Chemical potential of Bose–Einstein condensation

Question

A problem from Schroeder's Introduction to Thermal Physics [1]:

Problem 7.66. Consider a collection of $10\,000$ atoms of rubidium-87, conf‌ined inside a box of volume $(10^{-5}\;\mathrm{m})^3.$

(a) Calculate co, the energy of the ground state. (Express your answer in both joules and electron-volts.)

(b) Calculate the condensation temperature, and compare $kT_c$ to $\epsilon_0$.

(c) Suppose that $T = 0.9T_c.$ How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) f‌irst excited states?

(d) Repeat parts (b) and (c) for the case of $10^6$ atoms, conf‌ined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the f‌irst excited state.

(a) and (b) are easy to work out. In (c), I use the formula $\displaystyle N_\mathrm{excited}=\left(\frac{T}{T_c}\right)^{3/2}N, \ (T<T_c)$ to calculate how many atoms are there in the ground state. But I am completely lost about the difference between chemical potential and ground-state energy, because the book always assumes $\mu \approx 0$ when $T<T_c$. I don't think the answer should simply be $\epsilon_0$ but don't know to how to calculate.

I know this question seems to be silly, but please provide some hints about it.

References

1. Schroeder, D. V. An Introduction to Thermal Physics; Addison Wesley: San Francisco, CA, 2000. ISBN 978-0-201-38027-9.

Answer 1

It feels like you are going to fast in your way to think. The difference between the chemical potential and the ground state energy is clear :

1. The ground-state energy of your system is here $\epsilon_\textbf{k}=0$ correponding to the ground-state $|\textbf{k=0}\rangle$, which is macroscopically occupied in a BEC (i.e. $N\sim N_0$).

2. The chemical potential $\mu$ is the increment in energy that you give to your condensate by adding one particle. It is generally defined using free energy $F$ : $$\mu=\left(\frac{\partial F}{\partial N}\right)_{V,\;T}$$

If you want to evaluate $\mu$, considere the ground-state occupation : $$N_0=\frac{1}{e^{-\beta\mu}-1}$$

If you want $N_0$ to be the biggest possible (macroscopic occupation), you must fulfill the condition $e^{-\beta\mu}\rightarrow 1$, in other words $\mu\rightarrow 0$.

For larger resolution on $\mu$, one can do a Taylor expansion on the expression of $N_0$ : $$N_0\underset{\mu\rightarrow 0}{\sim}-\frac{1}{\beta\mu}$$ Then, $$\mu\simeq -\frac{1}{N_0\beta}=-\frac{k_\mathrm{B} T}{N_0}=-\frac{k_\mathrm{B} T}{N-N_\mathrm{exc}}$$