# Supersymmetric variation of $F$ term using super Jacobi identity [closed]

Basically, i am solving problems in note with a slightly different notation.

\begin{align} [Q_\alpha F] = -i \lambda_\alpha (x), \quad [ \bar{Q}_{\dot{\alpha}},F] = -i \bar{\chi}_{\dot{\alpha}} \end{align}

I obtain $\lambda_\alpha$ is vanishing. But i am not sure for obtaining $\chi$ as a total derivative of $\psi$.

Here i set \begin{align} [\bar{Q}_{\dot{\alpha}}, A(x)]=0, \quad [Q_\alpha, A(x)] = - i\sqrt{2} \psi_\alpha(x) \end{align} The super jacobi identity gives \begin{align} &0=[\psi_\alpha, \{ Q_\beta, \bar{Q}_{\dot{\gamma}} \} ] + [ Q_\beta, \{\bar{Q}_{\dot{\gamma}}, \psi_\alpha \} ] + [ \bar{Q}_{\dot{\gamma}}, \{ \psi_\alpha, Q_\beta \}] \\ &=2i\sigma^\mu_{\phantom{\s}\beta\dot{\gamma}}\partial_\mu \psi_\alpha + i [ Q_\beta, X_{\alpha\dot{\gamma}} ] + i [ \bar{Q}_{\dot{\gamma}}, F_{\beta\alpha}] \\ &= 2i\sigma^\mu_{\phantom{\s}\beta\dot{\gamma}}\partial_\mu \psi_\alpha + \sqrt{2} \sigma^\mu_{\phantom{\m}\alpha\dot{\gamma}} [ Q_\beta, \partial_\mu A] - i \sqrt{2} \epsilon_{\beta\alpha} [ \bar{Q}_{\dot{\gamma}},F] \\ &= 2i\sigma^\mu_{\phantom{\s}\beta\dot{\gamma}}\partial_\mu \psi_\alpha - i 2 \sigma^\mu_{\phantom{\mu}\alpha\dot{\gamma}} \partial_\mu \psi_{\beta} + \sqrt{2} \epsilon_{\beta\alpha} \bar{\chi}_\gamma \end{align}

what i want to obatin is $\bar{\chi} = - \sqrt{2} \sigma^\mu \partial_\mu \psi$

I think something must be wrong in my computation. Could you correct my computation? Is my procedure right?

## closed as off-topic by ACuriousMind♦, Daniel Griscom, Kyle Kanos, mpv, Sebastian RieseJan 20 '16 at 23:36

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