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A Basic questions on crystal lattice: k-vectors and Brillouin

  1. Jun 23, 2016 #1
    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
     
    Last edited: Jun 23, 2016
  2. jcsd
  3. Jun 28, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jun 28, 2016 #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
     
  5. Jun 30, 2016 #4
    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.
     
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