Basic questions on crystal lattice: k-vectors and Brillouin

In summary, the number of electrons in the first Brillouin zone is determined by the number of unit cells in the crystal and each k-point can have at most two electrons with different spins.
  • #1
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Hi all,

I’m brushing up my skills on solid state physics and I have a few questions about crystal lattices:

1. What’s the spacing between allowed kx, ky, and kz states for a lattice of dimensions La x Lb x Lc?


My attempt:

The spacing is kx, ky, and kz:

[tex]k_x = \frac{2\pi}{L_a}, \qquad k_y = \frac{2\pi}{L_b} , \qquad k_z = \frac{2\pi}{L_c}[/tex]

Assuming Born–von Karman periodic boundary conditions.

2. How many electrons fit in the first Brillouin zone?

My attempt:

[tex]N_a = \frac{L_a}{a},\qquad N_b = \frac{L_b}{b},\qquad N_c= \frac{L_c}{c}[/tex]

for an orthorhombic lattice of primitive direct lattice constants a, b, and c.

So, the total number of electrons is N = Na x Nb x Nc

Is this right?

Thanks
 
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  • #3
I have something to add. I think I figured out these questions; please confirm:

1. k-spacing:

Looks right on the initial post

2. Number of electrons:

[tex]N = (2N_a) (2 N_b)(2 N_c) = 8 N_a N_b N_c[/tex]

From the 1D analysis at the edge of the 1st Brillouin zone:

[tex] n_a \frac{2\pi}{L_a} = \frac{\pi}{a} \therefore n_a = \frac{L_a}{2a} = \frac{N_a a}{2a} = \frac{ N_a}{2} [/tex]

so for half of the the 1st Brillouin zone. For the whole thing, the number of k-states is 2 n_a = N_a. There are two electrons per k-state so (2N_a) electrons for dimension "a". For the other two, the analysis is identical, which again leads to:

[tex]N = (2N_a) (2 N_b)(2 N_c) = 8 N_a N_b N_c[/tex]

Are those correct? Thank you very much
 
  • #4
DivGradCurl said:
Hi all,

I’m brushing up my skills on solid state physics and I have a few questions about crystal lattices:


2. How many electrons fit in the first Brillouin zone?

My attempt:

[tex]N_a = \frac{L_a}{a},\qquad N_b = \frac{L_b}{b},\qquad N_c= \frac{L_c}{c}[/tex]

for an orthorhombic lattice of primitive direct lattice constants a, b, and c.

So, the total number of electrons is N = Na x Nb x Nc

Is this right?

Thanks
All the electrons are in FBZ distributed in several bands. If you mean how many electrons can settle in a specific band the answer is that for each k you can have at most two electrons with different spins and the number of k-points in FBZ is determined by the number of unit cells in the crystal.
 

1. What is a crystal lattice?

A crystal lattice is a three-dimensional arrangement of atoms, ions, or molecules in a crystalline solid. It is the repeating pattern that forms the basis of a crystal's structure.

2. What are k-vectors in relation to crystal lattices?

K-vectors, also known as wave vectors, are mathematical vectors that describe the direction and magnitude of the wave associated with a particle in a crystal lattice. They are used to calculate the energy and momentum of particles in the lattice.

3. How are k-vectors related to the Brillouin zone?

The Brillouin zone is a fundamental unit in the study of crystal lattices, representing the range of k-vectors that describe the energy and momentum of particles in the lattice. The k-vectors within the Brillouin zone are unique and can be used to understand the properties of the crystal.

4. What is the significance of the Brillouin zone in crystal lattices?

The Brillouin zone is significant because it allows for the simplification of calculations and analysis of crystal lattices. It also helps in understanding the behavior of particles and waves within the lattice, making it a crucial concept in solid state physics and materials science.

5. How are k-vectors and the Brillouin zone used in real-world applications?

K-vectors and the Brillouin zone are used in various fields such as materials science, condensed matter physics, and electronics. They are crucial in understanding the properties of materials and designing new materials for specific applications, such as in semiconductors, superconductors, and photonic crystals.

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