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Where do fermion properties come from?

  1. Mar 22, 2008 #1
    I'm still struggling to understand, that why precisely have I not understood what the Dirac's field is.

    hmhmh... okey, my question is this: Where does the anticommutation relation [itex]\{a_j,a_k^{\dagger}\}=\delta_{jk}[/itex] come from?

    Is there something better than: "because the commutation relation would give absurd results, hence the relation must be an anticommutation relation"?
     
  2. jcsd
  3. Mar 22, 2008 #2

    strangerep

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    If you're looking for some deeper theoretical reason, then,... no,
    that's about it.

    Although,... I vaguely recall a different argument involving spin-1/2
    Poincare unirreps and isotropy of space. It had something to do with
    the former being multiplied by -1 under a [itex]2\pi[/itex] rotation,
    combined with exchange of two particles at different points.
    But I suppose this is still just "...would give absurd results...".

    In the end, it all arises from Poincare/SR and positive-energy.
     
  4. Mar 29, 2008 #3
    The reason you mentioned, known as the spin-statistics theorem, is actually pretty good. Basically, mathematics gives you two options, and in order to describe nature, you have to pick one that works.

    You can think of the Dirac field first as having a classical limit field [itex]\psi_\alpha[/itex] where [itex]\alpha[/itex] is a spinor index. These fields are Grassmann numbers, and obey [itex]\{\psi_\alpha, \psi_\beta\} = 0[/itex], classically. When you quantize these fields using your favourite method (e.g. path integral), the anticommutation relations you mention pop out naturally.

    If you want to know where fermions "come from", well... that is a whole other ball of wax. In supersymmetry (which can be spontaneously or explicitly broken) they arise when spacetime has extra Grassmann dimensions (superspace). A (different) process, called Bosonization, they arise when you take a bosonic field theory and "change variables" and end up with a fermionic field theory. See for example, the Sine Gordon / Thirring duality.

    Finally, if you think that an anticommutation relation is unnatural, remember that you can always turn an anticommutator into a commutator by multiplying by grassmann parameters. So if your Lie algebra has Grassmann generators, then taking your parameters of your group to be Grassmann as well, you end up with good old fashioned commutators.
     
    Last edited: Mar 29, 2008
  5. Mar 30, 2008 #4
    I don't know if you know about representations of the Lorentz group. If you do, you will have noted that spin-1/2, 3/2, 5/2 etc. representations get a minus sign if we perform a pure rotation of [itex]2\pi[/itex] on them.

    Furthermore, you will know from exchange symmetry that a particle can have either a + sign or a - sign upon exchange.

    The spin statistics theorem essentially makes a connection between the two - if the particle obtains a minus sign under [itex]2\pi[/itex] rotations, then it obtains a minus sign upon exchange with an identical particle - we call these fermions. We call the other types bosons.
     
  6. Apr 19, 2008 #5

    samalkhaiat

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