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.

 

Nobody knows how to do this?

 

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?

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...

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.

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

 

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

\(\psi(0) = \frac{1}{\sqrt{L}\) = \(\psi(L) = \frac{1}{\sqrt{L}exp(ikL)\)

This yields

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

so

kL = n*2*pi

 

\(\psi(0) = \frac{1}{\sqrt{L}\) = \(\psi(L) = \frac{1}{\sqrt{L}exp(ikL)\)

This yields

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

so

kL = n*2*pi

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