# Extra Term in Super Particle Action

If you set up the first order formulation of a super particle $$S = \int dt \frac{1}{2}(\dot{x}^{\mu} \dot{x}_{\mu} - i \chi^{\mu} \dot{\chi}_{\mu})$$ by incorporating the constraints $$p^{\mu} p_{\mu} = 0, \chi^{\mu} p_{\mu} = 0$$ as a constrained Lagrangian $$S = \int dt ( \dot{x}^{\mu} p_{\mu} - \frac{e}{2}p^{\mu}p_{\mu} - \frac{i}{2} \chi_{\mu} \dot{\chi}^{\mu} - \frac{ik}{2} \psi \chi^{\mu} p_{\mu}),$$ eliminate the momentum from this via it's equations of motion $$\frac{\partial L}{\partial p_{\mu}} = 0,$$ to get $$p_{\mu} = e^{-1}(\dot{x}_{\mu} - \frac{ik}{2}\psi \chi_{\mu}),$$ and then plug this in to that first order action, you're supposed to get the action with super gravity $$S = \int d t \frac{1}{2}(e^{-1}\dot{x}^{\mu} \dot{x}_{\mu} - i \chi^{\mu} \dot{\chi}_{\mu} - i k e^{-1} \psi \dot{x}^{\mu} \dot{\chi}_{\mu})$$ yet I keep getting extra terms: \begin{align} S &= \int dt ( \dot{x}^{\mu} p_{\mu} - \frac{e}{2}p^{\mu}p_{\mu} - \frac{i}{2} \chi_{\mu} \dot{\chi}^{\mu} - \frac{ik}{2} \psi \chi^{\mu} p_{\mu}) \\ &= \int dt \{ \dot{x}^{\mu} e^{-1}(\dot{x}_{\mu} - \frac{ik}{2}\psi \chi_{\mu}) - \frac{e}{2}[e^{-1}(\dot{x}^{\mu} - \frac{ik}{2}\psi \chi^{\mu})][e^{-1}(\dot{x}_{\mu} - \frac{ik}{2}\psi \chi_{\mu})] \\ & \ \ \ \ \ \ \ \ - \frac{i}{2} \chi_{\mu} \dot{\chi}^{\mu} - \frac{ik}{2} \psi \chi^{\mu} (e^{-1}(\dot{x}_{\mu} - \frac{ik}{2}\psi \chi_{\mu}))\} \\ &= \int dt \{ e^{-1} \dot{x}^{\mu} \dot{x}_{\mu} - \frac{ik}{2}e^{-1} \dot{x}^{\mu} \psi \chi_{\mu} - \frac{e^{-1}}{2} \dot{x}^{\mu} \dot{x}_{\mu} + \frac{e^{-1}ik}{4}\psi \chi^{\mu} \dot{x}_{\mu} + \frac{e^{-1}ik}{4} \dot{x}^{\mu} \psi \chi_{\mu} \\ & \ \ \ \ \ \ \ \ \ + \frac{e^{-1}k^2}{8}\psi \chi^{\mu} \psi \chi_{\mu} - \frac{i}{2} \chi_{\mu} \dot{\chi}^{\mu} - \frac{e^{-1}ik}{2}\dot{x}^{\mu} \psi \chi_{\mu} - \frac{e^{-1}k^2}{4}\psi \chi^{\mu} \psi \chi_{\mu}\} \\ &= \int dt (\frac{1}{2}e^{-1}\dot{x}^{\mu} \dot{x}_{\mu} - \frac{1}{2} i \chi^{\mu} \dot{\chi}_{\mu} - \frac{1}{2} i k e^{-1} \psi \dot{x}^{\mu} \dot{\chi}_{\mu} + \frac{e^{-1}k^2}{8}\dot{x}^{\mu}\psi \chi_{\mu} \psi - \frac{e^{-1}k^2}{4}\dot{x}^{\mu}\psi \chi_{\mu} \psi) . \end{align} The last two terms should not be there - either missed something simple, or do you ignore order $k^2$ terms?

## 1 Answer

The "gravitino" $\psi$ is a Grassmann odd field, so it anticommutes with any other odd field. In particular, $\psi\psi = -\psi\psi$ so $\psi^2 = 0$.