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. Commented Jun 20, 2018 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. Commented Jun 20, 2018 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. Commented Jun 20, 2018 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? Commented Jun 20, 2018 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.