Bose-Einstein condensation of Helium

Question

So I'm recalling this derivation of Bose-Einstein condensation in helium in some Thermo-stat mech text book. I thought it was in Reif, but I couldn't find it. It calculates the condensation temperature in He, (with pure QM, no He-He interaction) to be something like $1.4 ~K$. (the number is not important, I just want to remember the calculation.) Does anyone know the reference I'm talking about, or another similar calculation?

Answer 1

There are numerous textbooks alongwith a lot of reliable online resources, where you can find the derivation. The famous some of them, which I've read includes,

1. Bose-Einstein Condensation by Lev. P. Pitaevskii & Sandro Stringari
2. Bose-Einstein Condensation by A. Griffin
3. Bose–Einstein Condensation in Dilute Gases by C. J. Pethick & H. Smith
4. Bose-Einstein Condensates: Theory, Characteristics & Current Research, editted by Paige W. Matthews
5. Universal Themes of Bose-Einstein Condensation by Nick P Proukakis, David W Snoke & Peter B Littlewood

But sometimes, it is better to have a overall idea about the derivation and then going through the textbooks (if further needed). So I'm giving here a short outline of the proof:

Let the density of states of the system be $D(\epsilon)$. So in D-dimensions (volume $V=L^D$) for infinite square well potential, $$D(\epsilon)~d\epsilon = \left(2S+1\right)\dfrac{L^D}{\Gamma\left(\frac{D}{2}\right)} \left(\dfrac{2m\pi}{\hbar^2}\right)^{D/2}~\epsilon^{\frac{D}{2}-1}~d\epsilon$$

and occupation number density, $n(\epsilon) = \dfrac{1}{exp\left[\beta\left(\epsilon-\mu\right)\right]-1}$ where $\beta = \dfrac{1}{k_B T}$

$\therefore$ Total number of particles in the system (ideal Bose gas), $$N = \displaystyle\int_0^\infty n(\epsilon)D(\epsilon)~d\epsilon = \left(2S+1\right)\dfrac{L^D}{\Gamma\left(\frac{D}{2}\right)} \left(\dfrac{2m\pi}{\hbar^2}\right)^{D/2}~\displaystyle\int_0^\infty \dfrac{\epsilon^{\frac{D}{2}-1}}{exp\left[\beta\left(\epsilon-\mu\right)\right]-1}~d\epsilon$$

In unit-less parameter, $x=\beta\epsilon$ and $\bar\mu = \mu\beta$, the integration becomes,

$$\Longrightarrow N = \left(2S+1\right)\dfrac{L^D}{\Gamma\left(\frac{D}{2}\right)} \left(\dfrac{2m\pi}{\hbar^2}\right)^{D/2}~\left(k_B T\right)^{\frac{D}{2}}~\displaystyle\int_0^\infty \dfrac{x^{\frac{D}{2}-1}}{e^{x-\bar{\mu}}-1}~d\epsilon$$

Let $T$ is being lowered, but $N$ is fixed. So the integral must increeased. Hence $\bar\mu$ must increase upto maximum valuie of zero$(0)$ (as $\mu \leq \epsilon_{min}$).

So it is possible that at a finite temperature $(T_C)$, we will arrive exactly at $\bar\mu=0$. Then, $$N = \left(2S+1\right)\dfrac{L^D}{\Gamma\left(\frac{D}{2}\right)} \left(\dfrac{2m\pi}{\hbar^2}\right)^{D/2}~\left(k_B T\right)^{\frac{D}{2}}~\displaystyle\int_0^\infty \dfrac{x^{\frac{D}{2}-1}}{e^{x}-1}~d\epsilon = \left(2S+1\right)L^D \left(\dfrac{2m\pi k_B T}{\hbar^2}\right)^{D/2}~\zeta\left(\dfrac{D}{2}\right)$$

It will converge, provided $D \gt 2$. Here $T_C$ is the condensate temperature.

At $T=T_C$, $\bar\mu=0$ is reached and at $T \lt T_C$, $\bar\mu$ cannot change. So the number of particles coming from the integral will be smaller than the actual number $N$.

$\Longrightarrow$ The remaining particles will settle in ground state with $\epsilon=0$ (this is Bose-Einstein condensate), let this number be $N_0$.

$\therefore N = N_0 + \left(2S+1\right)\dfrac{L^D}{\Gamma\left(\frac{D}{2}\right)} \left(\dfrac{2m\pi k_B T}{\hbar^2}\right)^{D/2}~\zeta\left(\dfrac{D}{2}\right) = N_0 + N~\left(\dfrac{T}{T_C}\right)^{\frac{D}{2}}$ for $T \lt T_C$

For $3D$, $$N_0 = N~\left[1-\left(\dfrac{T}{T_C}\right)^{\frac{3}{2}}\right]$$ $$\Longrightarrow \dfrac{N_0}{N} = 1-\left(\dfrac{T}{T_C}\right)^{\frac{3}{2}}$$

This is expression is so-called condensate fraction.