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How does a scalar transform under adjoint representation of SU(3)?

  1. Nov 19, 2007 #1
    I read this in a paper: Suppose there is a theory describing fermions transforming nontrivially under SU(3) gauge symmetry.
    L = \Psi^{\bar}(\gamma^A D_A+Y(\Phi))\Psi. The covariant derivative is: D_A\Psi=(\partial_A-i E_A^{\alpha}T_{\alpha})\Psi. Where E_A^{\alpha} are SU(3) gauge fields, T_{\alpha} are SU(3) generators and \alpha=1,2,...8. Then the author says \Phi=\Phi^{\alpha}T_{\alpha} is
    a scalar field that transforms in the adjoint representation
    of SU(3). I do not understand why should a scalar field transform that way. I thought scalars are invariant. Can one construct such a theory with one scalar instead of eight scalars? Can anybody explain? Since I do not know much of group theory, it may be helpful to refer me to some appropriate book also. What is meant by Cartan generators?
  2. jcsd
  3. Nov 19, 2007 #2
    It looks to me like it's saying that \Phi is a space-time scalar, not an SU(3) scalar. That is to say that it takes on a single value at each point in spacetime, which does not transform at all under Lorentz transformations, but that it carries SU(3) charge.
  4. Nov 23, 2007 #3


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    Parlyne is correct - when you say "scalar field", you are referring to the space-time transformation rules. A (complex) scalar field can transform any way it wants, but consider: if it is a *real* scalar field, it must transform under a *real* representation of the gauge group. Therefore it can be a singlet (doesn't transform at all) or an adjoint (like the gauge bosons). If [itex]\Psi[/itex] is in the fundamental of SU(3), then [itex]\bar{\Psi}\Psi[/itex] is already real, so therefore [itex]\phi[/itex] must also be real (since the action must be Hermitian). That's why they make it transform in the adjoint rep.

    "Cartan Generators" are the standard basis for the su(n) Lie Algebra. Check out Georgi's textbook for an in-depth explanation of this.
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