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\exnumber{5980}
\heading{BEC regarded as a phase transition} 
\auname{Edo Hogeg}


{\bf The problem:}

\Dn

Consider $N$ bosons that each have mass  $M$ in a box of volume $V$. 
The overall density of the particles is ${\rho=N/V}$.
The temperature is $T$. 
Denote by $m$ the number of particles that occupy the ground state orbital of the box. 
The canonical partition function of the system can be written as 
%
\[ Z \ \ = \ \ \sum_{m=0}^{N} Z_{N-m} 
\ \ = \ \ \sum_{m=0}^{N} e^{-\tilde{A}(m)} 
\ \ = \ \ \int d\varphi \ e^{-N\,A(\varphi) + \text{const}}  \]
%
In this question you are requested to regard the the Bose-Einstein condensation as phase transition that can be handled within the framework of the canonical formalism where $m$ is the order parameter. Whenever approximations are required assume that $1 \ll m \ll N $ such that ${\varphi=(m/N)}$ can be treated as a continuous variable. In the first part of the question assume that the gas is ideal, and that $Z_{N-m}$ can be calculated using the Gibbs prescription. In item~5 you are requested to take into account the interactions between the particles. Due to the interactions the dispersion relation in the presence of $m$ condensed bosons is modified as follows: 
%
\[ E_k \ = \sqrt{ \left(\epsilon_k + 2g\frac{m}{V} \right) \epsilon_k} \]  
%
where $ \epsilon_k $ are the one-particle energies in the absence of interaction, and ${ g }$ is the interaction strength. For the purpose of evaluating ${ Z_{N-m} }$ for large ${ m }$ assume that the above dispersion relation can be approximated by a linear function ${ E_k \propto k }$

\begin {itemize}

\item[(1)] Write an explicit expression for the probability ${ p_m }$ of finding ${ m }$ particles in the ground state orbital. Calculation of the overall normalization factor is not required.

\item[(2)] Find the most probable value ${ \bar{m} }$. Determine what is the condensation temperature ${T_c }$ below which the result is non-zero.    

\item[(3)] Assuming ${ T < T_c }$ write a Gaussian approximation for ${ p_m }$

\item[(4)] Using the Gaussian approximation determine the dispersion ${ \delta m }$ 

\item[(5)] Correct your answer for ${ p_m }$  in the large ${ m }$ range where the interactions dominate. 

\item[(6)] On the basis of your answer to item3, write an expression for ${ A(\varphi;f) }$ 
that involves a single parameter ${f}$ whose definition should be provided using ${\rho,M,T}$.  

\item[(7)] On the basis of your answer to item5, write an expression for ${ A(\varphi;a) }$ 
that involves a single parameter ${a}$ whose definition should be provided using ${\rho,M,T}$ and $g$.
  
\end {itemize}


\Dn


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{\bf The solution:}

\Dn

(1) 
By Gibbs prescription the probability $p_{m}$ of finding $m$ atoms in the ground state is given by

\[
p_{m}\propto \frac{1}{(N-m)!} z^{N-m}=\frac{1}{(N-m)!}\left(V\left(\frac{MT}{2\pi}\right)^{\frac{3}{2}}\right)^{N-m}
\]


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

(2)
To find the most probable $\bar{m}$ we donote $\tilde{A}(m)$ and find its extemum

\[
\tilde{A}(m)=\ln{((N-m)!)}-(N-m)\ln{z} 
\]

\[
\tilde{A}^{`}(m)=-\ln{(N-m)}+\ln{z}=0
\]

\[
\bar{m}=N-V\left(\frac{MT}{2\pi}\right)^{\frac{3}{2}}
\]

We find the condensation temperature by demanding $\bar{m}>0$ 

\[
N-V\left(\frac{MT}{2\pi}\right)^{\frac{3}{2}}>0
\]

\[
T_{c}=\frac{2\pi}{M} \left(\frac{N}{V}\right)^{\frac{2}{3}}
\]
and so if $T<T_{c}$ there will be $\bar{m}$ condensed atoms. 
%%%%%%%%%%%%%%
\Dn

(3)
Assuming $T<T_{c}$ we can rewrite the action $\tilde{A}(m)$ in the Gaussian approximation, to do that we'll make some approximations on $\tilde{A}(m)$:

\[
\tilde{A}(m)=\ln{((N-m)!)}-(N-m)\ln{z} 
\]

Using Stirling:

\[
\tilde{A}(m)\approx(N-m)\ln{(N-m)}-(N-m)-(N-m)\ln{z}
=(N-m)[\ln{N} + \ln{(1-\frac{m}{N})}]-(N-m)-(N-m)\ln{z} 
\]

Remembering that $N\gg m$ we can make the following approximation:

\[
\ln{(1-x)}\approx -x-\frac{1}{2}x^{2}
\]

Now we see that 

\[
\tilde{A}(m)\approx \frac{1}{2}\frac{m^2}{N} + (N-m)\ln{\left(\frac{N}{z}\right)}+O(m^3)
\]

 and see that the probability $p_{m}$ is:

\[
p_{m}\propto \exp{[-\frac{1}{2N}m^{2}-\ln{\left(\frac{N}{z}\right)}(N-m)]}
\]


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

(4)
From the result in item 3 we see immediately that the dispersion is 

\[
\delta m=\sqrt{N}
\] 

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

(5)
Due to the add interaction that dominates now, we can approximate the energy as follows:

\[
E_{k}=\sqrt{(\epsilon_k+2g\frac{m}{V})\epsilon_{k}} \approx \sqrt{2g\frac{m}{V}\epsilon_{k}}
=\sqrt{2g\frac{m}{V}\frac{k^2}{2M}}=\sqrt{\frac{gm}{VM}}k
\] 

Let us find the new one particle partition function:

\[
z'=\int \exp{(-\beta E_k)} \frac{d^3x d^3p}{(2\pi)^3}
=\frac{4\pi V}{(2\pi)^3} \int_{0}^{\infty} \exp{(-\beta\sqrt{\frac{gm}{VM}}k)}
\]

\[
z^{'}=\frac{VT^3}{\pi^2} \left(\frac{VM}{gm}\right)^\frac{3}{2}
\] 

Since the only thing in the probability that changes is $z'$ we can follow the previous procedure for the probability $p_{m}$:

\[
p_{m}\propto \exp{[-\frac{1}{2N}m^{2}+\frac{3}{2}(N-m)\ln{\frac{z^{'}}{N}}]}
\]

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

(6)
If we set $\ln{\frac{V}{N}\left(\frac{MT}{2\pi}\right)^{\frac{3}{2}}} = f(T)$ and recall $\varphi=\frac{m}{N}$ we can easly rewrite the answer in item 3 as

\[
A(\varphi;f)=-\frac{1}{2}\varphi^{2}-f(T)(1-\varphi)
\]

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

(7)
Again from the answer in item 5 we set $\frac{VT^3}{\pi^2} \left(\frac{VM}{gm}\right)^\frac{3}{2}=\left(\frac{a(T)}{\varphi}\right)^{\frac{3}{2}}$ we can rewrite to get:

\[
A(\varphi;a)=\frac{1}{2}\varphi^{2}+\frac{3}{2}(1-\varphi)\ln{\left(\frac{\varphi}{a(T)}\right)}
\]



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