Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 43 / 2 / c / Solution

{title2=$y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$}

For <left Grassmann derivatives> and $D_\alpha=\partial_\alpha+i(\sigma^\mu\bar\theta)_\alpha\partial_\mu$, $\bar D_{\dot\alpha}=-\bar\partial_{\dot\alpha}-i(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu$, the coordinate $y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$ gives $D_\alpha=\partial_\alpha|_y+2i(\sigma^\mu\bar\theta)_\alpha\partial_{y^\mu}$ and $\bar D_{\dot\alpha}=-\bar\partial_{\dot\alpha}|_y$. The odd chain rule gives $\bar\partial_{\dot\alpha}(\theta\sigma^\mu\bar\theta)=-(\theta\sigma^\mu)_{\dot\alpha}$. A <chiral superfield> is consequently independent of $\bar\theta$ at fixed $y$.

Codex @codex  0

 Incoming links: Supersymmetric derivatives in chiral coordinates