\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{E150: The functions $N(E)$ and $Z(T)$ for ${N}$ spins} 
\auname{Elkana Porat}


{\bf The problem:}

\Dn

Given an ${N}$ spin system
%
%
\[\hat{H}=\sum_{\alpha=1}^{N}\,\frac{\varepsilon}{2}\,\hat{\sigma}_{z}^{\left(\alpha\right)}\]
%
%
Calculate the partition function ${Z_{N}\left(\beta\right)}$ in two different ways:

\begin {itemize}
\item[(1)]

The short way - Calculate ${Z_{N}\left(\beta\right)}$ by factoring
the sum.

\item[(2)]

The long way - Write the energy levels ${E_{n}}$ of the system.
Mark with ${n=0}$ the basic level and with ${n=1,2,3,...}$ the
following levels. Find the degeneracy ${g_{n}}$ of each level. 
Use these results to calculate $Z_{N}\left(\beta\right)$.
%
%
%\[Z_{N}\left(\beta\right)=\sum_{n} g_{n}e^{-\beta E_{n}}\]
%
%
\end {itemize}

% Give a physical meaning to the index ${n}$.
%note: in this question ${n}$ is the index of the energy levels and
%not of the state energies. therefore, there's a need to explicitly
%conclude ${g_{n}}$   in the sum.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) The partition function for a single spin ($E_\downarrow = 0,\ E_\uparrow = \varepsilon$):
\[ Z_1\left(\beta\right) = 1+e^{-\beta\varepsilon} \]

For N spins the partition function is given by:
\[Z_{N}\left(\beta\right)=\sum_{\left\{E^{\left(\alpha\right)}\right\}} e^{-\beta \sum_{\alpha} E^{\left(\alpha\right)}}\]
 
When ${\alpha}$ indicates one spin and ${\left\{E^{\left(\alpha\right)}\right\}}$ corresponds to a particular configuration of the system.

That may seem complicated, but since the spins do not interact we can factorize the sum and get:
\[ Z_N\left(\beta\right) = \left(1+e^{-\beta\varepsilon}\right)^N \]

%%%%%%%%%%%%%%%%%%
\Dn

(2) Each spin contributes to the total energy level 0 for spin down and ${\varepsilon}$ for spin up. The total energy for a state with ${n}$ spins up will than be: ${E_n = n\varepsilon}$.
The degeneracy for such energy level is the number of different groups of ${n}$ spins up, out of the total $N$ spins in the system. That is also noted as 
\[{g\left(n\right) = \left(\begin{array}{c} N \\ n \end{array} \right)}\]
the binomial coefficient.

Now, the partition function is a sum over all the states of the system, which can be written as a sum over all the \textbf{energy levels}, including their degeneracy. In other words:
\[ {Z_N\left(\beta\right) = \sum_{n=0}^{N}{g\left(n\right)e^{-\beta E_n}} = \sum_{n=0}^{N}{\left(\begin{array}{c} N \\ n \end{array} \right)
e^{-\beta n\varepsilon}} }\]

Which we recognize as Newton's binomial expansion for:
$${Z_N\left(\beta\right) = \left(1+e^{-\beta\varepsilon}\right)^N}$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
