\documentclass[11pt,fleqn]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Page setup
\topmargin -1.5cm
\oddsidemargin -0.04cm
\evensidemargin -0.04cm
\textwidth 16.59cm
\textheight 24cm
\setlength{\parindent}{0cm}
\setlength{\parskip}{0cm}
% Fonts
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}
% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}
% Mass symbol
\newcommand{\mass}{\mathsf{m}}
\newcommand{\Mass}{\mathsf{M}}
% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}}
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}
% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}
% heading
\newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}}
\newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\heading{Bosons with Spin in magnetic field}
\auname{Shir Shenkar}
\exnumber{3021}
{\bf The problem:}
$N$ Bosons that have mass $m$ and spin 1 are placed
in a box that has volume $V$. A magnetic field $B$ is
applied, such that the interaction is $-\gamma B S_z$,
where $S_z=1,0,-1$ and $\gamma$ is the gyromagnetic ratio.
In items (c-f) assume the Boltzmann approximation
for the occupation of the $S_z\ne1$ states.
\Dn
(a) Find an equation for the condensation temperature $T_c$.
\Dn
(b) Find the condensation temperature $T_c(B)$ for $B=0$
and for $B\rightarrow \infty$.
\Dn
(c) Find the critical $B$ for condensation if $T$ is set
in the range of temperatures that has been defined in item(b).
\Dn
(d) Describe how $T_c(B)$ depends of $B$ in a qualitative manner.
Find approximate expressions for moderate and large fields.
\Dn
(e) Find the condensate fraction as a function of $T$ and $B$.
\Dn
(f) Find the heat capacity of the gas assuming large but finite field.
\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn
(a)
First, we can see that the Hamiltonian is $\mathcal{H}=\frac{p^2}{2m}-\gamma B S_z$, where $S_z=-1,0,1$. Therefore, we can treat the particles of the system like three different gasses.
\Dn
From the lecture we know that \[N=\sum_r f(\epsilon_r-\mu)\]
So for our case we get:
\[
N=\underbrace{\sum_p\frac{1}{e^{\frac{\beta p^2}{2m}-\beta\gamma B-\beta \mu}-1}}_{S_z=+1}
+\underbrace{\sum_p\frac{1}{e^{\frac{\beta p^2}{2m}-\beta \mu}-1}}_{S_z=0}
+\underbrace{\sum_p\frac{1}{e^{\frac{\beta p^2}{2m}+\beta\gamma B-\beta\mu}-1}}_{S_z=-1}
\]
So n equals to
\[
n=\frac{N}{V}=\frac{1}{\lambda_T ^{3}}\left[Li_{3/2}( e^{\beta \gamma B+\beta\mu})+Li_{3/2}(e^{\beta\mu})+Li_{3/2}(e^{-\beta \gamma B+\beta\mu})\right]
\]
The lowest energy state is for $S_z=+1$ particles. At the condensation we get
\[e^{\beta \gamma B+\beta\mu}=1\Rightarrow e^{\beta\mu} =e^{-\beta \gamma B} \Rightarrow \mu=-\gamma B.\]
At $T
\]
(b)
For $B=0$, n is
\[
n=\frac{3}{\lambda_T ^{3}}Li_{3/2}(1)=3Li_{3/2}(1)\cdot \left(\frac{m T_c}{2\pi}\right)^{3/2}
\]
So the condensation temperature is
\[
T_c= \left (\frac{n}{3\cdot 2.612} \right )^{2/3}\cdot \frac{2\pi}{m}
\]
*Note that $Li_{3/2}(1)\approx 2.612$.
\Dn
For $B\rightarrow \infty$:
\Dn
We have $\gamma B\gg T$, so the terms containing $e^{-2\beta \gamma B}$ are going to zero.
Hence, we get
\[
n\approx \frac{1}{\lambda_T ^{3}} Li_{3/2}(1)
\]
So the condensation temperature is
\[
T_{c}\approx \left ( \frac{n}{2.612} \right )^{2/3}\cdot \frac{2\pi}{m}
\]
\Dn
(c+d)
For the next items we can assume the Boltzmann approximation for the cases of $S_z\neq 1$: $\gamma B\gg T$.
\Dn
If we calculate the ratio between the condensation temperature $T_c(B)$ for $B=0$
and for $B\rightarrow \infty$, we get the following number
\[
\frac{T_{c}^{B=0}}{T_{c}^{B\rightarrow \infty}}\approx \frac{1}{3^{2/3}}
\]
As $B$ is increased, $T_c$ rises until $B_c$ is reached. At $B=B_c$, $T=T_c$ and the condensation occurs. Then we have from Boltzmann
\[
n\lambda_T ^{3}=Li_{3/2}(1)+Li{3/2}(e^{-\beta \gamma B_c})\approx Li_{3/2}(1)+e^{-\beta \gamma B_c}
\]
Notice we have neglected the third term in $n$, because it is of the second order in $e^{\beta \gamma B}$ and we only need the first.\Dn
Hence the critical magnetic field is
\[
B_c=\frac{-T}{\gamma }ln\left ( n\lambda_T ^{3}-Li_{3/2}(1) \right )\approx \frac{-T}{\gamma}ln \left(n\lambda_T^{3}-2.612 \right)
\]
(e)
We need to calculate $\frac{}{n}$. We have already showed that
\[
n=\frac{1}{\lambda_T ^{3}}\left ( Li_{3/2}(1)+e^{-\beta \gamma B} \right )+
\]
So the condensate fraction as function of $T$ and $B$ is
\[
\frac{}{n}=1-\frac{Li_{3/2}(1)+e^{-\beta \gamma B}}{n\lambda_T ^{3}}
\]
(f)
In order to calculate the heat capacity of the gas, we first need to write an expression for the energy. Once again, we continue to look at the system as 3 different gasses, and again we can assume $\gamma B\gg T$. We get
\[
\frac{E}{V}=\frac{3}{2}\cdot \frac{T}{\lambda_T^{3}}\left(Li_{5/2}(1)+e^{-\beta \gamma B}\right)=\frac{3}{2} \left(\frac{m}{2\pi}\right)^{3/2}\cdot T^{5/2}\left(Li_{5/2}(1)+e^{-\frac{\gamma B}{T}}\right)
\]
\Dn
Now, to get the heat capacity we differentiate with respect to $T$.
\[
C_V=\frac{\partial}{\partial T}\left( \frac{E}{V}\right)= \frac{15}{4}\cdot \frac{1}{\lambda_T ^{3}}\left(Li_{5/2}(1)+e^{-\beta \gamma B}\right)+\frac{3}{2}\cdot \frac{1}{\lambda ^{3}}\left ( \frac{\gamma B}{T} \right )e^{-\beta \gamma B}
\]
The third term is much more significant than the second term, therefore we can neglect the latter and get
\[
C_V\approx \frac{15}{4}\cdot \frac{1}{\lambda_T^{3}} Li_{5/2}(1)+\frac{3}{2}\cdot \frac{1}{\lambda_T^{3}} \left(\frac{\gamma B}{T}\right) e^{-\beta \gamma B}
\]
\Dn
\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}