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

1. Dec 16, 2014

### Nikitin

1. The problem statement, all variables and given/known data
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 .

Last edited: Dec 16, 2014
2. Dec 16, 2014

### Staff: Mentor

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. Dec 16, 2014

### Nikitin

wow. this is embarrassing. OK thanks, problem solved..