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Given a superfield (in 0+1 spacetime + 2 superspace coordinates)

$$X(t,\theta_1,\theta_2) = x(t) + \theta_i \psi_i(t) + \theta_1 \theta_2 F_{12}(t)\tag{1}$$ and given the standard supercharges represented as differential operators $$Q_i = \frac{\partial}{\partial \theta_i} + i \theta_i \frac{\partial}{\partial t},\tag{2}$$

an action

$$S = \int dt \int d\theta_1d\theta_2 \,\mathcal L\tag{3}$$ is supersymmetric if

$$[Q_i^{hilbert}, \mathcal L]\equiv Q_i( \mathcal L) = \frac{\partial}{\partial t} (\cdots)\tag{4}$$

i.e. if the (differential) action of the supercharge is a total derivative. I've heard that any action built out of superfields and superderivatives thereof is guaranteed to be supersymmetric. But from these definitions, I don't see why. In what way does the superspace formalism furnish actions which are guaranteed supersymmetric?

Qmechanic
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Dwagg
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1 Answers1

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Well, one may show that the infinitesimal SUSY transformation $$\delta {\cal L}~\propto~Q_i ({\cal L}), \tag{A}$$ cf. e.g. my Phys.SE answer here. Now $Q_i ({\cal L})$ is by construction (2) a sum of total derivative terms. (Note in particular that in superspace the total derivative can be wrt. a Grassmann-odd superspace coordinate $\theta_i$.) Hence the infinitesimal SUSY transformation (A) is a quasi-symmetry for the Lagrangian density ${\cal L}$.

Qmechanic
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