The shift of a scalar field $\Phi$: $$ \Phi \rightarrow \Phi'=\Phi - i \epsilon $$ is generated by $$ G = -i \frac{d}{d\Phi},$$ because $$ \mathrm{e}^{-i \epsilon \frac{d}{d\Phi} } \Phi = (1-i\epsilon \frac{d}{d\Phi} + \ldots ) \Phi \approx \Phi - i \epsilon .$$

What is the analogous generator for shifts of a Weyl spinor field $\Psi$ $$ \Psi \rightarrow \Psi'=\Psi - i \xi \quad ? $$ In index notation, we have $$ \Psi_\alpha \rightarrow \Psi_\alpha'=\Psi_\alpha - i \xi_\alpha .$$ Naively, one gets $$ G_\beta = -i \frac{d}{d\Psi_\beta},$$ $$ \mathrm{e}^{-i \xi_\beta \frac{d}{d\Psi_\beta} } \Psi_\alpha = (1-i\xi_\beta \frac{d}{d\Psi_\beta} + \ldots ) \Psi_\alpha \approx \Psi_\alpha - i \xi_\alpha .$$

where I used $\frac{d}{d\Psi_\beta} \Psi_\alpha = \delta_{\alpha \beta} $.

I'm really unsure about the spinor indices here. Shouldn't there be a spinor metric $i \sigma_2$ somewhere, because we are multiplying spinors? Does $\frac{d}{d\Psi_\beta} \Psi_\alpha = \delta_{\alpha \beta} $ make sense at all?

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    $\begingroup$ $\frac{\partial}{\partial\Psi_\beta}\Psi_\alpha=\delta^\beta_\alpha$ is a well-defined result of the basic algebra of spinors (but note the position of indices). $\endgroup$ Jan 4, 2017 at 11:10

1 Answer 1


As in my comment above, $$ \frac{\partial}{\partial\Psi_\beta}\Psi_\alpha=\delta^\beta_\alpha $$ is one of the basic results of the algebra of spinors (but notice that the index $\beta$ is an upper index).

The generator $$ G_\beta = -i \frac{d}{d\Psi_\beta}, $$ is used, for example, to define supercovariant derivatives (e.g., see Superspace, where you can find $D=\frac{\partial}{\partial\theta}+\cdots$).


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