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Pauli Exclusion across the three Generations

  1. Mar 11, 2012 #1
    Fermions are well known for NOT being able to exist in the same state, whereas bosons can. Hence why once an S orbital in an atom has two electrons (with opposite spins), that's it.

    But I've only ever seen this discussed for a single type of particle at a time. For instance, could a muon and an electron exist in the same state? Now, I realize this is a bit hand-wavy, since the orbital energies of a muon would (I suppose) be different than an electron... but imagine they weren't. Would there be any restriction to a muon and electron in an atom sharing all 4 quantum numbers?

    This would mean that so long as you had a bunch of different types of fermions (muons, taus...) they could 'overlap' like bosons. This seems strange to me. Or am I missing something?

    Thanks!
     
  2. jcsd
  3. Mar 11, 2012 #2
    Nope, no problem at all with particles of different flavour occupying the "same" state because as you say, they are different particles, and so they are NOT in the same state. The exclusion principle is only concerned with identical particles, and being of different flavours means two particles are not identical. So yes, in principle you could have a whole set of muon orbitals sitting alongside your electron orbitals in some atom, even if muons and electrons had the same masses and muons were stable.
     
  4. Mar 12, 2012 #3

    tom.stoer

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    Are you familiar with fermionic creation and annihilation operators?
     
  5. Mar 12, 2012 #4
    Thanks to both of you for clearing my head about this issue. I am familiar with the operators you speak of, and from a QM perspective it seems pretty clear. (You can swap two electrons and take into account what happens to the wave function, but not two distinct particles).

    I was doubting myself, though, because that flies in the face of those common (and oversimplified) adages. 'Fermions hate being together,' 'bosons can share a room, but fermions don't wanna' ...

    Of course, you can't expect total accuracy in such explanations, but I guess some part of my instinct was still based on those anthropomorphized particles I learned about as a kid.

    Thanks for setting me straight!
     
  6. Mar 13, 2012 #5

    tom.stoer

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    I asked for the creation and annihilation operators b/c of the following reason: you can create single-fermion states |S> using an operator

    [tex]|S\rangle = a^\dagger_S|0\rangle[/tex]

    where S means an index S = {momentum, spin, ...} with all other quantum numbers such as flavor and color.

    Now using these operators you can also create two-fermion states |SS'> for S≠S'

    [tex]|SS^\prime\rangle = a^\dagger_S\,a^\dagger_{S^\prime}|0\rangle[/tex]

    But for S=S' you get

    [tex]\left(a^\dagger_S\right)^2=0[/tex]

    which explains algenbraically why you can't create a two-fermion state with two identical particles S=S'. But b/c this holds only for S=S' the Pauli principle does not apply for different particles, i.e. S≠S', i.e. for an electron and a myon with identical momentum and spin.
     
  7. Mar 14, 2012 #6
    Thanks for the clear proof, Tom. That's one thing I've always enjoyed about Quantum Mechanics. In a topic where your intuition can so easily run you astray, the math is surprisingly clear.
     
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