What is the significance of singlets and gauge invariance in particle physics?

[ledamage]

Hi folks!

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

(2,-\frac{1}{2}) \otimes (2,-\frac{1}{2}) \otimes (1,1) = (1,0) \oplus (3,0)

imply that this term is gauge-invariant? What about the (3,0) part? I just can't see it.

 

[jdstokes]

Interesting question. I'm still learning this stuff so take this explanation with a grain of salt.

\varphi,\psi_1 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 \psi_2 because it is 1-dimensional.

Intuitively it makes sense that a doublet \varphi multiplied by the Dirac conjugate of another doublet is singlet, so \bar{\psi}_1\varphi 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 \bar{\psi}_1\varphi doesn't live in there.

 

[ledamage]

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?

 

[jdstokes]

I'd say that's correct yes.

 

[samalkhaiat]

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

<br /> \bar{\psi}_{2} \otimes \phi \otimes \psi_{1} \equiv (2,1) \otimes (2,1) \otimes (1,-2) = (1,0) \oplus (3,0)<br />

with the local product of the fields

\bar{\psi}_{2} \phi \psi_{1} \in (1,0)

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

\psi_{1} \rightarrow u_{1} \psi_{1}, \ \phi \rightarrow u_{2} \phi

\psi_{2} \rightarrow u_{1}u_{2} \psi_{2} = u_{2}u_{1} \psi_{2}

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

\bar{\psi}_{2} \phi \psi_{1} \rightarrow \bar{\psi}_{2} \phi \psi_{1}

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 q \equiv [3] and \bar{q} \equiv [\bar{3}];

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

This corresponds to the tensor identity

<br /> \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}<br />

So, while

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

is a reducible su(3)-tensor,

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

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

 

[ledamage]

Enlightenment! :)

 

[jdstokes]

A very nice explanation by samalkhaiat there.

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

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

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.

 

[jdstokes]

Suppose that I want to decompose \mathbf{3} \otimes \mathbf{6}. Is it sufficient to conclude that the symmetric tensors of the form T^{ijk}\equiv A^{ij}B^k+A^{ik}B^j+A^{kj}B^i form an irreducible subspace and that the remaining orthogonal subspace must be the adjoint rep because it has dimension 8?

Thus \mathbf{3}\otimes\mathbf{6} = \mathbf{10} \oplus \mathbf{8}.