Lagrangian and fields reprs

1. Jan 10, 2015

ChrisVer

I was wondering.
What's the reason for putting objects in low representations in the SM and not higher ones?
So, why fermions in a doublet of SU(2) and not a multiplet?

In analogy in SU(5) we put the particles in the 5-plet...

2. Jan 10, 2015

Staff Emeritus
You could do that. But it would predict particles that are unseen. If I put the left-handed electron and neutrino in an SU(2) triplet and not an SU(2) doublet, where is the third particle?

Last edited: Jan 10, 2015
3. Jan 10, 2015

ChrisVer

Hmmm I was thinking more the representations of the same gauge group.
For example the SU(2) has the 4-plet or triplet representation
(
$2 \otimes 2 = 3 \oplus 1$
$2 \otimes 3 = 4 \oplus 2$
)

and in a similar way I think you can work up to the 6plet.
In the 6plet one could put all the leptons.

4. Jan 10, 2015

Staff Emeritus
You can't do that and make the quantum numbers come out right. How do the e and mu get the same quantum numbers if they are in different positions in the multiplet?

5. Jan 10, 2015

ChrisVer

I think they would still be color neutral...
They would still have the same isospins and u(1) charges, and that's enough.
$[6] = \begin{pmatrix} e \\ \nu_e \\ \mu \\ \nu_\mu \\ \tau \\ \nu_\tau \end{pmatrix}$

The U(1) transformation matrix would have to be $Y_6 = \begin{pmatrix} Y_{el-flav} & 0 & 0 \\ 0 & Y_{mu-flav} & 0 \\ 0 & 0 & Y_{tau-flav} \end{pmatrix}$
(still traceless) with $Y_{i-flav}$ the same 2x2 matrices you have in the SM for the i-th flavor.
and similarily for the isospin
$T_6^i = diag ( \tau^i , \tau^i , tau^i )$
The only quantum number which I "feel" this would violate is the lepton number. But the lepton number is an accidental symmetry of SM.

Maybe I'm terribly wrong with the choices of Y and T matrices?

6. Jan 10, 2015