Why only first brillouin zone?

In summary, the justification for only considering the first Brillouin zone when calculating energy gaps 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. This means that only up to a reciprocal lattice vector is the plane wave unique, and this is seen by considering the reciprocal lattice vectors which in 1D are given by \mathbf{G}=\frac{2n\pi}{a}: e^{i G_{n1} r}=e^{i G_{n2} r}. For all values of n1 and n2, the interval is often chosen to be -pi/a to pi/a to get the band
  • #1
neu
230
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
 
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  • #2
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:

[tex]
\psi_{nk}(x) = e^{i k r} u_{nk}(x)
[/tex]

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 [tex]\mathbf{G}=\frac{2n\pi}{a}[/tex]:

[tex]
e^{i G_{n1} r}= e^{i G_{n2} r},
[/tex]

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.
 
  • #3
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!
 

1. Why is the first Brillouin zone important in solid state physics?

The first Brillouin zone is important in solid state physics because it is a fundamental concept that helps us understand the behavior of electrons in a crystalline lattice. It provides a way to map out the energy levels of electrons in a crystal and understand their interactions with the crystal lattice. This is crucial for understanding the properties of materials and designing new technologies.

2. How is the first Brillouin zone defined?

The first Brillouin zone is defined as the primitive unit cell in reciprocal space that contains all the points that can be reached by a wave vector perpendicular to a particular crystal plane. In other words, it is the region in reciprocal space that represents all possible momenta of electrons in a crystal.

3. Why is the first Brillouin zone a hexagon or square in 2D systems?

In 2D systems, the first Brillouin zone is a hexagon or square because these shapes allow for the most efficient packing of points in reciprocal space. This is due to the symmetry of the lattice in real space, which is reflected in the symmetry of the first Brillouin zone in reciprocal space.

4. How does the size of the first Brillouin zone relate to the lattice spacing?

The size of the first Brillouin zone is inversely proportional to the lattice spacing. This means that as the lattice spacing decreases, the first Brillouin zone becomes larger. This relationship is important for understanding the behavior of electrons in different materials, as the lattice spacing can affect the electronic properties of a material.

5. Can higher order Brillouin zones be observed in experiments?

Yes, higher order Brillouin zones can be observed in experiments, but they are usually not as important as the first Brillouin zone. This is because the higher order zones contain points that are already represented in the first Brillouin zone. However, in certain cases, such as when studying the electronic properties of very large crystals, higher order zones may need to be taken into account.

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