# Constraints on Chiral superfield

by ChrisVer
Tags: chiral, constraints, superfield
 P: 906 Suppose we have a superfield $\Phi(x,\theta,\bar{\theta})$ this can be expanded in component fields in the standard way as: $\Phi(x,\theta,\bar{\theta})= c(x) + \theta \psi(x) + \bar{\theta} \bar{ζ}(x) + \theta^{2} F(x) + \bar{\theta}^{2} Z(x) + \theta \sigma^{\mu} \bar{\theta} u_{\mu}(x) + \theta^{2} \bar{\theta} \bar{λ}(x) + \bar{\theta}^{2} \theta β(x) + \bar{\theta}^{2} \theta^{2} D(x)$ In case we want to say that this superfield is chiral, in general we have to impose the constraint that the supercovariant derivative on it will have to vanish (depends on if it's right or left which we choose).... Nevertheless. This will bring some contraints about the component fields in $\Phi$'s expansion above, which you can work by asking for the cov. derivative of it to vanish. However I'd like to ask if there's a faster way to do that, and avoid the tedious calculations.... For example I'd take a left chiral superfield: $\Phi_{L}( y, \theta) = \phi(y) + \theta \psi(y) + \theta^{2} F(y)$ and bring it from $S_{L}$ repr back to $S$, by doing a translation in the usual way: $\Phi_{L} (x+ i \theta \sigma \bar{\theta}, \theta)$ $= \phi(x) + i (\theta \sigma^{\mu} \bar{\theta}) \partial_{\mu} \phi(x) - \frac{1}{2} (\theta \sigma^{\mu} \bar{\theta})(\theta \sigma^{\nu} \bar{\theta}) \partial_{\mu} \partial_{\nu} \phi(x) + \theta \psi(x) + i (\theta \sigma^{\mu} \bar{\theta}) (\theta \partial_{\mu} \psi)+ \theta^{2} F(x)$ If I now try to compare the first superfield's components with the last one, shouldn't I get the constraints needed for it to be a chiral superfield? eg fastly: $\bar{ζ}(x)=0$ $Z(x)=0$ $D(x)= -\frac{1}{4} \partial_{\mu} \partial^{\mu} \phi$ ( because $(\theta \sigma^{\mu} \bar{\theta})(\theta \sigma^{\nu} \bar{\theta})=\frac{1}{2} n^{\mu \nu} \theta^{2} \bar{\theta}^{2}$) $u_{\mu}= i \partial_{\mu} \phi(x)$ $β(x)=0$ etc...
 P: 906 I don't really get the answer... A shift $x\rightarrow x + i \theta \sigma \bar{\theta}$ is going to change the cov.derivative of the Left repr to the Cov derivative of the S-repr... $\bar{D}_{\dot{A}}= - \frac{\partial}{\partial \bar{\theta}^{\dot{A}}}$ $= - \frac{\partial \bar{\theta}'^{\dot{B}}}{\partial \bar{\theta}^{\dot{A}}} \frac{\partial }{\partial \bar{\theta}'^{\dot{B}}}- \frac{\partial y^{\mu}}{\partial \bar{\theta}^{\dot{A}}} \partial_{\mu}^{(y)}$ $= - \frac{\partial}{\partial \bar{\theta}^{\dot{A}}} + i (θσ^{\mu})_{\dot{A}} \partial_{\mu}^{(y)}$ which isn't the same as the cov derivative in the general representation.... any idea? However I am sure after using it so many times that the coord change above is the one that's supposed to move me from $S_{L} \rightarrow S$ repr...