Is first Brillouin zone the same as Wigner-Seitz cell?

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SUMMARY

The first Brillouin zone is constructed using a Wigner-Seitz type cell in reciprocal space. For a hexagonal lattice with reciprocal lattice vectors A and B defined as $$A=2\pi\hat{x}+\frac{2\pi}{\sqrt{3}}\hat{y}$$ and $$B=\frac{4\pi}{\sqrt{3}}\hat{y}$$, the Brillouin zone is determined by the boundaries of the reciprocal lattice, specifically within the range of $$[-\frac{2\pi}{\sqrt{3}},\frac{2\pi}{\sqrt{3}}]$$. The construction involves bisecting the lines to the nearest neighbors, analogous to the Wigner-Seitz approach in real space. This method effectively captures the periodicity and symmetry of the lattice in reciprocal space.

PREREQUISITES
  • Understanding of reciprocal lattice vectors
  • Familiarity with Wigner-Seitz cell construction
  • Knowledge of hexagonal lattice structures
  • Basic concepts of solid state physics
NEXT STEPS
  • Study the construction of the first Brillouin zone for different lattice types
  • Learn about the relationship between reciprocal space and electronic band structure
  • Explore the implications of Brillouin zones in band theory
  • Investigate the mathematical derivation of reciprocal lattice vectors
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Students and researchers in solid state physics, materials science, and condensed matter physics who are looking to deepen their understanding of crystal structures and their electronic properties.

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Homework Statement


Not a homework question, but I am attempting to understand what exactly the first Brillouin zone is.

Homework Equations

The Attempt at a Solution


From my textbook, what I'm gathering is that one constructs the first Brillouin zone by constructing a "Wigner-Seitz" type cell in reciprocal space. My question is, is this how one constructs the first Brillouin zone, and if so, why/how does this work?
 
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Essentially, my question is, suppose we have a hexagonal lattice, such that the reciprocal lattice vectors are given by, $$A=2\pi\hat{x}+\frac{2\pi}{\sqrt{3}}\hat{y},$$ and $$B=\frac{4\pi}{\sqrt{3}}\hat{y}.$$ The magnitudes of these are, $$|A|=|B|=\frac{4\pi}{\sqrt{3}}.$$ Is the Brillouin zone just the bound between, $$[-\frac{2\pi}{\sqrt{3}},\frac{2\pi}{\sqrt{3}}].$$ Because, as I understand it, because the magnitudes of reciprocal square lattice vectors are, $$\frac{2\pi}{a},$$ the Brillouin zone is essentially bisection of this (i.e. bisection of line to nearest neighbors as it is with Wigner-Seitz construction).
 

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