Why does theta transform as exp(-i alpha)*theta under U_R(1)?

  • Context: Graduate 
  • Thread starter Thread starter alphaone
  • Start date Start date
  • Tags Tags
    Supersymmetry
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
1 reply · 2K views
alphaone
Messages
46
Reaction score
0
Hi
I have a question concerning the anticommuting variable theta of the superspace manifold: For a proof of some renormalisation theorems I have seen the author make use of the transformation properties of theta under the internal U_R(1) symmetry i.e. he said theta -> exp(-i alpha)*theta ,under this symmetry. Now I would like to know where this comes from. Is it coming from the fact that the SUSY generator Q does not commute with the R-symmetry and that we want theta*Q to be invariant under this symmetry? If so why do we need theta*Q to be invarint? I would really appreciate any response and if my question is not really clear please let me know as well.
 

Discussion

Physics news on Phys.org
alphaone said:
Hi
I have a question concerning the anticommuting variable theta of the superspace manifold: For a proof of some renormalisation theorems I have seen the author make use of the transformation properties of theta under the internal U_R(1) symmetry i.e. he said theta -> exp(-i alpha)*theta ,under this symmetry. Now I would like to know where this comes from. Is it coming from the fact that the SUSY generator Q does not commute with the R-symmetry and that we want theta*Q to be invariant under this symmetry? If so why do we need theta*Q to be invarint? I would really appreciate any response and if my question is not really clear please let me know as well.


I'm not an expert, but I think it goes something like this: Before we go to the superspace formalism, we know how the R-symmetry acts on the fields. To realize this symmetry in superspace, we look at for example the chiral superfield. We know how the R-symmetry acts on the components fields, the scalar and the fermion. If we look at the scalar, there are no thetas around, and so this will tell us how the entire superfield transforms.
But this transformation is only consistent with the transformation of the component fermions as long as the superspace coordinates, theta, also transforms.

I don't know if that made things clearer, I think this is explained in f.x the BUSSTEPP lectures on supersymmetry. You can find it here:
http://www.stringwiki.org/wiki/Supersymmetry_and_Supergravity

BTW: This question would probably be more at home in "beyond the standard model".

nonplus