# About the Fermi-Dirac distribution

In order for a function $$f$$ to be a probability density, it must satisfy $$\int f dx=1,$$ where $$x$$ spans the domain of the random variable. If we take the Fermi-Dirac distribution $$\tag{1}f(E)=\frac{1}{e^{(E-E_F)/kT}+1}$$ at $$T=0$$, we find a step function like this

which is described by $$f(E) = \left\{ \begin{array}{ll} 1 & \quad E < E_F \\ 0 & \quad E > E_F \end{array} \right.$$ So we clearly have $$\int f(E)dE= E_F\neq 1$$. So why is $$(1)$$ called a distribution?

• Strictly speking, Fermi-Dirac is mean occupation number $\langle n \rangle$.. See en.wikipedia.org/wiki/… Commented Jun 16, 2020 at 21:35
• Since f(E) is unitless, having the integral be equal to an energy is problematic... Commented Jun 16, 2020 at 21:49
• @JonCuster The units are OK because of.$dE$ contribution. Commented Jun 16, 2020 at 22:15
• @Alexander - sigh, coffee stopped working. More to the point, that integral determines where $E_{f}$ actually is in the band structure. Commented Jun 16, 2020 at 22:18
• Closely related: physics.stackexchange.com/q/275126/226902 Commented Sep 13, 2023 at 10:47

$$f(E)$$ is not a probability density function: it gives the probability that a state with energy $$E$$ is occupied (notice that it is a dimensionless quantity). As such, it does not need to integrate to $$1$$, and it doesn't.
• If each energy state has a probability of occupation (for $T > 0$), does this mean that at some instance maybe all the higher energy states are occupied? Commented Nov 4, 2021 at 20:03