# What is $z$ in the Bose gas?

I'm studying the ideal Bose gas, and I found this equation for the average occupation number of particles at the fundamental level: $$\langle n_0 \rangle = \frac{z}{1-z}$$ There's no degeneration due to spin, and $z$ is the fugacity defined as $$z = e^{\mu/kT}$$ But this is a problem because this makes $\langle n_0 \rangle$ negative.

Furthermore, I see that $z$ can take values from $0$ to $\infty$, but for $z<1$ that would mean either $\mu<0$ or $T<0$, and that doesn't seem physically possible. How do you define $z$ then?

• It is simply true that $\mu$ is always negative (or rather, smaller than the lowest lying single particle state) for massive Bose gases, since otherwise there had to be infinitely many particles. – Sebastian Riese Jun 20 '18 at 15:27

$\mu$ is the chemical potential and it is negative in the case of Bose gas.

• This should be a comment, not an answer. – myradio Jun 20 '18 at 15:35
• @myradio This IS an answer. It's making a physics claim that we should be able to upvote/downvote based on its truth/falsehood. And it fully answers the question (if $\mu$ is necessarily negative, problem solved!). It might not go into enough detail to explain WHY $\mu$ is necessarily negative, but it certainly gives a starting point for further research. – Jahan Claes Jun 20 '18 at 15:46
• It is posted as an answer and it certainly gives the important piece of information to solve the original doubt from the OP. Nevertheless, it does't answer the question strictly speaking. Just a brief explanation should have work. Concretely, the answer is mu is the chemical potential it is not answering the question What is z? – myradio Jun 20 '18 at 15:54

Let $\mathcal{Z}$ be the grand canonical partition function of your ideal quantum gas. Since particles do not interact with each other, you can factorize the partition function: $$\mathcal{Z} = \prod_{\lambda} \mathcal{z}_{\lambda}$$ where $\mathcal{z}_{\lambda}$ is the partition function for the $\lambda$-state, written as follows: $$\mathcal{z}_{\lambda} = \sum_{N_{\lambda}} \exp(-\beta(\epsilon_{\lambda}-\mu)N_{\lambda})$$ where $\epsilon_{\lambda}$ is the energy of the $\lambda$ state, $N_{\lambda}$ its occupation number and $\mu$ the chemical potential.

The above definitions stand for both fermion and boson gases. Now, since you are working with a Bose gas there is no limit concerning the occupation number, therefore the sum over $N_{\lambda}$ will go from zero to infinity. Making a change of variable we can rewrite the partition function as

$$\mathcal{z}_{\lambda} = \sum_{N_{\lambda} = 0}^{\infty} r^{N_{\lambda}}=\frac{1}{1-r}$$ which is verified only if $|r| = \exp(-\beta(\epsilon_{\lambda}-\mu)) < 1$ therefore $\mu < \epsilon_{\lambda}$

As we generally choose $\epsilon_{0} = 0$ for the ground level, the chemical potential is always negative.

Finally, the fugacity is defined as you pointed out, but for the choice of $\epsilon_0=0$ it takes values from 0 to 1.