Understanding Brillouin Zones - What do they represent physically?

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SUMMARY

The discussion focuses on the physical representation of Brillouin Zones (BZ) in solid-state physics, specifically how the volume of a BZ relates to the number of available k-states for charge carriers. It is established that the volume of the BZ divided by the volume of a k-state determines the total number of available k-states for a given band. If the volume enclosed by the Fermi surface occupies 5% of the BZ, this indicates that the valence band is 5% full, correlating directly to electron density. Recommended resources for further understanding include textbooks by Kittel and Ashcroft & Mermin.

PREREQUISITES
  • Understanding of solid-state physics concepts
  • Familiarity with Brillouin Zones and Fermi surfaces
  • Knowledge of k-space and k-states
  • Basic principles of electron density and band theory
NEXT STEPS
  • Study the concept of Brillouin Zones in detail using "Solid State Physics" by Kittel
  • Explore the relationship between Fermi surfaces and electron density in "Solid State Physics" by Ashcroft and Mermin
  • Learn about k-space and its implications in solid-state physics
  • Investigate the mathematical derivation of k-state volume and its significance in band theory
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Students and professionals in solid-state physics, particularly those studying electron behavior in materials, as well as educators seeking to clarify concepts related to Brillouin Zones and charge carriers.

F=qE
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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.
 
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F=qE said:
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?

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.
 
csmallw said:
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.

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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