Fermi-Dirac Statistics

  1. Hey kids,

    The question I'm having trouble with (this time) is as follows:

    Show that the Fermi-Dirac distribution function,

    [tex] f_{FD}(E)=\frac{1}{e^{(\frac{E-E_f}{kT})}+1} [/tex]

    Has the following functional form at T= 0K
    (see attachment)

    Now, the first thing that screamed at me was the division by T in the exponential bit. If T=0, what is going on!?

    The obvious things are:

    E>Ef then f(E) = 0


    E<Ef then f(E) = 1.

    I'm just really confused at how I can show that the function has that form at T=0K

    Any ideas?


    Attached Files:

  2. jcsd
  3. Galileo

    Galileo 1,999
    Science Advisor
    Homework Helper

    I don't understand your question. Didn't you just show the function has that form?
    [tex]\lim_{T \rightarrow 0}f_{FD}(E)=\left\{ \begin{array}{ll}1 & \mbox{if} E<E_f\\ \frac{1}{2} & \mbox{if} E=E_f \\0 & \mbox{if} E>E_f[/tex]
  4. I'm glad somebody else doesn't understand the question either.

    They want me to 'show' that the distribution has the (attached pic) form at T=0.

    The real problem I'm having is how do I "show" that it has that form? Via two lines of maths? That's it?
  5. Galileo

    Galileo 1,999
    Science Advisor
    Homework Helper

    You HAVE just shown it. By taking the limits.
    So yeah, that's it. :)
  6. Well, just use the definition of the Fermi-level... It is the the maximum energy-level at T = 0 K. Just fill up all the available energy-levels with all available electrons. The last electron is placed at the highest energylevel which is called the Fermi-level. Ofcourse all this is done at zero Kelvin. That is why the distribution function has the drawn form. All levels are filled up (probability one) till the Fermi-level. Above this level there are no filled levels since the Fermi-level is the highest. It is just by QM-definition of the Fermi-level

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