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Nearly free electron model - band gap

  1. Mar 4, 2015 #1
    For a wavefunction at the Brillouin boundary we have:

    [tex]\langle k|H|k\rangle = \epsilon_0 (\vec k) [/tex]
    [tex]\langle k'|H|k'\rangle = \epsilon_0 (\vec k+\vec G) [/tex]
    [tex]\langle k'|H|k\rangle = V_G = \frac{1}{L^3} \int e^{i(\vec k - \vec k') \cdot \vec r} V(r) d\vec r [/tex]
    [tex]\langle k|H|k'\rangle = V_G^* [/tex]

    Using degenerate perturbation theory, we diagonalize the hamiltonian and find the eigenvalues (energies) to be:
    [tex]E_{\pm} = \epsilon_0 (\vec k) \pm |V_g|[/tex]

    This would mean that for every ##k## right on the boundary we have a bandgap of ##2|V_G|##.


    Since ## V_G = \frac{1}{L^3} \int e^{i(\vec k - \vec k') \cdot \vec r} V(r) d\vec r##, shouldn't its magnitude be fixed at ##2|V_G##? However, in my notes for a periodic perturbation ##V_G = \tilde V cos(\frac{2n\pi}{a})## the gaps are increasing with ##k## (The bandgap at ##k=\frac{2\pi}{a}## seems to be twice as big as the bandgap at ##k=\frac{\pi}{a}##.

    freeelectron1.png
     
  2. jcsd
  3. Mar 9, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 13, 2015 #3
    Thanks for the bump greg. I think the potential here depends on ##k##, even though not stated. The constant band-gap thing only works when the potential is periodic like ##
    V_G = \tilde V cos(\frac{2n\pi}{a})##.
     
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