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Kahler potential for adjoint representation fields

+ 3 like - 0 dislike
158 views


In supersymmetric theories of fields in fundamental representation we write the Kahler interactions as 
\begin{equation} 
\Phi ^\dagger e ^{ 2 qV } \Phi 
\end{equation} 
where $V$ is the vector superfield in the fundamental representation. This is necessary to keep the fields which transform as,
\begin{align} 
& \Phi \rightarrow e ^{ i \Lambda } \Phi , \quad V \rightarrow V - \frac{ i }{ 2} \left( \Lambda - \Lambda ^\dagger \right) 
\end{align}
gauge invariant.

I would naively think that this requirement would transfer over to fields in other representations. However recently I reading a paper where they introduce fields in the adjoint representation, $\Phi _a$, and I believe they didn't include the gauge contribution and just wrote,
\begin{equation} 
\Phi _a ^\dagger \Phi _a 
\end{equation} 
(though they don't state or write this explicitly so I'm not sure). This doesn't make sense to me since adjoint representation fields still transform. Is there a reason why this would be justified, or did I misunderstand the paper?

asked May 16, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]

1 Answer

+ 3 like - 0 dislike

You need to check if the chiralilty condition on the  superfield -- if it involves the gauge covariant derivative, then $\Phi^\dagger \Phi$ will be locally gauge invariant.

A chiral superfield is usually defined by $\bar{D}_{\dot{\alpha}} \Phi=0$. The gauge covariant version is given by first defining $$\bar{\mathcal{D}}_{\dot{\alpha}}:= e^V \bar{D}_{\dot{\alpha}} e^{-V}\ ,$$ where $V$ is taken to be in a representation of the gauge group.Then, one defines a "covariant chiral super field" by $$\bar{\mathcal{D}}_{\dot{\alpha}} \Phi=0\ . $$ (see Wess & Bagger or section 2 of this paper).

(moving my comment into the answer)

answered May 17, 2014 by suresh (1,535 points) [ revision history ]
edited May 18, 2014 by suresh

By chirality condition do you mean to check that 

$$(\partial_\alpha - (\sigma \theta)_\alpha ) \Phi_a = 0 $$

for $\partial_\alpha$ or $D_\alpha$?

A chiral superfield is usually defined by $\bar{D}_{\dot{\alpha}} \Phi=0$. The gauge covariant version is given by first defining $\bar{\mathcal{D}}_{\dot{\alpha}}:= e^V \bar{D}_{\dot{\alpha}} e^{-V}$. Then, one defines a "covariant chiral super field" by $\bar{\mathcal{D}}_{\dot{\alpha}} \Phi=0$. (see Wess & Bagger or section 2 of this paper).

Thanks for your help. I think I misunderstood what they were doing in the paper. They had the correct Kahler term but were treating the $D$ terms of the gauge fields in a strange way. If you have time, I've added in a new related question here.

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