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Singlets and gauge invariance

  1. Jul 6, 2008 #1
    Hi folks!

    Another stupid question: Consider a Yukawa coupling [itex]\lambda \bar{\psi}_1 \psi_2 \phi[/itex] where [itex]\phi[/itex] is a scalar field in the [itex](2,-\frac{1}{2})[/itex] representation and [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are lh. Weyl fields in the [itex](2,-\frac{1}{2})[/itex] and [itex](1,1)[/itex] representation of [itex]\mathrm{SU}(2) \times \mathrm{U}(1)[/itex]. Why does the occurrence of the singlet [itex](1,0)[/itex] on the rhs of

    [tex](2,-\frac{1}{2}) \otimes (2,-\frac{1}{2}) \otimes (1,1) = (1,0) \oplus (3,0)[/tex]

    imply that this term is gauge-invariant? What about the [itex](3,0)[/itex] part? I just can't see it.
  2. jcsd
  3. Jul 8, 2008 #2
    Interesting question. I'm still learning this stuff so take this explanation with a grain of salt.

    [itex]\varphi,\psi_1[/itex] live in two-dimensional field vector spaces, so when we juxtapose them we must end up with a 4-dimensional space according to the tensor product rule for multiplication of vector spaces. We can ignore [itex]\psi_2[/itex] because it is 1-dimensional.

    Intuitively it makes sense that a doublet [itex]\varphi[/itex] multiplied by the Dirac conjugate of another doublet is singlet, so [itex]\bar{\psi}_1\varphi[/itex] must form a 1-dimensional subspace of our 4-dimensional tensor product space. The fact that we have a 3-dimensional subspace left over should not concern us much because [itex]\bar{\psi}_1\varphi[/itex] doesn't live in there.
  4. Jul 9, 2008 #3
    Yes, I think so. The appearance of a singlet in the above equation is just a necessary but not a sufficient condition for the term to be gauge invariant, right?
  5. Jul 9, 2008 #4
    I'd say that's correct yes.
  6. Jul 17, 2008 #5


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    Last edited: Jul 17, 2008
  7. Jul 18, 2008 #6
    Enlightenment! :)
  8. Jul 19, 2008 #7
    A very nice explanation by samalkhaiat there.

    Something I've been wondering is that given tensor components, e.g.,

    [itex]\left( \bar{q}_{i} q_{j} - \frac{1}{3} \delta_{ij} \bar{q}_{k}q_{k} \right)[/itex]

    how does one deduce that the corresponding subspace is irreducibly invariant?

    I believe a necessary and sufficient condition for this to hold is that it must not be possible to form a tensor of lower rank by contracting with the Kronecker delta and permutation symbol, although I don't quite understand why this works.
  9. Jul 19, 2008 #8
    Suppose that I want to decompose [itex]\mathbf{3} \otimes \mathbf{6}[/itex]. Is it sufficient to conclude that the symmetric tensors of the form [itex]T^{ijk}\equiv A^{ij}B^k+A^{ik}B^j+A^{kj}B^i[/itex] form an irreducible subspace and that the remaining orthogonal subspace must be the adjoint rep because it has dimension 8?

    Thus [itex]\mathbf{3}\otimes\mathbf{6} = \mathbf{10} \oplus \mathbf{8}[/itex].
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