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Density of States (Condensed Matter)

  1. Aug 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I need some help with the following problem:

    [Broken]

    2. Relevant equations

    ##\rho(k) dk = \frac{L}{\pi} dk##

    ##L=Na##

    ##\omega^2= \omega_m^2 \ sin^2 (qa/2)##

    3. The attempt at a solution

    The density of states is given by:

    ##g(\omega)= \rho (k) / \frac{dw}{dk}##​

    Where

    ##\frac{d\omega}{dk} = \frac{a}{2} \ cos \frac{qa}{2}##

    ##g(\omega) = \frac{L}{\pi} \frac{2}{a} \frac{1}{\omega_m \ cos (qa/2)}##

    Using the identity

    ##sin^2 x + cos^2 x =1 \implies cos x = \sqrt{1-sin^2 x} , \ cos(qa/2)=\sqrt{1-sin^2(qa/2)}##

    We get

    ##g(\omega)=\frac{L}{\pi} \frac{2}{a \omega_m \sqrt{1-sin^2(qa/2)}} = \frac{2Na}{a \pi \omega_m \sqrt{1-sin^2(qa/2)}}##

    ##\therefore g(\omega)= \frac{N}{\pi} \frac{2}{\sqrt{\omega_m^2 -\omega^2}}##​

    But this is not the correct answer. Why is there a "2" on the numerator, and how can we get rid of this factor of 2? :confused:

    Did I make a mistake, or is this a typo in the question?

    Any help is greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 19, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I believe your derivation is correct. The factor of 2 should be there as you have shown. See equation (35) here.

    Keeping track of factors of 2 can get confusing when comparing periodic boundary conditions and fixed boundary conditions. But the density of states should end up the same for either boundary condition.
     
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