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\exnumber{3341}
\heading{Bose in 2D harmonic trap} 
\auname{Yuval Edri}


{\bf The problem:}

\Dn

Consider a two dimensional bose gas in a harmonic potential
with energy eigenvalues $\hbar\omega(n_1+n_2+1)$ where $n_1,\,
n_2$ are integers. [This is how the recent cold atom experiments
realize condensation].

\Dn

(1) Find the number of particles in excited states $N-N_0$, where $N_0$ is the the ground state occupation. Assume $k_BT\gg \hbar\omega$ so that summations
on $n_1,\, n_2$ can be replaced by integrals.

(2) Find the Bose Einstein condensation
temperature $T_c$. [Note that $N$ here is not taken to its
thermodynamic limit; the transition is still fairly sharp if
$N>>1$.] 

 * use the integral -  $\int_{0}^{\infty}\frac{x}{ e^{x} -1} dx =\frac{\pi^2}{6}$ 

(3) Show that for $T<T_c$: $\,\,N_0=N[1-(T/T_c)^2].$


\Dn\Dn


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

\Dn

(1) 
The partition function in the Grand- canonical ensemble is: 

\[
Q=\sum_{n_1,n_2=1}^\infty\sum_{N_\epsilon}^\infty e^{-\beta N_\epsilon(\epsilon_{n_1,n_2}-\mu)}=\sum_{n_1,n_2=1}^\infty \frac{1}{1-e^{-\beta (\epsilon_{n_1,n_2}-\mu)}}
\]

The total occupation number is :   $ <n>=\frac{1}{\beta}\frac{\partial\ln{Q}}{\partial \mu}$ ,
\[
<n> =\sum_{n_1,n_2=1}^\infty \frac{e^{-\beta (\epsilon_{n_1,n_2}-\mu)}}{1-e^{-\beta (\epsilon_{n_1,n_2}-\mu)}}=\sum_{n_1,n_2=1}^\infty \frac{1}{e^{+\beta (\epsilon_{n_1,n_2}-\mu)}-1}
\]

In order to replace the summation to integral over the energies ,  we should find first the number of states up to energy E , and the density of states:\Dn

The relation between $n_1,n_2$ and the energy is describe by the equation: $\epsilon=\hbar\omega(n_1+n_2+1)$ 

Creating a right triangle at the plane of $n_1,n_2$ with leg length of $\frac{\epsilon}{\hbar\omega}$ so:
\[
N(\epsilon)=\frac{\epsilon^2}{2{\hbar\omega}^2}$  ,   $g(\epsilon)=\frac{\partial N(\epsilon)}{\partial \epsilon}=\frac{\epsilon}{(\hbar\omega)^2}
\] 

Having the density of states, we can replace the summation to an integral of $\epsilon$, we should notice that by writing it as an integral, we count only the occupation of the excited states.
\[
N_{excited}=\int_{0}^{\infty}\frac{1}{e^{+\beta (\epsilon-\mu)}-1} g(\epsilon) d\epsilon = \frac{1}{(\hbar\omega)^2} \int_{0}^{\infty}\frac{\epsilon}{e^{-\beta \mu}\cdot e^{\beta \epsilon}-1} d\epsilon 
\]

We shall write it in the manner of polylogarithm functions , which can be defined as : $ g_\nu(Z)=\frac{1}{\Gamma(\nu)} \int_{0}^{\infty}\frac{x}{z^{-1} e^{x} -1} dx $

Replacing $\beta\epsilon=x $ we get :
\[
N-N_0=N_{excited}=\frac{T^2}{(\hbar\omega)^2}\cdot g_2(e^{\beta\mu})\cdot \Gamma(2)=\frac{T^2}{(\hbar\omega)^2}\cdot g_2(e^{\beta\mu})
\]

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

(2)
We are asked to find the critical temperature of the Bose -Einstein condensation , so we just have to ask on which temperature the number of particles occupied on the excited states is equal to the total number of particles N,  $g_2(e^{\beta\mu}) $ can be taken to be $ g_2(1)=\int_{0}^{\infty}\frac{x}{e^{x} -1} dx =\frac{\pi^2}{6}$  when $\mu<<T$ which is something seems to characterize the condensation - lower $\mu$ means there is no energy cost in adding another particle to the system

\[
N_{excited}=\frac{\pi^2\cdot T^2}{6\cdot(\hbar\omega)^2}=N  \rightarrow   T_c=\pi\hbar\omega\cdot \sqrt{\frac{N}{6}}
\]

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

(3)
Beneath the critical temperature, the occupation of the ground state is equal to  $N-N_{excited}$ :

\[
N_0=N-N_{excited}=N-\frac{\pi^2\cdot T^2}{6\cdot(\hbar\omega)^2}=N-\frac{\pi^2} {6\cdot(\hbar\omega)^2} (T/T_c)^2 \cdot T_c^2=N(1-(T/T_c)^2)
\]


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\Dn

 


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