Singlets and gauge invariance

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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.
 

Answers and Replies

523
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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.
 
36
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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?
 
523
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I'd say that's correct yes.
 
samalkhaiat
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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,
Do not confuse the REDUCIBLE tensor

[tex]
\bar{\psi}_{2} \otimes \phi \otimes \psi_{1} \equiv (2,1) \otimes (2,1) \otimes (1,-2) = (1,0) \oplus (3,0)
[/tex]

with the local product of the fields

[tex]\bar{\psi}_{2} \phi \psi_{1} \in (1,0)[/tex]

which forms the su(2)Xu(1)-invariant Yukawa coupling; anything belongs to a singlet is a scalar (invariant).
Indeed, using the transformations

[tex]\psi_{1} \rightarrow u_{1} \psi_{1}, \ \phi \rightarrow u_{2} \phi[/tex]

[tex]\psi_{2} \rightarrow u_{1}u_{2} \psi_{2} = u_{2}u_{1} \psi_{2}[/tex]

where [itex]u_{1} \in u(1)[/itex] and [itex]u_{2} \in su(2)[/itex], it is easy to show that

[tex]\bar{\psi}_{2} \phi \psi_{1} \rightarrow \bar{\psi}_{2} \phi \psi_{1}[/tex]

We build up higher-dimensional irreducible representations by taking tensor pruducts of the fundamental representations and projecting out the invariant subspaces. Take the simple case of su(3), the fundamental representations are [itex]q \equiv [3][/itex] and [itex]\bar{q} \equiv [\bar{3}][/itex];

[tex]\bar{q} \otimes q \equiv [\bar{3}] \otimes [3] = [8] \oplus [1][/tex]

This corresponds to the tensor identity

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

So, while

[tex]\{\bar{q} \otimes q \equiv \bar{q}_{i}q_{j}\} \in [\bar{3}]\otimes [3][/tex]

is a reducible su(3)-tensor,

[tex]\bar{q}_{k}q_{k} \in [1][/tex]

is an invariant ( su(3)-scalar).

The other object in the bracket(...) is an irreducible, traceless tensor with 8 components, i.e., it belongs to the octet [8] ( mesons or gluons matrix).

regards

sam
 
Last edited:
36
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Enlightenment! :)
 
523
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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.
 
523
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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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