Why only first brillouin zone?

Click For Summary
SUMMARY

The discussion centers on the justification for only considering the first Brillouin zone when analyzing energy gaps in crystalline materials. Participants explain that Bragg reflection at the zone boundary causes electron wavevectors to be diffracted back into the first Brillouin zone. The periodicity of the wave function in a crystal, represented as a function of lattice periodicity times a plane wave, leads to the conclusion that all Brillouin zones are equivalent. Therefore, the analysis can be simplified to focus solely on the first Brillouin zone without loss of generality.

PREREQUISITES
  • Understanding of Bragg reflection and its role in diffraction.
  • Familiarity with Bloch waves and their properties in crystal lattices.
  • Knowledge of reciprocal lattice vectors and their significance in solid-state physics.
  • Concept of translational symmetry in crystalline structures.
NEXT STEPS
  • Study the implications of Bragg reflection in solid-state physics.
  • Explore the mathematical formulation of Bloch's theorem in more detail.
  • Investigate the role of Umklapp processes in thermal conductivity.
  • Learn about the band structure of materials and how it relates to Brillouin zones.
USEFUL FOR

Physicists, materials scientists, and students studying solid-state physics, particularly those interested in the behavior of electrons in crystalline structures and the implications of Brillouin zones on electronic properties.

neu
Messages
228
Reaction score
3
I realize this a fundamental flaw in my understanding but I can't seem to get my head around it.

I understand that bragg reflection occurs at the zone boundary and so electron wavevectors are diffracted back into the 1st brillouin zone, but what is the justification for only considering the 1st brillouin zone, for example, when considering energy gaps?

This is also connected with my earlier question about why Umklapp processes are required, where large k is transmitted back into 1st brillouin zone by G:

K1+K2=K3+G
 
Physics news on Phys.org
The reason why it is only necessary to consider the first Brillouin zone lies in the fact that the wave function in a crystal is written as a function with the periodicity of the lattice times a plane wave, so that in one dimension:

<br /> \psi_{nk}(x) = e^{i k r} u_{nk}(x)<br />

The plane wave is unique only up to a reciprocal lattice vector. This is seen by considering the reciprocal lattice vectors which in 1D are given by \mathbf{G}=\frac{2n\pi}{a}:

<br /> e^{i G_{n1} r}= e^{i G_{n2} r},<br />

for all values of n1 and n2. The interval is often chosen to be -pi/a to pi/a to get the band structure around a symmetric interval.
 
Ok, thanks for that

I understand Bloch waves. So are you saying that a bloch wave traveling through the crystal is the same in all brillouin zones so consideration of which one is arbitary?

So what about the electrons which remain within the 1st brillouin zone as their wavevector K is large enough to experience diffraction (laue condition: delta K=G & at zone boundary min kin for diffraction = (delta k/2) = G/2) ?

I think i understand. The pure isotropic crystal has translational symetry, any position has same properties as that position translated by the lattice vectors. So every brillouin zone is equivalent and choice between any 1st zone, or any 2nd zone is arbitrary. So electrons which are bound behave in the same way in every brillouin zone, as do electrons passing through the crystal.

so may only consider 1st brilluoin zone.

is that the reasoning or am I missing something? Sorry to go on about it!
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
8K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K