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I Fermi Energy Calculations About Non Parabolic Dispersions

  1. Sep 10, 2018 #1
    It is easy to understand that for a free electron, we can easily define the energy state density, and by doing the integration of the State density* Fermi-Dirac distribution we will be able to figure out the chemical potential at zero kelvin, which is the Fermi-Energy. Hence, we can define the Fermi Momentum correspondingly.

    However, I feel rather struggling to define the fermi-momentum and Fermi energy for a non-parabolic dispersion.

    Say, For example, A 2D system, I have a dispersion
    E = A*k^2 +B*k^3

    So k(E) will be in general complicated due to its cubic relations.

    So If we still want to evaluate the Fermi Energy, Do still solve the integration
    int _0 ^E_f state density*Fermi-Dirac Distribution dE = Number of Particles
    to evaluate E_f?

    And I think this integration will be very complicated.

    Also, Under this circumstance, how do we define the Ferim-Momentum? Do we still solve
    E_f = A k_F^2 + B k_F^3 to find k_F?

  2. jcsd
  3. Sep 10, 2018 #2


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    Non-quadratic dispersion is not a big deal as long as the system is isotropic. You don't need to integrate anything!
    All you need is a volume of the occupied state. For 3D case, the Fermi surface is a sphere whose volume is equal to ## V_F = \frac 4 3 \pi k^3_F##. If you have one electron per primitive unit cell, the Fermi sphere volume is 1/2 the volume of the first Brillouin zone. That gives you the Fermi momentum using elementary math. Then the Fermi energy can be easily obtained by inserting the Fermi momentum value into the dispersion formula. For the 2-D case, the Fermi surface is a cylinder. Project that onto a plane and use planar geometry formulae.
  4. Sep 10, 2018 #3
    Thanks for your reply!

    I think I somehow understand the idea of this. Would you like to recommend some literature on these materials?
  5. Sep 10, 2018 #4


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    I don't know anything specific about non-quadratic dispersion. But this should be intuitive. You fill up the band with all available electrons and as long as the dispersion is isotropic, the shape of the Fermi surface is spherical. From Born-van Karman boundary conditions, you will get that number of allowed states in each band is equal to 2 x number of Bravais cells in the crystal (2 because of spin degeneracy). This point is easy to see in the case of a cubic lattice.
    If the bands are not symmetric, then the situation gets a bit more complicated. In simpler case, they could be ellipsoidal. These could happen in semi-metals when the bands are much less than 1/2 full. In general, the shape can be quite complex. See for example Ashcroft and Mermin for some examples. I did Fermi surface calculations for intercalated graphite and had to do numerical calculations.
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