I'm studying statistical mechanics and band theory, with two different professors. My statistical mechanical teacher defines the Fermi Occupation function this way: \begin{equation} F(E,T)=\frac{1}{1+e^{\beta({E-\mu})}} \end{equation} where $\beta=kT$. He this defines the Fermi level as $E_F=\mu(T)$ as T tends to zero. enter image description here

So, about this I have some questions because even at this point I'm very confused. What does F(E,T) represent? Is it a probability distribution? And if not, what does it represent? I got very confused because, as $E_F$ is defined as being independent on the temperature, thus a constant in the picture (depending on the solid considered I suppose) I see from the image that there are no fermions over $E_F$, even when the temperature increases. Why is that? If, for example, I consider a semiconductor I have that the Fermi level lies in the middle of an energy gap, so at very low temperatures I have no conduction. But, as the temperature increases, I need some electron over $E_F$ for the conduction to take place.

With the other professor is a different story. He defines \begin{equation} F(E,T)=\frac{1}{1+e^{\beta({E-E_F})}} \end{equation} and he says it's a distribution probability. He uses a picture like this one: enter image description here

The two pictures obviously can't refer to the same thing. Can anyone give me a clue?

  • $\begingroup$ In your first graph, the functions are not normalized so that the area under the curve is constant. For a given material, there are only so many electrons in the conduction band, after all. $\endgroup$
    – Jon Custer
    Aug 18, 2016 at 23:20
  • $\begingroup$ You almost certainly meant to write $\beta = 1/kT$. $\endgroup$
    – DanielSank
    Aug 19, 2016 at 4:42
  • $\begingroup$ Closely related; physics.stackexchange.com/q/559824/226902 $\endgroup$
    – Quillo
    Sep 13, 2023 at 10:47

1 Answer 1


The two pictures obviously can't refer to the same thing.

Actually, they do.

You see, $FE(E,T)$, the Fermi function, is the mean number of fermions in the state of given energy E at temperature T. Hopefuly, it is bounded between 0 and 1 so only zero or one fermion in these quantum states. In this case, $\mu$ is the limit of energy at wich you have one half of a chance to found a fermion in you gas. Lower energy levels will mostly be filled and higher ones will be empty.

Obviously, in order to generalise the distribution it is common to normalize regarding $k_bT$ i.e $\beta = \frac{1}{k_bT}$. $\mu$ actually depends on your system and mostly on your system's reservoir and so is fixed with regards to N and T (else the distribution isn't valid, meaning you don't have enough particles or the temperature is too high).

I did a plot of two characterising limit $\nu$ (sorry for notation change, it's old) relatively different to the temperature of their systems as $\nu = 10k_bT$ and $\nu = 50k_bT$. e is in $k_bT$ units. Fermi-Dirac distribution for $\nu = 10k_bT$ and $\nu = 50k_bT$. e is in $k_bT$ unit.

As you can see, as the distribution is exponentially decaying between $\nu-5k_bT$ and $\nu+5k_bT$. This gives you the density of distribution of your fermions in the different energy states around.

If you keep an eye on a constant $\mu$ (or $\nu$ in the case of my plot) you will see the distribution sharpen as you decrease the temperature. This defines the so called "Fermi sphere" in the p space containing all the energy levels of your fermions at zero temperature. As such you define it's radius as $p_F$ giving you a total number of quantic states (for electrons): $$\frac{4}{3}\pi p_F^3*\frac{V}{h^3}*2=N$$ (2 is due to the 1/2 spin of electrons) As such, you can define the Fermi energy as $\epsilon_F = \frac{p_F^2}{2m}$ giving you the fermi level at $0\ K$: $\mu_F=\epsilon_F$.

Remember that $\mu$ depends on the system bath and that the distribution along the energy states varies with $T$. Hope this helps a bit.


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