261 lines
11 KiB
TeX
261 lines
11 KiB
TeX
\documentclass[../ising_model.tex]{subfiles}
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\begin{document}
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\section{Methods}\label{sec:methods}
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\subsection{The Ising model}\label{subsec:ising_model}
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The Ising model consist of a lattice of spins, which can be though of as atoms
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in a grid. In two dimensions, the length of the lattice is given by $L$, and the
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number of spins within a lattice is given by $N = L^{2}$. When we consider the
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entire lattice, the system spin configuration is denoted as a vector $\mathbf{s}
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= [s_{1}, s_{2}, \dots, s_{N}]$. The total number of spin configurations, or
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microstates, is $2^{N}$.
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A given spin $i$ can take one of two possible discrete values $s_{i} \in \{-1, +1\}$,
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either spin down or spin up. The spins interact with its nearest neighbors, and
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in a two-dimensional lattice each spin has up to four nearest neighbors. However,
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the model is not restricted to this dimentionality \cite[p. 3]{obermeyer:2020:ising}.
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In our experiment we will use periodic boundary conditions, meaning all spins
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have exactly four nearest neighbors. To find the analytical expressions necessary
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for validating our model implementation, we will assume a $2 \times 2$ lattice.
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We count the neighboring spins as visualized in Fig. \ref{fig:tikz_boundary},
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\begin{figure}
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\centering
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\begin{tikzpicture}
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\draw[help lines] (0, 0) grid (2, 2);
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% \node[inner] (s1) at (0.5, 0.5) {s};
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\node (s1) at (0.5, 1.5) {$s_1$};
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\node (s2) at (1.5, 1.5) {$s_2$};
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\node (s3) at (0.5, 0.5) {$s_3$};
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\node (s4) at (1.5, 0.5) {$s_4$};
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\node[gray] (s12) at (2.5, 1.5) {$s_1$};
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\node[gray] (s34) at (2.5, 0.5) {$s_3$};
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\node[gray] (s13) at (0.5, -0.5) {$s_1$};
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\node[gray] (s24) at (1.5, -0.5) {$s_2$};
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\draw[red, ->] (s1.east) -- (s2.west);
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\draw[red, ->] (s2.east) -- (s12.west);
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\draw[red, ->] (s1.south) -- (s3.north);
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\draw[red, ->] (s2.south) -- (s4.north);
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\draw[red, ->] (s3.east) -- (s4.west);
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\draw[red, ->] (s4.east) -- (s34.west);
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\draw[red, ->] (s3.south) -- (s13.north);
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\draw[red, ->] (s4.south) -- (s24.north);
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\end{tikzpicture}
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\caption{Visualization of counting rules for each pair of spins in a
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$2 \times 2$ lattice, where periodic boundary conditions are applied.}
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\label{fig:tikz_boundary}
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\end{figure}
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and find the total system energy given by % Eq. \eqref{eq:energy_total}
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\begin{equation}
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E(\mathbf{s}) = -J \sum_{\langle k l \rangle}^{N} s_{k} s_{l} \ .
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\label{eq:energy_total}
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\end{equation}
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$\langle k l \rangle$ denote a pair of spins, and $J$ is the coupling constant.
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We also find the total magnetization of the system, which is given by % Eq. \eqref{eq:magnetization_total}
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\begin{equation}
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M(\mathbf{s}) = \sum_{i}^{N} s_{i} \ .
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\label{eq:magnetization_total}
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\end{equation}
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In addition, we have to consider the state degeneracy, the number of different
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microstates sharing the same value of total magnetization. The derivation of the
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analytical values can be found in Appendix % \ref{}
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\begin{table}[H]
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\centering
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\begin{tabular}{cccc} % @{\extracolsep{\fill}}
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\hline
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Spins up & $E(\mathbf{s})$ & $M(\mathbf{s})$ & Degeneracy \\
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\hline
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4 & -8J & 4 & 1 \\
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3 & 0 & 2 & 4 \\
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\multirow{2}*{2} & 0 & 0 & 4 \\
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& 8J & 0 & 2 \\
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1 & 0 & -2 & 4 \\
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0 & -8J & -4 & 1 \\
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\hline
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\end{tabular}
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\caption{Values of total energy, total magnetization, and degeneracy for the
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possible states of a system for a $2 \times 2$ lattice, with periodic boundary
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conditions.}
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\label{tab:lattice_config}
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\end{table}
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We use the analytical values, found in Table for both for lattices where $L = 2$ and $L > 2$.
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However, to compare the quantities for lattices where $L > 2$, we find energy
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per spin given by
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\begin{equation}
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\epsilon (\mathbf{s}) = \frac{E(\mathbf{s})}{N} \ ,
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\label{eq:energy_spin}
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\end{equation}
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and magnetization per spin given by
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\begin{equation}
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m (\mathbf{s}) = \frac{M(\mathbf{s})}{N} \ .
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\label{eq:magnetization_spin}
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\end{equation}
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\subsection{Statistical mechanics}\label{subsec:statistical_mechanics}
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When we study ferromagnetism, we have to consider the probability for a microstate
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$\mathbf{s}$ at a fixed temperature $T$. The probability distribution function
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(pdf) is given by
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\begin{equation}
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p(\mathbf{s} \ | \ T) = \frac{1}{Z} \exp^{-\beta E(\mathbf{s})},
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\label{eq:boltzmann_distribution}
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\end{equation}
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known as the Boltzmann distribution. This is an exponential distribution, where
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$\beta$ and $Z$ are given by
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\begin{align*}
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\beta =& \frac{1}{k_{B}} \ , &
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Z &= \sum_{\text{all possible } \mathbf{s}} e^{-\beta E(\mathbf{s})} \ , \\
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\end{align*}
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and $k_{B}$ is the Boltzmann constant. $Z$ is a normalizing factor of the pdf,
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known as the partition function, which we derive in Appendix \ref{sec:partition_function}
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\begin{equation*}
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Z = 4 \cosh (8 \beta J) + 12 \ .
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\end{equation*}
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Using the partition function and Eq. \eqref{eq:boltzmann_distribution}, the pdf
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of a microstate at a fixed temperature is given by
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\begin{equation}
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p(\mathbf{s} \ | \ T) = \frac{1}{4 \cosh (8 \beta J) + 12} e^{-\beta E(\mathbf{s})} \ .
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\end{equation}
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% Add something about why we use the expectation values?
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We derive the analytical expressions for expectation values in Appendix.
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\ref{sec:expectation_values}. We find the expected total energy
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\begin{equation*} %\label{eq:energy_total_first}
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\langle E \rangle = -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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\end{equation*}
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and the expected energy per spin
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\begin{equation*} %\label{eq:energy_spin_first}
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\langle \epsilon \rangle = \frac{-2J \sinh(8 \beta J)}{ \cosh(8 \beta J) + 3} \ .
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\end{equation*}
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We find the expected absolute total magnetization
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\begin{equation*} %\label{eq:magnetization_total_first}
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\langle |M| \rangle = \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ ,
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\end{equation*}
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and the expected magnetization per spin
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\begin{equation*} %\label{eq:magnetization_spin_first}
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\langle |m| \rangle = \frac{e^{8 \beta J} + 1}{2( \cosh(8 \beta J) + 3)} \ .
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\end{equation*}
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We will also determine the heat capacity
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\begin{equation}
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C_{V} = \frac{1}{k_{B} T^{2}} (\mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2}) \ ,
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\label{eq:heat_capacity}
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\end{equation}
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and the susceptibility
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\begin{align*}
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\chi = \frac{1}{k_{\text{B} T}} (\mathbb{E}(M^{2}) - [\mathbb{E}(M)]^{2}) \ .
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\label{eq:susceptibility}
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\end{align*}
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In Appendix \ref{sec:heat_susceptibility} we derive expressions for
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the heat capacity and the susceptibility, given by
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\begin{align*}
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\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) \ ,
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\end{align*}
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\begin{align*}
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\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) \ .
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\end{align*}
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For scalability of our model, we want to work with unitless spins. To obtain this,
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we introduce the coupling constant $J$ as the base unit for energy. With the
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Boltzmann constant we derive the remaining units, which can be found in Table
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\ref{tab:units}.
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\begin{table}[H]
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\centering
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\begin{tabular}{ccc} % @{\extracolsep{\fill}}
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\hline
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Value & Unit & \\
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\hline
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$[E]$ & $J$ & \\
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$[M]$ & $1$ & (unitless) \\
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$[T]$ & $J / k_{B}$ & \\
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$[C_{V}]$ & $k_{B}$ & \\
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$[\chi]$ & $1 / J$ & \\
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\hline
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\end{tabular}
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\caption{Values of total energy, total magnetization, and degeneracy for the possible states of a system for a $2 \times 2$ lattice, with periodic boundary conditions.}
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\label{tab:units}
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\end{table}
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\subsection{The Markov chain Monte Carlo method}\label{subsec:mcmc_method}
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Markov chains consist of a sequence of samples, where the probability of the next
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sample depend on the probability of the current sample. Whereas the Monte Carlo
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method introduces randomness to the sampling, which allows us to approximate
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statistical quantities \cite{tds:2021:mcmc} without using the pdf directly. We
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generate samples of spin configuration, to approximate $\langle \epsilon \rangle$,
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$\langle |m| \rangle$, $C_{V}$, and $\chi$, using the Markov chain Monte Carlo
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(MCMC) method.
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The Ising model is initialized in a random state, and the Markov chain is evolved
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until the model reaches an equilibrium state. However, generating new random states
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require ergodicity and detailed balance. A Markov chain is ergoditic when all system
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states can be reached at every current state, whereas detaild balance implies no
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net flux of probability. To satisfy these criterias we use the Metropolis-Hastings
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algorithm, found in Fig. \ref{algo:metropolis}, to implement the MCMC method.
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The Markov process will reach an equilibrium, reflecting the state of a real system.
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% Add something about burn-in time
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At the point of equilibrium we start sampling, as the distribution of the set of
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samples will tend toward the actual distribution.
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% Add something about spin flipping
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At each step of flipping a spin, the change in energy is evaluated as
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\begin{align*}
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\Delta E &= E_{\text{after}} - E_{\text{before}} \ .
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\end{align*}
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We have to evaluate
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\begin{figure}[H]
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\begin{algorithm}[H]
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\caption{Metropolis-Hastings Algorithm}
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\label{algo:metropolis}
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\begin{algorithmic}
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\Procedure{Metropolis}{$lattice, energy, magnetization$}
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\For{ Each spin in the lattice }
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\State $ri \leftarrow \text{random index of the lattice}$
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\State $rj \leftarrow \text{random index of the lattice}$
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\State $dE \leftarrow 2 \cdot lattice_{ri,rj} \cdot (lattice_{ri,rj-1} + lattice_{ri,rj+1}
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+ lattice_{ri-1,rj} + lattice_{ri+1,rj} )$
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\State $rn \leftarrow \text{random number} \in [0,1)$
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\If{$rn \leq e^{- \frac{dE}{T}}$}
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\State $lattice_{ri,rj} = -1 \cdot lattice_{ri,rj}$
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\State $magnetization = magnetization + 2 \cdot lattice_{ri,rj}$
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\State $energy = energy + dE$
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\EndIf
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\EndFor
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\caption{Algo}
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\end{figure}
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\subsection{Phase transition and critical temperature}\label{subsec:phase_critical}
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\subsection{Implementation}\label{subsec:implementation}
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% P2
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\subsection{Tools}\label{subsec:tools}
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The Ising model and MCMC methods are implemented in C++, and parallelized using
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both \verb|MPI| and \verb|OpenMP| \cite{openmp:2018}. We used the Python library
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\verb|matplotlib| \cite{hunter:2007:matplotlib} to produce all the plots, and
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\verb|seaborn| \cite{waskom:2021:seaborn} to set the theme in the figures. In
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addition, while optimizing our implementation we used the profiler
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% Add profiler
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\end{document}
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% Method
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% Problem 1: OK
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% Problem 2:
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% Problem 3:
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% Results
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% Problem 4:
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% Problem 5:
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% Problem 6:
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% Problem 7:
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% Problem 8:
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% Problem 9:
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%
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%
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