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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?

Beom.Jean
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1 Answers1

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For a superfield $S$, an infinitesimal supersymmetric transformation gives :

$\delta_\epsilon S = -i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S$

Here $\mathcal{Q}_a$ and ${ \mathcal{\bar Q}}^{\dot a}$ are differential operators applying on the superfield:

$\mathcal{Q}_a = i \partial_a - (\sigma^\mu \bar \theta)_a \partial_\mu$,

${ \mathcal{\bar Q}}^{\dot a} = i \partial^{\dot a} - (\bar \sigma^\mu \theta)^{\dot a} \partial_\mu$

(with notations $\partial_a = \frac{\partial}{\partial \theta_a}, \partial^{\dot a} = \frac{\partial}{\partial \theta^{\dot a}}$)

So :

$\delta_\epsilon S = (\epsilon.\partial+ \bar\epsilon.\bar \partial + i (\epsilon \sigma^\mu \bar \theta + \bar \epsilon \sigma^\mu \theta ) \partial_\mu)S $

So, you have :

$\partial_a \delta_\epsilon S = \delta_\epsilon \partial_aS-i ( \bar \epsilon \sigma^\mu)_a \partial_\mu S$

Now, looking for "covariant" derivatives means that we are looking for a definition of derivatives which are consistent with the supersymmetry transformations, so we want that the derivative operation and the (infinitesimal) supersymmetry operation, applying on a superfield, commute.

This is clearly not the case with the derivatives $\partial_a$ (or $\partial^{\dot a}$)

Trimok
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  • Thank you for the answer, but it seems that my problem has not been resolved yet. As you explained in detail, it is obvious that $\partial_{\alpha}$ doesn't commute with $\delta$, and $D_{\alpha}$ does. And I think it's OK if we want to define the covariant derivative as those which commute with $\delta$. But What I'm wondering is that, according to Quevedo, the condition looks effective only when it is used for $D_{\alpha}S$ to satisfy the first eq. I wrote above. And my problem is, as I mentioned in my question. I can't see the relation clearly. – Beom.Jean Oct 03 '14 at 13:54
  • @sbthesy : I begin to think that there is an error in the text (beginning of the page $40$): I would have written : $i\partial_\alpha (\delta_\epsilon S) =i\partial_{\alpha}[S, \epsilon Q+\bar{\epsilon}\bar{Q}] = i\partial_\alpha(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S \neq i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S) = \delta_\epsilon (\partial_\alpha S)$ – Trimok Oct 04 '14 at 15:17