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  Supersymmetry invariants

+ 2 like - 0 dislike

On page 158 of Fields, the supersymmetry algebra is represented in terms of the action on supercoordinates as

$$\delta \theta^\alpha = \epsilon^\alpha$$ $$\delta\bar{\theta}^{\dot{\alpha}} = \bar{\epsilon}^{\dot{\alpha}}$$ $$\delta x^{\alpha\dot{\beta}} = \frac{1}{2}i(\epsilon^\alpha \bar{\theta}^{\dot{\beta}} + \bar{\epsilon}^{\dot{\beta}}\theta^\alpha)$$

Further down on the page, the following statement is made:

In the supersymmetric case the infinitesimal invariants under the q's (and therefore p) are

$$d\theta^\alpha, \qquad d\bar{\theta}^{\dot{\alpha}}, \qquad dx^{\alpha\dot{\beta}} + \frac{1}{2}i(d\theta^\alpha)\bar{\theta}^{\dot{\beta}} + \frac{1}{2}i(d\bar{\theta}^{\dot{\beta}})\theta^\alpha$$

I get a sense that these should follow from the definition of the supersymmetry transformations, but what exactly does the quoted statement mean?

Thinking of the SUSY variation $\delta_\epsilon$ as something similar to a BRST variation, is this statement about nilpotency of the algebra in form notation or something?

This post imported from StackExchange Physics at 2015-05-22 21:05 (UTC), posted by SE-user leastaction

asked May 20, 2015 in Theoretical Physics by leastaction (425 points) [ revision history ]
edited Sep 29, 2015 by leastaction

It looks like $d\theta$ is simply the Berezin measure for Grassmann integration. 

Yes, that may as well be the case. I can't seem to edit my original post right now, but there's a distinction between $d\theta$ and $\delta \theta$ of course which I seem to have overlooked in my post. I suppose the claim the author is making is that in superspace, the stated quantities are in fact invariants under the supersymmetry transformations.

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