Talking about the free electron approximation the Fermi-surface is a circle. However, sometimes it fits into the first BZ and sometimes it does not. I read that, in general, it does not fit into the first BZ if the metal has 2 or more valence electrons.

Why is this?

  • $\begingroup$ It is because if you have many valence electrons, the density of electrons in the metal N is high. You can easily figure out the fermi energy of the corresponding gas of electrons and assuming the Fermi surface circle in k-space, you can find out what is the maximum wave vector (or fermi wave vector) $k_F$ (see en.wikipedia.org/wiki/Fermi_energy). if $k_F$ is larger than the first Brillouin zone, the circle is larger than the first BZ. $\endgroup$ Commented Feb 11, 2017 at 13:49
  • $\begingroup$ Thanks, But isnt there some obvious explanation why 2 valence electrons to overfill the first BZ, why not 3 for instance? $\endgroup$
    – Marsl
    Commented Feb 11, 2017 at 14:02
  • $\begingroup$ I think this is a general rule of thumb. Usually more than 2 valence bands mean transition metals so first BZ automatically full which explains why they tend to be good conductors. However, all crystals must be studied in details to get a more accurate description of their properties, so I am not sure if a straight forward calculation could show what you are looking for. $\endgroup$ Commented Feb 11, 2017 at 14:21

1 Answer 1


You can understand this in the following oversimplified model. Let us consider periodic 1D chain of $N$ atoms with lattice constant $a$. Brillouin zone boundaries are located then at $\pm n\pi/a$. Let us consider three configurations:

  1. Atoms have 1 valence electron on s-type orbital. In this case the energy level of this atomic orbital is split into $N$-level band. Only $N/2$ levels are occupied due to spin degeneracy. Fermi momentum is equal to $k_F = \pm\pi/2a$, hence Fermi surface is fully inside the 1st Brillouin zone.
  2. Atoms have 2 valence electrons on s-type orbital. Now all $N$ levels are occupied and Fermi momentum $k_F = \pm\pi/a$, so that Fermi surface touches the edges of the 1st BZ.
  3. Now let us switch to atoms with p-type orbitals occupied. If the orbitals are oriented like shown in figure below, then two energy bands of different width would emerge: one is with low mass (from two orbitals touching each other like $\infty - \infty$) and one is with large mass (from two orbitals oriented like $8 - 8$), each having $N$ levels. They have to be occupied up to the same energy level. Now, if there are 3 valence electrons, then the high mass band will be fully occupied by 2 valence electrons, whereas the third electron will be distributed among the low mass band in the 1st BZ and high mass in the 2nd BZ, as shown schematically in the figure below.

The general idea is that 2 electrons would fill one band completely and others can stay either in the first, or in the second BZ. It does not mean, that 3 valence electrons must lead to the Fermi surface protracting to the 2nd BZ. In fact, if you rotate p-orbitals in this model 45 degrees, then you will end up with the single band, where each level is degenerate not only in spin, but also in angular momentum. In this case all 3 electrons will stay in the 1st BZ.

Real band structures are quite complicated due to many factors involved, and for complex materials it often occurs that Fermi surfaces touch the BZ boundaries.

Two atoms with p-orbitals. Overlap occurs between horizontal and vertical orbitals but the magnitude is different, resulting in different transfer integrals (bandwidths)

Band structure for the 1D atomic chain with p-type valence electrons

  • $\begingroup$ Thanks, could you please write a few words about the mass you are referring to. Is it the effective mass and how can you assign a mass to a band? I know that the effective mass is the curvature of a band but how does it relate to the orbitals drawn? Why is it low for the $\sigma$-bond and high for the $\pi$-bond? $\endgroup$
    – Marsl
    Commented Feb 12, 2017 at 13:26
  • $\begingroup$ That's right, mass is the curvature of the band. In this picture high or low mass is not the feature of the orbital type of the band, but rather the feature of overlap between orbitals. Simply speaking, the stronger the overlap is the wider the band is and, hence, the higher kinetic energy an electron can acquire in such band. It is the same, as saying that mass is low. In this picture, that $\infty-\infty$ overlap is strong and $8-8$ overlap is weak, so that $m_\infty < m_8$. $\endgroup$
    – drYG
    Commented Feb 12, 2017 at 16:02

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