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Using the covariant derivative when building a superpotential

+ 7 like - 0 dislike
3778 views

When building a supersymmetric theory people typically write the superpotential as a products of fields. However, there is another ingredient that could in principle be used - the covariant derivative. Since it still transforms properly under supersymmetry it seems like a perfectly valid operator. Furthermore such terms will be SUSY invariant since we integrate over superspace.

As an example I've seen it written that the most general renormalizable superpotential for a single chargeless field is given by,
\begin{equation} 
W = \alpha + \beta \Phi + m\Phi ^2 + \lambda \Phi ^3 
\end{equation} 
However, there could in principle also be a term,
\begin{equation} 
{\cal D} ^\alpha \Phi {\cal D} _\alpha \Phi 
\end{equation} 
where $ {\cal D} ^\alpha \equiv \partial ^\alpha + i ( \sigma ^\mu \bar{\theta} ) _\alpha \partial _\mu $. This would give Lagrangian contribution:
\begin{equation} 
{\cal L} \supset \int \,d^2\theta {\cal D} ^\alpha \Phi {\cal D} _\alpha \Phi + h.c. = F ( y ) F  ( y ) + h.c. 
\end{equation} 
and since $ F $ is real this would potentially wreck havoc on the scalar potential. 

Is there some reason they can be omitted?

asked Jun 14, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
retagged Jun 18, 2014 by Dilaton

3 Answers

+ 6 like - 0 dislike

In 4d, with $\mathcal{N}=1$ supersymmetry, there are two kinds of superspace integrals -- one over the full superspace $d^4\theta$ and the other over chiral superspace $d^2\theta$. Integrals over chiral superspace are automatically invariant under supersymmetry if the integrand is a chiral. If $\Phi$ is a chiral super field, then any power or function of it remains chiral. However after  taking a derivative, a term like $D_\alpha \Phi$  is no longer chiral as one can easily check. That is the reason one never sees terms like the one Jeff suggests. That leaves behind the possibility of the integral over an arbitrary function of chiral superfields and this is called the superpotential. Adding a vector field makes the story more interesting. The kinetic term for a vector field appears through a chiral integral over a chiral field created from derivatives of the vector field. The most general kinetic term can include $\mathcal{W}^2$ multiplied by an arbitrary function of chiral superfields. Such terms do indeed appear in low-energy effective actions.

answered Jun 17, 2014 by suresh (1,535 points) [ revision history ]

This is true when you take 1 superderivative, but when you take 2 superderivatives, you are back to normal. The real reason is explained e.g. in Weinberg's QFTIII book on page 72, terms which involve 2 superderivatives in the superpotential are really secretly doing an integral over the other 2 fermionic coordinates in superspace, and can be moved over to the Kahler potential. The superpotential by convention omits all terms with two derivatives, which are moved to the D term by convention. Although they could have been written in the superpotential, this is never done (and it is the right convention, because these terms really are doing 4 grassman integrals inside, not 2).

Taking two derivative doesn't make it chiral either. I mean $\bar{D}_{\dot{\alpha}} D_\beta D_\gamma \Phi\neq 0$.  Please check! What Weinberg (and Ron) are saying is that a term like the Kahler potential (call it $K(\Phi,\bar{\Phi})$) which is written as a integral over the full superspace can be written as a integral over chiral superspace by $\bar{D}^2 ( K)$ plus the complex conjugate term. The key is that  $\bar{D}^2 ( K)$ is chiral and is permitted as per my argument above. However, Jeff is talking about the other derivative $D_\alpha$. So it doesn't matter if there are one or two, the answer will remain non-chiral.

Whoa, whoa, you are giving me too much credit! I just made a mistake. That's what happens when you read books, instead of working it out.

+ 2 like - 0 dislike

I found why you can omit superderivative terms from the superpotential in Weinberg QFTIII p.72. The double super-derivative is equivalent to integrating over the other dimensions of superspace, and any derivative term which is still a chiral superfield is such that it can be written as a D-term, and is taken to contribute to the Kahler potential, not the superpotential.

answered Jun 18, 2014 by Ron Maimon (7,535 points) [ no revision ]

Dear Ron, good to see you here. Note that Weinberg's page III/72 isn't addressing the same terms as the OP. The OP talks about $(D\Phi)(D\Phi)$ while Weinberg talks about $DD\Phi$. It's a different thing. The OP doesn't discuss second D-derivatives.

Thanks Lubos, sorry for being an idiot, I'll fix the answer. To be overly-generous to myself, however, the general question is why you can omit superderivative terms in general from the superpotential, not just the specific one in the question, and Weinberg's correct answer is that the properly chiral superderivatives are always of the form that can be rewritten as a full integral over superspace, and moved to the Kahler potential. This is true, but I should explain which terms are properly chiral, and can contribute to the superpotential, and which are not (like OP's). Again, not making excuses for the stupidity above, just explaining why I am not deleting the answer immediately.

+ 2 like - 1 dislike

The spinor superfields are also built from contractions over covariant derivatives of the vector superfield. I'm not sure this helps your question, but you are correct that the covariant derivative plays a role in the construction of superfield Lagrangians. In fact, when we look at a chiral superfield coupled to a vector superfield the term $\bar{\Phi} V \Phi|_{\theta \theta \bar{\theta} \bar{\theta}}$ contains the product of $\bar{F}$ and $F$ where $F$ appears as the $\theta \theta$ component of $\Phi$.

Of course, when a term is not included in a Lagrangian, sometimes the reason is simply, because we can.
 

answered Jun 15, 2014 by James S. Cook (95 points) [ no revision ]

We can't leave out terms from a Lagrangian which is supposed to be consistent, the terms that need to be included are determined from renormalizability, and the only way to leave out something is to forbid it by symmetry. The renormalizability constrains the terms to those of low dimension, and this is why you get simple superpotentials.

@JamesS.Cook: Thanks, good point. I forgot that we do in fact use the covariant derivative when building $W_\alpha W^\alpha$ with $W_\alpha\sim {\bar {\cal D}} ^2 {\cal D} ^\alpha V $. This makes the suggestion of disallowing other terms with  ${\cal D} ^\alpha$ even stranger. I starting to suspect that the reason we forget about it is just because its typically forbidden by gauge invariance and renormalizability...

@Ron Maimon the dimension of the terms $D^{\alpha} \Phi D_{\alpha} \Phi$ is higher than those in the standard cubic potential? My understanding of the quantum mechanical side of all this is a bit fuzzy.

@JamesS.Cook: The term ${\cal D}^\alpha \Phi {\cal D}_\alpha \Phi $ is of dimension 3 (dimension 4 after integration over $d^2\theta$) which means its still renormalizable.

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