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Exclusion principle question.

  1. Dec 6, 2008 #1
    In terms of the Pauli exclusion principle, what does it mean, quantitatively, to possess the same location in space?

    Is there a definite distance which Fermions with the same states will simply not get closer than? Or is it some force of repulsion?
     
  2. jcsd
  3. Dec 7, 2008 #2

    malawi_glenn

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    No not "same location in space" but occupying the same Quantum State. Two identical fermions can not occupy the same quantum state.

    Due to this, fermionic gas will constitute a pressure. Even when the fermions are free (not bound to an atom or similar) the states are still DISCRETE! In fact, they are DISCRETE in momentum space as well, so there will exists momentum greater than zero in the fermionic gas. And pressure is fore/area = (change in momentum / time)/are... c.f. classical gas molecules in a container, they will give rise to pressure at container walls.

    The limit of what Pressure a collection of fermions can withstand are related first to the chandrasekhar limit - electrons gas withstanding gravity. Then you have the maximum mass of a neutron star, before it collapses to a black hole.

    Those are some "applications" of pauli principle ;)
     
  4. Dec 7, 2008 #3
    Two different fermions in two different atoms can occupy the same energy state.
     
  5. Dec 7, 2008 #4

    malawi_glenn

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    A state is NOT specified by its energy!

    The state in an atom is specified by n,L, m_L and S.

    So there can be several eletrons having same energy, but they then differ in S (spin) or L (m_L) (angular momentum). Hence only one fermion/state.

    In hydrogentic atoms, L can take vaules n-1,n-2,..,0
    n can take 0,1,...
    etc.

    Basic exercise in intro QM
     
  6. Dec 7, 2008 #5
    As malawi_glenn wrote the energy eigenvalue does not specify the quantum state uniquely, not even in an atom. In this case, the relevant thing is that the wavefunctions of the "same state" would still be different because the atoms are located at different places. So, if one wavefunction is psi(x), the other wavefunction is psi(x-a), if the other atom is at position a.
     
  7. Dec 7, 2008 #6
    OK, 'quantum states'. But these atoms are two identical bosons...
     
  8. Dec 7, 2008 #7

    malawi_glenn

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    and??
     
  9. Dec 8, 2008 #8
    I thought you could tell me. If you have two identically prepared, neutral atoms described by a single wave equation, does it make sense to talk about whether a particular electron is a constituent of one atom or the other?
     
  10. Dec 8, 2008 #9

    malawi_glenn

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    Look at the H2 molecule for instance, there this effect occurs. The overlapping of each atoms electron state forces the electrons to "be away from each other" and thus makes each atom to work as a weak dipole -> which will make the atoms bind together. So to answer your question: This was an example where the electrons "can't decide" on which atom they belong to.
     
  11. Dec 8, 2008 #10
    Yes. I was thinking more along the lines of unbound atoms. So if the atoms themselves can only be described by a single wave equation, what of their constituent electrons?
     
  12. Dec 8, 2008 #11

    malawi_glenn

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    Then the electrons "know" what atom they belong to. You have the "same" pheneomenon in solids, when putting many atoms close toghter, this force the energies of atoms to form a band.
     
  13. Dec 8, 2008 #12
    Another practice problem for our QM students: Suppose we have N noninteracting Bosons in a box. Then it is easy to find the ground state of this system: all the Bosons occupy the single particle ground state. Now, suppose that the Bosons are really bound states of two Fermions. Let's say that the Fermons are bound by a harmonic potential. We assume that the average distance of the bound Fermions in a boson, which is of order
    sqrt[hbar/(m omega)], is much less than the dimension (L) of the box.

    What is the ground state of this system?
     
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