Lagrangian and fields reprs

[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...

 

[Vanadium 50]

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?

 

[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.

 

[Vanadium 50]

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?

 

[ChrisVer]

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

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?

 

[Vanadium 50]

You can't have them in a 6-plet of isospin and have the same isospins.

Consider angular momentum, also an SU(2). A 6-plet corresponds to J=5/2 which has m = +/- 5/2, +/- 3/2 and +/- 1/2. You can't declare it to have three +/- 1/2 and no +/- 3/2 or +/- 5/2. That's not a J=5/2 state and it's not a 6-plet.

 

[vanhees71]

There's no way (yet?) to derive the particle content of the Standard Model theoretically. It's just empirical input to the model. The same holds for the many free parameters (coupling constants/masses).