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On SUSY irreps: eigenvalues of Pauli-Lubanski operator?

  1. Oct 31, 2013 #1
    Hi everyone,

    Just an easy question that came to my mind while studying basics of SUSY.
    Consider in N=1, D=4 a massive clifford vacuum [itex]|m,s,s_3\rangle[/itex], and for cconcreteness take its spin to be s=1/2.
    Now, acting with the four supercharges on both the [itex]|m,1/2,1/2\rangle[/itex] and [itex]|m,1/2,-1/2\rangle[/itex] states of the cliffor vacuum i get a total of 8 states, namely
    [tex]
    \begin{align}
    &|m,1/2,1/2\rangle\\
    &\overline Q_{\dot 1}|m,1/2,1/2\rangle \text{ with }s_3=0\\[2mm]
    &\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=1\\[2mm]
    &\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=1/2\\[2mm]
    \end{align}
    [/tex]
    and
    [tex]
    \begin{align}
    &|m,1/2,-1/2\rangle\\
    &\overline Q_{\dot 1}|m,1/2,-1/2\rangle \text{ with }s_3=-1\\[2mm]
    &\overline Q_{\dot 2}|m,1/2,-1/2\rangle \text{ with }s_3=0\\[2mm]
    &\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle \text{ with }s_3=-1/2\\[2mm]
    &\end{align}
    [/tex]

    Now, as far as I understood, these should form a massive vector, which I'd say then is set up by ([itex]\overline Q_{\dot 2}|m,1/2,1/2\rangle\,,\, \overline Q_{\dot 1}|m,1/2,-1/2\rangle\,,\,\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle+\overline Q_{\dot 1}|m,1/2,1/2\rangle)[/itex]), two weyl fermions of opposite chirality (thus a dirac spinor) ([itex]|m,1/2,1/2\rangle\,,|m,1/2,-1/2\rangle [/itex]) and ([itex]\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,1/2\rangle\,,\overline Q_{\dot 1}\overline Q_{\dot 2}|m,1/2,-1/2\rangle [/itex]) and at last one real scalar [itex]\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle-\overline Q_{\dot 1}|m,1/2,1/2\rangle) [/itex].

    Now, my questions:

    1) How do I know that the state [itex]\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle-\overline Q_{\dot 1}|m,1/2,1/2\rangle)[/itex] is indeed a scalar? I mean, I only know that its [itex]s_3[/itex] component is zero, but I don't know whether ist spin is! Equivalently, how do I know that the state [itex]\frac{1}{\sqrt{2}}(\overline Q_{\dot 2}|m,1/2,-1/2\rangle+\overline Q_{\dot 1}|m,1/2,1/2\rangle)[/itex] is part of a vector and not just another scalar (or maybe that the plus sign linear combination above is a scalar and the minus sign combination is part of the vector field)?
    To answer this question I thought of letting the square of the Pauli-Lubanski operator W act on these states: even if this is no longer a casimir it should nevertheless give me the spin of the state, right? Anyway, to this end i tried to compute the commutator [itex][W^2,Q] [/itex] (where Q here is just any of the supercharges), but I got a non-sense answer, ([itex][W^2,Q] =\frac{-3m^2}{4}Q[/itex], which seems reasonable but actually isn't - i probably messed up somewhere) so that I'really missing something here...could you please help me on this matter?


    2) Also, if you still have time you could also explain me why in the multiplet above the scalar is real (without referring to the fact that bosonic modes must be equal in number to fermionic ones) and why the two weyl fermions have opposite chirality?



    Cheers!
     
    Last edited: Oct 31, 2013
  2. jcsd
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