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Pauli's exclusion principle

  1. Oct 30, 2011 #1
    I know the principle says that no two fermions canNot have the same atomic number, but could you explain that in detail?
     
  2. jcsd
  3. Oct 30, 2011 #2

    Ken G

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    It means two identical fermions (say, any two electrons) cannot be in exactly the same state. So if you have a discrete way of labeling the states, say by atomic quantum numbers, then you can only have one electron in each of those states (or two if you are not counting the spin in the quantum numbers). One reason this is really important is that it gives us the periodic table-- atoms get a lot larger when you "break into a new valence shell" as you add electrons, and that's only because the new electrons can't go into the states that the previous electrons are already in.
     
  4. Oct 30, 2011 #3

    tom.stoer

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    Are you familiar with the harmonic oscillator?

    Fermionic creation and annihilation operators are constructed using anti-commutators (instead of commutators

    [tex]\{b_s,b^\dagger_{s^\prime}\} = \delta_{s\,s^\prime}[/tex]

    [tex]\{b_s,b_{s^\prime}\} = \{b^\dagger_s,b^\dagger_{s^\prime}\} = 0[/tex]

    The last equation is important b/c for s=s' it can be rewritten as :

    [tex](b^\dagger_s)^2 = 0[/tex]

    s,s' represent the quantum numbers

    You can construct a states where each s, s', s'', ... is occupied by either zero or one particle. But if you want to construct a state with two particles sitting in state s you get

    [tex]|2s\rangle = (b^\dagger_s)^2|0\rangle = 0[/tex]

    due to the anticommutator.
     
    Last edited: Oct 30, 2011
  5. Oct 30, 2011 #4
    They can be in the same state as long as they have different positions though right?
     
  6. Oct 30, 2011 #5

    tom.stoer

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    Sorry, I forgot to explain that creation and annihilation operators act in momentum space and that this means that s is a collection of momentum, spin, isospin, etc.}. Therefore they are not located anywhere in position space.
     
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