\chapter{Statistical Physics-Pset1 Solutions}
\begin{abox}
	Practise set-01
\end{abox}
\begin{enumerate}
	\item A particle is confined to the region $x \geq 0$ by a potential which increases linearly as $u(x)=u_{0} x$. The mean position of the particle at temperature $T$ is
	{	\exyear{NET/JRF(JUNE-2011)}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $\frac{k_{B} T}{u_{0}}$
		\task[\textbf{b.}]$\left(k_{B} T\right)^{2} / u_{0}$
		\task[\textbf{c.}]$\sqrt{\frac{k_{B} T}{u_{0}}}$
		\task[\textbf{d.}]  $u_{0} k_{B} T$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text{Partition function }Z&=\frac{1}{h} \int e^{-\frac{p^{2}}{2 m k_{B} T}} d p \int e^{-\frac{u_{0} x}{k_{B} T}} d x\text{ and }\langle x\rangle=\int x p(x) d x d p_{x}\\
		\Rightarrow\langle x\rangle&=\frac{\iint x e^{-\frac{p^{2}}{2 m k_{B} T}} d p \cdot e^{\frac{u_{0} x}{k_{B} T}} d x}{\iint e^{\frac{p^{2}}{2 m k_{B} T}} d p \cdot e^{\frac{u_{0} x}{k_{B} T}} d x}=\frac{\int_{0}^{\infty} x e^{\frac{\mu_{0} x}{k_{B} T}} d x}{\int_{0}^{\infty} e^{-\frac{\mu_{0} x}{k_{B} T}} d x}=\frac{\left(\frac{k_{B} T}{u_{0}}\right)^{2}}{\left(\frac{k_{B} T}{u_{0}}\right)} \frac{\int_{0}^{\infty} t e^{-t} d t}{\int_{0}^{\infty} e^{-t} d t}=\frac{k_{B} T}{u_{0}}
		\end{align*}
		So the correct answer is \textbf{Option (a)}
	\end{answer}
	\item 	Consider a system of $N$ non-interacting spins, each of which has classical magnetic moment of magnitude $\mu$. The Hamiltonian of this system in an external magnetic field $\vec{H}$ is $\sum_{i=1}^{N} \vec{\mu}_{i} \cdot \vec{H}$, where $\vec{\mu}_{i}$ is the magnetic moment of the $i^{\text {th }}$ spin. The magnetization per spin at temperature $T$ is
	{	\exyear{NET/JRF(JUNE-2011)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{\mu^{2} H}{k_{B} T}$
		\task[\textbf{b.}]$\mu\left[\operatorname{coth}\left(\frac{\mu H}{k_{B} T}\right)-\frac{k_{B} T}{\mu H}\right]$
		\task[\textbf{c.}] $\mu \sinh \left(\frac{\mu H}{k_{B} T}\right)$
		\task[\textbf{d.}] $\mu \tanh \left(\frac{\mu H}{k_{B} T}\right)$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text { For classical limit } M&=\frac{\int_{0}^{2 \pi} \int_{0}^{\pi} \mu \cos \theta \exp \frac{\mu H \cos \theta}{k T} \sin \theta d \theta d \phi}{\iint \exp \frac{\mu H \cos \theta}{k_{B} T} \sin \theta d \theta d \phi} \\
		M&=\mu\left[\operatorname{coth}\left(\frac{\mu H}{k_{B} T}\right)-\frac{k_{B} T}{\mu H}\right]
		\end{align*}
		So the correct answer is \textbf{Option (b)}
	\end{answer}
	\item 	The internal energy $E$ of a system is given by $E=\frac{b S^{3}}{V N}$, where $b$ is a constant and other symbols have their usual meaning. The temperature of this system is equal to
	{	\exyear{NET/JRF(DEC-2011)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{b S^{2}}{V N}$
		\task[\textbf{b.}]$\frac{3 b S^{2}}{V N}$
		\task[\textbf{c.}]$\frac{b S^{3}}{V^{2} N}$
		\task[\textbf{d.}] $\left(\frac{S}{N}\right)^{2}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		T d S=d E+P d V \Rightarrow d E=T d S-P d V \Rightarrow\left(\frac{\partial E}{\partial S}\right)_{V}=T \Rightarrow T=\frac{3 b S^{2}}{V N}
		\end{align*}
		So the correct answer is \textbf{Option (b)}
	\end{answer}
	\item A gas of $N$ non-interacting particles is in thermal equilibrium at temperature $T$. Each particle can be in any of the possible non-degenerate states of energy $0,2 \varepsilon$ and $4 \varepsilon$. The average energy per particle of the gas, when $\beta \varepsilon<<1$, is
	{	\exyear{NET/JRF(DEC-2011)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$2 \varepsilon$
		\task[\textbf{b.}] $3 \varepsilon$
		\task[\textbf{c.}]$2 \varepsilon / 3$
		\task[\textbf{d.}]  $\varepsilon$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		E_{1}&=0, E_{2}=2 \varepsilon, E_{3}=4 \varepsilon, Z=e^{-0 \beta}+e^{-2 \varepsilon \beta}+e^{-4 \varepsilon \beta} \Rightarrow\langle E\rangle=\frac{0 \cdot e^{-o \beta}+2 \varepsilon e^{-2 \varepsilon \beta}+4 \varepsilon e^{-4 \varepsilon \beta}}{e^{-0 \beta}+e^{-2 \varepsilon \beta}+e^{-4 \varepsilon 1 \beta}} \\
		\Rightarrow\langle E\rangle&=\frac{2 \varepsilon e^{-2 \varepsilon \beta}+4 \varepsilon e^{-4 \varepsilon \beta}}{1+e^{-2 \varepsilon \beta}+e^{-4 \varepsilon \beta}}=\frac{2 \varepsilon(1-2 \varepsilon \beta \ldots .)+4 \varepsilon(1-(4 \varepsilon \beta) \ldots .)}{1+(1-2 \varepsilon \beta \ldots .)+(1-4 \varepsilon \beta \ldots . .)}=\frac{2 \varepsilon+4 \varepsilon}{1+1+1}=\frac{6 \varepsilon}{3}=2 \varepsilon\\
		&\text{	where }\beta \varepsilon<<1.
		\end{align*}
		So the correct answer is \textbf{Option (a)}
	\end{answer}
	\item 	Gas molecules of mass $m$ are confined in a cylinder of radius $R$ and height $L$ (with $R>L$ ) kept vertically in the Earth's gravitational field. The average energy of the gas at low temperatures (such that $m g L \gg k_{B} T$ ) is given by
	{	\exyear{NET/JRF(DEC-2011)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$N k_{B} T / 2$
		\task[\textbf{b.}]$3 N k_{B} T / 2$
		\task[\textbf{c.}]$2 N k_{B} T$
		\task[\textbf{d.}] $5 N k_{B} T / 2$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		Z&=\frac{1}{h^{3}} \int e^{-\beta H} d p_{x} d p_{y} d p_{z} d x d y d z\\
		Z&=\int_{-\infty}^{\infty} e^{\frac{-p_{x}^{2}}{2 m k_{B} T}} d p_{x} \int_{-\infty}^{\infty} e^{\frac{-p_{y}^{2}}{2 m k_{B} T}} d p_{y} \int_{-\infty}^{\infty} e^{\frac{-p_{z}^{2}}{2 m k_{B} T}} d p_{z} \int d x d y \int_{0}^{L} e^{-\frac{m g z}{k_{B} T}} d z\\
		Z&=\pi R^{2}\left(\frac{m k_{B} T}{2 \pi \hbar^{2}}\right)^{\frac{3}{2}} \int_{0}^{L} e^{-\frac{m g z}{k_{B} T}} d z \Rightarrow Z=\pi R^{2}\left(\frac{m k_{B} T}{2 \pi \hbar^{2}}\right)^{\frac{3}{2}}\left(\frac{1-e^{-\frac{m g L}{k_{B} T}}}{\frac{m g}{k_{B} T}}\right)\\
		Z_{N}&=Z^{N}\\
		\Rightarrow\langle E\rangle&=k_{B} T^{2} \frac{\partial \ln z}{\partial T}=\frac{5 N k_{B} T}{2} \text {, since } m g L>>k_{B} T
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item 	The free energy of the gas of $N$ particles in a volume $V$ and at a temperature $T$ is $F=N k_{B} T \ln \left[a_{0} V\left(k_{B} T\right)^{5 / 2} / N\right]$, where $a_{0}$ is a constant and $k_{B}$ denotes the Boltzmann constant. The internal energy of the gas is
	{	\exyear{NET/JRF(JUNE-2012)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{3}{2} N k_{B} T$
		\task[\textbf{b.}]$\frac{5}{2} N k_{B} T$
		\task[\textbf{c.}]$N k_{B} T \ln \left[a_{0} V\left(k_{B} T\right)^{5 / 2} / N\right]-\frac{3}{2} N k_{B} T$
		\task[\textbf{d.}]  $N k_{B} T \ln \left[a_{0} V /\left(k_{B} T\right)^{5 / 2}\right]$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		F&=N k_{B} T \ln \left[a_{0} V\left(k_{B} T\right)^{5 / 2} / N\right], F=U-T S, U=F+T S\\
		d F&=-S d T-P d V \Rightarrow\left(\frac{\partial F}{\partial T}\right)_{V}=-S \text { or } S=-\left(\frac{\partial F}{\partial T}\right)_{V} \Rightarrow U=F-T\left(\frac{\partial F}{\partial T}\right)_{V} \\
		F&=N k_{B} T \ln \left(C T^{5 / 2}\right) \text { where } C=\frac{a_{0} V k_{B}^{5 / 2}}{N} \\
		\left(\frac{\partial F}{\partial T}\right)_{V}&=N k_{B} \ln \left(C T^{5 / 2}\right)+N k_{B} T \frac{C}{C T^{5 / 2}} \frac{5}{2} T^{3 / 2} \Rightarrow T\left(\frac{\partial F}{\partial T}\right)_{V}=N k_{B} T \ln \left(C T^{5 / 2}\right)+\frac{5}{2} N k_{B} T \\
		T\left(\frac{\partial F}{\partial T}\right)_{V}&=F+\frac{5}{2} N k_{B} T \Rightarrow U=F-T\left(\frac{\partial F}{\partial T}\right)_{V}=-\frac{5}{2} N k_{B} T .\\
		T\left(\frac{\partial F}{\partial T}\right)_{V}&=F+\frac{5}{2} N k_{B} T \Rightarrow U=F-T\left(\frac{\partial F}{\partial T}\right)_{V}=-\frac{5}{2} N k_{B} T .
		\end{align*}
		So the correct answer is \textbf{Option (b)}
	\end{answer}
	\item A system has two normal modes of vibration, with frequencies $\omega_{1}$ and $\omega_{2}=2 \omega_{1}$. What is the probability that at temperature $T$, the system has an energy less than $4 \hbar \omega_{1}$ ?
	[In the following $x=e^{-\beta \hbar \omega_{1}}$ and $Z$ is the partition function of the system.]
	{	\exyear{NET/JRF(JUNE-2012)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$x^{3 / 2}\left(x+2 x^{2}\right) / Z$
		\task[\textbf{b.}]$x^{3 / 2}\left(1+x+x^{2}\right) / Z$
		\task[\textbf{c.}]$x^{3 / 2}\left(1+2 x^{2}\right) / Z$
		\task[\textbf{d.}] $x^{3 / 2}\left(1+x+2 x^{2}\right) / Z$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\intertext{There is two normal mode so there is two degree of freedom.}
		\intertext{	Energy of harmonic oscillator is $E=\left(n_{1}+\frac{1}{2}\right) \hbar \omega_{1}+\left(n_{2}+\frac{1}{2}\right) \hbar \omega_{2}$.} E&=\left(n_{1}+\frac{1}{2}\right) \hbar \omega_{1}+\left(n_{2}+\frac{1}{2}\right) \hbar 2 \omega_{1}\text{ where } n_{1}=0,1,2,3 \ldots .
		\text{ and }n_{2}=0,1,2,3 \ldots .\\
		\intertext{	Ground state energy $E=\frac{3 \hbar \omega_{1}}{2}$, first excited state energy $E=\frac{5 \hbar \omega_{1}}{2}$. Second excited state energy $E=\frac{7 \hbar \omega_{1}}{2}$ which is doubly degenerate state so $g=2$, other state have more energy than $4 \hbar \omega_{1}$.}
		P\left(E<4 \hbar \omega_{1}\right)&=\frac{e^{-\frac{3 \beta \hbar \omega_{1}}{2}}+e^{-\frac{5 \beta \hbar \omega_{1}}{2}}+2 e^{-\frac{7 \beta \hbar \omega_{1}}{2}}}{Z}=\frac{x^{3 / 2}\left(1+x+2 x^{2}\right)}{Z}\text{ where }x=e^{-\beta \hbar \omega_{1}} .
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item 	The entropy of a system, $(S)$, is related to the accessible phase space volume $\Gamma$ by $S=k_{B} \ln \Gamma(E, N, V)$ where $E, N$ and $V$ are the energy, number of particles and volume respectively. From this one can conclude that $\Gamma$
	{	\exyear{NET/JRF(DEC-2012)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]does not change during evolution to equilibrium
		\task[\textbf{b.}]oscillates during evolution to equilibrium
		\task[\textbf{c.}]is a maximum at equilibrium
		\task[\textbf{d.}]is a minimum at equilibrium
	\end{tasks}
	\begin{answer}
		Entropy is maximum at equilibrium.\\
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item 	Consider a one-dimensional Ising model with $N$ spins, at very low temperatures when almost all spins are aligned parallel to each other. There will be a few spin flips with each flip costing an energy $2 J$. In a configuration with $r$ spin flips, the energy of the system is $E=-N J+2 r J$ and the number of configuration is ${ }^{N} C_{r} ; r$ varies from 0 to $N$. The partition function is
	{	\exyear{NET/JRF(DEC-2012)}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $\left(\frac{J}{k_{B} T}\right)^{N}$
		\task[\textbf{b.}]$e^{-N J / k_{B} T}$
		\task[\textbf{c.}] $\left(\sinh \frac{J}{k_{B} T}\right)^{N}$
		\task[\textbf{d.}] $\left(\cosh \frac{J}{k_{B} T}\right)^{N}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text { Let us}&\text{ consider only three energy levels, } E_{r}=-2 J+2 r J \text { i.e. } E_{0}=-2 J, E_{1}=0 \text { and }\\
		E_{2}&=2 J.\\
		Q_{2}&=\frac{\left({ }^{2} C_{0} e^{-\beta E_{0}}+{ }^{2} C_{1} e^{-\beta E_{1}}+{ }^{2} C_{2} e^{-\beta E_{2}}\right)}{\sum_{r=0}^{2}{ }^{2} C_{r}}=\frac{\left(e^{\beta 2 J}+2 e^{0}+e^{\beta 2 J}\right)}{4}=\frac{\left(e^{\beta J}+e^{\beta J}\right)^{2}}{4} \\
		Q_{2}&=\left(\frac{e^{\beta J}+e^{\beta J}}{2}\right)^{2}=(\cosh \beta J)^{2} \Rightarrow(\cosh \beta J)^{2} \Rightarrow Q_{N}=(\cosh \beta J)^{N}
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item Consider a system of three spins $S_{1}, S_{2}$ and $S_{3}$ each of which can take values $+1$ and $-1$. The energy of the system is given by $E=-J\left[S_{1} S_{2}+S_{2} S_{3}+S_{3} S_{1}\right]$ where $J$ is a positive constant. The minimum energy and the corresponding number of spin configuration are, respectively,
	{	\exyear{NET/JRF(DEC-2012)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$J$ and 1
		\task[\textbf{b.}]$-3 J$ and 1
		\task[\textbf{c.}]$-3 J$ and 2
		\task[\textbf{d.}]  $-6 J$ and 2
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text{If we take }S_{1}&=S_{2}=S_{3}=+1\text{ i.e. }\hspace{3cm}\stackrel{\uparrow}{s_1}\qquad \stackrel{\uparrow}{s_2}\qquad\stackrel{\uparrow}{s_3}\\
		\text { Then energy, } E&=-J[1 \times 1+1 \times 1+1 \times 1]=-3 J\\
		\text { Again } S_{1}&=S_{2}=S_{3}=-1 \text {, then }\hspace{2.8cm}\downarrow\qquad \ \downarrow\qquad \ \downarrow\\
		\text{Energy }(E)&=-3 J
		\intertext{So, minimum energy is $(-3 J)$ and there are two spin configuration.}\\
		\text { If we take }&\stackrel{\uparrow}{s_1}\qquad \stackrel{\downarrow}{s_2}\qquad\stackrel{\uparrow}{s_3}\\
		\text { Then we get Maximum energy } E&=J \text {. }
		\end{align*}
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item 	Consider a system of two Ising spins $S_{1}$ and $S_{2}$ taking values $\pm 1$ with interaction energy given by $\varepsilon=-J S_{1} S_{2}$, when it is in thermal equilibrium at temperature $T$. For large $T$, the average energy of the system varies as $C / k_{B} T$, with $C$ given by
	{	\exyear{NET/JRF(JUNE-2013)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$-2 J^{2}$
		\task[\textbf{b.}] $-J^{2}$
		\task[\textbf{c.}]$J^{2}$
		\task[\textbf{d.}] $4 J$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\intertext{The interaction energy is given by $E=-J S_{1} S_{2}$ where $S_{1}$ and $S_{2}$ taking values $\pm 1$.}
		\text { Possible }&\text{values of the Energy of the system are }\\
		E_{1}&=-J 1 \cdot 1=-J, \quad E_{2}=-J(-1) \cdot(1)=+J\\
		E_{3}&=-J(1) \cdot(-1)=+J, \quad E_{4}=-J(-1) \cdot(-1)=-J\\
		\langle U\rangle&=\frac{\sum_{r} E_{r} g_{r} e^{-\frac{E_{r}}{k T}}}{\sum_{r} g_{r} e^{\frac{E_{r}}{k T}}}=\frac{-2 J e^{\frac{J}{k T}}+2 J e^{\frac{J}{k T}}}{2 e^{\frac{J}{k T}}+2 e^{-\frac{J}{k T}}}=-J\left(\frac{e^{\frac{J}{k T}}-e^{-\frac{J}{k T}}}{e^{\frac{J}{k T}}+e^{\frac{J}{k T}}}\right)=-J \frac{\left(1+\frac{J}{k T}-\left(1-\frac{J}{k T}\right)\right)}{1+\left(\frac{J}{k T}\right)+1-\left(\frac{J}{k T}\right)}\\
		\Rightarrow\langle U\rangle&=-\frac{J^{2}}{k T} \Rightarrow C=-J^{2} \quad \text { (For large } T, \frac{J}{k T}<<1 \text { ) }
		\end{align*}
		So the correct answer is \textbf{Option (b)}
	\end{answer}
	\item A collection $N$ of non-interacting spins $S_{i}, i=1,2, \ldots \ldots, N,\left(S_{i}=\pm 1\right)$ is kept in an external magnetic field $B$ at a temperature $T$. The Hamiltonian of the system is $H=-\mu B \Sigma_{i} S_{i}$. What should be the minimum value of $\frac{\mu B}{k_{B} T}$ for which the mean value $\left\langle S_{i}\right\rangle \geq \frac{1}{3}$ ?
	{	\exyear{NET/JRF(DEC-2014)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{1}{2} N \ln 2$
		\task[\textbf{b.}] $2 \ln 2$
		\task[\textbf{c.}]$\frac{1}{2} \ln 2$
		\task[\textbf{d.}] $N \ln 2$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		P\left(S_{i}=+1\right)&=\frac{e^{-\frac{\mu B}{k T}}}{e^{-\frac{\mu B}{k T}}+e^{\frac{\mu B}{k T}}}, P\left(S_{i}=-1\right)=\frac{e^{\frac{\mu B}{k T}}}{e^{-\frac{\mu B}{k T}}+e^{\frac{\mu B}{k T}}}\\
		\left\langle S_{i}\right\rangle&=\frac{+1 e^{-\frac{\mu B}{k T}}-e^{+\frac{\mu B}{k T}}}{e^{-\frac{\mu B}{k T}}+e^{\frac{\mu B}{k T}}} \Rightarrow\left\langle S_{i}\right\rangle=-\left(\tanh \frac{\mu B}{k T}\right)\\
		\text{For $N$ particle }&\left\langle S_{i}\right\rangle=-N \tanh \frac{\mu B}{k T}\\
		\text{	According to question, }&\frac{\left\langle S_{i}\right\rangle}{N} \geq \frac{1}{3} \Rightarrow-\tanh \left(\frac{\mu B}{k T}\right)=\frac{1}{3} \Rightarrow \frac{\mu B}{k T}=\frac{1}{2} \ln 2
		\end{align*}
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item A system of $N$ distinguishable particles, each of which can be in one of the two energy levels 0 and $\in$, has a total energy $n \in$, where $n$ is an integer. The entropy of the system is proportional to
	{	\exyear{NET/JRF(JUNE-2015)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$N \ln n$
		\task[\textbf{b.}]$n \ln N$
		\task[\textbf{c.}] $\ln \left(\frac{N !}{n !}\right)$
		\task[\textbf{d.}] $\ln \left(\frac{N !}{n !(N-n) !}\right)$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		??
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item Consider three Ising spins at the vertices of a triangle which interact with each other with a ferromagnetic Ising interaction of strength $J$. The partition function of the system at temperature $T$ is given by $\left(\beta=\frac{1}{k_{B} T}\right)$ :
	{	\exyear{NET/JRF(JUNE-2015)}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $2 e^{3 \beta J}+6 e^{-\beta J}$
		\task[\textbf{b.}] $2 e^{-3 \beta J}+6 e^{\beta J}$
		\task[\textbf{c.}]$2 e^{3 \beta J}+6 e^{-3 \beta J}+3 e^{\beta J}+3 e^{-\beta J}$
		\task[\textbf{d.}] $(2 \cosh \beta J)^{3}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		&H=J\left(S_{1} S_{2}+S_{1} S_{3}+S_{2} S_{3}\right)\\
		&\begin{array}{llll}
		S_{1} & S_{2} & S_{3} & E
		\end{array}\\
		&\begin{array}{llll}
		1 & 1 & 1 & 3 J
		\end{array}\\
		&\left.\begin{array}{ccc}
		1 & 1 & -1 \\
		1 & -1 & 1 \\
		-1 & 1 & 1
		\end{array}\right\}-J\\
		&\begin{array}{lll}
		-1 & -1 & 1
		\end{array}\\
		&\left.\begin{array}{ccc}
		-1 & 1 & -1 \\
		1 & -1 & -1
		\end{array}\right\}-J\\
		&\left.\begin{array}{lll}
		-1 & -1 & -1
		\end{array}\right\} 3 J\\
		\Rightarrow&z=2 e^{-3 \beta J}+6 e^{\beta J}
		\end{align*}
		So the correct answer is \textbf{Option (b)}
	\end{answer}
	\item The partition function of a system of $N$ Ising spins is $Z=\lambda_{1}^{N}+\lambda_{2}^{N}$ where $\lambda_{1}$ and $\lambda_{2}$ are functions of temperature, but are independent of $N$. If $\lambda_{1}>\lambda_{2}$, the free energy per spin in the limit $N \rightarrow \infty$ is
	{	\exyear{NET/JRF(DEC-2015)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$-k_{B} T \ln \left(\frac{\lambda_{1}}{\lambda_{2}}\right)$
		\task[\textbf{b.}] $-k_{B} T \ln \lambda_{2}$
		\task[\textbf{c.}]$-k_{B} T \ln \left(\lambda_{1} \lambda_{2}\right)$
		\task[\textbf{d.}] $-k_{B} T \ln \lambda_{1}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		Z&=\lambda_{1}^{N}+\lambda_{2}^{N}, F=-k T \ln \left(\lambda_{1}^{N}+\lambda_{2}^{N}\right)\text{, it is given } \lambda_{1}>>\lambda_{2}\\
		\Rightarrow F&=-k T \ln \left[\lambda_{1}^{N}\left(1+\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{N}\right], \quad \frac{\lambda_{2}}{\lambda_{1}} \approx 0\right. \\
		F&=-k T \ln \lambda_{1}^{N}=-N k T \ln \lambda_{1} \Rightarrow \frac{F}{N}=-k T \ln \lambda_{1}
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item The Hamiltonian of a system of $N$ non interacting spin $\frac{1}{2}$ particles is $H=-\mu_{0} B \sum_{i} S_{i}^{z}$, where $S_{i}^{z}=\pm 1$ are components of $i^{\text {th }}$ spin along an external magnetic field $B$. At a temperature $T$ such that $e^{\frac{\mu_{0} B}{k_{B} T}}=2$. the specific heat per particle is
	{	\exyear{NET/JRF(DEC-2015)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{16}{25} k_{B}$
		\task[\textbf{b.}] $\frac{8}{25} k_{B} \ln 2$
		\task[\textbf{c.}]$k_{B}(\ln 2)^{2}$
		\task[\textbf{d.}]  $\frac{16}{25} k_{B}(\ln 2)^{2}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text {  For the given system } E&=-\mu_{0} B \tanh \frac{\mu_{0} B}{k T}\\
		C_{V}&=\left(\frac{\mu_{0} B}{k T}\right)^{2} N k \sec ^{2} h \frac{\mu_{0} B}{k T}\\
		\frac{C_{V}}{N}&=\left(\frac{\mu_{0} B}{k T}\right)^{2} k \frac{4}{\left(e^{\frac{\mu_{0} B}{k T}}+e^{-\frac{\mu_{0} B}{k T}}\right)^{2}} \quad \because e^{\frac{\mu_{0} B}{k T}}=2 \Rightarrow \frac{\mu_{0} B}{k T}=\ln 2\\
		\frac{C_{V}}{N}&=\left(\frac{\mu_{0} B}{k T}\right)^{2} k \frac{4}{\left(2+\frac{1}{2}\right)^{2}}=\left(\frac{\mu_{0} B}{k T}\right)^{2} k \frac{16}{25}=(\ln 2)^{2} \frac{16 k}{25}=\frac{16}{25} k(\ln 2)^{2}
		\end{align*}
		So the correct answer is \textbf{Option (d)}
	\end{answer}
	\item A gas of non-relativistic classical particles in one dimension is subjected to a potential $V(x)=\alpha|x|$ (where $\alpha$ is a constant). The partition function is $\left(\beta=\frac{1}{k_{B} T}\right)$
	{	\exyear{NET/JRF(JUNE-2016)}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $\sqrt{\frac{4 m \pi}{\beta^{3} \alpha^{2} h^{2}}}$
		\task[\textbf{b.}] $\sqrt{\frac{2 m \pi}{\beta^{3} \alpha^{2} h^{2}}}$
		\task[\textbf{c.}]$\sqrt{\frac{8 m \pi}{\beta^{3} \alpha^{2} h^{2}}}$
		\task[\textbf{d.}] $\sqrt{\frac{3 m \pi}{\beta^{3} \alpha^{2} h^{2}}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		z&=\frac{1}{h} \int_{-\infty}^{\infty} e^{-\frac{p_{x}^{2}}{2 m k T}} d p_{x} \int_{-\infty}^{\infty} e^{-\frac{\alpha|x|}{k T}} d x=\frac{1}{h}(2 \pi m k T)^{1 / 2} \int_{-\infty}^{\infty} e^{\frac{\alpha|x|}{k T}} d x\\
		\Rightarrow z&=\left(\frac{2 \pi m k T}{h^{2}}\right)^{1 / 2} \int_{-\infty}^{\infty} e^{-\frac{\alpha|x|}{k T}} d x\\
		\because \int_{-\infty}^{\infty} e^{-\frac{\alpha|x|}{k T}} d x&=\int_{-\infty}^{0} e^{+\frac{\alpha x}{k T}} d x+\int_{0}^{\infty} e^{-\frac{\alpha x}{k T}} d x=\frac{k T}{\alpha}+\frac{k T}{\alpha}=\frac{2 k T}{\alpha}\\
		z&=\left(\frac{2 \pi m k T}{h^{2}}\right)^{1 / 2}\left(\frac{2 k T}{\alpha}\right)=\left(\frac{8 \pi m}{h^{2} \beta^{3} \alpha^{2}}\right)^{\frac{1}{2}}\qquad
		\text{	put }\beta=\frac{1}{k T}
		\end{align*}
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item Consider a random walk on an infinite two-dimensional triangular lattice, a part of which is shown in the figure below.
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=4cm]{SP-11}
	\end{figure}
	If the probabilities of moving to any of the nearest neighbour sites are equal, what is the probability that the walker returns to the starting position at the end of exactly three steps?
	{	\exyear{NET/JRF(DEC-2016)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{1}{36}$
		\task[\textbf{b.}]$\frac{1}{216}$
		\task[\textbf{c.}] $\frac{1}{18}$
		\task[\textbf{d.}] $\frac{1}{12}$
	\end{tasks}
	\begin{answer}
		Solution: For walker to return to starting position it must move along an equivalent triangle in three steps.\\
		For steps one any movement can result in equilateral triangle.\\
		For step two, two out of six options will form equilateral triangle.\\
		For step three, only one out of six options will form equilateral\\ triangle Total probability $=\frac{6}{6} \times \frac{2}{6} \times \frac{1}{6}=\frac{1}{18}$\\
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item 	An atom has a non-degenerate ground-state and a doubly-degenerate excited state. The energy difference between the two states is $\varepsilon$. The specific heat at very low temperatures $(\beta \varepsilon \gg 1)$ is given by
	{	\exyear{NET/JRF(DEC-2016)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$k_{B}(\beta \varepsilon)$
		\task[\textbf{b.}]$k_{B} e^{-\beta \varepsilon}$
		\task[\textbf{c.}] $2 k_{B}(\beta \varepsilon)^{2} e^{-\beta \varepsilon}$
		\task[\textbf{d.}]  $k_{B}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\intertext{Assume energy at ground state is 0 and energy at first excited state is $\in$. The partition function is $Z=1+2 e^{-\beta \epsilon}$}
		\text{	Energy }&=\frac{2 \in e^{-\beta \epsilon}}{\left(1+2 e^{-\beta \epsilon}\right)}\\
		\text{Specific heat, }C_{V}&=\left(\frac{\partial U}{\partial T}\right)_{V}=\frac{2 \in e^{-\frac{\epsilon}{k T}}(-\in) \frac{-1}{k T^{2}}}{\left(1+2 e^{\frac{-\epsilon}{k T}}\right)}+\frac{2 \in e^{\frac{-\epsilon}{k T}} \in \in \frac{2}{k T^{2}}}{\left(1+2 e^{\frac{-\epsilon}{k T}}\right)^{2}}\\
		&=2 k\left(\frac{\epsilon}{k T}\right)^{2} e^{\frac{-\epsilon}{k T}} \frac{\left(1+2 e^{\frac{-\epsilon}{k T}}\right)}{\left(1+2 e^{\frac{-\epsilon}{k T}}\right)^{2}}=2 k(\beta \in)^{2} e^{-\beta \in} \frac{\left(1+2 e^{-\beta \epsilon}\right)}{\left(1+2 e^{-\beta \epsilon}\right)^{2}}\\
		C_{V} &\simeq 2 k(\beta \in)^{2} e^{-\beta \epsilon}, \quad \beta \in \rightarrow \infty
		\end{align*}
		So the correct answer is \textbf{Option (c)}
	\end{answer}
	\item In a thermodynamic system in equilibrium, each molecule can exist in three possible states with probabilities $1 / 2,1 / 3$ and $1 / 6$ respectively. The entropy per molecule is
	{	\exyear{NET/JRF(JUNE-2017)}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $k_{B} \ln 3$
		\task[\textbf{b.}]$\frac{1}{2} k_{B} \ln 2+\frac{2}{3} k_{B} \ln 3$
		\task[\textbf{c.}]$\frac{2}{3} k_{B} \ln 2+\frac{1}{2} k_{B} \ln 3$
		\task[\textbf{d.}] $\frac{1}{2} k_{B} \ln 2+\frac{1}{6} k_{B} \ln 3$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		S&=-k_{B} \sum_{i} P i \ln P i\\
		P_{1}&=\frac{1}{2}, P_{2}=1 / 3 \text { and } P_{3}=1 / 6 \\
		S&=-k_{B}\left(\frac{1}{2} \ln 1 / 2+1 / 3 \ln 1 / 3+1 / 6 \ln 1 / 6\right) . \\
		&=-k_{b}\left(\frac{1}{2}(\ln 1-\ln 2)+\frac{1}{3}(\ln 1-\ln 3)+\frac{1}{6}(\ln 1-\ln 6)\right. \\
		&=k_{B}\left[\frac{1}{2} \ln 2+\frac{1}{3} \ln 3+\frac{1}{6} \ln 2+\frac{1}{6} \ln 3\right]=k_{B}\left[\frac{1}{2} \ln 2+\frac{1}{6} \ln 2+\frac{1}{3} \ln 3+\frac{1}{6} \ln 3\right] \\
		S&=k_{B}\left[\frac{3 \ln 2+\ln 2}{6}+\frac{2 \ln 3+\ln 3}{6}\right]=k_{B}\left(\frac{4 \ln 2}{6}+\frac{3 \ln 3}{6}\right)=k_{B}\left[\frac{2}{3} \ln 2+\frac{1}{2} \ln 3\right]
		\end{align*}
		So the correct answer is \textbf{Option (c)}
	\end{answer}
\end{enumerate}
\newpage
\begin{abox}
	Practise set-2
\end{abox}
\begin{enumerate}
	\item A system of $\mathrm{N}$ non-interacting classical point particles is constrained to move on the twodimensional surface of a sphere. The internal energy of the system is
	{\exyear{GATE 2010}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{3}{2} N k_{B} T$
		\task[\textbf{B.}] $\frac{1}{2} N k_{B} T$
		\task[\textbf{C.}] $N k_{B} T$
		\task[\textbf{D.}] $\frac{5}{2} N k_{B} T$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text{There are $2 \mathrm{~N}$ degree of freedom.}\\
		\text{The internal energy of the system is}\frac{N k_{B} T}{2}+\frac{N k_{B} T}{2}&=N k_{B} T
		\end{align*}
		So the correct answer is \textbf{Option (C)}
	\end{answer}
	\item Partition function for a gas of photons is given as, $\ln Z=\frac{\pi^{2} V\left(k_{B} T\right)^{3}}{45 \hbar^{3} C^{3}}$. The specific heat of the photon gas varies with temperature as
	{\exyear{GATE 2010}}
	\begin{tasks}(2)
		\task[\textbf{A.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4.5cm]{diagram-20210910(14)-crop}
		\end{figure}
		\task[\textbf{B.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4.5cm]{diagram-20210910(15)-crop}
		\end{figure}
		\task[\textbf{C.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4.5cm]{diagram-20210910(16)-crop}
		\end{figure}
		\task[\textbf{D.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4.5cm]{diagram-20210910(17)-crop}
		\end{figure}
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\mathrm{U}=\mathrm{K}_{\mathrm{B}} \mathrm{T}^{2} \frac{\partial \ln \mathrm{z}}{\partial \mathrm{T}}, \quad \mathrm{C}_{\mathrm{v}}=\left(\frac{\partial \mathrm{U}}{\partial \mathrm{T}}\right)_{\mathrm{v}} \Rightarrow \mathrm{C}_{\mathrm{v}} \propto \mathrm{T}^{3}
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item From Q. no. 2, the pressure of the photon gas is
	{\exyear{GATE 2010}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{\pi^{2}\left(k_{B} T\right)^{3}}{15 \hbar^{3} C^{3}}$
		\task[\textbf{B.}]  $\frac{\pi^{2}\left(k_{B} T\right)^{4}}{8 \hbar^{3} C^{3}}$
		\task[\textbf{C.}] $\frac{\pi^{2}\left(k_{B} T\right)^{4}}{45 \hbar^{3} C^{3}}$
		\task[\textbf{D.}] $\frac{\pi^{2}\left(k_{B} T\right)^{3 / 2}}{45 \hbar^{3} C^{3}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text{Since, }P&=-\frac{\partial F}{\partial V} \Rightarrow P=K T\left(\frac{\partial \ln z}{\partial V}\right)_{T}\\&=\frac{\pi^{2}\left(k_{0} T\right)^{4}}{45 \hbar^{3} C^{3}}
		\end{align*}
		So the correct answer is \textbf{Option (C)}
	\end{answer}
	\item A system of $N$ non-interacting and distinguishable particle of spin 1 is in thermodynamic equilibrium. The entropy of the system is
	{\exyear{GATE 2011}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $2 k_{B} \ln N$
		\task[\textbf{B.}] $3 k_{B} \ln N$
		\task[\textbf{C.}] $N k_{B} \ln 2$
		\task[\textbf{D.}] $N k_{B} \ln 3$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\mathrm{S}=\mathrm{k}_{\mathrm{B}} \sum_{\mathrm{i}} \ln \Omega, \Omega&=3 \text{is number of microstate. }\mathrm{S}=1 ; \quad \mathrm{S}_{\mathrm{z}}=-1, \quad 0, \quad 1\\
		\text{	The entropy of the system is }& N k_{B} \ln 3
		\end{align*}
		So the correct answer is \textbf{Option (D)}
	\end{answer}
	\item A system has two energy levels with energies $\varepsilon$ and $2 \varepsilon .$ The lower level is 4 -fold degenerate while the upper level is doubly degenerate. If there are $N$ non-interacting classical particles in the system, which is in thermodynamic equilibrium at a temperature $T$, the fraction of particles in the upper level is
	{\exyear{GATE 2011}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{1}{1+e^{\varepsilon / k_{B} T}}$
		\task[\textbf{B.}] $\frac{1}{1+2 e^{\varepsilon / k_{B} T}}$
		\task[\textbf{C.}] $\frac{1}{2 e^{\varepsilon / k_{B} T}+4 e^{2 \varepsilon / k_{B} T}}$
		\task[\textbf{D.}] $\frac{1}{2 e^{\varepsilon / k_{B} T}-4 e^{2 \varepsilon / k_{B} T}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\text{Partition function }Z&=4 e^{-\epsilon / k T}+2 e^{-\epsilon / k T} \Rightarrow P(2 \varepsilon)\\&=\frac{2 e^{-2 \in / k T}}{4 e^{-\epsilon / k T}+2 e^{-2 \varepsilon / k T}}=\frac{1}{1+2 e^{\epsilon / k T}}
		\end{align*}
		So the correct answer is \textbf{Option (B)}
	\end{answer}
	\item Consider a system whose three energy levels are given by $0, \varepsilon$ and $2 \varepsilon$. The energy level $\varepsilon$ is two-fold degenerate and the other two are non-degenerate. The partition function of the system with $\beta=\frac{1}{k_{B} T}$ is given by
	{\exyear{GATE 2012}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $1+2 e^{-\beta \varepsilon}$
		\task[\textbf{B.}] $2 e^{-\beta \varepsilon}+e^{-2 \beta \varepsilon}$
		\task[\textbf{C.}] $\left(1+e^{-\beta \varepsilon}\right)^{2}$
		\task[\textbf{D.}] $1+e^{-\beta \varepsilon}+e^{-2 \beta \varepsilon}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		E_{1}&=0, E_{2}=\varepsilon, E_{3}=2 \varepsilon ; g_{1}=1, g_{2}=2, g_{3}=1\\&\text{ where} g_{1}, g_{2}\text{ and }g_{3}\text{ are degeneracy}\\
		\text{	The partition function }Z&=g_{1} e^{-\beta \cdot E_{1}}+g_{2} e^{-\beta \cdot E_{2}}+g_{3} e^{-\beta \cdot E_{3}}\\&=1+2 e^{-\beta c}+e^{-\beta 2 \varepsilon}=\left(1+e^{-\beta c}\right)^{2}
		\end{align*}
		So the correct answer is \textbf{Option (C)}
	\end{answer}
	\item Consider a linear collection of $N$ independent spin $1 / 2$ particles, each at a fixed location. The entropy of this system is $(k$ is the Boltzmann constant)
	{\exyear{GATE 2013}}
	\begin{tasks}(4)
		\task[\textbf{A.}] Zero
		\task[\textbf{B.}]  $N k$
		\task[\textbf{C.}]  $\frac{1}{2} N k$
		\task[\textbf{D.}] $N k \ln (2)$
	\end{tasks}
	\begin{answer}
		There are two microstates possible for spin $\frac{1}{2}$ particle, so entropy is given by $N k \ln (2)$\\\\
		So the correct answer is \textbf{Option (D)}
	\end{answer}
	\item Consider a system of $N$ non-interacting spin $-\frac{1}{2}$ particles, each having a magnetic moment $\mu$, is in a magnetic field $\vec{B}=B \hat{z} .$ If $E$ is the total energy of the system, then number of accessible microstates $\Omega$ is given by
	{\exyear{GATE 2015}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\Omega=\frac{N !}{\frac{1}{2}\left(N-\frac{E}{\mu B}\right) ! \frac{1}{2}\left(N+\frac{E}{\mu B}\right) !}$
		\task[\textbf{B.}] $\Omega=\frac{\left(N-\frac{E}{\mu B}\right) !}{\left(N+\frac{E}{\mu B}\right) !}$
		\task[\textbf{C.}] $\Omega=\frac{1}{2}\left(N-\frac{E}{\mu B}\right) ! \frac{1}{2}\left(N+\frac{E}{\mu B}\right) !$
		\task[\textbf{D.}] $\Omega=\frac{N !}{\left(N+\frac{E}{\mu B}\right) !}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\intertext{Number of microstate is ${ }^{N} C_{n_{1}}$, where $n_{1}$ is number of particle in $+\frac{1}{2}$ state and}
		n_{2}&=\left(N-n_{1}\right)\text{ is number of state in }-\frac{1}{2}\text{ state.}\\
		\text{	where }n_{1}&=\frac{1}{2}\left(N-\frac{E}{\mu B}\right), n_{2}=\frac{1}{2}\left(N+\frac{E}{\mu B}\right)\\
		\text{So, number of microstate}=&\Omega=\frac{1}{2}\left(N-\frac{E}{\mu B}\right) ! \frac{1}{2}\left(N+\frac{E}{\mu B}\right) !
		\end{align*}
	\end{answer}
	\item The average energy $U$ of a one dimensional quantum oscillator of frequency $\omega$ and in contact with a heat bath at temperature $T$ is given by
	{\exyear{GATE 2015}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $U=\frac{1}{2} \hbar \omega \operatorname{coth}\left(\frac{1}{2} \beta \hbar \omega\right)$
		\task[\textbf{B.}] $U=\frac{1}{2} \hbar \omega \sinh \left(\frac{1}{2} \beta \hbar \omega\right)$
		\task[\textbf{C.}] $U=\frac{1}{2} \hbar \omega \tanh \left(\frac{1}{2} \beta \hbar \omega\right)$
		\task[\textbf{D.}] $U=\frac{1}{2} \hbar \omega \cosh \left(\frac{1}{2} \beta \hbar \omega\right)$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\because Z&=\sum e^{-\beta E_{i}}=\sum_{i=0}^{\infty} e^{-\beta\left(n+\frac{1}{2}\right) \hbar \omega}\text{ where }E\\&=\left(n+\frac{1}{2}\right) \hbar \omega \Rightarrow Z=\frac{1}{2 \sinh \left(\frac{\beta \hbar \omega}{2}\right)}\\
		\because U&=\frac{-\partial}{\partial \beta} \ln Z=-\frac{\partial}{\partial \beta} \ln \left[\frac{1}{2 \sinh \left(\frac{\beta \hbar \omega}{2}\right)}\right]\\&=\frac{\hbar \omega}{2} \operatorname{coth}\left(\frac{\beta \hbar \omega}{2}\right)
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item  The entropy of a gas containing $N$ particles enclosed in a volume $V$ is given by $S=N k_{B} \ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)$, where $E$ is the total energy, $a$ is a constant and $k_{B}$ is the Boltzmann constant. The chemical potential $\mu$ of the system at a temperature $T$ is given by
	{\exyear{GATE 2015}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\mu=-k_{B} T\left[\ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)-\frac{5}{2}\right]$
		\task[\textbf{B.}] $\mu=-k_{B} T\left[\ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)-\frac{3}{2}\right]$
		\task[\textbf{C.}] $\mu=-k_{B} T\left[\ln \left(\frac{a V E^{3 / 2}}{N^{3 / 2}}\right)-\frac{5}{2}\right]$
		\task[\textbf{D.}] $\mu=-k_{B} T\left[\ln \left(\frac{a V E^{3 / 2}}{N^{3 / 2}}\right)-\frac{3}{2}\right]$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\left(\frac{\partial G}{\partial T}\right)_{P}&=-S=-N k_{B} \ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)\\ \because S&=N k_{B} \ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)\\
		\Rightarrow G&=-N k_{B} T \ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)+\ln A\\
		\Rightarrow \mu&=\left(\frac{\partial G}{\partial N}\right)=-\left[k_{B} T \ln \left(\frac{a V E^{3 / 2}}{N^{5 / 2}}\right)+N k_{B} T \frac{N^{5 / 2}}{a V E^{3 / 2}} \cdot \frac{(-5 / 2)}{N^{7 / 2}} a V E^{3 / 2}\right]\\&=-k_{B} T\left[\ln \left(\frac{a V E^{3 / 2}}{N^{\frac{5}{2}}}\right)-\frac{5}{2}\right]
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item entropy $S$ of a system of $N$ spins, which may align either in the upward or in the downward direction, is given by $S=-k_{B} N[p \ln p+(1-p) \operatorname{In}(1-p)] .$ Here $k_{B}$ is the Boltzmann constant. The probability of alignment in the upward direction is $p$. The value of $p$, at which the entropy is maximum, is---------- (Give your answer upto one decimal place)
	{\exyear{GATE 2016}}
	\begin{answer}
		\begin{align*}
		S&=-k_{B} N[p \ln p+(1-p) \operatorname{In}(1-p)]\\
		\text{	For maximum entropy, }\frac{d S}{d p}&=0 \Rightarrow \ln p+p \times \frac{1}{p}-\ln (1-p)+(1-p) \times \frac{1}{1-p}(-1)=0\\
		\ln p+1-\ln (1-p)-1&=0 \Rightarrow \ln \left(\frac{p}{1-p}\right)\\&=0 \Rightarrow p=1-p \Rightarrow p=0.5
		\end{align*}
	\end{answer}
	\item 	$N$ atoms of an ideal gas are enclosed in a container of volume $V$. The volume of the container is changed to $4 V$, while keeping the total energy constant. The change in the entropy of the gas, in units of $N k_{B} \ln 2$, is------ where $k_{B}$ is the Boltzmann constant.
	{\exyear{GATE 2016}}
	\begin{answer}
		\begin{align*}
		S_{1}&=-N k_{B} \ln 1, S_{2}=-N k_{B} \ln \frac{1}{4} \Rightarrow \Delta S\\&=S_{2}-S_{1}=N k_{B} \ln 4=2 N k_{B} \ln 2
		\end{align*}
	\end{answer}
	\item A two-level system has energies zero and $E$. The level with zero energy is nondegenerate, while the level with energy $E$ is triply degenerate. The mean energy of a classical particle in this system at a temperature $T$ is
	{\exyear{GATE 2016}}
	\begin{tasks}(4)
		\task[\textbf{A.}] (a) $\frac{E e^{\frac{-E}{k_{B} T}}}{1+3 e^{\frac{-E}{k_{B} T}}}$
		\task[\textbf{B.}] $\frac{E e^{\frac{-E}{k_{B} T}}}{1+e^{\frac{-E}{k_{s} T}}}$
		\task[\textbf{C.}] $\frac{3 E e^{\frac{-E}{k_{B} T}}}{1+e^{\frac{-E}{k_{B} T}}}$
		\task[\textbf{D.}] $\frac{3 E e^{\frac{-E}{k_{B} T}}}{1+3 e^{\frac{-E}{k_{s} T}}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\langle E\rangle&=\frac{\sum_{i} g_{i} E_{i} e^{-\frac{E_{i}}{k T}}}{\sum_{i} g_{i} e^{-\frac{E_{i}}{k T}}}=\frac{0 \times e^{-\frac{0}{k T}}+3 \times E \times e^{-\frac{E}{k T}}}{e^{-\frac{0}{k T}}+3 \times e^{-\frac{E}{k T}}}=\frac{3 E e^{\frac{-E}{k_{B} T}}}{1+3 e^{\frac{-E}{k_{B} T}}}
		\end{align*}
		So the correct answer is \textbf{Option (D)}
	\end{answer}
	\item Consider a triatomic molecule of the shape shown in the figure in three dimensions. The heat capacity of this molecule at high temperature (temperature much higher than the vibrational and rotational energy scales of the molecule but lower than its bond dissociation energies) is:
	{\exyear{GATE 2017}}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=4.5cm]{diagram-20210911(13)-crop}
	\end{figure}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{3}{2} k_{B}$
		\task[\textbf{B.}] $3 k_{B}$
		\task[\textbf{C.}] $\frac{9}{2} k_{B}$
		\task[\textbf{D.}] $6 k_{B}$
	\end{tasks}
	\begin{answer}
		If given molecules are at lower temperature i.e. atoms are attached to rigid rod then degree of freedom is 6 , so internal energy is $\frac{6 k_{B} T}{2}$, but at high temperature, vibration mode will active, so there are three extra vibration mode will active, so total energy
		\begin{align*}
		U&=3 k_{B} T+3 k_{B} T=6 k_{B} T\\
		C_{V}&=\left(\frac{\partial U}{\partial T}\right)_{V}=6 k_{B}
		\end{align*}
		So the correct answer is \textbf{Option (D)}
	\end{answer}
	\item Consider $N$ non- interacting, distinguishable particles in a two-level system at temperature $T$. The energies of the levels are 0 and $\varepsilon$, where $\varepsilon>0$. In the high temperature limit $\left(k_{B} T>\varepsilon\right)$, what is the population of particles in the level with energy $\varepsilon ?$
	{\exyear{GATE 2017}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{N}{2}$
		\task[\textbf{B.}] $N$
		\task[\textbf{C.}] $\frac{N}{4}$
		\task[\textbf{D.}] $\frac{3 N}{4}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		P(\varepsilon)&=\frac{\exp -\frac{\varepsilon}{k T}}{1+\exp -\frac{\varepsilon}{k T}},\text{ population of particle in the level with energy $\varepsilon$ is}\\
		N P(\varepsilon)&=N \frac{\exp -\frac{\varepsilon}{k T}}{1+\exp -\frac{\varepsilon}{k T}},\text{ for }\left(k_{B} T>\varepsilon\right), N P(\varepsilon)\\&=N \frac{\exp -\frac{\varepsilon}{k T}}{1+\exp -\frac{\varepsilon}{k T}}=N \frac{1}{1+1}=\frac{N}{2}
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item  A microcanonical ensemble consists of 12 atoms with each taking either energy 0 state, or energy $\in$ state. Both states are non-degenerate. If the total energy of this ensemble is $4 \in$, its entropy will be----------- $k_{B}$ (up to one decimal place), where $k_{B}$ is the Boltzmann constant.
	{\exyear{GATE 2018}}
	\begin{answer}
		\begin{align*}
		\intertext{The number of ways having total energy $4 \in$, out of 12 atom is}
		&={ }^{12} C_{4}=\frac{12}{\lfloor 48}=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2}=495\\
		\text{Hence, entropy, }S&=k_{B} \ln w=k_{B} \ln (495)=k_{B}(6.204)=6.204 k_{B}
		\end{align*}
	\end{answer}
	\item  The partition function of an ensemble at a temperature $T$ is
	$$
	Z=\left(2 \cosh \frac{\varepsilon}{k_{B} T}\right)^{N}
	$$
	where $k_{B}$ is the Boltzmann constant. The heat capacity of this ensemble at $T=\frac{\varepsilon}{k_{B}}$ is $X N k_{B}$, where the value of $X$ is (up to two decimal places).
	{\exyear{GATE 2018}}
	\begin{answer}
		\begin{align*}
		\text{	The partition function, }z&=\left[2 \cosh \left(\frac{\varepsilon}{k_{B} T}\right)\right]^{N}\\
		\text{The average energy, }\langle E\rangle&=k_{B} T^{2} \frac{\partial(\ln z)}{\partial T}\\
		&=\frac{N k_{B} T^{2}\left[2 \sinh \left(\frac{\varepsilon}{k_{B} T}\right)\right]\left(\frac{-\varepsilon}{k_{B} T^{2}}\right)}{2 \cosh \left(\frac{\varepsilon}{k_{B} T}\right)}\\&=-N \varepsilon \tanh \left(\frac{\varepsilon}{k_{B} T}\right)\\
		C&=\frac{d\langle E\rangle}{d T}=-N \varepsilon \sec h^{2}\left(\frac{\varepsilon}{k_{B} T}\right) \cdot\left(\frac{-\varepsilon}{k_{B} T^{2}}\right)\\
		\text{At }T&=\frac{\varepsilon}{k}, C=\frac{N \varepsilon^{2}}{k \cdot\left(\varepsilon^{2} / k^{2}\right)} \sec h^{2}(1)\\&=N k \sec h^{2}(1)=0.42 N k_{B}
		\end{align*}
	\end{answer}
	\item A collection of $N$ two-level systems with energies 0 and $E>0$ is in thermal
	equilibrium at temperature $T$. For $T \rightarrow \infty$, the specific heat approaches to,
	{\exyear{JEST 2012}}
	\begin{tasks}(4)
		\task[\textbf{A.}] 0
		\task[\textbf{B.}] $N k_{B}$
		\task[\textbf{C.}] $\frac{3 N k_{B}}{2}$
		\task[\textbf{D.}] $\infty$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		Z&=\sum e^{-\beta E_{i}}=e^{-\beta \times 0}+e^{-\beta E_{i}} \Rightarrow Z\\&=1+e^{-\beta E} \Rightarrow \ln z\\&=\ln \left(1+e^{-\beta E}\right)\\
		U&=\langle E\rangle=-\frac{\partial}{\partial \beta} \ln z\\&=-\frac{\partial}{\partial \beta} \ln \left(1+e^{-\beta E}\right)\\&=-\frac{1}{1+e^{-\beta E}} \times e^{-\beta E}(-E)\\&=\frac{E e^{-\beta E}}{1+e^{-\beta E}}\\
		\text{Now},\left(\frac{\partial U}{\partial T}\right)_{V}&=C_{V}\\
		&=\frac{\partial}{\partial T}\left(\frac{E e^{-\frac{E}{k T}}}{1+e^{-\frac{E}{k T}}}\right)\\
		\Rightarrow C_{V}&=\frac{\left(\frac{E^{2}}{k T^{2}} e^{\frac{-E}{k T}}+\frac{E^{2}}{k T^{2}} e^{\frac{-2 E}{k T}}-\frac{E^{2}}{k T^{2}} e^{\frac{-2 E}{k T}}\right)}{\left(1+e^{\frac{-E}{k T}}\right)^{2}} \Rightarrow C_{V}\\
		&=\left.\frac{\frac{E^{2}}{k T^{2}} e^{\frac{-E}{k T}}}{\left(1+e^{\frac{-E}{k T}}\right)^{2}} \Rightarrow C_{V}\right|_{T \rightarrow \infty}=0
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item  A monoatomic gas consists of atoms with two internal energy levels, ground state $E_{0}=0$ and an excited state $E_{1}=E$. The specific heat of the gas is given by
	{\exyear{JEST 2014}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\frac{3}{2} k$
		\task[\textbf{B.}] $\frac{E^{2} e^{E / k T}}{k T^{2}\left(1+e^{E / k T}\right)^{2}}$
		\task[\textbf{C.}] $\frac{3}{2} k+\frac{E^{2} e^{E / k T}}{k T^{2}\left(1+e^{E / k T}\right)^{2}}$
		\task[\textbf{D.}] $\frac{3}{2} k-\frac{E^{2} e^{E / k T}}{k T^{2}\left(1+e^{E / k T}\right)^{2}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		E_{0}&=0, \quad E_{1}=E\\
		\text{Then partition function is}\\
		z&=\sum e^{-\beta E_{i}} \Rightarrow z\\&=e^{-\beta \times 0}+e^{-\beta E} \Rightarrow \ln z\\&=\ln \left(1+e^{-\beta E_{1}}\right)\\
		U&=\langle E\rangle=\frac{-\partial}{\partial \beta} \ln z\\&=-\frac{\partial}{\partial \beta} \ln \left(1+e^{-\beta E}\right)\\&=-\frac{1}{\left(1+e^{-\beta E}\right)}(-E) e^{-\beta E}\\&=\frac{E e^{-\beta E}}{1+e^{-\beta E}} \quad\left[\because \beta=k_{B} T\right]\\
		\left(\frac{\partial U}{\partial T}\right)_{v}=C_{V}&=\frac{\left(1+e^{-\frac{E}{k_{B} T}}\right) E . e^{-\frac{E}{k_{B} T}} \cdot\left(\frac{E}{k_{B} T^{2}}\right)-E e^{-\frac{E}{k_{B} T}} \cdot e^{-\frac{E}{k_{B} T}}\left(\frac{E}{k_{B} T^{2}}\right)}{\left(1+e^{-\frac{E}{k_{B} T}}\right)^{2}}\\
		C_{V}&=\frac{\frac{E^{2}}{k_{B} T^{2}} e^{-\frac{E}{k_{\mathrm{B}} T}}+\frac{E^{2}}{k_{B} T^{2}} e^{-\frac{2 E}{k_{\mathrm{B}} T}}-\frac{E^{2}}{k_{B} T^{2}} e^{-\frac{2 E}{k_{\mathrm{B}} T}}}{\left(1+e^{-\frac{E}{k_{\mathrm{B}} T}}\right)^{2}}\\&=\frac{E^{2} e^{-\frac{E}{k_{\mathrm{B}} T}}}{k_{B} T^{2}\left(1+e^{-\frac{E}{k_{\mathrm{B}} T}}\right)^{2}}\\&=\frac{E^{2} e^{\frac{E}{k_{\mathrm{B}} T}}}{k_{B} T^{2}\left(1+e^{\frac{E}{k_{\mathrm{B}} T}}\right)^{2}}\\
		\text{If gas will classically allowed, then }C_{V}&=\frac{3}{2} k_{B}\\
		\text{	and quantum mechanically, }C_{V}&=\frac{E^{2} e^{\frac{E}{k_{B} T}}}{k_{B} T^{2}\left(1+e^{\frac{E}{k_{B} T}}\right)^{2}}\\
		\therefore \quad C_{V}&=\frac{3}{2} k_{B}+\frac{E^{2} e^{E / k T}}{k T^{2}\left(1+e^{E / k T}\right)^{2}}
		\end{align*}
		So the correct answer is \textbf{Option (C)}
	\end{answer}
	\item Consider a system of $2 N$ non-interacting spin $1 / 2$ particles each fixed in position and carrying a magnetic moment $\mu$. The system is immersed in a uniform magnetic field $B$. The number of spin up particles for which the entropy of the system will be maximum is
	{\exyear{JEST 2014}}
	\begin{tasks}(4)
		\task[\textbf{A.}]  0
		\task[\textbf{B.}] $N$
		\task[\textbf{C.}] $2 N$
		\task[\textbf{D.}] $N / 2$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\intertext{Let us consider $n$ number of spin out of $2 N$ particle have spin up remaining $2 N-n$ is down.}
		\text{Number of ways, }\omega&=\left\{\begin{array}{ll}2^{N} C_{n} & \text { for spin } \frac{1}{2}(\text { up }) \\ 2^{N} C_{2 N-n} & \text { for spin } \frac{1}{2}(\text { down })\end{array}\right.\\
		\text{Entropy, }S&=k \ln \omega \Rightarrow S\\&=k \ln { }^{2 N} C_{2 N-n}+k \ln { }^{2 N} C_{n}\\
		S&=k\left\{\left[\ln \frac{2 N !}{(n !)(2 N-n) !}\right]+\left[\ln \frac{2 N !}{(n !)(2 N-n) !}\right]\right\}\\
		S&=2 k[(\ln 2 N !-\ln n !-\ln (2 N-n) !)]\\
		S&=2 k[2 N \ln 2 N-2 N-n \ln n+n-\{(2 N-n) \ln (2 N-n)-(2 N-n)\}]\\
		[\because \ln N !&=N \ln N-N !]\\
		S&=2 k[2 N \ln 2 N-2 N-n \ln n+n-2 N \ln (2 N-n)+n \ln (2 N-n)+(2 N-n)]\\
		S&=2 k[2 N \ln 2 N-n \ln n-2 N \ln (2 N-n)+n \ln (2 N-n)]
		\intertext{Now for maximum entropy at equilibrium for spin $\frac{1}{2}$ up particle,}
		\frac{d S}{d n}&=0\\
		\frac{d S}{d n}&=2 k\left[-\frac{n}{n} \cdot 1-\ln n-\frac{2 N}{2 N-n}(-1)+\frac{n}{2 N-n}(-1)+\ln (2 N-n)\right]\\
		&=2 k\left[-1-\ln n+\frac{2 N}{2 N-n}-\frac{n}{2 N-n}+\ln (2 N-n)\right]\\
		&=2 k\left[-1+\frac{2 N-n}{2 N-n}+\ln (2 N-n)-\ln n\right]\\& \Rightarrow 2 k\left[-1+1+\ln \frac{(2 N-n)}{n}\right]=0\\
		\because \quad 2 k &\neq 0\\
		\therefore \ln \left(\frac{2 N-n}{n}\right)&=0 \Rightarrow \frac{2 N-n}{n}\\&=1 \Rightarrow 2 N=2 n \\
		\Rightarrow n&=N
		\end{align*}
		So the correct answer is \textbf{Option (B)}
	\end{answer}
	\item For a system in thermal equilibrium with a heat bath at temperature $T$, which one of the following equalities is correct? $\left(\beta=\frac{1}{k_{B} T}\right)$
	{\exyear{JEST 2015}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\frac{\partial}{\partial \beta}\langle E\rangle=\langle E\rangle^{2}-\left\langle E^{2}\right\rangle$
		\task[\textbf{B.}] $\frac{\partial}{\partial \beta}\langle E\rangle=\left\langle E^{2}\right\rangle-\langle E\rangle^{2}$
		\task[\textbf{C.}] $\frac{\partial}{\partial \beta}\langle E\rangle=\left\langle E^{2}\right\rangle+\langle E\rangle^{2}$
		\task[\textbf{D.}] $\frac{\partial}{\partial \beta}\langle E\rangle=-\left(\left\langle E^{2}\right\rangle+\langle E\rangle^{2}\right)$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\because\langle E\rangle&=\frac{\sum_{i} E_{i} e^{-\beta E_{i}}}{\sum_{i} e^{-\beta E_{i}}}\\
		\frac{\partial\langle E\rangle}{\partial \beta}&=-\frac{\sum_{i} E_{i}^{2} e^{-\beta E_{i}}}{\sum_{i} e^{-\beta E_{i}}}+\frac{\sum_{i} E_{i}^{2} e^{-\beta E_{i}} \cdot e^{-\beta E_{i}}}{\left(\sum_{i} e^{-\beta E_{i}}\right)^{2}}\\&=-\frac{\sum_{i} E_{i}^{2} e^{-\beta E_{i}}}{\sum_{i} e^{-\beta E_{i}}}+\frac{\sum_{i} E_{i}^{2} e^{-2 \beta E_{i}}}{\left(\sum_{i} e^{-\beta E_{i}}\right)^{2}}\\
		\Rightarrow \frac{\partial\langle E\rangle}{\partial \beta}&=\langle E\rangle^{2}-\left\langle E^{2}\right\rangle
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item A particle in thermal equilibrium has only 3 possible states with energies $-\in, 0, \in .$ If the system is maintained at a temperature, $T>>\frac{\epsilon}{k_{B}}$, then the average energy of the particle can be approximated to,
	{\exyear{JEST 2015}}
	\begin{tasks}(4)
		\task[\textbf{A.}] $\frac{2 \in^{2}}{3 k_{B} T}$
		\task[\textbf{B.}] $\frac{-2 \in^{2}}{3 k_{B} T}$
		\task[\textbf{C.}] $\frac{-\epsilon^{2}}{k_{B} T}$
		\task[\textbf{D.}]  0
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\langle E\rangle&=\frac{-\in e^{+\frac{\epsilon}{k T}}+0+\in e^{-\frac{\epsilon}{k T}}}{e^{\frac{\epsilon}{k T}}+1+e^{-\frac{\epsilon}{k T}}}\\&=\epsilon\left(\frac{e^{-\frac{\epsilon}{k T}}-e^{\frac{\epsilon}{k T}}}{1+e^{-\frac{\epsilon}{k T}}+e^{\frac{\epsilon}{k T}}}\right)\\
		\Rightarrow\langle E\rangle&=\frac{\left[\left(1-\frac{\epsilon}{k T}\right)-\left(1+\frac{\epsilon}{k T}\right)\right]}{1+\left(1-\frac{\epsilon}{k T}\right)+\left(1+\frac{\epsilon}{k T}\right)}\\&=\frac{-2 \epsilon^{2}}{3 k T}
		\end{align*}
		So the correct answer is \textbf{Option (B)}
	\end{answer}
	\item A collection of $N$ interacting magnetic moments, each of magnitude $\mu$, is subjected to a magnetic field $\mathrm{H}$ along the $\mathrm{z}$ direction. Each magnetic moment has a doubly degenerate level of energy zero and two non-degenerate levels of energies $-\mu H$ and $\mu H$ respectively.The collection is in thermal equilibrium at temperature $T$. The total energy $E(T, H)$ of the collection is
	{\exyear{JEST 2018}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $-\frac{\mu H N \sinh \left(\frac{\mu H}{k_{B} T}\right)}{1+\cosh \left(\frac{\mu H}{k_{b} T}\right)}$
		\task[\textbf{B.}] $-\frac{\mu H N}{2\left(1+\cosh \left(\frac{\mu H}{k_{b} T}\right)\right)}$
		\task[\textbf{C.}] $-\frac{\mu H N \cosh \left(\frac{\mu H}{k_{B} T}\right)}{1+\cosh \left(\frac{\mu H}{k_{b} T}\right)}$
		\task[\textbf{D.}] $-\mu H N \frac{\sinh \left(\frac{\mu H}{k_{B} T}\right)}{\cosh \left(\frac{\mu H}{k_{b} T}\right)}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		Z_{1}&=\left(2 \times \exp \left(-\frac{0}{k_{B} T}\right)+\exp \left(-\frac{-\mu H}{k_{B} T}\right)+\exp \left(-\frac{\mu H}{k_{B} T}\right)\right) \\\Rightarrow Z_{1}&=\left(2+2 \cosh \frac{\mu H}{k_{B} T}\right)\\
		Z_{N}&=\left(2+2 \cosh \frac{\mu H}{k_{B} T}\right)^{N}\\U&=k_{B} T^{2}\left(\frac{\partial \ln Z_{N}}{\partial T}\right)_{N, V}\\&=-\frac{N \mu H \sinh \left(\frac{\mu H}{k_{B} T}\right)}{1+\cos \frac{\mu H}{k_{B} T}}
		\end{align*}
		So the correct answer is \textbf{Option (A)}
	\end{answer}
	\item 	Consider a system of $N$ distinguishable particles with two energy levels for each particle, a ground state with energy zero and an excited state with energy $\varepsilon>0$. What is the average energy per particle as the system temperature $T \rightarrow \infty$ ?
	{\exyear{JEST 2019}}
	\begin{tasks}(4)
		\task[\textbf{A.}] 0
		\task[\textbf{B.}]  $\frac{\varepsilon}{2}$
		\task[\textbf{C.}] $\varepsilon$
		\task[\textbf{D.}] $\infty$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		\langle E\rangle&=\sum_{i} P_{i} E_{i} \Rightarrow P_{i}=\frac{e^{\beta E_{i}}}{z}\\
		\langle E\rangle&=0 \times \frac{01}{1+e^{-\beta \varepsilon}}+\varepsilon \times \frac{1}{1+e^{-\beta \varepsilon}}\\
		&=\frac{\varepsilon}{1+e^{-\varepsilon / k_{B} T}}=\frac{\varepsilon}{2}\text{`}T \rightarrow \infty
		\end{align*}
		So the correct answer is \textbf{Option (B)}
	\end{answer}
	\item Consider a diatomic molecule with an infinite number of equally spaced non-degenerate energy levels. The spacing between any two adjacent levels is $\varepsilon$ and the ground state energy is zero. What is the single particle partition function $Z$ ?
	{\exyear{JEST 2019}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $Z=\frac{1}{1-\frac{\varepsilon}{k_{B} T}}$
		\task[\textbf{B.}] $Z=\frac{1}{1-e^{\frac{\varepsilon}{k_{\mathrm{B}} T}}}$
		\task[\textbf{C.}] $Z=\frac{1}{1-e^{\frac{2 \varepsilon}{k_{B} T}}}$
		\task[\textbf{D.}] $Z=\frac{1-\frac{\varepsilon}{k_{B} T}}{1+\frac{\varepsilon}{k_{B} T}}$
	\end{tasks}
	\begin{answer}
		\begin{align*}
		Z&=\sum_{i} g_{i} e^{-\beta \varepsilon_{i}}\\
		g_{i}&=1\\
		Z&=1+e^{-\beta \varepsilon}+e^{-2 \beta \varepsilon}+\ldots \ldots\\
		Z&=\frac{1}{1-e^{-\beta \varepsilon}}
		\end{align*}
		No option is matched
	\end{answer}
\end{enumerate}

Statistical Mechanics/chapter/Statistical Physics-Pset1 Solutions.tex


\documentclass[12pt,thmsa]{article}

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\usepackage{graphicx} % Allows including images
\usepackage{booktabs} % Allows the use of \toprule, \midrule and \bottomrule in tables
\usepackage{cancel}
\usepackage{color}    % color
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\usepackage{amssymb}  %  \varnothing
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% Linhui added for newly defined color
\definecolor{forestgreen}{RGB}{34,139,34}

% Linhui added for Expectation and Variance
\newcommand{\Exp}{{\mathbb E}\! }
\newcommand{\Var}{\mbox{Var}\! }

% Linhui added for rename the command for empty set.
\let\oldemptyset\emptyset
\let\emptyset\varnothing

%------------------------------------------------------%
\makeatletter
\def\maketitle{%
	\par
	\hrule height 1.5pt\vspace{1ex}
	\par\noindent

	\begin{minipage}{0.5\textwidth}
		\scshape
		Purdue University $\cdot$ ece 60000 \\[1ex]
		Random Variables and Signals \\
		Prof. Bell, Prof. Salama
	\end{minipage}
	\begin{minipage}{0.45\textwidth}
		\raggedleft
		\MakeTextUppercase{{\@title}}\\[0.3ex] % 0.2ex height space between two line
		\textit{\@author}\\[0.2ex]
		\textit{August 31, 2021}
	\end{minipage}
	\par\vspace{1ex}
	\hrule height 1.5pt\vspace{1ex}
	\par
}
\makeatother

\author{Linhui Xie}
\title{Lecture Note 1}
%------------------------------------------------------%

\begin{document}
\maketitle

\section{Lecture 1\medskip}

%------------------------------------------------------%
\subsection{Set theory} % 1.1
A simple random experiment - roll a fair die,
\[\Omega=\mathcal{S}=\{1,2,3,4,5,6\}.\]
Also, \(\Omega, \mathcal{S}\) is \underline{sample space}.
\begin{itemize}
	\item Each event of interest can be describe by a subset of \(\Omega=\{1,2,3,4,5,6\}\).\\
    For example, \\
	\[\begin{aligned}
	A_{1} &= \text{ the outcome is odd } = \{1, 3, 5\}, \\
	A_{2} &= \text{ the outcome is divisible by 3 } = \{3, 6\}, \\
	A_{3} &= \text{ the outcome is prime } = \{2, 3, 5\}.
	\end{aligned}\]
	\item There are \(C_{6}^{0} + C_{6}^{1} + C_{6}^{2} + C_{6}^{3} + C_{6}^{4} + C_{6}^{5} + C_{6}^{6}=2^6=64\) distinct subsets of \(\Omega\).
	\item In order to fully characterize a random experiment, we must know the probability of each of these sets.
\end{itemize}

%------------------------------------------------------%
\subsection{Events} % 1.2
\begin{itemize}
	\item Events are subsets of \(\Omega\).
	\item The collection of all events is called the \underline{event space},
\[\mathcal{F}(\Omega)=\left\{A_{1}, {A}_{2}, \ldots, A_{64}\right\}.\]
	\item Our random experiment is completely characterized by
\[
\{\Omega, \mathcal{F}(\Omega), P(\cdot)\},
\]where\[
P(\cdot): \mathcal{F}(\Omega) \rightarrow[0,1],
\] and assigns probability to each of the events.
\end{itemize}

%------------------------------------------------------%
\subsection{Basic Set Theory} % 1.3
A set is simply a collection of objects. In any given problem, the set containing all possible elements of interest is called the universe, universal set, or space. We typically denote the space by \(\Omega\) or \(\mathcal{S}\).
\begin{itemize}
	\item The \underline{union} of two sets \(A\) and \(B\), denoted \(A \cup B\), is defined as \[
A \cup B =\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A {\color{blue}{\text { or }}} \omega \in B\}.
\]
    \begin{figure}[H] % The h stands for "here." You could also specify the bottom of the page by saying [b]
    \centering
    \includegraphics[width=.4\linewidth]{Figure/Note01_Figure_1.png}
    \end{figure}

	\item  The \underline{intersection} of two sets \(A\) and \(B\), denoted \(A \cap B\), is defined as \[
A \cap B=\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A {\color{blue}{\text { and }}} \omega \in B\}.
\]
    \begin{figure}[H] % The h stands for "here." You could also specify the bottom of the page by saying [b]
    \centering
    \includegraphics[width=.4\linewidth]{Figure/Note01_Figure_2.png}
    \end{figure}

	\item The \underline{complement} of a set \(A\) (with respect to \(\Omega\) ), denoted \(\overline{A}, A^{''}, A^{c}\), is defined as \[
\overline{A}=\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \notin  A\}.
\]
    \begin{figure}[H] % The h stands for "here." You could also specify the bottom of the page by saying [b]
    \centering
    \includegraphics[width=.4\linewidth]{Figure/Note01_Figure_3.png}
    \end{figure}

	\item The set containing no elements is called the \underline{empty set} or \underline{null set}, and is denoted by \(\emptyset\) or \(\{ \}\). (n.b. \(\{\emptyset\}\) is not correct notation.)

	\item  If two sets \(A\) and \(B\) have no elements in common, then \(A \cap B = \emptyset\), and \(A\) and \(B\) are said to be \underline{disjoint}.

	\item  The \underline{set difference} of two sets \(A\) and \(B\) is defined as
\[\begin{aligned}
A-B &=\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A {\color{blue}{\text { and }}} \omega \notin B\}\\
&= A \cap \overline{B}.
\end{aligned}
\]
    \begin{figure}[H] % The h stands for "here." You could also specify the bottom of the page by saying [b]
    \centering
    \includegraphics[width=.4\linewidth]{Figure/Note01_Figure_4.png}
    \end{figure}

	\item  The \underline{symmetric difference} of two sets \(A\) and \(B\), is defined as
\[\begin{aligned}
A \Delta B & =\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A {\color{blue}{\text { or }}} \omega \in B, \text { but not both. }\} \\
& = \left( A \cup B \right) - \left( A \cap B \right) \\
& = \left( A \cap \overline{B} \right) \cup \left( \overline{A} \cap B \right).
\end{aligned}
\]
    \begin{figure}[H] % The h stands for "here." You could also specify the bottom of the page by saying [b]
    \centering
    \includegraphics[width=.4\linewidth]{Figure/Note01_Figure_5.png}
    \end{figure}

	\item  Two sets \(A\) and \(B\) are  \underline{equal} if and only if they contain exactly the same elements. \\
	\(\Longleftrightarrow A=B \Leftrightarrow A \subset B \text{ and } B \subset A.\) \\
	Proof, \\
	\(\Longleftarrow\) \\
	If \(B \subset A\), then \(\omega \in B \Rightarrow \omega \in A \), and if \(A \subset B\), then \(\omega \in A \Rightarrow \omega \in B \). \\
	If both conditions hold, that means every element of \(B\) is an element of \(A\), and every element of \(A\) is an element of \(B\). (or say, \(A\) has no element that are not in \(B\) and \(B\) has no elements that are not in \(A\). ) \\
	\(A\) and \(B\) have exactly the same elements, \(A = B\).\\
	\(\Longrightarrow\) \\
	From the definition of the subset,
	\(A =\{ \omega \in \Omega: \omega \in A \}\),  \(B =\{ \omega \in \Omega: \omega \in B \}\). \\
	If \(A=B\), every element of \(A\) is an element of \(B\), \(A \subset B\). Similarly, every element of \(B\) is an element of \(A\), \(B \subset A\).
\end{itemize}

%------------------------------------------------------%
\subsection{Index sets \(\mathcal{I}\)} % 1.4
\begin{itemize}
	\item   Indexed collections of sets \( \{A_{i} ; i \in \mathcal{I}\} \), where \(\mathcal{I} \) is an index set.
	\item   So \( \{A_{i} ; i \in \mathcal{I}\} \) is a ``set of sets", or a \underline{family of sets}, or a \underline{collection of sets}.
	\item  There is one and only one set \( A_{i} \), for each \( i \in \mathcal{I} \).

For example,
\[\begin{array}{l}  \mathcal{I} = \{ 1,2,3\} , \\A_{1}=[0,1]=\{x \in \mathbb{R} ; 0 \leq x \leq 1\}, \\  A_{2}=[1,2] , \\  A_{3}=[2,3], \end{array}\]
\[ \{A_{i} ; i \in \mathcal{I}\} =\{ A_1, A_2, A_3  \} = \left\{ [0,1], [1,2], [2,3] \right\}, \]
\[\bigcup_{i \in \mathcal{I}} A_{i}=[0,1] \cup[1,2] \cup[2,3].\]
Typical Index sets:
\[\begin{array}{l}
	\mathbb{N}=\{1,2,3, \ldots\}=\text { natural numbers }, \\
	\mathbb{Z}=\{\ldots-1,0,1,2, \ldots\}=\text { integers }, \\
	\mathbb{Z}_{+}=\{0,1,2, \ldots\}=\text { non-negative integers }, \\
	\mathbb{R}=(-\infty, \infty), \\
	\mathbb{I}_{n}=\{0,1,2, \ldots, n-1\}.
\end{array}\]

For another example,
\[\mathcal{S}=\{1,2,3,4,5,6\}.\]
There are 64 subsets, \(A_{1}, A_{2}, \cdots, A_{64}\),
\[\left\{ A_{1}, A_{2}, \cdots, A_{64} \right\} \text{ is an indexed collection of sets.}\]

	\item  An index set \(\mathcal{I}\) is countable if it has an infinite number of elements and they can be put in one-to-one correspondance with the natural numbers \(\mathbb{I}\).

	\item  Given an indexed collection of sets \( \{A_{i} ; i \in \mathcal{I}\} \), the union of the family is \[ {\color{blue}{\bigcup_{i \in I}}} A_{i} \triangleq\left\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A_{i} \text { for } {\color{blue}{ \text{at least}}} \text{ one } i \in \mathcal{I} \right\}. \]
	the intersection of the family is
	\[ {\color{blue}{\bigcap_{i \in I}}} A_{i} \triangleq\left\{ {\color{forestgreen}{\omega \in \Omega}}: \omega \in A_{i} \text { for } {\color{blue}{ \text{all } }} i \in \mathcal{I} \right\}. \]
	For example, \\
	\(\text {Let } F_{r}=[0,1 / r), r \in(0,1] . \text { Find }\)
    \[ \bigcup_{r \in(0,1]} F_{r} \quad \text { and } \quad \bigcap_{r \in(0,1]} F_{r}.\]
    Short answer,
	\[ \begin{aligned}\bigcup_{r \in (0,1]} F_r
    &= \bigcup_{r \in (0,1]}[0,1 / r)
      =\big\{\omega: \omega \in[0,1 / r) \text { for } {\color{blue}{\text{at least}}} \text{ one } r \in(0,1]\big\}
      = [0,\infty). \\  \\
    \bigcap_{r \in (0,1]} F_r
    &= \bigcap_{r \in (0,1]}[0,1 / r)
      =\big\{\omega: \omega \in[0,1 / r) \text { for } {\color{blue}{\text{all }}} r \in(0,1]\big\}
      = [0,1).
    \end{aligned} \]
\end{itemize}

%------------------------------------------------------%
\subsection{Algebra of Set Theory} % 1.5
There are 16 rules,

\(\left.\begin{array}{l}\text { 1. } A \cup B=B \cup A \text { . } \\ \text { 2. } A \cap B=B \cap A \text { . }\end{array} \qquad \qquad \qquad \; \;  \right\}\) Commutative laws

\(\left.\begin{array}{l}\text { 3. } A \cup(B \cup C)=(A \cup B) \cup C. \\ \text { 4. } A \cap(B \cap C)=(A \cap B) \cap C .\end{array} \quad \quad  \right\}\) Associative laws

\(\left.\begin{array}{l}\text { 5. } A \cap(B \boldsymbol{\color{blue}{\cup}} C)=(A \cap B) \boldsymbol{\color{blue}{\cup}} (A \cap C). \\ \text { 6. } A \cup(B \boldsymbol{\color{blue}{\cap}} C)=(A \cup B) \boldsymbol{\color{blue}{\cap}} (A \cup C) .\end{array} \right\}\) Distributive laws

\( \begin{array}{l} \text{ 7. } \overline{\overline{A}}=A\end{array}. \)

\(\left.\begin{array}{l}
	\text { 8. } \overline{ {\color{black}{A}} \cap {\color{black}{B}} }=\overline{\color{black}{A}} \cup \overline{\color{black}{B}}. \\
	\text { 9. } \overline{ {\color{black}{A}} \cup {\color{black}{B}} }=\overline{\color{black}{A}} \cap \overline{\color{black}{B}}.\end{array} \qquad \qquad \qquad \quad \; \;\right\}\) De Morgan's Laws

 \(	\text { 10. }\overline{\mathcal{S}}=\varnothing. \)

 \(	\text { 11. }A \cap \mathcal{S}=A.\)

 \(	\text { 12. }A \cap \varnothing=\varnothing.\)

 \(	\text { 13. }A \cup \mathcal{S}= \mathcal{S}.\)

 \(	\text { 14. }A \cup \varnothing=A.\)

 \(	\text { 15. }A \cup \bar{A}= \mathcal{S}.\)

 \(	\text { 16. }A \cap \bar{A}=\varnothing.\)

\noindent
Proof:

1) Union is commutative.
\[
\begin{aligned}
	A \cup B & \triangleq\{x \in \Omega: \quad x \in A \text { or } x \in B\} \\
	&=\{x \in \Omega: \quad x \in B \text { or } x \in A\} \\
	&=B \cup A.
\end{aligned}
\]

\medskip

2) Intersection is commutative.
\[
\begin{aligned}
	A \cap B & \triangleq\{x \in \Omega: \quad x \in A \text { and } x \in B\} \\
	&=\{x \in \Omega: \quad x \in B \text { and } x \in A\} \\
	&=B \cap A.
\end{aligned}
\]

\medskip

3)  Union is associative.
\[
\begin{aligned}
	A \cup(B \cup C) &\triangleq \{x \in \Omega: \qquad \quad x \in A \text { or } x \in(B \cup C)\} \\
	&=\{x \in \Omega: \qquad \quad x \in A \text { or } x \in B \text { or } x \in C\} \\
	&=\{x \in \Omega: \; x \in(A \cup B) \text {  or } x \in C)\} \\
	&=(A \cup B) \cup C.
\end{aligned}
\]

\medskip

4) Intersection is associative.
\[
\begin{aligned}
	A \cap(B \cap C) &\triangleq \{x \in \Omega: \qquad \quad x \in A \text { and } x \in(B \cap C)\} \\
	&=\{x \in \Omega: \qquad \quad x \in A \text { and } x \in B \text { and } x \in C\} \\
	&=\{x \in \Omega: \; x \in(A \cap B) \text{  and } x \in C)\} \\
	&=(A \cap B) \cap C.
\end{aligned}
\]

\medskip

5) Intersection is distributive over union.

Let \(x \in A \cap(B \cup C) \). Then, \(x \in A\) and \(x \in(B \cup C)\),
\[ \Rightarrow x \in A \text{ and at the same time, } x \in B \text{ or } x \in C \text{, possibly both}, \] \[\Rightarrow \text{ either } x \in A \text{ and } x \in B \text{, or } x \in A  \text{ and } x \in C \text{ (possibly both).} \]

Hence, \[x \in(A \cap B) \text{ or } x \in(A \cap C),\]

i.e. \(x \in(A \cap B) \cup(A \cap C)\). So we have that \[
\begin{aligned}
x \in A \cap(B \cup C) & \Rightarrow x \in(A \cap B) \cup(A \cap C), \\
&\Rightarrow A \cap(B \cup C) \subset(A \cap B) \cup(A \cap C).
\end{aligned}\]

Next we assume that \(x \in(A \cap B) \cup(A \cap C) \). Then, \(x \in(A \cap B)\) or \(x \in(A \cap C)\),
\[ \begin{aligned} &\Rightarrow x \in A \text{ in addition to being in } B \text{ or } C \text{, or both,} \\
	&\Rightarrow x \in A \cap(B \cup C). \end{aligned}\]

This gives us that \[\begin{aligned}
x \in(A \cap B) \cup(A \cap C) &\Rightarrow x \in A \cap(B \cup C),\\
&\Rightarrow (A \cap B) \cup(A \cap C) \subset A \cap(B \cup C).
\end{aligned}\]

Combining the two results we have
\[A \cap(B \cup C) \subset(A \cap B) \cup(A \cap C) \text{ and } (A \cap B) \cup(A \cap C) \subset A \cap(B \cup C) \]
\[\Rightarrow A \cap(B \cup C)=(A \cap B) \cup(A \cap C).\]

\medskip

6)  Union is distributive over intersection.

Let \(x \in A \cup(B \cap C) \). Then, \(x \in A\) or \(x \in(B \cap C)\),
\[ \Rightarrow \text{ either }  x \in A \text{, or } x \in B \text{ and } x \in C \text{, possibly both}. \]

If \(x \in A\), we have \(x \in A \cup B \) and \(x \in A \cup C \) both satisfied. If \(x \in B \cap C\), we have \(x \in B \) and \(x \in C\). Then \(x \in A \cup B \) and \(x \in A \cup C \) both hold,
\[ \Rightarrow \text{ either }  x \in A \text{ or } x \in B \text{ and at the same time, either } x \in A \text{ or } x \in C, \]
\[\Rightarrow  x \in (A \cup B) \text{, and } x \in (A \cup C), \]

i.e. \(x \in(A \cup B) \cap(A \cup C)\). So we have that \[
\begin{aligned}
x \in A \cup(B \cap C) & \Rightarrow x \in(A \cup B) \cap(A \cup C), \\
&\Rightarrow A \cup(B \cap C) \subset(A \cup B) \cap(A \cup C).
\end{aligned}\]

Next we assume that \(x \in(A \cup B) \cap(A \cup C) \). Then, \(x \in(A \cup B)\) and \(x \in(A \cup C)\). If \(x \in A\), \(x \in A \cup (B \cap C)\) holds. If \(x \notin A\), since \(x \in(A \cup B) \) and \(x \in(A \cup C) \), we have \(x \in B\) and \(x \in C\). It gives \(x \in(B \cap C)\), related on \(x \in A \cup (B \cap C)\).

\[\begin{aligned}
x \in(A \cup B) \cap(A \cup C) &\Rightarrow x \in A \cup(B \cap C),\\
&\Rightarrow (A \cup B) \cap(A \cup C) \subset A \cup(B \cap C).
\end{aligned}\]

Combining the two results we have
\[A \cup(B \cap C) \subset(A \cup B) \cap(A \cup C) \text{ and } (A \cup B) \cap(A \cup C) \subset A \cup(B \cap C) \]
\[\Rightarrow A \cup(B \cap C)=(A \cup B) \cap(A \cup C).\]

\medskip

7)
\[\begin{aligned}
\overline{\overline{A}} &=\{ x \in \Omega: \omega \notin  \overline{A}\} \\
& = \{ x \in \Omega: \omega \in A\} \\
& = A
\end{aligned}
\]

\medskip

8) De Morgan's Laws
\[
\begin{aligned}
    x \in \overline{A \cap B} &\Leftrightarrow x \notin A \cap B, & & \text{ By definition of complement.}\\
    &\Leftrightarrow x \in \overline{A} \text { or } x \in \overline{B}, & & \text{ By definition of intersection and complement.}\\
    &\Leftrightarrow  x \in \overline{A} \cup \overline{B},  & & \text{ By definition of union.}\\
    \overline{A \cap B} & = \overline{A} \cup \overline{B} .
\end{aligned}
\]

\medskip

9) De Morgan's Laws
\[
\begin{aligned}
    x \in \overline{A \cup B} &\Leftrightarrow x \notin A \text{ and } x \notin B, & & \text{ By definition of union and complement.}\\
    &\Leftrightarrow x \in \overline{A} \text { and } x \in \overline{B}, & & \text{ By definition of complement.}\\
    &\Leftrightarrow  x \in \overline{A} \cap \overline{B}, & & \text{ By definition of intersection.} \\
    \overline{A \cup B} & = \overline{A} \cap \overline{B}.
\end{aligned}
\]

\medskip

10)Here \(\mathcal{S}\) is the universal set. Based on the definition,
\[
\overline{\mathcal{S}} =\{ \omega \in \Omega ( \Omega= \mathcal{S}): \omega \notin  \mathcal{S} \}.
\]
The complement of the set \(\mathcal{S}\) is the difference between the universal set \(\mathcal{S}\) and set \(\mathcal{S}\) itself. Obviously, no element satisfies this statement. The set containing no elements is empty set. Hence, \(\overline{\mathcal{S}} = \varnothing\).

\medskip

11)
Here \(\mathcal{S}\) is the universal set.
First we show that \(A \cap  \mathcal{S} \subset A\). In general, the set resulting from the intersection of two sets is a subset of both of the sets. Let \(x \in A \cap  B\), \(x\) is an element of \(A \cap  B\).
\[\begin{aligned}
 x \in A \cap  B & \Rightarrow x \in A \text{ and } x \in B, \qquad \text{ By definition of intersection.}\\
 & \Rightarrow x \in A, \\
 & \Rightarrow  A \cap B \subset A.
\end{aligned}\]
Next, we want to show that \( A \subset A \cap  \mathcal{S} \).
Let \(x \in A\), then \(x \in  \mathcal{S}\) such that \(x \in A \). Therefore, \(x \in A \Rightarrow x \in(A \cap \mathcal{S}) \Rightarrow A \subset A \cap  \mathcal{S} \).
Since \(A \cap \mathcal{S} \subset \mathrm{A}\) and \(\mathrm{A} \subset \mathrm{A} \cap \mathcal{S},\) we have that \(\mathrm{A} \cap \mathcal{S}=\mathrm{A}\).

\medskip

12) First we show that \(A \cap  \varnothing \subset \varnothing\). We have shown in 11) that the set resulting from the intersection of two sets is a subset of both of the sets. Hence, \(A \cap  \varnothing \subset \varnothing\) holds.

Next, we want to show that \( \varnothing \subset A \cap  \varnothing \). Since every set contains the set itself and the empty set, the \( \varnothing \) is subset of a set resulting from \(A \cap  \varnothing \), i.e., \( \varnothing \subset A \cap  \varnothing \).

Since \(A \cap  \varnothing \subset \varnothing\) and \( \varnothing \subset A \cap  \varnothing \), we have \( A \cap  \varnothing = \varnothing \).

\medskip

13)
Here \(\mathcal{S}\) is the universal set.
First we show that \(\mathcal{S} \subset A \cup  \mathcal{S}\). In general, both of the sets are subsets of the set resulting from the union of two sets. Let \(x \in B \), \(x\) is an element of \(B \).
\[\begin{aligned}
 x \in B & \Rightarrow x \in A \text{ or } x \in B, \\
 & \Rightarrow x \in A \cup B, \qquad \quad \text{ By definition of union.}\\
 & \Rightarrow B \subset A \cup B.
\end{aligned}\]

Next, we want to show that \( A \cup  \mathcal{S} \subset \mathcal{S} \).
Let \(x \in A \cup  \mathcal{S}\), then \(x \in  \mathcal{S}\) such that \(x \in A \text{ or } x \in \mathcal{S} \).
\[\begin{aligned}
 x \in A \cup  \mathcal{S} & \Rightarrow x \in A \text{ or } x \in \mathcal{S}, & & \text{ By definition of union.}\\
 & \Rightarrow x \in \mathcal{S},  & & \text{ By the case that } A \text{ is subset of } \mathcal{S}.\\
 & \Rightarrow A \cup  \mathcal{S} \subset \mathcal{S}.
\end{aligned}\]

Since \(\mathcal{S} \subset \mathrm{A} \cup \mathcal{S}\) and \(A \cup \mathcal{S} \subset \mathcal{S}\), we have that \(\mathrm{A} \cup \mathcal{S}=\mathcal{S}\).

\medskip

14)
First we show that \(A \subset A \cup \varnothing\). We have shown in 13) that both of the sets are subsets of the set resulting from the union of two sets. Hence, \(A \subset A \cup \varnothing\) holds.

Next, we want to show that \( A \cup \varnothing \subset A \). Let \( x \in A \cup \varnothing\), we have \( x \in A\) or \( x \in \varnothing\). Since no element exists in empty set, \( x \in \varnothing\) is logically false. Hence, only \( x \in A\) holds. It results in \( x \in A \cup \varnothing \Rightarrow x \in A\).

Since \(A \subset A \cup \varnothing\) and \( A \cup \varnothing \subset A \), we have \( A \cup  \varnothing = A \).

\medskip

15)
First we show that \(A \cup \overline{A} \subset \mathcal{S}\). We have shown in 13) that both of the sets are subsets of the set resulting from the union of two sets. Hence, \(A \subset A \cup \overline{A} \Rightarrow A \subset \mathcal{S}\), and \(\overline{A} \subset A \cup \overline{A} \Rightarrow \overline{A} \subset \mathcal{S}\). Let \(x \in A \cup \overline{A} \), \(x \in A\) or \( x \in \overline{A} \), we all have \(x \in \mathcal{S}\). \(A \cup \overline{A} \subset \mathcal{S}\) holds.

Next, let \(x \in \mathcal{S}\), for the statements \(x\) is an element of \(A\), and \(x\) is not an element of \(A\), only one statement is true. Hence, either \(x \in A\) is true or \(x \notin A\) is true, i.e.,
\(x \in \mathcal{S} \Rightarrow x \in A  \text{ or } x \notin A\). Hence, \(\mathcal{S} \subset A \cup \overline{A} \).

Since \(A \cup \overline{A} \subset \mathcal{S}\) and \(\mathcal{S} \subset A \cup \overline{A} \), we have \( A \cup \overline{A} = \mathcal{S} \).

\medskip

16)
First we show that \(A \cap  \overline{A} \subset \varnothing\). Let \(x \in A \cap  \overline{A}\),
\[\begin{aligned}
x \in A \cap  \overline{A} & \Rightarrow x \in A \text{ and } x \in  \overline{A} \\
& \Rightarrow x \in A \text{ and } x \notin A \\
& \Rightarrow \text{no element satisfies both statement. }
\end{aligned}\]
Since no element exists in empty set, such statement result in a set contain no element, i.e., \(A \cap  \overline{A} \subset \varnothing\) holds.

Next, we want to show that \( \varnothing \subset A \cap  \overline{A} \). Since every set contains the set itself and the empty set, the \( \varnothing \) is subset of a set resulting from \(A \cap  \overline{A} \), i.e., \( \varnothing \subset A \cap  \overline{A} \).

Since \(A \cap  \overline{A} \subset \varnothing\) and \( \varnothing \subset A \cap \overline{A} \), we have \( A \cap  \overline{A} = \varnothing \).
\bigskip

\noindent
[Ref]:

R. G. Bartle, D. R. Sherbert, ``Sets and Functions" in ``Introduction to Real Analysis", 3rd Edition, John Wiley and Sons, Inc. 2000. ch 1, pp 3.

W. Rudin, ``Basic Topology" in ``Principles of Mathematical Analysis", 3rd Edition, McGraw-Hill Inc. ch 2, pp 28.

\end{document}


\providecommand{\main}{.}
\input{\main/../../general/packages.tex}
\usepackage{booktabs}
\begin{document}

\input{\main/../../general/title.tex}

{\hypersetup{hidelinks}
	\tableofcontents  % Generates the table of contents
}
\part{Dataset}
\chapter{Data points}
First we begin looking at the dataset, the distributions of the given metrics and the statistical analysis of these data points.

\section{Retrieving the dataset}
The dataset can be downloaded from \url{https://homes.di.unimi.it/valentini/ProgettoBioinformatica1718/data/}.

\section{Composition}
\subsection{Training dataset}
In the training dataset there are 981388 data points, each one comprised of 26 metrics. The first 356 are pathogenic and all the others are negative.

\subsection{Testing dataset}
In the test dataset there are 19018 data points, still each one comprised of 26 metrics. The first 40 are pathogenic and the following are negative.

\chapter{Metrics}
\section{How the graphs are realized}
All the graphs are in triples: positives, negatives and mixed.

The normalization is done, as usual, in the following way:

\begin{figure}
	\[
		m' = \frac{\text{metric} - \mean{\text{metric values}}}{\max\crl{\text{metric values}}- \min\crl{\text{metric values}}}
	\]
	\caption{Input normalization}
\end{figure}

\subsection{Metric sample distribution}
Are realized by calculating the frequencies and estimating the density distributions parameters via MLE.

\subsection{Plot graphs}
Plot graphs are realized by sorting the values of the single metrics.

\subsection{Normalized plot graphs}
Are realized by sorting the values of the metric, with the domain and codomain normalized.

\subfile{\main/chapters/datapoints_plots}
\subfile{\main/chapters/metric_distributions}

\chapter{Data correlation}
We now proceed to try and identify eventual data correlations.

\subfile{\main/chapters/scatter_plot}
\clearpage
\subfile{\main/chapters/correlation_coefficient}
\clearpage
\subfile{\main/chapters/correlations}
\subfile{\main/chapters/data_correlations_table}
\clearpage
\subfile{\main/chapters/less_correlated_correlation_matrix}

\chapter{Dataset visualization}
After having removed the correlated metrics we can proceed to use techniques of dimensionality reduction for visualization to see if the dataset is valid for a clustering or machine learning approach:

\section{PCA}
\subsection{Training dataset visualization}
The positive data are clustered inside the negative data: a simple clustering approach would, most probably, not be enough for separating the two classes, but easily separable for any multi-parameters ML approach like networks.

The points isolated on the right are most probably errors in the realization of the dataset.

\begin{figure}
	\includegraphics[width=0.5\textwidth]{pca/training}
	\caption{Training dataset visualization}
\end{figure}

\subsection{Testing dataset visualization}
The testing dataset is probably malformed: all the positives point are visibly clustered. A simple clustering approach such as K-Means could probably separate most of them successfully.

An ML approach with a network will probably reach sooner an high accuracy on the testing dataset than on the training dataset for this reason, being able to classify most of the positive points immediately.

\begin{figure}
	\includegraphics[width=0.5\textwidth]{pca/testing}
	\caption{Testing dataset visualization}
\end{figure}

\subsection{Mixed dataset visualization}
The two datasets overlap correctly.
\begin{figure}
	\includegraphics[width=0.5\textwidth]{pca/mixed}
	\caption{Mixed dataset visualization}
\end{figure}

\section{TSNE}

\subsection{Training dataset visualization}
Implementation exists but it is not runnable on the current machine in a decent time.

\subsection{Testing dataset visualization}
With TSNE visualization, after 1000 iterations, the training dataset positive class appears on the border right of the cluster.

\begin{figure}
	\includegraphics[width=0.5\textwidth]{tsne/testing}
	\caption{Testing dataset visualization}
\end{figure}

\subsection{Mixed dataset visualization}
Implementation exists but it is not runnable on the current machine in a decent time.

\chapter{Dataset issues}

\section{Possible dataset errors}
The PCA visualization shows points extremely out of the dataset cluster. It is possible that these points are errors.

\section{Biased testing dataset}
The testing dataset does not seem to reflect the training dataset distribution, but agglomerates the two classes in two separated clusters. It is therefore unhelpful when calculating the classification success in networks, as the training dataset result harder to classify that the testing one.

\subsection{Experimental proof}
To test if the testing dataset really is unhelpful in validating the network results I have tried to train an extremely small network (2 layers, 3 neurons each). The results after 300 epochs, trained on the usual training set, show a considerable success on the testing set while still low success on the training set.

\begin{figure}
	\includegraphics[width=\textwidth]{doubts.png}
	\caption{Quick ``overfitting'' on test set}
\end{figure}

\part{Network implementation}
\chapter{Model architecture}
Based on the informations extracted from the dataset structure, a small network should work best.

\begin{table}
	\begin{tabular}{|c|c|c|}
		\hline
		\textbf{Layer (type)}      & \textbf{Output Shape} & \textbf{Parameters number} \\
		\hline
		Input dense layer (Dense)  & (None, 22)            & 506                        \\
		\hline
		Input SeLU (Activation)    & (None, 22)            & 0                          \\
		\hline
		Hidden dense layer (Dense) & (None, 22)            & 506                        \\
		\hline
		Hidden SeLU (Activation)   & (None, 22)            & 0                          \\
		\hline
		Output dense layer (Dense) & (None, 1)             & 23                         \\
		\hline
		Sigmoid (Activation)       & (None, 1)             & 0                          \\
		\hline
	\end{tabular}
	\caption{Model summary}
\end{table}

\section{Activation function}
We'll be using the \textbf{SeLU} (\ref{selu}) activation function for most of the layers for its property of auto-normalization.

\begin{figure}
	\[
		\text{SeLU}(x) = \lambda \begin{cases}
			x                 & x > 0    \\
			\alpha e^x-\alpha & x \leq 0
		\end{cases}
	\]
	\caption{SeLU}
\end{figure}

\section{Dense layers}
Dense layers will be used through all the network:
\begin{figure}
	\includegraphics[width=0.3\textwidth]{dense_connected}
	\caption{Dense connected layer}
\end{figure}

\section{Input}
The input is a dense layer with 22 neurons, expecting as input the normalized metrics. As activation function \textbf{SeLU} will be used.

\section{Hidden layer}
A single hidden dense hidden layer with 22 neurons will be used, with \textbf{SeLU} as activation function.

\section{Drop out}
For the small dimension of the network, no drop out will probably be necessary.

\section{Output}
The output layer of the network is a dense layer with one neuron and a \textbf{sigmoid} as activation function. When \textbf{active} models the positive class, and when \textbf{inactive} models the negative class.

\begin{figure}
	\[
		\sigma(x) = \text{sigmoid}(x) = \frac{e^x}{e^x + 1}
	\]
	\caption{Sigmoid}
\end{figure}

\section{Batch Size}
Since the positive class is extremely smaller of the negative class a large batch size will be necessary: 4096 data points per batch allow for a probability that at least one of the data points is positive in the training set of the \(\approx0.77\).

\section{Loss function}
Since the task assigned to the network is a binary classification the loss function will be the \textbf{cross entropy}.

\section{Update policy}
As update policy we are going to use a form of gradient \textbf{back propagation} with \textbf{Adam} optimizer (\ref{adam}).

\section{Mathematical representation}
Since the network is extremely simple a short mathematical representation is possible:

\begin{figure}
	\[
		\text{prediction}(\bmx) = \sigma(\matr{W}_{output}\cdot\text{SeLU}(\matr{W}_{hidden}\cdot\text{SeLU}(\matr{W}_{input}\cdot\bmx)))
	\]
	\caption{Mathematical model}
\end{figure}

\section{Network model representation}
Graphical model of the network.
\begin{figure}
	\includegraphics[width=0.7\textwidth]{network}
	\caption{Model of the network}
\end{figure}

\chapter{Results}

These are the results obtained after training the network for 100 epochs.

\section{Accuracy and Loss}
\begin{table}
	\begin{tabular}{|c|c|c|}
		\hline
		\textbf{Metric} & \textbf{Training}     & \textbf{Testing}      \\
		\hline
		Accuracy        & 0.9998970845374103    & 0.9995267641182038    \\
		\hline
		Loss            & 0.0004865645350134752 & 0.0018388369186131115 \\
		\hline
	\end{tabular}
	\caption{Accuracy and Loss after 100 epochs}
\end{table}

\begin{figure}
	\includegraphics[width=0.7\textwidth]{results/accuracy_loss}
	\caption{Accuracy and Loss after 100 epochs}
\end{figure}

\section{ROC, AUROC, PRC and AUPRC}
\begin{table}
	\begin{tabular}{|c|c|c|}
		\hline
		\textbf{Metric} & \textbf{Training}  & \textbf{Testing}   \\
		\hline
		AUROC           & 0.9982423662593878 & 0.9944672779007271 \\
		\hline
		AUPRC           & 0.8780726019818784 & 0.9337718051372246 \\
		\hline
	\end{tabular}
	\caption{Accuracy and Loss after 100 epochs}
\end{table}

\begin{figure}
	\includegraphics[width=0.7\textwidth]{results/ROC_PRC}
	\caption{ROC, AUROC, PRC and AUPRC after 100 epochs}
\end{figure}

\section{Confusion matrices}

\subsection{Training confusion matrix}
\begin{figure}
	\includegraphics[width=0.7\textwidth]{results/training_confusion}
	\caption{Training confusion matrix after 100 epochs}
\end{figure}

\subsection{Testing confusion matrix}
\begin{figure}
	\includegraphics[width=0.7\textwidth]{results/testing_confusion}
	\caption{Testing confusion matrix after 100 epochs}
\end{figure}

\section{Prediction speed}
The network classifies a data point in \(375 \mu s \pm 108 \mu s\).

This value was computer on a machine with the following carattestics:

\begin{table}
	\begin{tabular}{|c|c|}
		\hline
		\textbf{Computer Specification}       \\
		\hline
		Model Name            & MacBook Pro   \\
		\hline
		Processor Name        & Intel Core i7 \\
		\hline
		Processor Speed       & 2.3 GHz       \\
		\hline
		Number of Processors  & 1             \\
		\hline
		Total Number of Cores & 4             \\
		\hline
		L2 Cache (per Core)   & 256 KB        \\
		\hline
		L3 Cache              & 6 MB          \\
		\hline
		Memory                & 16 GB         \\
		\hline
	\end{tabular}
	\caption{Computer specifications}
\end{table}

\part{References}

\chapter{References}

\begin{description}
	\item[Adam] Adam: A Method for Stochastic Optimization: \url{https://arxiv.org/abs/1412.6980} \label{adam}
	\item[SeLU] Self-Normalizing Neural Networks: \url{https://arxiv.org/abs/1706.02515} \label{selu}
\end{description}

\input{\main/../../general/footer.tex}

\end{document}