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Degenrate fermi gas

  1. Mar 5, 2010 #1
    Why is it called degenerate ??

    Is it because all levels upto fermi level are filled or all degeneracies of the energy levels that are present occupied at T = 0k .

    In deriving the average occupation no. for a deg fermi gas , we have used classical idea of momentum that is integrating over a sphere of p^2 dp , why not qntm mechanical operator form since this a quantum gas.

    Also in calculating average N we have an integral of (E)^1/2 dE within limit 0 to Ef where Ef is fermi energy level , now while evaluating this we don't consider the E= 0 state i.e ground state because if we put 0 then avg N = 0 , what's the thought behind this ?

    I had posted this in general section , got no reply , atleast someone reply here
  2. jcsd
  3. Mar 5, 2010 #2


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    Degenerate just means all the energy levels below a certain level (the fermi energy) are taken. I don't know exactly why we use the term degenerate...but we do.

    We use classical-style calculations for simplicity. You can do a fully quantum mechanical treatment of a degenerate gas; however, it's a lot more complicated. You can get a lot of insights by doing the classical treatment, and the approximations aren't all that bad.

    It's analogous to using Newton's laws for dynamics instead of Einstein's special relativity. It's simpler and the answers aren't that bad.
  4. Mar 5, 2010 #3
    Where are the supposed classical bits? It all seems fully quantum mechanical to me.
  5. Mar 5, 2010 #4


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    Perhaps you are right and they are all QM. I may have been confused since I saw none of the regular QM "stuff" (e.g. wavefunctions, operators, bra-ket notation, etc) when dealing with degenerate Fermi gasses.
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