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Density of states, volume

  1. Apr 17, 2013 #1
    In my text:

    The number of states per unit volume of the real space & the reciprocal space is given by


    1 / (4∏³)

    No further explanation is given.

    How do you get to this 4∏³

    And how come the density of states is the same in real space & reciprocal space?

    I think this is incorrect, they should be the reversed version of each other.

    Thx in advance
     
  2. jcsd
  3. Apr 17, 2013 #2
    Consider a cubic box of size L. Then periodic boundary conditions imply that kx=nx*2pi/L where nx=0,1,2,3,... The same follows for ky and kz.
    Now you can see that in k-space a state with a given k occupies a k-space volume of (2pi/L)^3. Then the number of states per unit volume is (1/V) L^3 * 2 /(2pi)^3 and you get that result. The factor of 2 comes from the fact that the electron has 2 distinct values for the z component of the spin. Note that if you consider a rectangular box you would obtain an identical conclusion.
     
  4. Apr 24, 2013 #3
    I have attached a figure from Ashcroft and Mermin.

    Do this. In the case of 2D, take the reciprocal lattice and you draw a circle or radius k from the center.

    How many electrons states you would have in that area encircled in k-space? If you can figure this out, you can easily find the answer you are looking for. Also, in the case of electrons, there are 2 electrons allowed for each state (spin).

    If all else fails, you can use this formula. [itex]g(E)dE = 2g(k)dk/\nabla_k E(k)[/itex]
     

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    Last edited: Apr 24, 2013
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