per spin
\begin{align*}
    \frac{C_{V}}{N} &= \frac{1}{N} \frac{1}{k_{B} T^{2}} (\mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2}) \\
    &= \frac{1}{N k_{B} T^{2}} \mathbb{V}(E) \ .
\end{align*}
Using Eq. \eqref{eq:susceptibility}, we find he susceptibility per spin
\begin{align*}
    \frac{\chi}{N} &= \frac{1}{N} \frac{1}{k_{B} T} (\mathbb{E}(M^{2}) - [\mathbb{E}(M)]^{2}) \\
    &= \frac{1}{N k_{B} T} \mathbb{V}(M) \ .
\end{align*}
We now have to find the variance of both total energy and total magnetizaation. 
We obtain this using the definition 
\begin{align*}
    \mathbb{V}(X) &= \sum_{x \in D} [(x - \mathbb{E}(x))^2 \cdot p(x)] & \text{\cite[p. 132]{springer:2012:modernstat}} \\
    &= \mathbb{E}(X^{2}) - [\mathbb{E}(X)]^{2} \ .
\end{align*}
The variance of the total energy is then given by
\begin{align*}
    \mathbb{V}(E) &= \mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2} \\
    &= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} - \frac{64J^{2} \sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \\
    &= 64J^{2} \bigg( \frac{\cosh(8 \beta J)}{\cosh(8 \beta J) + 3} - \frac{\sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \\
    &= 64J^{2} \bigg( \frac{(\cosh(8 \beta J)) \cdot (\cosh(8 \beta J) + 3)}{(\cosh(8 \beta J) + 3)^{2}} \\
    & \quad - \frac{\sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \\
    &= 64J^{2} \bigg( \frac{\cosh^{2}(8 \beta J) + 3\cosh(8 \beta J) - \sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \\
    &= 64J^{2} \bigg( \frac{3\cosh(8 \beta J) + 1}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \ ,
\end{align*}
and the variance of the total magnetization is given by
\begin{align*}
    \mathbb{V}(M) &= \mathbb{E}(M^{2}) - [\mathbb{E}(|M|)]^{2} \\
    &= \frac{8e^{8 \beta J} + 8}{\cosh(8 \beta J) + 3} - \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\
    &= \frac{(8(e^{8 \beta J} + 1)) \cdot (\cosh(8 \beta J) + 3) - 4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\
    &= \frac{4(e^{8 \beta J} + 1) \cdot (e^{8 \beta J} + e^{-8 \beta J}) + 24(e^{8 \beta J} + 1) - 4(e^{8 \beta J} + 1)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\
    &= \frac{4e^{2(8 \beta J)} + 4e^{8 \beta J} 4e^{0} + 4e^{-8 \beta J} 24e^{8 \beta J} + 24 - 4e^{2(8 \beta J)} - 16e^{8 \beta J} - 16}{(\cosh(8 \beta J) + 3)^{2}} \\
    &= \frac{4(3e^{8 \beta J} + e^{-8 \beta J} + 3)}{(\cosh(8 \beta J) + 3)^{2}} \ .
\end{align*}
We find the heat capacity 
\begin{align*}
    \frac{C_{V}}{N} &= \frac{64J^{2} }{N k_{B} T^{2}} \bigg( \frac{3\cosh(8 \beta J) + 1}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \ ,
\end{align*}
and susceptibility
\begin{align*}
    \frac{\chi}{N} &= \frac{4}{N k_{B} T} \bigg( \frac{3e^{8 \beta J} + e^{-8 \beta J} + 3}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \ .
\end{align*}

\section{Change in total system energy}\label{sec:delta_energy}
When we consider the change in energy after flipping a single spin, we evaluate 
$\Delta E = E_{\text{after}} - E_{\text{before}}$. We find the $3^{2}$ values as
\begin{align*}
    \Delta E = -8J - (-8J) = 0 \\
    \Delta E = -8J - 0 = -8J \\
    \Delta E = -8J - 8J = -16J \\
    \Delta E = 0 - (-8J) = 8J \\
    \Delta E = 0 - 0 = 0 \\
    \Delta E = 0 - 8J = -8J \\
    \Delta E = 8J - (-8J) = 16J \\
    \Delta E = 8J - 0 = 8J \\
    \Delta E = 8J - 8J = 0,
\end{align*}
where the five distinct values are $\Delta E = \{-16J, -8J, 0, 8J, 16J\}$.

\end{document}