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  The criterion to obtain covariant spinor derivatives in superspace

+ 4 like - 0 dislike

It can be simply said that covariant derivatives in 4d superspace are given by $\mathcal{D}_\mu $, $\mathcal{D}_{\alpha}$ and $\mathcal{D}_{\dot{\alpha}}$, so that they commute with the representation of the supercharge generator $\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}}$, which means that the covariant derivatives of any superfield are again superfields.

But while following the detailed argument to derive the result (Quevedo {arxiv:1011.1491}), I found I don't understand the logic at the very critical point. According to him, the criterion for a field to be a superfield is \begin{equation} i[S, \epsilon Q+\bar{\epsilon}\bar{Q}]=i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S=\delta S, \end{equation}

And $\partial_{\alpha}S$ is not a superfield because

$\delta(\partial_{\alpha}S)=i[\partial_{\alpha}S, \epsilon Q+\bar{\epsilon}\bar{Q}] =i\partial_{\alpha}[S, \epsilon Q+\bar{\epsilon}\bar{Q}] = i\partial_\alpha(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S \neq i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S)$.

What I don't get is the second equality. It seems not trivial at all for me that $\partial_{\alpha}$ can be simply extracted outside (and this happens again when $\mathcal{D}$ is substituted for $\partial$), and I feel that we can jump directly to the equality

$ i[\partial_{\alpha}S, \epsilon Q+\bar{\epsilon}\bar{Q}] =i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S) $

from the superfield transformation rule. Can anybody explain what the matter is with my understanding?

This post imported from StackExchange Physics at 2014-10-03 11:12 (UTC), posted by SE-user sbthesy
asked Oct 3, 2014 in Theoretical Physics by sbthesy (20 points) [ no revision ]

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