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A Spontaneous breaking of supersymmetry

  1. Apr 4, 2016 #1
    Dear All
    I am studying spontaneous breaking of supersymmetry but i am confused about something. Why should the goldstone be a spin 1/2, why it is not a boson? I have read that this is due to supersymmetry is a fermionic symmetry but i am not convinced with this answer!
    Thank you
     
  2. jcsd
  3. Apr 6, 2016 #2

    haushofer

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    I'm not really into this stuff, but you should at least say why you are not convinced. See e.g. page 60-61, eqn. (4.23) of

    http://thep.housing.rug.nl/sites/default/files/theses/PhD thesis_Andrea Borghese.pdf

    I'd say you mimick the usual Goldstone theorem, but now for spontaneous breaking of SUSY and see how you construct the accompanying Goldstone particles. For the bosonic case, I really like Zee's treatment in his QFT book (chapter 4.1). There you can see that the Goldstone bosons are explicitly constructed using the conserved charges Q of the corresponding symmetry. In SUSY, these charges are fermionic. So I'd would be highly surprised if you wouldn't get fermionic Goldstone particles for spontaneous broken SUSY.
     
  4. May 28, 2016 #3
    The Goldstone particle from SUSY breaking is indeed a fermion, and it's called a "goldstino". So search for "goldstino" to find more on it.
     
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