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
     
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