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Kahler potential with multiple gauge groups

+ 2 like - 0 dislike
169 views

The Kahler potential for a gauge theory is given by
\begin{equation} 
\Phi ^\dagger e ^{ V} \Phi 
\end{equation} 
where $ V $ is the vector superfield corresponding to the gauge group. 

This doesn't seem to extend as smoothly to multiple gauge groups as I would have thought. Suppose we have two gauge superfields, $ V $ and $  V ' $. Since they live in different spaces I would think we would just have,
\begin{equation} 
\Phi ^\dagger e ^{ V} e ^{ V ' }  \Phi 
\end{equation} 
If we work in the Wess-Zumino gauge,
\begin{equation} 
V = ( \theta \sigma ^\mu \bar{\theta} ) V _\mu ^a T ^a 
\end{equation} 
then we have,
\begin{equation} 
{\cal L} \supset \int \,d^4\theta \Phi ^\dagger \left( 1 + V + \frac{1}{2} V ^2 \right) \left( 1 + V ' + \frac{1}{2}V ^{ \prime 2} \right) \Phi 
\end{equation} 
Since the generators in $ V $ and $ V ' $ act on different spaces typically the exponentials are independent of one another as you'd expect. 

However, there is one term that I wouldn't have expected which couples the different gauge fields,
\begin{equation} 
{\cal L} \supset \int \,d^4\theta \Phi ^\dagger V V' \Phi = \phi  ^\dagger t _a t _b ' V _\mu ^a V  ^{ \prime \, \mu b } \phi  
\end{equation} 
where $ t _a , t _b ' $ are the generators and $ \phi $ are the scalar components of the chiral superfields. 

I can't think of such an analogue of such terms in non-supersymmetric theories. Are these special to supersymmetric gauge theories or do they not exist for some reason?

asked Jun 11, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]

1 Answer

+ 3 like - 0 dislike

When you have a field charged under two gauge groups, the covariant derivative has three terms, the ordinary derivative, the first gauge field, and the second gauge field. When you square it to make the action, you get a mixed term in the product. It's not surprising--- it just means that in the contact-diagram involving two separate gauge bosons impinging at the same point on the scalar, the type of the gauge boson is freely variable, and it's required for gauge invariance. It's the same in supersymmetric and non-supersymmetric theories, the gauge interactions are determined from the free action and minimal coupling in both cases.

answered Jun 11, 2014 by Ron Maimon (7,535 points) [ revision history ]

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