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Density of states, and integral of the Sommerfeld type

  1. Feb 5, 2016 #1
    It is easy to show that when you have a quantum system, lets think for example in electrons in a metal, then there appears summation over electron states of the form, e.g. for the energy for a free electron gas at T=0K:

    ##E=2 \sum_{k\leq k_f} \frac{\hbar^2}{2m}k^2##

    Where ##k_f## denotes the Fermi wave vector.

    In general one has sums of the form:

    ##\sum_{k} F(\vec{k})##

    Now, for a free electorn gas one has that the volume in k space per allowed k value is ##\Delta \vec{k}=\frac{8\pi^3}{V}## where V is the volume of the solid (I'm following Aschroft and Mermin, solid state physics, page 37). So that in the limit of large V, one has:

    ##\displaystyle \lim_{V\rightarrow \infty} \frac{1}{V} \sum_{k} F(\vec{k}) \Delta k= \int \frac{d^3k}{8\pi^3} F(k).##

    For the energy at non zero temperature one has

    ##E=2 \sum_k \epsilon(k) f(\epsilon_k)##

    With:
    ##\epsilon_k=\frac{\hbar^2}{2m}k^2##

    In general one has a dependence on energy ##\epsilon## per electron on the wave vector k, and f denotes the Fermi-Dirac distribution.

    In the limit of large V, one then has:

    ##\lim \frac{E}{V}=u=\frac{1}{4\pi} \int \epsilon(k)f(\epsilon(k))d^3k##

    Now, for the arbitrary function ##F(k)##, one can solve the integral in spherical coordinates:
    ##\int \frac{d^3k}{8\pi^3} F(k)=\frac{1}{\pi^2}\int dk k^2 F(\epsilon_k)##

    So, thats ok, now the next step the book does is the one that is not clear to me how to get. In this step, this integral is taken to the one that is commonly used to evaluate the internal energy, or the number of electrons, etc. its the integral which I call as the "Sommerfeld type" because is that Sommerfeld used to make the "Sommerfeld expansion"

    In this step, it is introduced the density of states

    ##\frac{1}{\pi^2}\int dk k^2 f(\epsilon_k)=\int_{-\infty}^{\infty} d\epsilon g(\epsilon) F(\epsilon)##

    Where is used the fact that the integrand depends on ##\vec k## only through the electronic energy ##\epsilon =\frac{\hbar^2}{2m}k^2##

    I hoped to derive this result just by calculation of the differentials, to get something like ##d\epsilon g(\epsilon)=\frac{k^2 dk}{\pi^2}##

    The density of states is defined as ##g(\epsilon)=\frac{dn}{d\epsilon}##

    Where ##n=\frac{N}{V}## is the number of electrons per unit volume, the electronic density.
     
  2. jcsd
  3. Feb 10, 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. Feb 11, 2016 #3
    Hi Telemachus! I don't quite understand the question here - is it the change of variable from [itex]k[/itex] to [itex]\epsilon[/itex] that's causing confusion?
     
  5. Feb 12, 2016 #4
    Yes, thats it.
     
  6. Feb 12, 2016 #5
    Yes, thats it, I dont understand how to get the density of states in that step. I know the integral makes perfect sense, I have used it, but I don't know how to derive the result rigorously.

    PS: sorry for the double post.

    PS2: I saw a mistake in the formula I've posted, it is obvious, but anyway it should read:

    ##\frac{1}{\pi^2}\int dk k^2 F(\epsilon_k)=\int_{-\infty}^{\infty} d\epsilon g(\epsilon) F(\epsilon)##
     
    Last edited: Feb 12, 2016
  7. Feb 12, 2016 #6
    The change of variables in the integral should be straightforward:
    [tex]\epsilon = \frac{\hbar^2 k^2}{2m}\\

    k = \sqrt{\frac{2m}{\hbar^2}} \; \epsilon^{1/2}\\

    dk = \frac{1}{2} \sqrt{\frac{2m}{\hbar^2}} \; \epsilon^{-1/2} \; d\epsilon
    [/tex]
     
  8. Feb 12, 2016 #7
    Great, thanks! thats exactly what I wanted, I don't know why I couldn't see that the density of states emerged naturally from there!
     
  9. Feb 12, 2016 #8
    No problem!

    Btw your integrals should go from [itex]0[/itex] to [itex]\infty[/itex], not from [itex]-\infty[/itex] to [itex]\infty[/itex], since [itex]k[/itex] and [itex]\epsilon[/itex] are both positive.
     
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