Basic questions on crystal lattice: k-vectors and Brillouin

[DivGradCurl]

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:

$$k_x = \frac{2\pi}{L_a}, \qquad k_y = \frac{2\pi}{L_b} , \qquad k_z = \frac{2\pi}{L_c}$$

Assuming Born–von Karman periodic boundary conditions.

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

My attempt:

$$N_a = \frac{L_a}{a},\qquad N_b = \frac{L_b}{b},\qquad N_c= \frac{L_c}{c}$$

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

 

[DivGradCurl]

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:

$$N = (2N_a) (2 N_b)(2 N_c) = 8 N_a N_b N_c$$

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

$$n_a \frac{2\pi}{L_a} = \frac{\pi}{a} \therefore n_a = \frac{L_a}{2a} = \frac{N_a a}{2a} = \frac{ N_a}{2}$$

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:

$$N = (2N_a) (2 N_b)(2 N_c) = 8 N_a N_b N_c$$

Are those correct? Thank you very much

 

[hokhani]

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:

$$N_a = \frac{L_a}{a},\qquad N_b = \frac{L_b}{b},\qquad N_c= \frac{L_c}{c}$$

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.