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BLRPV, a new SUSY, it seems.

  1. Sep 22, 2014 #1

    arivero

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    Interesting paper:

    http://arxiv.org/abs/1409.5438 Simultaneous B and L Violation: New Signatures from RPV-SUSY

    The Superpotential has terms UDD and QLD.

    Does it mean that a Diquark can decay to an antiquark, and a Meson can decay to a Lepton? I am not sure of the second point. The paper show Neutron decaying to leptons, but this is simultaneous violation of B and L, while Meson to Lepton only violates L.
     
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  3. Sep 25, 2014 #2

    ChrisVer

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    when you say diquark to an antiquark what do you mean?
    If I recall well U and D are both singlets of SU(2), and they are in the [itex]\textbf{3}[/itex] repr of SU(3) - so they are quarks..
    http://prntscr.com/4qbbst
    From M.Drees book- since the Ubar and Dbar are written as antiquarks, I suppose U,D are quarks... but nevertheless you can change my 3 above to 3bar and "quarks" to "antiquarks" without messing the reasoning...the main point is that it doesn't contain anything like [itex] 3 \otimes 3 \otimes \bar{3} [/itex] which wouldn't only violate B or L numbers, but also the local SU(3) symmetry [on which the MSSM is built on]

    The meson decaying to lepton is true from the RPV terms. In fact this is already known and that's why people imposed R parity in the first place [to kill those terms which are SU[3]xSU[2]xU[1] invariant but violate L,B numbers - the coupling constants are then unnaturally small, so you impose a symmetry U[1]_R to explain their smallness. Then you break it to discrete symmetry]
     
    Last edited: Sep 25, 2014
  4. Sep 25, 2014 #3

    arivero

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    in the diagram a) in page 3, you can see a diquark-antiquark vertex, or it seems so.

    SU(3) is not violated. For sure UDD, seen as an interaction term, can be a SU(3) singlet. Or, from group theory,
    [itex] 3 \otimes 3 = 6 \oplus \bar 3 [/itex] we can see that a diquark (or a disquark) can be in a colour triplet, just as if it were an antiquark.

    For SU(2) you are right that something is going on. UDD is not a SU(2) singlet, if all the quarks or squarks are of the left chirality.
     
  5. Sep 25, 2014 #4

    ChrisVer

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    Yes I understand your confusion with the diagram, but in fact that's not an antiquark there... it's a squark [scalar]. The thing above is not a bar [indicating anti- ] but a tilde [indicating super-]. You can zoom in and see it by yourself, at some point it becomes pretty obvious :)
    The coupling is then with quark-quark-squark.

    If you allowed a vertex with quark-quark-antiquark then your Lagrangian would need to have a term with fields containing those 3 particles- to make it simple let's say it has the fields [itex]\mathcal{L}= q_1 q_2 \bar{q}_3 [/itex] so that's a [itex] 3 \otimes 3 \otimes \bar{3}[/itex] term which violates the SU[3] symmetry. SU[3] permits only [itex] 3 \otimes \bar{3} [/itex] (eg in a mass term) or [itex] 3 \otimes 3 \otimes 3[/itex] and of course [itex] \bar{3} \otimes \bar{3} \otimes \bar{3}[/itex] , or in general words: configurations which have the singlet representation as a result , so they can be SU[3] invariant. That's how you build the MSSM: send your SM fields to chiral superfields existing in the same representations as in SM, take SU(3)xSU(2)xU(1) and start building up your model's lagrangian which has those invariant terms.
    The RPV terms then naturally appear in the model (so they are singlets) as well as the rest MSSM interaction terms...

    For SU(2):
    It is a singlet, since U,U and D are singlets. So you have [itex] 1 \otimes 1 \otimes 1 [/itex]...
    In case you have the one singlet and put a doublet like for example the Q, you also need to put a 2nd doublet (like L) to keep SU(2) invariance- in fact the one is bared but it doesn't make a big difference for SU(2) since the fundamental and antifundamental reprs are the same- : [itex] 2 \otimes 2 \otimes 1 = ( 3 \oplus 1 ) \otimes 1 = 3 \oplus 1[/itex] which can be SU(2) invariant.

    (Sad personal fact:
    :( this way of configurations of representations costed me a whole question to my SUSY exams and lost a very good grade, where I had to find those RPV terms and explain the need for R-parity afterwards with diagrams, and I messed up the fields resulting in non-SU(3) x SU(2) x U(1) invariant quantities... when I asked for some compassion especially because I couldn't see why 3x3x3 is a singlet, I was told "you shouldn't have tried SUSY if you don't know this trivial stuff" - of course I did know, but during the exam I had forgotten. Also during the exam they also tried to help us saying "thing of protons and pions" but it was more confusing than helpful during the exam time. Well although the mark was "bad" for me, I learned out of it.
    )
     
    Last edited: Sep 25, 2014
  6. Sep 26, 2014 #5

    arivero

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    Hey, yes, it is true, they are the singlets. I noticed it in my first reading but somehow I forgot.


    Here I also concede the point, but I am not so sure about it not being an antiparticle (respective of the other two), because the squark is outgoing from the vertex, while the quarks are incoming.

    Also I would expect that the vertex violates barion number by one, while the next vertex keeps barion number and violates lepton number by one. If it is so, and reading the process in the diagram as going from left to right, I would say that it describes two (conjugate-)quarks that collide to form a (conjugate-)anti-s-quark, which in turn decays to a lepton plus a quark. In the first vertex, the incoming particles have barion number (-1/3)x2 and the outgoing squark should have barion number (+1/3). Then in the second vertex the outcoming quark should have still barion number (+1/3) and the lepton has of course lepton number (+1) So the first vertex violates B (and B-L), the second vertex violates L (and B-L) but the four fermion effective vertex magically preserves B-L, while violating both B and L.
     
    Last edited: Sep 26, 2014
  7. Sep 26, 2014 #6

    ChrisVer

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    and what does outgoing actually mean? the squarks are scalar fields. Also in this case, they are the propagators...
     
  8. Sep 27, 2014 #7

    arivero

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    I have seen some books that, in order to have a criteria both for internal and external lines, refer to the "flowing of charge" and the "flowing of momentum". Depending if charge flows in the same direction or contrary to momentum, you name particle or antiparticle. Here by outgoing I was implying that the momentum was flowing from left to right, describing a process of decay from Nucleon to (Meson + Lepton).

    By the way, now that we have agreed that the internal line is an scalar, it is actually the same diagram you were calculating in https://www.physicsforums.com/threads/amplitude-for-fermion-fermion-yukawa-scattering.772742/ isn't it? Or a pure chiral variant of it; in the other thread the fermions were Dirac and here I guess they are Weyl, are there?
     
  9. Sep 27, 2014 #8

    ChrisVer

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    It depends. I think you can as well combine your weyl spinors in the MSSM lagrangian into dirac, no? The only thing that would change is that you could then avoid in refering to R or L quarks as well as to R and L squarks.But I am not really sure about that.. What I am sure about is that in the end you can sum everything up in the diagram...
    As for the diagram. Yes it is the same. You can apply the same Feynman rules to it and get the amplitude. The only things that can change are the parameters and the coupling constants at the vertices...the free theory propagator will remain the same:
    [itex] \frac{i}{p^2 - m^2} [/itex]
     
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