120 lines
4.7 KiB
TeX
120 lines
4.7 KiB
TeX
\documentclass{article}
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% set up telugu
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\usepackage{fontspec}
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\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
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\usepackage{polyglossia}
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\setdefaultlanguage{english}
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\setotherlanguage{telugu}
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%other packages
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{physics}
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\usepackage{siunitx}
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\usepackage{todonotes}
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\usepackage{luacode}
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\usepackage{titling}
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\usepackage{enumerate}
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% custom deepak packages
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\usepackage{luatrivially}
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\usepackage{subtitling}
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\usepackage{cleveref}
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\begin{luacode*}
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math.randomseed(31415926)
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\end{luacode*}
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\newcommand{\kb}{k_{\mathrm{B}}}
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\title{Problem 3.19}
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\subtitle{Random energy model}
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\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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% want empty date
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\predate{}
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\date{}
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\postdate{}
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% !TeX spellcheck = en_GB
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\begin{document}
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\maketitle
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Given $N$ spins (so $M = 2^N$ possible states), assign a random energy to each.
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Given $j \in \left{1, 2, \ldots, 2^N \right}$, assume each energy $E_j$ is selected with probability $P(E) = \frac{1}{\sqrt{\pi N}} e^{\flatfrac{-E^2}{N}}$.
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Gaussian with mean $0$ and standard deviation $\sqrt{\flatfrac{N}{2}}$
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\subsubsection*{(a) Microcanonical ensemble}
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Consider the states in a small range $E < E_j < E+\deltaE$.
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Let the number of such states in this range be $\Omega(E) \delta E$.
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Calculate the average
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\begin{equation}
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\left< \Omega(N\epsilon) \right>_{REM}
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\end{equation}
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over the ensemble of REM systems, in terms of energy per particle $\epsilon$.
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For energies per particle near zero, show that this average density of states grows exponentially as the system size $N$ grows.
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In contrast, show that $\left< \Omega(N\epsilon) \right>_{REM}$ decreases exponentially for $\epsilon = \flatfrac{E}{N} < -\epsilon_\ast$ and for $\epsilon > \epsilon_\ast$, where the limiting energy per particle
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\begin{equation}
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\epsilon_\ast = \sqrt{\log 2}.
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\end{equation}
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(Hint: as $N$ grows, the probability density $P(N \epsilon)$ decreases exponentially, which the total number of states $2^N$ grows exponentially.
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Which one wins?)
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\subsubsection*{(b) Entropy}
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What does an exponentially growing number of states mean?
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Let the entropy per particle be $s(\epsilon) = \flatfrac{S(N\epsilon)}{N}$.
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Then (setting $\kb = 1$) $\Omega(E) = \exp(S(E)) = \exp(N s(\epsilon))$ grows exponentially whenever the entropy per particle is positive.
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How do we calculate the entropy per particle $s(\epsilon)$ of a typical REM?
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Can we just use the annealed average
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\begin{equation}
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s_{\mathrm{annealed}}(\epsilon) = \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}}?
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\end{equation}
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Show that $s_{\mathrm{annealed}} = \log2 - \epsilon^2$.
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\subsubsection*{Not a part just an answer to that question}
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Sethna says no, because that limit doesn't work in the glassy state below $\epsilon_\ast$.
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\section{Solution} \label{sec:solution}
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\subsection*{(a)}
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We want $\left< \Omega(N\epsilon) \right>_{REM}$.
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Pretty simply, $\Omega(N \epsilon)$ is the total number of states with a given energy, and the average of that over possible configurations is just the number of possible states $M$ times the probability of any individual state having energy $N \epsilon$, $P(N \epsilon)$:
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\begin{align}
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\left< \Omega(N\epsilon) \right>_{REM} &= M P(N \epsilon) \\
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&= 2^N \frac{1}{\sqrt{\pi N}} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
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&= \frac{1}{\sqrt{\pi N}} e^{N \log 2} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
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&= \frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}
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\end{align}
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This grows exponentially if the argument of the exponential is positive, so
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\begin{align}
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N \log 2 &> \frac{(N \epsilon)^2}{N} \\
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N^2 \log 2 &> N^2 \epsilon^2 \\
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\sqrt{\log 2} &> \abs{\epsilon}
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\end{align}
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This is the condition we were asked to show.
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\subsection*{(b)}
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We can just jump straight in.
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\begin{align}
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s_{\mathrm{annealed}}(\epsilon) &= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}} \\
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&= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log(\frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}) \\
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&= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \left( \log \frac{1}{\sqrt{\pi N}} + N \log 2 - \flatfrac{(N \epsilon)^2}{N} \right) \\
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&= \lim_{N \rightarrow \infty} \frac{1}{N} \log \frac{1}{\sqrt{\pi N}} + \log 2 - \epsilon^2 \\
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\end{align}
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The first term clearly goes to $0$ for $N \rightarrow \infty$, because it ends up proportional to $\flatfrac{\log{N}}{N}$.
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So that just leaves the last two terms which are $N$-independent, giving us
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\begin{equation}
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s_{\mathrm{annealed}} = \log2 - \epsilon^2.
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\end{equation}
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\newpage
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\listoftodos
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\end{document}
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