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Number of states in a band

  1. Sep 23, 2009 #1
    Hello all. I'm having trouble finding anywhere but my (unclear) lecture notes about the total number of states in a band. I'm getting a little muddled between my different models and am just wondering if the number of states in a band is equal to the number of atoms in the solid. In this case, a set of monovalent atoms would fill half of the band and divalent atoms would fill the whole band. This would explain why Lithium, Sodium, Potassium etc all conduct, but I still don't understand why (e.g.) phosphorus (valency 5) is an insulator. Any ideas would be much appreciated!!
    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 23, 2009 #2
    For a monovalent atomic crystal, the total number of states will be equal to the number of atoms, that's correct.

    But conduction is not just related to the number of states in a crystal. The band structure of different crystals are all different and depend on many details, usually there is a splitting between lower and upper bands and these splittings, band-gaps, at the relevant energy scale (at around Fermi Level) decide whether the crystal is a conductor or an insulator.
  4. Sep 23, 2009 #3
    As I had it, in a filled band, there is no "room" for electrons who want to gain momentum in a certain direction, so they can't flow (i.e. no conduction). In a partially filled band, this is not the case (hence conduction). However, if the bands overlap (there is no band gap) then the electrons are free to gain momentum without worrying about a gap of forbidden energy and the electrons can flow (hence conduction).

    I guess my question could be rephrased as "Does the fermi level always lie either: between two bands; or at the exact energy level where the band is exactly half filled?"
  5. Sep 23, 2009 #4
    There is also an issue of geometry. In 1D, you're used to just drawing some curves, and a horizontal line representing the Fermi energy (or really, chemical potential); in general, you don't draw multiple bands in the same energy. In 2D and 3D, it's common that higher bands lowest energies are lower than the lowest bands highest energies, but with different momenta. Thus it may be that a naive counting argument suggests a filled band, but actually some of the electrons end up in a higher band, thus giving you an extended Fermi surface.
  6. Sep 24, 2009 #5
    So when my lecturers are on about the fermi surface overlapping the walls of the brillouin zone, this is what they mean. The electrons are, in a way, bent away (due to satisfaction of the Bragg law) from overlapping the zone by the lattice potential, so they have values of k less than pi/a (if a is the lattice spacing in that direction), i.e. less than the brilloin zone. So, in the free electron model, the surface is a sphere and so it just overlaps the zone without quarrel (if the brillouin zone exists?). Then, in a strong potential, the fermi surface is warped (so that it assumes a shape similar to that of the brillouin zone?) and so, if the occupation of states is large, the fermi surface will just fill the brillouin zone. If the occupation of states is not large, then the fermi surface can be moved by an electric field and induce a current. Am I on the right track here?

    Edit: In summary, does a "full" brillouin zone correspond to an insulator and a non full one correspond to a conductor? Is this why a material which is an insulator can become conductive when a dopant releases an electron into the lattice, extending the boundaries of the fermi surface into the next brillouin zone? If so, is a nearly full brillouin zone the same as one containing holes and can therefore conduct as well? Then, in a p-n junction, the electrons which are in the overflow zone of the n-doped brillouin zone could flow into the not-quite-full brillouin zone of the p-doped zone, "filling up" the holes. Then, there would be an excess of electrons in the p-doped zone and it would be negatively charged and there would be an electric field set up. To place electrons into the p-zone would require a voltage high enough to overcome the field set up and I'm guessing the electrons would have to tunnel through it, requiring a large voltage that would break down similar to an insulator and thus create a spark, destroying the diode? In the opposite direction, the electrons would be taken from the p zone, recreating the holes and then a current of electrons would flow into the p zone from the n zone? Is this right?
    Last edited: Sep 24, 2009
  7. Sep 25, 2009 #6
    Basically, the issue of what makes things insulating or conducting is very complex; see http://nanoscale.blogspot.com/2008/12/more-about-insulators.html

    At the end of the day, the issue is settled by the existence of low energy states which have macroscopic transport. For situations where the Laudau (or more accurate Born-Sommerfeld) model applies, this boils down to whether the Fermi surface exists as an extended surface. In 1D, you can then reliably do counting arguments. Beyond 1D, these counting arguments tend to get very unreliable. Then there are all the places where the Landau model does not apply at all...
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