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Fermi-Dirac statistics valid for electron gas in metals?

  1. Feb 2, 2008 #1

    In my course of solid states physics we use the fermi-dirac statistics for a free electron gas in metals. The fermi wave length of the electrons is about 1 Angström. Now, the wavelength may be intepreted as something as a coherence range - the electron should forget about the state of the other electrons on a scale of about 1 Angström, right? But then, how can we use FD-statistics if there is no coherence of the electron gas along the whole metal (or crystal)?

    Thanks for all answers!

  2. jcsd
  3. Feb 2, 2008 #2


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    FD statistics has nothing to do with the coherence length.
    Two electrons can not occupy the same quantum state, i.e. they obey the exclusion principle which is why they obey FD statistics.
    Mathematically this is related to the the symmetry of the total wavefunction (more specifically if it changes sign when you swap the particles around) which in turn is related to the spin of the electron; electrons are fermions since they have half-integer spin (1/2)
  4. Feb 2, 2008 #3
    Yeah, ok, but how do I know if electrons are to be described by the same wave function? I mean, taking two seperated crystals would give seperate FD statistics. But bringing them together would yield all electrons to be described by one wavefunction thus obeying one FD statistic.

    So, the question is: When is the distance between two electrons big enough so that the two can be in the same quantum numbers? Any relation to the coherence length?

    Thanks again!

  5. Feb 2, 2008 #4


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    It is not the same quantum number; it is the same quantum state. It is not the same thing.
    Two electrons that are spatially separated are not in the same state.
  6. Feb 2, 2008 #5
    Hmmm... ? But isn't quantum state und the set of all quantum numbers the same? I mean, the state of an electron is determined by its quantum numbers, right? And so the state of two electrons is the same if they have the same quantum number? I am confused...

    Sure this only holds if the two electrons are not seperated spatially. But so are electrons in a crystal? Could you explain your statement in more detail, f95toli? I have the feeling to be one the wrong path...

    Thank you!

  7. Feb 3, 2008 #6


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    Not quite. it is true that if the electron is bound to an atom we can use quantum numbers to do the "bookkeeping", two electrons with the same numers would indeed be in the same state which is forbidden. However, electrons in a metal are free to move so you can't assign numbers like n,l,m etc to them. Instead we have to start from the beginning and solve the SE for electrons in a periodic potential (in the simplest case), we then get new numbers that I guess you could call "quantum numbers" that are associated with Bloch states which in turn depend on the wavevector k. This rather simple description still give rise to quite complicated physics, specficially a very large number of energy bands in k-space that can be occuiped by electrons (the bands are very densily packed, which is why we talk about the density of states as a continuum). The main point here is that whereas we can still only have one electron per state there is a huge number of states available and I suppose two electrons could at least in principle be in the "same place" (whatever that means) as long as they had different momentum.
    Note that "real space" descriptions become rather complicated, it is much easier to describe what is going on in k-space.
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