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Fermi gas density of states

  1. May 21, 2014 #1
    1. The problem statement, all variables and given/known data

    20fwcc7.png

    Part (a): Plot fermi energy as a function of N
    Part (b): Derive the density of states and find its value
    Part (c): How many atoms reside at 20% of fermi energy? Estimate diameter of cloud
    Part (d): For the same atoms without spin, why is the cloud much smaller? Estimate the transition temperature.

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    The Fermi Energy is the highest energy level occupied by the atoms at T = 0.

    We know that ##\epsilon_f \propto \left( \frac{N}{V}\right)^{\frac{2}{3}}## so the graph looks like:
    29c1ma8.png

    Part (b)

    I'm not sure why there is a factor of ##\omega^3##, as the density of states seem to be independent:

    [tex]g_{(k)} dk = (2S+1) \frac{1}{8} \times \frac{4\pi k^2 dk}{(\frac{\pi}{L})^3}[/tex]
    [tex]g_{(k)} dk = \frac{V}{\pi^2} k^2 dk[/tex]

    Using the substitution ##E = \frac{\hbar^2 k^2}{2m}## still doesn't produce any ##\omega## What does it mean when they say ##\epsilon >> \hbar \omega##?
     
  2. jcsd
  3. May 21, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    That equation for the energy is valid for a free particle, not for one confined to a harmonic oscillator.

    It means you can neglect the discreteness of the energy levels and consider them to be continuously distributed.
     
  4. May 21, 2014 #3
    Part (b)
    For one confined to a harmonic oscillator, ##E = (n+\frac{1}{2})\hbar \omega##. I've been thinking what the density of states would look like in n-space. Since ##\epsilon>>\hbar \omega##, I can simply take ##\epsilon \approx n\hbar \omega##.

    Would it be something like:
    [tex]g_{(n)} dn = (2S+1) \frac{1}{8} \frac{4 \pi n^2 dn}{1}[/tex]
    [tex]g_{(\epsilon)} d\epsilon= \frac{\pi}{\hbar^3} \frac{\epsilon^2}{\omega^3} d\epsilon[/tex]

    Therefore ##\alpha = \frac{\pi}{\hbar^3}##.

    At T = 0 K, the occupation number up till fermi energy is 1. After fermi energy, it is 0. It is a heavyside function.

    [tex]N = \sum n_{(\epsilon)} = \int g_{(\epsilon)} d\epsilon[/tex]
    [tex]N = \int_0^{\epsilon_F}\frac{\pi}{\hbar^3} \frac{\epsilon^2}{\omega^3} d\epsilon[/tex]
    [tex]\epsilon_F = \hbar \omega\left(\frac{3N}{\pi}\right)^{\frac{1}{3}}[/tex]

    For ##N = 10^6## and ##\omega = 2\pi \times 10^5##,
    [tex]\epsilon_F = 6.52 \times 10^{-27} J = 41 \space neV[/tex]

    Checking online, the fermi energy of Lithium is 4.7 eV. I'm off by 10 orders of magnitude..
     
    Last edited: May 21, 2014
  5. May 22, 2014 #4

    DrClaude

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    Staff: Mentor

    I guess you stumbled upon the Fermi energy for electrons in solid lithium. For a degenerate Fermi gas of atoms, that's indeed the order of magnitude. The corresponding Fermi temperature is ##T_\mathrm{F} \approx 0.5\ \mathrm{mK}##, which is why advanced cooling techniques are required!
     
  6. May 22, 2014 #5
    Thanks alot, that does make sense. Where does the degeneracy of ##\frac{(n+1)(n+2)}{2}## come into play here? Should it go into the density of states or the partition function?
     
  7. May 24, 2014 #6
    Any input on whether my density of states is right?

    [Edit]I'm concerned about the missing degeneracy given in the question of ##\frac{(n+1)(n+2)}{2}##
     
    Last edited: May 24, 2014
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