[QUE/SM-10001] SM-PPROBEM

1. {Show that for ideal bosons in an open system, the total number of particles is given by $$N =\frac{\lambda}{1-\lambda}+\frac{V}{\Lambda^3}\ g_{3/2}(\lambda).$$ In the above $\lambda=\exp\left(\frac{\mu}{k_BT}\right)$ is fugacity. $\mu$ is chemical potential. $\Lambda$ is thermal/quantum wavelength given by $\Lambda = {\displaystyle \frac{h}{\sqrt{2\pi mk_BT}}.}$ Also, $g_{3/2}(\lambda)={\displaystyle \sum_{k=1}^\infty\ \frac{\lambda^k}{k^{3/2}}.}$
2. Postulate that $0\ \le\ \lambda\ \le 1$ can be as large as $1-{\displaystyle\frac{a}{N}}$ where $a$ is constant. 

1. Show that $$a=\frac{\rho\Lambda^3}{\rho\Lambda^3-g_{3/3}(\lambda)}.$$ The number of bosons in the ground state is denoted by $N_0=\lambda/(1-\lambda)$.
2.  Derive an expression for $N_0/N$ in terms of $a$ and show that \begin{eqnarray*} \frac{N_0}{N}=\left\{\begin{array}{lll} 1-\left(\frac{T}{T_C}\right)^{3/2} & \ {\rm for}\ T\ \le \ T_C\ ,\\[10mm] \approx 0 & \ {\rm for}\ T\ > T_C\ , \end{array}\right. \end{eqnarray*} where the Bose condensation temperature, $T_C$ is given by the relation, $$\frac{N}{V}\left(\frac{h}{\sqrt{2\pi mk_BT_C}}\right)^3=g_{3/2}(\lambda=1)=\zeta(3/2)\approx 2.612.$$

KPN