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Brillouin Zone Confusion

  1. May 10, 2014 #1
    Hey all,

    Quick question about BZ's and it's probably a really dumb one. Solid State isn't my favorite class...

    What does it physically represent? Like is the area of the BZ divided by the area of a k-state (2π/L) equal to the number of charge carriers in the system? So if there was an electron fermi surface area that takes up 5% of the Brillouin Zone, does that relate to the electron density by just being 5% of the number of charge carriers from the above division?

    Thanks for the explanation. It will help me out in understanding this.
     
  2. jcsd
  3. May 10, 2014 #2
    Not exactly. The volume of the BZ divided by the volume of a k-state is equal to the total number of available k-states for a given band in the system, i.e., the number of slots in the band that are available for charge carriers (electrons) to inhabit. Thus, if the volume enclosed by the Fermi surface takes up 5% of the volume of the Brillouin zone, then electrons occupy 5% of the valence band. Alternatively, we could say that the valence band is 5% full.

    There are also many other ways in which the concept of Brillouin zones can be physically understood. You might want to consider consulting an undergraduate-level textbook like Kittel or Ashcroft and Mermin.
     
  4. May 10, 2014 #3
    Yeah I plan on looking at a Kittel book. That made sense though - thank you for clearing that up for me :)
     
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