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Massless 3D Density of States

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the single particle density of states for massless particles with dispersion E=h_bar ck for a 3D cube of volume V

    2. Relevant equations
    E=pc, p=E/c,
    dp=dE/c, d^3p = 4pi*p^2 dp
    k=sqrt(k_x^2+k_y^2+k_z^2)
    k_j = 2pi/L l_j (j=x,y,z)

    3. The attempt at a solution
    I have tried calculating the density of states in the exact same way as I do for a massive particle but using different energy relations.

    Sum(all K)
    = sum(all kx,ky,kz)
    = int(dl_x dl_y dl_z)
    = int ((2pi/Lh_bar)^3 d^3p)
    = int ((2pi/Lh_bar)^3 4pi*p^2 dp)
    = int ((2pi/Lh_bar)^3 4pi*E^2/c^3 dE)

    D(E) = (2pi/Lh_bar)^3 4pi*E^2/c^3

    The powers are correct for E and C however I seem to have a dependce on the volume (1/L^3 = 1/V) which I should not be getting - does anyone know why I am getting this?

    Thanks,
    James
     
  2. jcsd
  3. Apr 5, 2014 #2

    TSny

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    From ##k_j = \frac{2\pi}{L}l_j##, what do you get for ##dl_j## in terms of ##dk_j##?
     
  4. Apr 6, 2014 #3
    I get:

    dl_j = L/2pi dk_j = L/(2pi*h_bar) dp_j

    Sorry that must have been a typo, I still seem to have an L^3 term which won't go away though.

    I now get:
    = int(dl_x dl_y dl_z)
    = int ((L/2pi*h_bar)^3 d^3p)
    = int ((L/2pi*h_bar)^3 4pi*p^2 dp)
    = int ((L/2pi*h_bar)^3 4pi*E^2/c^3 dE)

    D(E) = (L/2pi*h_bar)^3 4pi*E^2/c^3
     
  5. Apr 6, 2014 #4

    TSny

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    OK, I think that's corect. But the density of states is often defined on a "per unit volume" basis. If that's what you need, how would you fix your result so that D(E)dE represents the number of states per unit volume with energy between E and E+dE?
     
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