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

[zellwwf]

Hello PF :)

Homework Statement

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) ...

Homework 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)

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 believe 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
----------------------------------------------------------------------------------------------------
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 :)

 

[TSny]

Hello PF :)

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 believe it's correct):
D(k) = \frac{2V}{(2 \pi)^3}

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.

\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}

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 $d\vec{p}$?

Now I will attempt two:
\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

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