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Can i use the $y^*$ representation of the chiral covariant derivatives on a superfield that contains both $y$ and $y^*$?

+ 2 like - 0 dislike
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Imagine I want to compute this

$$D^{\dagger2}D_{\alpha}(\Phi^*\Phi)$$

where the $D$-s are super-covariant derivatives and $\Phi$ is a chiral superfield. Following the notation of this review http://arxiv.org/abs/hep-ph/9709356 on supersymmetry, the chiral covariant derivatives are (page 33)

$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{\mu}\theta^{\dagger})_{\alpha}\partial_{\mu}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}+i(\theta^{\dagger}\bar{\sigma}^{\mu})^{\alpha}\partial_{\mu}$$

$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}-i(\bar{\sigma}^{\mu}\theta)^{\dot{\alpha}}\partial_{\mu}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}+i(\theta\sigma^{\mu})_{\dot{\alpha}}\partial_{\mu}$$

let's now consider the following coordinate change

$$x^{\mu}\to{}y^{\mu}=x^{\mu}+i\theta^{\dagger}\bar{\sigma}^{\mu}\theta$$
  
the review claims (page 34) that in these $y,\theta,\theta^{\dagger}$ coordinates the chiral covariant derivative are

$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-2i(\sigma^{\mu}\theta^{\dagger})_{\alpha}\frac{\partial}{\partial{}y^{\mu}}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}+2i(\theta^{\dagger}\bar{\sigma}^{\mu})^{\alpha}\frac{\partial}{\partial{}y^{\mu}}$$

$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}$$

we can as well go to coordinates

$$x^{\mu}\to{}\bar{y}^{\mu}=x^{\mu}-i\theta^{\dagger}\bar{\sigma}^{\mu}\theta$$

where in page 35 it is claimed that the chiral covariant derivatives take the form

$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}$$

$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}-2i(\bar{\sigma}^{\mu}\theta)^{\dot{\alpha}}\frac{\partial}{\partial\bar{y}^{\mu}}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}+2i(\theta\sigma^{\mu})_{\dot{\alpha}}\frac{\partial}{\partial\bar{y}^{\mu}}$$

so far so fine. It is obvious that for some computations the choice of the clever coordinates will make computations considerably less cumbersome. This is where my problems arise. The computation I want to do contains a chiral superfield, which can be expressed most compactly using the $y,\theta$ coordinates, whilst the antichiral superfield is better written using $\bar{y},\theta^{\dagger}$. My question is, is itlegitimate to use the chiral covariant derivatives is the $\bar{y},\bar{\theta}$ form, even though what i wanna take the derivative of contains all $y\bar{y},\theta,\theta^{\dagger}$? even if the answer is no, what i the wisest way to perform my computation?

asked Jul 17, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]

1 Answer

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I still am unsure about wether it is legitimate to use one representation of the super-covariant derivatives on something that is $y$ and $y^*$ dependent (even though I suspect that the answeris no). Nonetheless, I have found that an efficient way of making the computation not so cumbersome is writing the chiral superfield in $y$ coordinates, where it is very simple, and writing the antichiral superfield also in $y$ coordinates. This one will be longer. The reason for this choice is that this makes the last two super covariant derivatives trivial since they only involve a derivative over the Grasmann variable $\theta^{\dagger}$. 

Thus, we expand both chiral and antichiral superfields in $y$ coordinates, we multiply them, we take the first super-covariat derivative, and we keep only terms with $\theta^{\dagger{}2}$, since only these will survive the last two derivatives.

answered Jul 18, 2015 by Dmitry hand me the Kalashnikov (720 points) [ revision history ]

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