Thermal Properties of the Free Electron Gas: Fermi-Dirac Distribution

[matteoargos]

TL;DR Summary

I have I have some doubts about the relationship between the probability of occupying an electronic state defined by the Fermi Dirac distribution and the relationship to the total number N of free electrons in a solid at equilibrium.

Hello

I have some doubts about 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

Where: f_i 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

where we neglect the superscript N.

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

In this section is explained that
"f_i 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) x 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.

 

[Matthew_]

The first formulation gives the total number of electrons, the reasoning behind that sum is exactly what you reported. The expression $n(\mathcal{E})=g(\mathcal{E}) f_D(\mathcal{E})$, with the same logic, actually represents the density of particles with energy $\mathcal{E}$, being $g(\mathcal{E})$ the density of states (i.e. the number of states per unit volume with energy $\mathcal{E}$) and $f_D(\mathcal{E})$ the occupation probability of such states. Therefore, one should write the total number of electrons as: $$N=Vn=V \int \mathrm{d}\mathcal{E}\: n(E)=V \int \mathrm{d}\mathcal{E}\: g(\mathcal{E}) f_D(\mathcal{E}), $$
where I assumed a continous spectrum.
To get an explicit connection between the two formulations of $N$, one needs a dispersion relation, as $N=\sum_i f_i$ is meant as a sum over all single-particle states and not as a sum over energy. More explicitly, one has $N=\sum_i f_D(\mathcal{E}_i)$ where $f_D$ is the Fermi-Dirac distribution and $i$ is generally a multi-index.
It is customary to consider bloch states, so that each one-particle state is fully described by the band index $n$, the bloch wavevector $\mathbf{k}$ and spin $\sigma$. The number of electrons associated to each band (suppressing band index for convenience, as the total number of electrons is just the sum over all $N_n$): $$N=2\sum_\mathbf{k}f_D(\mathcal{E_\mathbf{k}}).$$ Note that the additional factor 2 emerges as two electrons with opposite spin can be found with the same $\mathbf{k}$. For sums in k-space, one usually uses the continuum limit often exploited in A. & M. $$\sum_\mathbf{k}\mapsto \dfrac{V}{(2\pi)^3}\int \mathrm{d}^3k.$$ Under the right assumptions and given that the constant energy surfaces are "regular enough", one can manipulate the integral to find the definition of the density of states associated to each band. The end result is exactly what you were searching for: $$N=\dfrac{2V}{(2\pi)^3}\int \mathrm{d}^3k\:f_D(\mathcal{E_\mathbf{k}}) = V\int\mathrm{d}\mathcal{E}\left( \iint_{S_\mathcal{E}}\dfrac{\mathrm{d}S_\mathcal{E}}{4\pi^3}\dfrac{1}{\|\nabla_\mathbf{k} \mathcal{E}(\mathbf{k})\|}\right)f_D(\mathcal{E_\mathbf{k}}) \equiv V \int \mathrm{d}\mathcal{E}\: g(\mathcal{E}) f_D(\mathcal{E}),$$
where we first integrate in k-space over the surface $S_\mathcal{E}$ with constant energy $\mathcal{E}$. As a side note, there are different possibilities for the definition of the density of states, which may rely on the symmetries or regularities of constant energy surfaces in k-space (the simplest case being the spherical surfaces of the free-electron model. In that context, one can easly get the density of states by using spherical coordinates and exploiting the simmetry of the dispersion relation). A more general formulation is actually: $$g(\mathcal{E})=\dfrac{1}{4\pi^3}\int \mathrm{d}^3k\:\delta^3(\mathcal{E}-\mathcal{E}(\mathbf{k})).$$