I have some difficulties to understand the relationship between the probability of occupying an electronic state defined by the Fermi Dirac distribution and the relationship to the number $N$ of free electrons in a solid of $N$ atoms. In particular I refer to the Section 2.2 of Solid State Physics (Ashcroft, Mermin) formula (2.49)

$$ N = \sum_i f_i = \sum_i \frac{1}{e^{(\varepsilon_i-\mu)/k_bT}+1} \tag{2.49}$$

Where: fi is the probability of the one-electron level i being occupied at temperature $T$ in an $N$-electron system which is the well known Fermi-Dirac distribution

$$ f_i^N = \frac{1}{e^{(\varepsilon_i-\mu)/k_bT}+1},\tag{2.48}$$

where we neglect the superscript $N$.

I do not arrive to understand conceptually the (2.49) as $N$ to be summatory of fi contributions. Being f a probability.

In this section is explained that "fi is the mean number of electrons in the one-electron level"

as proof is indicated that

"[...]A level can contain either 0 or 1 electron (more than one being prohibited by the exclusion principle). The mean number of electros is therefore 1 times the probability of 1 electron plus 0 times the probability of 0 electrons. Thus the mean number of electrons in theleve is numerically equal to the probability of its being occupied. Note that this would not be so if multiple occupation of levels were permitted".

If we think this of course the (2.49) is evident.

Nevertheless I'm used to think to the number of electrons for a certain energy level for a solid like

$$ n(E)=g(E)f(E),$$

where g(E) is the density of levels , f(E) is the Fermi distribution

so that N is the integral on the space of the whole $E$ of $g(E)f(E)dE$ (or as the limit of the summatory $g_i(E) f_i(E)$ )

I miss $g_i(E)$ in the first part on the left of the (2.49)....

Someone can explain me conceptually this point? Thank you in advance.


1 Answer 1


You use $n(E)=g(E)f(E)$ when you perform an integral over energy. For example, the number of electrons is given by $$ N = \int dE \, g(E) f(E)$$

However, if you are summing over each state, you should use $$ N = \sum_i f_i(E)$$

The difference is that when you sum (integrate) over energy, the density of states $g(E)$ tells you how many states you have at that energy (you should consider this factor because you are not summing over each state!). However, you don't need to use the density of states if you are summing over each state!

  • $\begingroup$ thank you. In effect without g(E) the integral return 1 over all the E space. Where in the summatory of states i do not need an extra function. Obvious. Thanks and sorry for the stupid question. $\endgroup$
    – Argos
    Nov 18, 2023 at 17:48
  • $\begingroup$ Don't be too hard on yourself this is complicated stuff! $\endgroup$
    – dllahr
    Nov 19, 2023 at 12:42
  • $\begingroup$ @Argos don't be sorry! It is not easy to understand everything on a first read and this post will help others having the same question! $\endgroup$
    – Ivan
    Nov 19, 2023 at 19:06

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