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Fermi Surface squashed by potentials

  1. Apr 1, 2015 #1
    Taken from my textbook:


    My understanding is that:

    • One valence electron, 2 spin states -> Half-filled Brillouin zone
    • Seeking inspiration from "Nearly Free Electron Model": gaps open up at zone boundaries
    • States nearer to zone boundaries get pushed down in energy further

    Since a fermi surface is a surface of constant energy in k-space, shouldn't the surfaces nearer to the zone boundaries that get pushed down in energies get repelled even more? It seems that surfaces nearer to the boundaries get closer even! Why are the fermi seas like these below?

  2. jcsd
  3. Apr 2, 2015 #2


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    Take a simple potential with this symmetry as an example: ##V(x,y)=A \cos(x)\cos(y)##. Can you find a direction along which the potential is constant? To which lines do these directions in reciprocal space? What happens at the cell boundary?
  4. Apr 4, 2015 #3

    The surfaces of constant potential are the circles* about (0,0). Given this is the fermi surface, these surfaces are the reciprocal space. Still doesn't answer my question about "pushed down in energy -> decrease in radius, squashed inwards"?
  5. Apr 8, 2015 #4
  6. Apr 10, 2015 #5


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    Sorry, I was on the wrong track and had no time to think about your problem recently. Now I think I understand the behaviour. Near the BZ boundary, the lower energy band will be lowered as compared to the free electron case, so the Fermi surface will be shifted to higher k values. At stronger potentials, it will even protrude into the second BZ or even farther.
  7. Apr 10, 2015 #6
    I get the downward shift in energy due to a perturbation. Why will the fermi surface be shifted to higher k values?
  8. Apr 11, 2015 #7


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    The fermi surface is a surface of constant energy. If the levels split, this constant energy value will be reached at higher values of k. Think of an irregularly shaped plate: where the rim is lower, soup will ooze out more.
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