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\title{Problem 3.19}
\subtitle{Random energy model}
\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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\begin{document}
	\maketitle
	
	Given $N$ spins (so $M = 2^N$ possible states), assign a random energy to each.
	Given $j \in \left{1, 2, \ldots, 2^N \right}$, assume each energy $E_j$ is selected with probability $P(E) = \frac{1}{\sqrt{\pi N}} e^{\flatfrac{-E^2}{N}}$.
	Gaussian with mean $0$ and standard deviation $\sqrt{\flatfrac{N}{2}}$

	\subsubsection*{(a) Microcanonical ensemble}
	Consider the states in a small range $E < E_j < E+\deltaE$.
	Let the number of such states in this range be $\Omega(E) \delta E$.
	
	Calculate the average
	\begin{equation}
		\left< \Omega(N\epsilon) \right>_{REM}
	\end{equation}
	over the ensemble of REM systems, in terms of energy per particle $\epsilon$.
	For energies per particle near zero, show that this average density of states grows exponentially as the system size $N$ grows.
	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
	\begin{equation}
		\epsilon_\ast = \sqrt{\log 2}.
	\end{equation}
	(Hint: as $N$ grows, the probability density $P(N \epsilon)$ decreases exponentially, which the total number of states $2^N$ grows exponentially.
	Which one wins?)

	\subsubsection*{(b) Entropy}
	What does an exponentially growing number of states mean?
	Let the entropy per particle be $s(\epsilon) = \flatfrac{S(N\epsilon)}{N}$.
	Then (setting $\kb = 1$) $\Omega(E) = \exp(S(E)) = \exp(N s(\epsilon))$ grows exponentially whenever the entropy per particle is positive.
	
	How do we calculate the entropy per particle $s(\epsilon)$ of a typical REM?
	Can we just use the annealed average
	\begin{equation}
		s_{\mathrm{annealed}}(\epsilon) = \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}}?
	\end{equation}
	
	Show that $s_{\mathrm{annealed}} = \log2 - \epsilon^2$.

	\subsubsection*{Not a part just an answer to that question}
	Sethna says no, because that limit doesn't work in the glassy state below $\epsilon_\ast$.

	\section{Solution} \label{sec:solution}
	
	\subsection*{(a)}
	We want $\left< \Omega(N\epsilon) \right>_{REM}$.

	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 $N \epsilon$, $P(N \epsilon)$:
	\begin{align}
		\left< \Omega(N\epsilon) \right>_{REM} &= M P(N \epsilon) \\
		&= 2^N \frac{1}{\sqrt{\pi N}} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
		&= \frac{1}{\sqrt{\pi N}} e^{N \log 2} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
		&= \frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}
	\end{align}
	This grows exponentially if the argument of the exponential is positive, so
	\begin{align}
		N \log 2 &> \frac{(N \epsilon)^2}{N} \\
		N^2 \log 2 &> N^2 \epsilon^2 \\
		\sqrt{\log 2} &> \abs{\epsilon}
	\end{align}
	This is the condition we were asked to show.

	\subsection*{(b)}
	We can just jump straight in.
	
	\begin{align}
		s_{\mathrm{annealed}}(\epsilon) &= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}} \\
	&= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log(\frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}) \\
	&= \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) \\
	&= \lim_{N \rightarrow \infty} \frac{1}{N} \log \frac{1}{\sqrt{\pi N}} + \log 2 - \epsilon^2 \\
	\end{align}

	The first term clearly goes to $0$ for $N \rightarrow \infty$, because it ends up proportional to $\flatfrac{\log{N}}{N}$.
	So that just leaves the last two terms which are $N$-independent, giving us
	\begin{equation}
		s_{\mathrm{annealed}} = \log2 - \epsilon^2.
	\end{equation}
	\newpage
	\listoftodos

\end{document}