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?

  • 4
    $\begingroup$ Strictly speking, Fermi-Dirac is mean occupation number $\langle n \rangle$.. See en.wikipedia.org/wiki/… $\endgroup$ – Alexander Jun 16 '20 at 21:35
  • $\begingroup$ Since f(E) is unitless, having the integral be equal to an energy is problematic... $\endgroup$ – Jon Custer Jun 16 '20 at 21:49
  • 1
    $\begingroup$ @JonCuster The units are OK because of.$dE$ contribution. $\endgroup$ – Alexander Jun 16 '20 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 Jun 16 '20 at 22:18

$f(E)$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.