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Spontaneusly broken R-symmetry

  1. Mar 10, 2012 #1
    Spontaneously broken R-symmetry is a sufficient condition for spontaneously supersymmetry breaking.

    My question is: What is a spontaneously R-symmetry breaking?
     
  2. jcsd
  3. Mar 12, 2012 #2

    blechman

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    This is technical. Since I have no idea what your level is, I'll give you the full answer. But if you don't understand any of this, then the short answer is, "this is a technical thing that theoretical physicists consider when constructing mathematical models of supersymmetry."

    An R symmetry in SUSY is a symmetry that treats the supersymmetric partners differently. The most famous example is "R-parity" where the supersymmetric particles are "odd" and the ordinary particles are "even".

    When you have a SUSY theory, you often (perhaps ALWAYS - I can't think of a counterexample) have an "R symmetry" as well. For so called "N=1 SUSY" the R symmetry is U(1). For "N=2 SUSY" the R symmetry is SU(2), etc. The statement you are making is that if you spontaneously break the R symmetry, you also tend to break SUSY as well. This is very important if you are trying to build a model of SUSY, for example, and you are trying to figure out how SUSY is broken - if you can find a way to break the R symmetry, then you also break SUSY while you're at it!

    The minimal supersymmetric standard model (MSSM) is an example of N=1 SUSY whose U(1) R symmetry was broken down to a Z_2 parity. I have written papers that consider N=2 SUSY that does NOT break the R symmetry (we called it Minimal-R-symmetric-Supersymmetric-Standard-Model, or MRSSM).

    The above statement you made is also not true in general - you need something called "generic conditions". But that's another technical point...

    Hope that helps!
     
  4. Mar 16, 2012 #3
    Hi Blechman,

    Thank for your reply.

    First of all, I was studying and thinking about my initial question and maybe I got the answer.

    May I say that R-symmetry is spontaneously broken if and only if any R-charged superfield get a non-vanishing VEV?

    Another question:

    In fact spontaneously broken R-sym is a sufficient cond. for susy breaking only if the superpotential is generic. I gave a look at the paper of Nelson, Seiberg, but I didn't understand exactly what is a generic superpotential. Would you give me some hint?

    Tank you.
     
  5. Mar 16, 2012 #4

    blechman

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    That is correct.

    That is right. When you build models (or do any kind of "effective field theory", or EFT), the idea is that you have a list of fields and symmetries, and you write down every operator you can that is made of products of the fields and is consistent with those symmetries. In ordinary EFT, you are REQUIRED to do this, since if you miss an operator, quantum corrections will generate those operators upon renormalization. This is why, for example, you can never set the Higgs mass to zero in the Standard Model.

    But SUSY is different: thanks to non-renormalization theorems, even if you don't include all the allowed operators, you will NOT generate them from quantum corrections. This is very special to supersymmetric theories and is not true in general.

    So when building SUSY models, you can write down "incomplete" superpotentials that are consistent even at the quantum level. These superpotentials are called "non-generic". A "generic" superpotential would be one that includes every operator allowed by all the symmetries. If you had such a superpotential, and it is arranged so that there is a spontaneously broken R symmetry, then Nelson-Seiberg point out that (at least in N=1) SUSY is also broken.

    Note that broken R symmetry is sufficient but **NOT** necessary.
     
  6. Apr 10, 2012 #5
  7. Apr 21, 2012 #6
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