# Fermi distribution and ideal gas

I was wondering about the following:

If we have ideal gas particles, then $E \ge 0$, so one would expect that the state $E=0$ is occupied with probability one for low temperatures, but this is not the case, actually we have:

$$\langle n_{E=0} \rangle = \frac{1}{e^{\beta \cdot 0}+1} = \frac{1}{2}.$$

Somehow I find this result very counterintuitive, as I would have expected this state to be occupied with probability $1$ (at least for low temperatures). Why is this not the case?

## 1 Answer

You have forgotten to include the chemical potential $\mu$, which enforces the constraint of a fixed average number of particles in the system. The correct result for the thermal occupation of an ideal Fermi gas at temperature $T = 1/k_B \beta$ is $$n(E) = \frac{1}{e^{\beta(E-\mu)} + 1}.$$ At zero temperature, you have $\beta\to\infty$, so that for $E=0$ $$\lim_{\beta\to\infty}n(E=0) = \lim_{\beta\to\infty} \frac{1}{e^{-\beta\mu} + 1} = 1.$$ Indeed, this holds at zero temperature for all $E<\mu$, since $(E-\mu)$ is negative. If $E>\mu$ then $(E-\mu)$ is positive and the exponential washes out the denominator, so that $n(E>\mu)\to 0$. Only when $E=\mu$ exactly do you have $n(E=\mu) = 1/2$ (this is true at all temperatures).

• mhmm, the meaning of the chemical potential is something like: How much energy do we have to add to the system in order to put one more particle in. So if $E < \mu$, this means that the state I am considering has a lower energy than the chemical potential, so this state is likely to be occupied. Yeah, thank you, that makes perfect sense. – Xin Wang Feb 12 '15 at 20:15
• Indeed, at $T=0$ the chemical potential is exactly the energy required to add one particle to the system. Therefore $E=\mu$ is the energy of the highest occupied state (Fermi energy), and in order to add another particle you must place it in the next highest energy state. – Mark Mitchison Feb 12 '15 at 21:47