Fermi-dirac statistics, Griffiths 5.28

  1. 1. The problem statement, all variables and given/known data
    Evaluate the integrals (eqns 5.108 and 5.109) for the case of identical fermions at absolute zero.


    2. Relevant equations

    5.108
    [tex]N=\frac{V}{2\pi^{2}}\int_{0}^{\infty}\frac{k^2}{e^{[(\hbar^{2}k^{2}/2m)-\mu]/kT}+1}dk[/tex]

    5.109
    [tex]E=\frac{V}{2\pi^2}\frac{\hbar^2}{2m}\int_0^{\infty}\frac{k^4}{e^{[(\hbar^{2}k^{2}/2m)-\mu]/kT}+1}dk[/tex]
    3. The attempt at a solution

    Ok so at absolute zero, the chemical potential is equal to the fermi energy E_f. I'm not sure how to approach either integral because of the T dependence in the denominator in the argument of the exponential.
    Aren't there two cases, one for the energy of the state being above the chemical potential, and another for it being less than.
    If the energy is less, then the argument goes to - infinite, and the integral is just of k^2, from 0 to infinite... that doesn't seem right.
    If the energy is greater than mu, then the argument goes to positive infinite, and the integrand goes to 0. Fantastic.

    There's got to be something going on with the expressions in the argument of hte exponential to give a reasonable integrand for T=0.
    I think I'm missing some crucial observation.
     
  2. jcsd
  3. George Jones

    George Jones 6,396
    Staff Emeritus
    Science Advisor
    Gold Member

    Chegg
    Look at equations [5.103] and [5.104].
     
  4. OK, so for the first integral. I have a feeling that I need to change the limits of integration because once I get to k values that exceed the Fermi-Energy, the integrand goes to zero, so setting the upper limit to a specific k value will ensure convergence as well.

    Oh ok, this is the probably the same method for the second integral as well.
     
  5. nrqed

    nrqed 3,077
    Science Advisor
    Homework Helper

    Yes, in both cases you must integrate up to [itex] k_F[/itex] only (the k value at the Fermi energy).
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?