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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

enter image description here

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?

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

1 Answer 1

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$f(E)$ 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.

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  • $\begingroup$ 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? $\endgroup$
    – Anton
    Commented Nov 4, 2021 at 20:03

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