Density of states

Discussion in 'Physics' started by boks, Nov 14, 2008.

1. boks Thread Starter Active Member

Oct 10, 2008
218
0
1. The problem statement, all variables and given/known data

We study a one dimensional metal with length L at 0 K, and ignore the electron spin. Assume that the electrons do not interact with each other. The electron states are given by

$\psi(x) = \frac{1}{\sqrt{L}}exp(ikx), \psi(x) = \psi(x + L)$

$\psi(x) = \psi(x + L)$

What is the density of states at the Fermi level for this metal?

3. The attempt at a solution

According to my book, the total energy of the system is

$E = \frac{\hbar^{2}\pi^{2}n^{2}}{2mL^{2}}$

why is this?

It's evident that k = n*2*pi because of the boundary contidions. I don't know what to do next.

Last edited: Dec 7, 2008
2. boks Thread Starter Active Member

Oct 10, 2008
218
0
Nobody knows how to do this?

3. blazedaces Active Member

Jul 24, 2008
130
0
Are you sure this is the correct equation? According to those two equations, k must equal 0. That doesn't sound right...

And why isn't this wave equation dependent on time? I'll just ignore that i guess...

Explain to me why k = n * 2 * pi according to the information you have provided.

$E = \frac{\hbar^{2}\pi^{2}n^{2}}{2mL^{2}}$ has nothing to do with your wave equation. Look up the proof, as I'm sure it's in your book as well as all over the internet.

It has to do with schrodinger's wave equation and its time-independant solution.

-blazed

4. boks Thread Starter Active Member

Oct 10, 2008
218
0
$\psi(0) = \frac{1}{\sqrt{L}$ = $\psi(L) = \frac{1}{\sqrt{L}exp(ikL)$

This yields

exp(ikL) = cos(kL) + i sin(kL) = 1

so

kL = n*2*pi

5. blazedaces Active Member

Jul 24, 2008
130
0
Ah, I see. Point taken.

So, is it exp(ikL) or exp(-ikL)? Otherwise, why is exp(ikL) in the denominator of the fraction?

Well, if this helps, the density of quantum states per unit volume of a crystal is given by (again, you can look up the proof yourself if you would like):

$g(E)=\frac{4\pi(2m)^{3/2}}{h^3}\sqrt{E}$

So for the fermi energy level you have to plug in the fermi energy for E.

Good luck,
-blazed