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Infinite effective mass in metals

  1. Jun 2, 2012 #1
    Hello, I have a question about conduction in metals.

    I guess you all know a common pedagogical picture where an electron bands are drawn as ~ cosine curves in 1D or E_0 - cos(k_x) - cos(k_y) in 2D.

    Now, in metals , we were told that the Fermi surface passes through the band. Therefore, electrons "under" Fermi surface can go to states above Fermi surface and thus conduct electricity. That sounds good so far.

    We also defined an effective mass as m* ~ [itex](\frac{d^2E}{dk^2})^{-1}[/itex]. But what happens if Fermi surface crosses the cosine band exactly in the middle (or where it is linear locally), then the double derivative is zero, and m* = infinity. How can these electrons move anywhere or even conduct? I see possible resolutions to this question:

    1) electrons further away from such surface can still move to other states, but this will be a bad conductor (because m* for those nearby electrons will be huge)

    2) in real metals Fermi surface doesn't cross bands where there is no curvature?

    3) maybe conduction in not related to m*, but to how many electrons go above the Fermi surface ?

    4) maybe m* definition breaks down far away from Brillouin zone boundaries?

    5) maybe we don't define effective mass for metals? only for semiconductors?

    6) something else?

    Thanks in advance
    Last edited: Jun 2, 2012
  2. jcsd
  3. Jun 3, 2012 #2
    I found the answer: effective mass is only useful near the top or bottom of the band, where the dispersion is nearly quadratic, and hence this can be thought of as a free particle with effective mass ( E_free = h^2 k^2/(2m*) ).

    After reading another book, I think I understood how conduction really happens in metals. I will try to explain what I understood here.

    Suppose Fermi surface is somewhere close to the bottom of a band ( in a perfect metal, where there is NO scattering), and so conduction is done by electrons. Consider an electron at k=0 (bottom of the band, not too far from Fermi surface). If we apply an electric field to this electron, it will start to move in the direction opposite to the force (because q<0, m*>0). However h_bar* dk/dt = F, so the force will start to change k uniformly, so the electron will move up the band E(k). As this happens d^E/dk^2 decreases, so the effective mass increases and becomes infinite where E(k) is linear. m* = inf means zero acceleration. After the linear part of E(k), the effective mass flips sign, so that m*<0 and |m*| decreases, thus the particle (hole -like) now experiences deceleration while it gets to the top of the band and back to the linear interval. And so on.

    The overall result is that the velocity of the charge carrier oscillates about v=0, and no current is transferred !

    However, charge carriers in metals constantly experience scattering, which effectively resets k value to zero. So, overall there is conduction now, because there is some average velocity due to the force applied. And so there is current.

    (this is what is essentially written in Singleton chapter 9.1)
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