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Suppose we have a field theory with a single complex scalar field $\phi$ and a single Dirac Fermion $\psi$, both massless. Let us write $\psi _L=\frac{1}{2}(1-\gamma ^5)\psi$. Then, the Yukawa coupling of the scalar field to the left-handed Fermion field should be of the form $$ g\overline{\psi}\psi \phi, $$ where $g$ is the coupling constant. So far, so good (at least I think; please correct me if this is incorrect).

Now, we introduce gauge invariance into the theory and demand that $\phi$ transform as a triplet and $\psi _L$ transform as a doublet under the gauge group $SU(2)$. What form does the Lagrangian take now? My confusion is arising because now, in particular, although $\phi$ transforms as a scalar under the Lorentz group, it must be described by $3$-components so that it can transform as a triplet under $SU(2)$. But then, the Yukawa coupling term listed above, as written, is not a number! I know this has something to do with the fact (I think) that $\overline{2}\otimes 2$ decomposes into something involving the triplet representation of $SU(2)$. Unfortunately, I don't know enough about the representation theory of $SU(2)$ to turn this into a Yukawa coupling term that makes sense.

Once again, because of my lack of knowledge of the representation theory of $SU(2)$, I don't know how to write the gauge covariant derivative corresponding to the triplet representation of $SU(2)$. If we use the Pauli matrices as a basis for $\mathfrak{su}(2)$, how is the triplet representation of $SU(2)$ described in terms of the Pauli matrices acting on a three-dimensional complex vector space?

I am also unsure as to exactly what should happen to the kinetic term for the Fermion field. Before insisting upon gauge invariance, this term should be of the form $$ i\overline{\psi}\gamma ^\mu \partial _\mu \psi . $$ However, because (I believe) gauge invariance is only being demanded for $\psi _L$, presumably something more complex happens to this term than just $i\overline{\psi}\gamma ^\mu D_\mu \psi$, where $D_\mu$ is the appropriate gauge covariant derivative. Instead, should this kinetic term be written in the form $$ i\overline{\psi_L}\gamma ^\mu D_\mu \psi _L+i\overline{\psi _R}\gamma ^\mu \partial _\mu \psi _R? $$

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You seem want to introduce gauge invariance into a theory that doesn't appear to have the global symmetry need in the first place. One way to think of gauge invariance is that you 'gauge' the global symmetry, then you just change your derivative terms to covariant derivatives like you mentioned. In other words, we can only concern ourselves with the global symmetry for now and gauge it at the very end if we wish. Now, at the very least in the term you wrote down

$$\phi \bar{\psi} \psi$$

it's not clear how the indices are contracted. Do the $\phi$ and $\psi$ have indices? For example I could make the $\psi$ transform in under $SU(2)$ by introducing a second copy of the $\psi$ and sum over them:

$$ \phi \bar{\psi^a} \psi^a $$

but now I can't make the $\phi $ transform because there isn't anything left to contract the $\phi$ index with. That is,

$$ \phi^b \bar{\psi^a} \psi^a $$

isn't a singlet (a singlet doesn't transform under the symmetry ) so it doesn't make any sense as a term in your lagrangian. Or you could introduce a second spinor that is a singlet under the global symmetry so you have something to contract your scalar indices with:

$$ \phi^a \bar{\psi^a} \eta $$

Finally, if you want a spinor to transform as a triplet, or in the adjoint of $SU(2)$ you could introduce the generator of SU(2) and contract indices as follows:

$$ \phi^a (t^a)^{bc} \bar{\psi^b} \psi^c = \phi^a \bar{\psi} t^a \psi$$

where the $t^a$ is in the fundamental (or doublet) representation so that we can properly trade the adjoint index for two doublet indices and get an overall singlet for the lagrangian.

Now, as for the kinetic terms, like you mentioned, if you want to introduce gauge symmetry trade the regular derivative for a covariant derivative.

$$ \partial_\mu \phi^a \rightarrow D_\mu \phi^a =\partial_\mu \phi^a + i {A_\mu}^b (t^b)^{ac}\phi^c $$

where the $t^a$ is in whatever rep the $\phi^a$ transforms in. The same holds for the fermion fields.

There is one caveat to all this however - this prescription of naively gauging a global symmetry that I have outlined breaks down if the global symmetry is 'anomalous'. That is, quantum mechanical effects break the naive, classical global symmetry. I'm not going to get into what that is, but keep it in the back of your mind for the time being and read about it when you have a chance.

I have a feeling you might want more info than this, but I'll stop here for now and if you edit I will clarify/ add.

EDIT: In retrospect this seems to work more easily for $SU(2)$ reps easier than other groups since for $SU(2)$ the adjoint rep is the same as the triplet rep so I can trade triplet rep indices for doublet indices using the generators $(t^a)^{ij}$. I am not sure if you can do these sort of things for groups in general.

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This actually came from a homework problem (there was a five tag limit). The problem reads: "One method of generating neutrino masses that we did not discuss in class involves adding a Higgs field $T$ which is a triplet under $SU(2)_L$ with a Yukawa coupling to the left-handed lepton doublets. If this triplet is a complex field, then it is possible to assign a lepton number to this new scalar field such that the lepton number is conserved.". He doesn't actually write out the Yukawa coupling term, so we are left to figure this out on our own. –  Jonathan Gleason May 30 '12 at 0:16
    
This part of my question can be phrased as: given this statement of the problem, what is the correct form of the Yukawa coupling term? (I realize what I wrote down didn't make sense. The problem is, I don't know how to modify it so that it does make sense). –  Jonathan Gleason May 30 '12 at 0:17
    
Say we have a Higgs doublet, and a Higgs triplet. Is it possible to produce a Higgs triplet out of Higgs doublet? I mean such that we have a term like $\sim D D T$? –  stupidity Jun 6 '12 at 14:28
    
JG - I'm sorry, didn't I answer this above? Stupidity - To get a triplet you can get use the regular decomposition $|1/2> \times |1/2> = | 1> + |0>$. For the term in your lagrangian, I believe $D^i (t^a)^{ij} D^j T^a$ does the job. –  DJBunk Jun 6 '12 at 15:02
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