# Three dimensional representation of ##U(1)\times SU(2)##

• A
Consider a three dimensional representation of ##U(1)\times SU(2)## with zero hypercharge ##Y=0##:

$$L= \begin{pmatrix} L^+ \\ L^0 \\ L^- \end{pmatrix}$$

Then the mass term is given by [1]:

$$\mathcal{L} \supset -\frac m 2 \left( 2 L^+ L^- +L^0 L^0 \right)$$

I am wondering where the mass term is coming from.

I know that in the Standard Model the mass term for a doublet which is in two dimensional representation, is given by ## m \bar L L##.

Reference:

1. Eqn (4) in arXiv:0710.1668v2 [hep-ph]

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
It is just a parameter of the Lagrangian. Since ##L## is a full non-chiral Dirac fermion, there is no problem in introducing that mass term, just as there is no problem in introducing a mass term in QED. The problem in introducing fermion masses in the SM is that the SM is chiral and left- and right-handed fields transform differently under SU(2). In turn, this means that the mass term needs to be generated through something like the Higgs mechanism, but this is not an issue here.

Thanks Orodruin . So, in this case mass term should read:

$$-\frac m 2 \bar L L = -\frac m 2 \left( L^{+†} L^+ + L^{0†} L^0 + L^{-†} L^- \right)$$

So, why should this expression be the same as the expression above in eqn (4) of arXiv:0710.1668v2 [hep-ph] ?

vanhees71
Gold Member
Isn't this the same up to a constant?

##L^+##, ##L^0## and ##L^-## are independent fields. Let's call them ##\psi^+##, ##\phi^0## and ##\chi^-##.
So I am wondering why ##\psi^{+†}=\chi^-## or ##\chi^{-†}=\psi^+## ?

vanhees71