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I work on a introduction into Super-symmetry. In the course we define \begin{equation} D_{\alpha} = \frac{\partial}{\partial \theta^{\alpha}} - i \sigma^{\mu}_{\alpha \dot{\alpha}} \bar{\theta}^{\dot{\alpha}} \partial_{\mu} \end{equation} and \begin{equation} \bar{D}_{\dot{\alpha}} = \frac{\partial}{\partial \bar{\theta}^{\dot{\alpha}}} - i \bar{\sigma}^{\mu}_{\dot{\alpha} \alpha} \theta^{\alpha} \partial_{\mu}. \end{equation} The super field strength is defined as \begin{equation} W_{\alpha} = - \frac{1}{4} \bar{D}^2 D_{\alpha} V \end{equation} and \begin{equation} \bar{W}_{\dot{\alpha}} = - \frac{1}{4} D^2 \bar{D}_{\dot{\alpha}} V \end{equation} where $V$ is a real superfield.

Now I wanna show the following identity \begin{equation} D^{\alpha}W_{\alpha} = \bar{D}^{\dot{\alpha}} \bar{W}_{\dot{\alpha}}. \end{equation}

But when I start calculating I never end up with the given identity. I guess this is a well known identity and can be found in some textbooks or easy to prove in a few lines when one is very familiar with spinor notation. My question is if someone can prove this (explaining steps) here.

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We need to prove the operator identity $$ D^\alpha \bar{D}^2 D_\alpha = \bar{D}^{\dot{\alpha}} D^2 \bar{D}_{\dot{\alpha}} \,. $$ To prove this we first note the anti-commutation relation $$ \{D_\alpha,\bar{D}_{\dot{\beta}}\} = 2i\sigma^\mu_{\alpha\dot{\beta}}\partial_\mu \,, $$ and then use this relation twice as follows, \begin{align} D^\alpha \bar{D}^2 D_\alpha &= -\bar{D}_{\dot\beta}D_\alpha\bar{D}^{\dot{\beta}}D^\alpha +2i\sigma^\mu_{\alpha\dot{\beta}}\partial_\mu\bar{D}^{\dot{\beta}}D^\alpha \\&= \bar{D}_{\dot{\beta}}\bar{D}^{\dot{\beta}}D_\alpha D^\alpha -2i\bar{D}^{\dot{\beta}}\sigma^\mu_{\alpha\dot{\beta}}\partial_\mu D^\alpha +2i\sigma^\mu_{\alpha\dot{\beta}}\partial_\mu\bar{D}^{\dot{\beta}}D^\alpha \,. \end{align} (Recall we are allowed to move one $\alpha$ index up if we move the other one down, and similarly for the $\dot{\alpha}$ indices.) The last two terms cancel, as $\partial_\mu$ commutes with any of the $D,\bar{D}$ operators, and hence we just have $$ D^\alpha \bar{D}^2 D_\alpha = \bar{D}^2 D^2 \,. $$ It should be clear that we could have moved the $D,\bar{D}$ operators around differently to get instead $D^2 \bar{D}^2$. It should also be clear that we can make a similar computation for $\bar{D}^{\dot{\alpha}} D^2 \bar{D}_{\dot{\alpha}}$. With these results we have established the required identity.

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