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Fermi-dirac statistics, Griffiths 5.28

by Elvex
Tags: fermidirac, griffiths, statistics
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Elvex
#1
Mar19-07, 08:28 PM
P: 11
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.
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George Jones
#2
Mar20-07, 07:27 AM
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Look at equations [5.103] and [5.104].
Elvex
#3
Mar20-07, 09:32 AM
P: 11
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.

nrqed
#4
Mar20-07, 09:36 AM
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P: 2,986
Fermi-dirac statistics, Griffiths 5.28

Quote Quote by Elvex View Post
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.
Yes, in both cases you must integrate up to [itex] k_F[/itex] only (the k value at the Fermi energy).


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