# Filling of energy bands

In the context of electron energy bands for crystals, I have heard the phrase

There are $$x$$ electrons per band, therefore the electrons fill $$y$$ bands.

I am trying to understand the 'therefore' in that sentence. I believe it could be related to the number of atoms per unit cell, and the number of valence electrons per atom.

1. What is the rationale behind the statement? How can electrons fill a band? I thought the bands were the eigenvalues of the Bloch wave?
2. Can you relate the number of valence electrons of Silicon to the valence bands? Here is a plot of the bands for fcc silicon in case it is helpful:

• That question requires a quite long answer I guess.. I would recommend reading CH15 of Ashcroft & Mermin 'solid state physics' of CH9 of C. Kittel 'Introduction to solid state physics'. If you are familiar with the concepts of Fermi surfaces and Brillouin zones, these references might be perfect you! Commented Jan 7, 2020 at 19:43
• The quote is without source and too vague. Commented Jan 8, 2020 at 0:19

Bloch's theorem states that the eigenvalues and eigenvectors of a periodic one-electron Hamiltonian can be labeled by Bloch's wavevectors $$\bf k$$ lying in the first Brillouin zone. For each $$\bf k$$ there is an infinite set of solutions of the eigenvalue problem, labeled by another quantum number, $$n$$, the band index. The eigenvalues $$E_n({\bf k})$$ depend on both $$n$$ and $$\bf k$$, being continuous functions of $$\bf k$$. Each function $$E_n({\bf k})$$ , at fixed $$n$$, and for $$\bf k$$ in the first Brillouin zone, is an energy band.

The one-electron eigenstates $$\psi^{(n)}_{\bf k}({\bf r})$$ can be used to build a totally antisymmetric $$N_e$$-electron ground state wavefunction by using the lowest lying eigenvectors in a Slater determinant where each spatial state enters twice, once multiplied by a spin up and then multiplied by a spin down wavefunction.

Filling the n-th band is a pictorial expression equivalent to say that all the $$\bf k$$ states of the $$n$$-th band appear in the Slater determinant with double occupation (to take into account the spin)

In a finite periodic crystal (it can be obtained by using 3D periodic boundary conditions (pbc)), there is a finite number of $${\bf k}$$ vectors in the first Brilluoin zone, exactly equal to the total number $$N$$ of cells in the pbc crystal. So, all the $$N$$ orbital $${\bf k}$$ states of a band can accomodate a maximum of $$N_e=2N$$ electrons.

In order to have the lowest possible energy (the ground state), one has to "fill", i.e. to use, the one-electron states, sorting the values of $$E_n({\bf k})$$, starting from the lowest values and using all the states in increasing order of energy until the number of used values of $$n$$ and $${\bf k}$$ is exactly equal to $$\frac{N_e}{2}$$ (remember that there are two spin values for each $$n$$ and $${\bf k}$$).

For insulators and semiconductors, such an ordered filling of one-electron states stops when all the $${\bf k}$$ values of a finite number of bands have been used. The first empty state is separated from the highest filled state by a finite gap of energy. Things are more complex in the case of metals because in that case there is no gap neatly separating filled and empty bands and partially filled bands appear.

About question n.2, Silicon, in its equilibrium crystalline structure at normal conditions is a semiconductor with the diamond structure. The diamond structure is not a pure Bravais lattice but it can described as an fcc lattice with a 2 atoms base. If the pbc finite size crystal contains $$N$$ cells, there will be $$2N$$ atoms. Each atom will contribute to the valence bands with $$4$$ electrons ( $$3s^23p^2$$ electronic configuration of the atom). In total, the finite size crystal will have $$N_e=8N$$ valence electrons. Consequently, the number of filled bands will be $$4$$.

• Dear GiorgioP, thank you for your kind answer. Could you please see if I now understand you correctly. I have written an explanation in my own words based on your answer: (1) Assume the crystal has $N$ primitive cells and that $M$ are the number of valence electrons in each primitive cell. Assume that we ignore correlation, then the many-body ground state is a slater determinant of bloch waves. It is well defined after having made a choice of quantum numbers $(n,\vec{k})$ for each bloch wave that appear in the slater determinant. The band corresponding to Commented Jan 8, 2020 at 15:01
• quantum number $n$ is called filled if there is a bloch wave in the many-body ground state with quantum numbers $(n,\vec{k})$ for each $\vec{k}$ in the first Brillouin zone. (2) Because there are exactly $N$ allowable values of $\vec{k}$ in the first Brillouin zone and using the Pauli principle, the band corresponding to quantum number $n$ is filled if and only if there are exactly $2N$ bloch waves with quantum numbers $(n,\vec{k})$ in the many body wave function. The aufbau principle says that the ground state is acquired by picking $n$ as small as possible for each bloch wave in the Commented Jan 8, 2020 at 15:02
• slater determinant. This ensures that that the number of bands that are filled is $\lfloor(MN)/(2N) \rfloor$. If $M$ is even then $(MN)/(2N)$ bands are filled. Commented Jan 8, 2020 at 15:02
• @MikkelRev I agree with all you wrote, with an additional clarification: the aufbau principle in general works by looking at the total ordering of energies as function of both $n$ and $\bf k$. In the case of insulators or semiconductors, it doesn't make difference with respect to picking bands in the order of increasing $n$, but it makes a huge difference in the case of metals. Commented Jan 8, 2020 at 18:05