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Fermi distribution: Sum over states --> integral over states

  • Thread starter Nikitin
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  • #1
726
27

Homework Statement


http://web.phys.ntnu.no/~kolausen/TFY4230/.oldExams/17_eksdes12.en.pdf
solution: http://web.phys.ntnu.no/~kolausen/TFY4230/.oldExams/18_losdes12.en.pdf

Look at problem 4a, formula (27) or the expression between (29) and (30).

My professor keeps converting sums into integrals in a manner I don't understand. In fact, I think he might be wrong and miss a factor of 1/8. OK let me explain what I mean:

For instance, when calculating the total number of particles ##N## by summing the number of particles per state, ##N_k##, at ##T=0## he does this:

$$N=\sum_k N_k = \sum_k ln(\frac{1}{1+e^{\beta( \mu + E_k}}) \rightarrow \int_0^{\infty} \frac{dk}{(2 \pi)^3/L^3} 4 \pi k^2 ln(\frac{1}{1+e^{\beta( \mu + E_k}})$$, where ##k_x = 2 \pi n_x/L##, ##k_y = 2 \pi n_y/L## and ##k_z = 2 \pi n_z/L##

OK, so he inserts a factor ## (\Delta k)^3 / (\Delta k)^3 ##, where ##\Delta k = 2 \pi /L##, into the sum, and since ##L## is very large the sum becomes a Riemann sum and thus can be rewritten into a volume-integral over the k-space.

BUT: my professor integrates over ALL of the volume in k-space, even for negative ##k_x##, ##k_y## and ##k_z##! That is unphysical, since neither ##n_x,n_y## or ##n_z## can be negative. In fact, the only legal way to integrate this is by integrating over the octant of space where all three k-axises are positive. Hence you must multiply the integral with a factor of ##1/8##, which my professor doesn't do .

Please help I have my exam on friday!!!
 
Last edited:

Answers and Replies

  • #2
DrClaude
Mentor
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That is unphysical, since neither ##n_x,n_y## or ##n_z## can be negative.
That is not correct. The problem statement itself mentions that the n's can be positive and negative. Note that the starting point is a particle in a box with periodic boundary conditions.
 
  • #3
726
27
wow. this is embarrassing. OK thanks, problem solved..
 

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