Bose–Einstein statistics exercise

Question

I've a basic Bose–Einstein statistics exercise. I've tried to solve it in two ways, but each way gives a different result.

We have $n$ identical bosons without interactions at temperature $T$. There are two states, of energies $\epsilon > 0$ and $0$. The question is: What is the value of $U$?

• The first way is to say that the number of particles of the system is the sum of the average number of bosons for each energy :

$$n = \frac{1}{e^{\beta\epsilon-\alpha}-1}+\frac{1}{e^{-\alpha}-1}$$

Given $n$, we can obtain $\alpha = f(n)$ and finally say that

$$U = \frac{\epsilon}{e^{\beta\epsilon-f(n)}-1}$$

I've calculated $f$ (it is quite long...), and it doesn't give the same results than the second method.

• The second way is to say that

$$Z(\beta) = \sum_{n_0 + n_\epsilon = n} e^{-\beta n_\epsilon \epsilon} = \frac{1 - e^{-\beta\epsilon (n+1)}}{1-e^{-\beta \epsilon}}$$

and then

$$U = -\frac{\partial \ln Z}{\partial \beta}$$

In fact, I've seen the solution of the exercise and it is the second answer. But why is the first wrong?

Answer 1

Actually, the chemical potential is zero.

Consider that, at all temperatures, there is a macroscopic population in the ground state, i.e., the system is in a mixed phase with a BEC at all temperatures. The criteria for a BEC to form is $\mu=0$ ($\alpha/\beta$ in your notation). Thus, at all temperatures, $\alpha=0$.

Now, if you take the thermodynamic limit viz. $n\rightarrow\infty$, both your answers coincide. Your first answer remains unaltered save for setting $f(n)=0$, yielding $$ U = \frac{\epsilon}{e^{\beta\epsilon}-1}. $$

In the thermodynamic limit, your second answer becomes

$$ \displaystyle\lim_{n\rightarrow\infty} Z(\beta) = \frac{1}{1-e^{-\beta\epsilon}}. $$

Using $U=-\frac{\partial}{\partial\beta}\ln{Z}$ yields

$$ \displaystyle\lim_{n\rightarrow\infty}U = \frac{\partial}{\partial\beta}\ln{\left[1-e^{-\beta\epsilon}\right]}=\frac{\epsilon}{e^{\beta\epsilon}-1}, $$ the same as your first answer. QED

I think the discrepancy for finite $n$ arises due to inequivalency in particle number fluctuations between canonical and grand canonical ensembles for for finite $n$ ...

Answer 2

In the first way, you are using grand-canonical ensemble (http://en.wikipedia.org/wiki/Grand_canonical_ensemble), with fixed temperature and chemical potential.

In this situation (grand-canonical ensemble), the number of particles is not fixed.

So it is in contradiction with you assertion of $n$ identical bosons, because you suppose here that the number of particles is fixed.