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Spin of sfermions

  1. Oct 30, 2013 #1
    I understand that the degrees of freedom must match.

    A Weyl spinor has 2 d.o.f (spin up and spin down), thus the superpartner must also have 2 d.o.f
    Is there a reason why it is a complex scalar and not of spin-1?
     
  2. jcsd
  3. Oct 30, 2013 #2

    fzero

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    The left-handed fermions in the Standard Model are doublets under the electroweak group, while the right-handed fermions are singlets. The only way to do this in am ##N=1## SUSY theory is if we put them in chiral multiplets, which only have scalars and fermions. Conversely, a spin 1 field must be in the adjoint representation of a gauge group for the quantum theory to make sense. Since there are no SM fermions in an adjoint representation, the fermions in any vector multiplets must be superpartners (gauginos), not SM particles.
     
  4. Oct 30, 2013 #3
    Hi fzero, thanks for your answer.

    Can you check if my chain of reasoning is correct?

    - Spin 1 fields must lie in the adjoint representation of any quantum theory (I actually didn't know this was a requirement)
    - SM Fermions live in the fundamental representation => When extending SM with SUSY, the superpartners of the fermions must also be in the fundamental rep.
    - Thus sfermions are of spin 0, not spin 1.
     
  5. Oct 31, 2013 #4

    fzero

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    Yes, to flesh this out, the longitudinal component of a spin 1 field with no gauge invariance would have negative norm once the theory is quantized. However, if the spin 1 field transforms in the familiar way under a gauge invariance (##A_\mu \rightarrow A_\mu + \partial_\mu\alpha##), this longitudinal component becomes an unphysical, pure gauge degree of freedom. This gauge transformation corresponds to the one in which ##A_\mu## is the connection of a gauge-covariant derivative, as is familiar from demonstrating that the matter kinetic terms are gauge-invariant. This identification with the connection on the gauge group is precisely the one which requires that the gauge field be in the adjoint representation.

    Yes, this is true that the left-handed fermions turn out to be in fundamental representations and so the entire supermultiplets must transform the same way.
     
  6. Oct 31, 2013 #5
    I don't think that logic is as tight as it seems because of the possibility of having a massless spin 1 particle in a representation other than the adjoint representation. Massless particles don't have longitudinal components.
     
  7. Oct 31, 2013 #6

    fzero

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    To be complete I should have referred to both timelike and longitudinal components. But that is for a formalism where we start with a theory of a 4-vector and attempt to quantize. You suggest the equally appropriate starting point of quantizing a massless particle with helicity ##\pm 1## components. But then you run into difficulty because if we try to assemble these components into a 4-vector, you find that a generic Lorentz transformation generates a timelike component. So the theory is not Lorentz-invariant unless we admit the ##A_\mu \rightarrow A_\mu + \partial_\mu \alpha## gauge symmetry. This is discussed in Weinberg.

    One could certainly produce a much more rigorous argument than I have, but one should always be led to the conclusion that consistent quantization of a spin 1 field requires a specific gauge-invariance. I am not sure if one could add an additional representation under a 2nd gauge group in which a gauge field transforms in a representation other than the adjoint.
     
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