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R-parity and conservation of angular momentum

  1. Aug 24, 2015 #1
    Assuming R-pairity and thus the creation/destruction of supersymmetric particles happens in pairs,
    how is angular momentum conserved when a particle and its supersymmetric partner have different spin by 1/2?
     
  2. jcsd
  3. Aug 24, 2015 #2

    fzero

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    It depends on the process, but angular momentum will be conserved in ways that are already familiar from SM processes. For example, in the SM, we can have processes like ##gg\rightarrow q \bar{q} ##. At tree-level, there is a diagram

    ##g## ^^^^^^^^^^^^^^^^---------------------- ##q##
    ##\hspace{3.65cm}## |
    ##\hspace{3.65cm}## |
    ##\hspace{3.65cm}## |
    ##g## ^^^^^^^^^^^^^^^^---------------------- ##\bar{q}##

    Angular momentum is conserved because the initial state can have ##J = 0,1,2,\ldots## and so can the final state.

    In the MSSM, the same diagram can describe gluino production ##gg\rightarrow \tilde{g} \bar{\tilde{g}}##. It is also easy to add additional vertices to produce a 1-loop diagram with squarks or a squark + ##q\bar{q}## in the final state
     
  4. Aug 24, 2015 #3
    Thanks for responding! Is it obvious why in your example, gg→g̃ barg̃ , ( where NN->SS) , the reverse reaction (SS->NN) would not be allowed? The R- parity would not change, so CPT stays the same...
     
  5. Aug 24, 2015 #4

    fzero

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    The reverse reaction would be allowed. The R-parity of ##\tilde{g}## is ##-1##, but the R-partity is multiplicative, so for a system of two of them it is ##1##. In general, R-parity is conserved by particle-antiparticle annihilation for all SUSY particles.
     
  6. Aug 25, 2015 #5

    ohwilleke

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    More generally, any Feynman diagram's truth or falsity is never changed by rotating the diagram.
     
  7. Aug 29, 2015 #6
    Do you know this document: "mT2: the truth behind the glamour", arXiv:hep-ph/0304226v1, 23 April 2003? Perhaps you might be interested by it.
     
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