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Questions about SUSY

  1. Aug 13, 2013 #1
    For N=1 SUSY, the SUSY algebra is closed as long as Bianchi identity [itex]\epsilon^{\mu\nu\rho\sigma}D_\nu F^a_{\rho\sigma}=0[/itex] is valid, that is [itex][\delta_{\epsilon_1},\delta_{\epsilon_2}] \propto P [/itex]. What if there is magnetic monopole when Bianchi identity doesn't hold?
    When N=2, there is central charge. But for N=1, there is no central charge. SO how to solve this problem?
     
  2. jcsd
  3. Aug 13, 2013 #2

    fzero

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    For any N=1 theory, the monopole solutions break SUSY spontaneously. So any state with a monopole present is not a supersymmetric state, but the Lagrangian is still supersymmetric.

    Even for the N=2 theory with central charge, the BPS monopole only preserves 1/2 of the supersymmetries. These preserved supersymmetries satisfy the algebra where the central charges are identified in a particular way with the electric and magnetic charge of the (dyonic) state. There is a discussion of this around page 41 of Harvey's lectures. You should get the impression that the fact that any SUSY at all is preserved is very special and has huge consequences. These include the BPS bound on the mass and charge of the monopole for theories with at least N=2 SUSY, and further the exact Montonen-Olive duality of the N=4 SYM theory.
     
  4. Aug 14, 2013 #3
    Why do monopole solutions break SUSY? I am totally new in this topic and it seems there are so many new ideas to understand. If a vacuum breaks SUSY, I know the potential energy will be nonzero. This condition is usually satisfied in monopole solutions. Could you please explain more about the SUSY breaking of monopole solution? Thanks,
    Lagrangian preserves SUSY ???:confused:
     
  5. Aug 14, 2013 #4
    Coming back to the SUSY algebra, do you mean we don't need to discuss whether it's closed or not since SUSY is broken for such solutions?
     
  6. Aug 15, 2013 #5

    fzero

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    You might want to review the subject of spontaneously broken global symmetries in ordinary QFT. It is rather generic that we can have a symmetry present in the Lagrangian, but an individual state is generally not invariant under that symmetry. Typically one asks about whether the vacuum state preserves the symmetry, because, if it does not, then the low-energy effective Lagrangian describing the physics below the symmetry breaking scale no longer has the larger symmetry. But you can ask about multiparticle states as well.

    For example, Poincare-invariant field theories have translational invariance, but if we consider a state consisting of a particle located at a definite point in space, then translation invariance is broken. This doesn't change the importance of translational invariance in the theory, the Lagrangian still has the symmetry, even though individual states do not.

    In the case of SUSY theories, we recall that SUSY is a symmetry between bosons and fermions. Any state consisting of an unequal number of bosons and fermions is no longer supersymmetric. So already from this, we can understand why the monopole in an N=1 theory is not supersymmetric. In a theory with extended SUSY, there are composite SUSY transformations which relate scalars to vectors. I haven't personally worked out the details, but I would suspect that these are important in the discussion of what part of the SUSY is preserved by the BPS monopoles. If you look at what these are, you'll see that by construction, it is a background for the scalar fields of the theory that provides the magnetic source term in the equation of motion for the gauge field.

    Monopole are also finite energy states, so a state containing a monopole will have higher energy than a SUSY vacuum.

    If the Lagrangian is SUSY, then the SUSY algebra closes for the fields, viewed as perturbations of a trivial background. In a sense, this is true in the limit where we can take the strength of the interactions to zero and we have free fields. In the interacting theory, we have to be able to say something about the vacuum state. If we are in a SUSY vacuum, we will be able to show that the SUSY algebra closes. If some scalar fields pick up VEVs, then the vacuum might not be SUSY (or may not have as much SUSY as the free theory).

    Now, as you've pointed out, the SUSY algebra closes on the gauge multiplet when the Bianchi identity is satisfied. This is always the case in the free theory, since there are no magnetic sources, only electric ones. When we include interacting scalar fields, as in the BPS case, we can generate magnetic sources in certain vacua where the gauge symmetry is spontaneously broken to a smaller group. So, in this sense, the spontaneous breaking of SUSY is directly related to the spontaneous breaking of the gauge symmetry. It is not any more or less special than standard examples of symmetry breaking in QFT.
     
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