# Solid State - Sommerfeld Drude Model, Calculating N of Elect.

1. Mar 24, 2013

### zellwwf

Hello PF :)

1. The problem statement, all variables and given/known data
Consider a system of N electrons at temperature T = 0, each having a mass of m, confined to volume V. Find the number of electrons that:

a) have momentum $p<p_f/2$
b) ...

2. Relevant equations
The relevant equations can be derived below:
but here is a sample of them
$\int F(k)D(k)d\vec{k}$
where F: How many electrons per k-state. function of magnitute of K
where D: density in k-space. (since we turned a discrete sum into an integral), function of magnitude of K too (although it's usually a constant)

3. The attempt at a solution

I have two attempted solutions for A, and they are NOT equal, they differ by a factor of h:

Attempt 1:
note, i will be using the reduced planck's constant, but i can't find the bar, so i will denote it by H.
$\vec{p}=h \vec{k}$
$\frac{\vec{p}}{h}=\vec{k}$
thus:
$d\vec{p}=h d\vec{k}$
$\frac{1}{h}d\vec{p}=d\vec{k}$

Now we need to calculate the density of states in K-space (too early in the morning to write it down how i got it, but i belive it's correct):
$D(k) = \frac{2V}{(2 \pi)^3}$

and since:
$F(k) = 1$ at T = 0;
disregarding spin, as we just multiply by two if we need to 'account' for it in this question

now we can replace these results in our integral above:
$\int^{\frac{p_f}{2}}_{0}F(k)D(k)d\vec{k} = \frac{V}{4h\pi^3}\int^{\frac{p_f}{2h}}_{0}d\vec{p}$
transforming to spherical coordinates reduces the integral to:
$=\frac{V}{4h\pi^3}\int^{\frac{p_f}{2h}}_{0}4\pi p^2dp$
$=\frac{V}{h\pi^2}\int^{\frac{p_f}{2h}}_{0}p^2dp$
$=\frac{V}{24h^4\pi^2}p_f^3$
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Now I will attempt two:
i will try to derive the integral itself for P-space, not K-Space

the only thing that i believe changes is the Density function, D(p)
since:
$\vec{k}=\frac{\vec{p}}{h}=\frac{2\pi}{L}*(\vec{i}+\vec{j}+\vec{k})$
thus:
$\vec{p}=\frac{2h\pi}{L}*(\vec{i}+\vec{j}+\vec{k})$
thus D(p) can be obtained:
$D(p) = \frac{2L^3}{8h^3\pi^3}=\frac{V}{4h^3\pi^3}$

F(p) = 1 also, at T = 0; (again, disregarding spin)
so now we need to only evaluate the integral:

$\int F(p)D(p)d\vec{p}$
$=\frac{V}{4h^3\pi^3}\int^{\frac{p_f}{2}}_{0}d\vec{p}$
$=\frac{V}{4h^3\pi^3}\int^{\frac{p_f}{2}}_{0}4\pi p^2dp$
$=\frac{V}{3h^3\pi^2}p_f^3$

-----
Apparently they are not equal,
i know i made a mistake, or two, or three..
can someone show me :)

2. Mar 25, 2013

### TSny

If I recall correctly, the factor of 2 in the numerator above is to take care of the spin. If so, you are not disregarding the spin.

Note that the upper limit in the integral on the left should be a value of k rather than p. Correcting this will change the upper limit in the integral on right (which should be a value of p). Also, the integrals are over 3 dimensions in k or p space. Did you include enough factors of h when changing the $d\vec{k}$ to $d\vec{p}$?

When evaluating the integral at the upper limit, it appears that you used $p_f$ for the upper limit instead of $p_f/2$