I am reading "Electronic Transport in Mesoscopic Systems" by Supriyo Datta and am a little confused by his discussion of the Fermi Dirac distribution function, which is given by:


where $E$ is the energy of a state, $T$ is temperature, $E_f$ is the Fermi energy (the highest occupied state at $T=0$) and $k_B$ is Boltzmann's constant.

The author claims that in the "...high temperature or the non-degenerate limit $(\exp\left[E-E_f\right]/k_BT\gg1)$" it has the following simple form:


Question: I agree that when the exponential dominates the denominator this is the result, but is there any justification in calling it a high-temperature limit?

Indeed, I would think that if the temperature is high, it would make sense to say that $\frac{E-E_f}{k_BT}\ll 1$ in which case the exponential does not dominate the denominator at all!

I would brush it off as a typo, but when I look at a graph of Fermi-Dirac distributions for different temperatures it does seem that they take on the form of an exponentially decaying function when the temperature is high:

enter image description here


1 Answer 1


Unless you are in the low temperature limit, then the F-D distribution should be written as $$ F(E) = [\exp (E-\mu)/kT +1]^{-1},$$ where $\mu$ is the chemical potential.

I think what you are missing is that the chemical potential is not a constant, it is temperature-dependent. At high temperatures then $\mu < 0$ and $\exp(-\mu/kT) \gg 1$. You could have a look at this answer to another question for an explanation of why that is.

Thus $$[\exp (E-\mu)/kT +1]^{-1} \simeq [\exp(-\mu/kT)\exp(E/kT)]^{-1} =\exp(\mu/kT)\exp(-E/kT) $$ which is the Boltzmann distribution and is $\ll 1$ for all $E$.

  • 2
    $\begingroup$ Okay thanks for your response. This difference between chemical potential and Fermi energy was known to me, but I was following the textbook's author rather naively. Just a note that without reading the link to a different question that you provided, your answer is incomplete. It is not enough for $\mu<0$ to give $\exp(-\mu/kT)\gg 1$, we also require $|\mu|>>kT$ to get the desired behaviour. Again: it's in the link, but perhaps if you would be so kind as to append this one sentence to your answer for future readers it would make things easier for them. Thanks very much for the response! $\endgroup$
    – Lachy
    Commented Nov 1, 2016 at 17:38
  • 1
    $\begingroup$ @Lachy That's why I said "$\mu<0$ and $\exp(-\mu/kT)\gg 1$. $\endgroup$
    – ProfRob
    Commented Nov 1, 2016 at 18:05

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