131 lines
6.9 KiB
TeX
131 lines
6.9 KiB
TeX
\documentclass[../ising_model.tex]{subfiles}
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\begin{document}
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\appendix
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\section{Partition function}\label{sec:partition_function}
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Using the values estimated for the $2 \times 2$ case, found in \ref{tab:lattice_config},
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we find the partition function
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\begin{align*}
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Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} + 2 \cdot e^{-\beta (8J)} \\
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& \quad + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\
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&= 2e^{8 \beta J} + 2e^{-8 \beta J} + 12.
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\end{align*}
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We rewrite the expression using the identity $\cosh(8 \beta J) = 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})$,
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and get
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\begin{align*}
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z &= 4 \cosh (8 \beta J) + 12 \ .
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\end{align*}
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\section{Expectation values}\label{sec:expectation_values}
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For a linear function of a stochastic random variable $X$, with a known probability
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distribution, the expected value of $x$ is given by
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\begin{align*}
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\mathbb{E}(aX + b) &= a \cdot \mathbb{E}(X) + b & \text{\cite[p. 131]{springer:2012:modernstat}}
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\end{align*}
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In our case the discrete random variable is the spin configuration, and we want
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to find the expected value of the function $E(\mathbf{s})$. Continuing, we will
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use the notation $\langle E \rangle$ for the expectation value of a given variable,
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in this case $E$.
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Both energy per spin and magnetization per spin are functions of $\mathbf{s}$.
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In addition, the number of spins is given as a constant for each lattice. We can
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use the expression for $\langle E \rangle$ and $\langle M \rangle$ to find the
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expectation values per spin. For energy per spin
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\begin{align*}
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\langle \epsilon \rangle &= \sum_{i=1}^{N} \epsilon(s_{i}) p(s_{i} \ | \ T) \\
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&= \sum_{i=1}^{N} \frac{E(\mathbf{s})}{N} p(s_{i} \ | \ T) \\
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&= \frac{1}{N} \sum_{i=1}^{N} E(\mathbf{s}) p(s_{i} \ | \ T)
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\end{align*}
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The same applies to magnetization per spin
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\begin{align*}
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\langle |m| \rangle = \frac{1}{N} \sum_{i=1}^{N} |M(s_{i})| p(s_{i} \ | \ T) \ .
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\end{align*}
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Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which give zero we get
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\begin{align*}
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\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} \\
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&= \frac{16J}{Z} \big(e^{-8 \beta J} - e^{8 \beta J}) \\
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&= -\frac{32J \sinh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
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&= -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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\end{align*}
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and
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\begin{align*}
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\langle |M| \rangle &= 4 \cdot \frac{1}{Z} \cdot e^{8 \beta J} + 4 \cdot 2 \cdot \frac{1}{Z} \cdot e^{0} \\
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& \quad + 4 \cdot | -2| \cdot \frac{1}{Z} \cdot e^{0} + | -4| \cdot e^{8 \beta J} \\
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&= \frac{8 e^{8 \beta J} + 16}{Z} \\
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&= \frac{4 (2e^{8 \beta J} + 4)}{4(\cosh(8 \beta J) + 3)} \\
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&= \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ .
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\end{align*}
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The squared function
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\begin{align*}
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\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} \\
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&= \frac{128J^{2}}{Z} \big(e^{8 \beta J} + e^{-8 \beta J} \big) \\
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&= \frac{128J^{2} \cosh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
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&= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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\end{align*}
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and
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\begin{align*}
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\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} \\
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& \quad + 4 \cdot (-2)^{2} \cdot \frac{1}{Z} \cdot e^{0} + (-4)^{2}\cdot e^{8 \beta J} \\
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&= \frac{32e^{8 \beta J} + 32}{Z} \\
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&= \frac{4 (8e^{8 \beta J} + 8)}{4(\cosh(8 \beta J) + 3)} \\
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&= \frac{8e^{8 \beta J} + 8}{\cosh(8 \beta J) + 3} \ .
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\end{align*}
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The squared expectation value is given by
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\begin{align*}
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\langle E \rangle^{2} &= \bigg(-\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \bigg)^{2} \\
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&= \frac{64J^{2} \sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \ ,
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\end{align*}
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and
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\begin{align*}
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\langle |M| \rangle^{2} &= \Big( \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \Big)^{2} \\
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&= \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \ .
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\end{align*}
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\section{Heat capacity and magnetic susceptibility}\label{sec:heat_susceptibility}
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To find the heat capacity in Eq. \ref{eq:heat_capacity}, we normalize to heat
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capacity per spin
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\begin{align*}
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\frac{C_{V}}{N} &= \frac{1}{N} \frac{1}{k_{B} T^{2}} (\mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2}) \\
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&= \frac{1}{N k_{B} T^{2}} \mathbb{V}(E) \ .
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\end{align*}
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Using Eq. \eqref{eq:susceptibility}, we find he susceptibility per spin
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\begin{align*}
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\frac{\chi}{N} &= \frac{1}{N} \frac{1}{k_{B} T} (\mathbb{E}(M^{2}) - [\mathbb{E}(M)]^{2}) \\
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&= \frac{1}{N k_{B} T} \mathbb{V}(M) \ .
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\end{align*}
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We now have to find the variance of both total energy and total magnetizaation.
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We obtain this using the definition
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\begin{align*}
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\mathbb{V}(X) &= \sum_{x \in D} [(x - \mathbb{E}(x))^2 \cdot p(x)] & \text{\cite[p. 132]{springer:2012:modernstat}} \\
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&= \mathbb{E}(X^{2}) - [\mathbb{E}(X)]^{2} \ .
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\end{align*}
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The variance of the total energy is then given by
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\begin{align*}
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\mathbb{V}(E) &= \mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2} \\
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&= \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}} \\
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&= 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) \\
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&= 64J^{2} \bigg( \frac{(\cosh(8 \beta J)) \cdot (\cosh(8 \beta J) + 3)}{(\cosh(8 \beta J) + 3)^{2}} \\
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& \quad - \frac{\sinh^{2}(8 \beta J)}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \\
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&= 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) \\
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&= 64J^{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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and the variance of the total magnetization is given by
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\begin{align*}
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\mathbb{V}(M) &= \mathbb{E}(M^{2}) - [\mathbb{E}(|M|)]^{2} \\
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&= \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}} \\
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&= \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}} \\
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&= \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}} \\
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&= \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}} \\
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&= \frac{4(3e^{8 \beta J} + e^{-8 \beta J} + 3)}{(\cosh(8 \beta J) + 3)^{2}} \ .
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\end{align*}
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We find the heat capacity
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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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and susceptibility
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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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\end{document} |