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My reference is section 5.2 of BUSSTEPP Lectures on Supersymmetry, page 46.

A Vector superfield $V$ is a scalar superfield which satisfies the reality condition

$$V=\overline{V},$$

where the bar means complex conjugation.

A generic vector superfield has the following component expansion

\begin{align} V(x,\theta,\bar{\theta})=& C(x)+\theta\xi(x)+\bar{\theta}\bar{\xi}(x)+\bar{\theta}\bar{\sigma}^{\mu}\theta v_{\mu}(x)+\theta^{2}G(x)+\bar{\theta}^{2}\bar{G}(x) \\ & +\bar{\theta}^{2}\theta\eta(x)+\theta^{2}\bar{\theta}\bar{\eta}(x)+\theta^{2}\bar{\theta}^{2}E(x), \end{align}

where $C$, $v_{\mu}$ and $E$ are real fields.

Let $\Lambda$ be a chiral superfield. Then, it has the following component expansion

$$\Lambda(x,\theta,\bar{\theta})=\phi(x)+\theta\chi(x)+\theta^{2}F(x)+i\bar{\theta}\bar{\sigma}^{\mu}\theta\partial_{\mu}\phi(x)-\frac{i}{2}\theta^{2}\bar{\theta}\bar{\sigma}^{\mu}\partial_{\mu}\chi+\frac{1}{4}\theta^{2}\bar{\theta}^{2}\Box\phi(x).$$

Its real part is a particular vector superfield. The $U(1)$-gauge transformation is given by

$$V\rightarrow V-(\Lambda+\bar{\Lambda}).$$

Or more specifically,

\begin{align} C & \rightarrow C-(\phi+\bar{\phi}) \\ \xi & \rightarrow\xi-\chi \\ G & \rightarrow G-F \\ v_{\mu} & \rightarrow v_{\mu}-i\partial_{mu}(\phi-\bar{\phi}) \\ \eta_{\alpha} & \rightarrow\eta_{\alpha}+\frac{i}{2}(\sigma^{\mu})_{\alpha\dot{\beta}}\partial_{\mu}\bar{\chi}^{\dot{\beta}} \\ E & \rightarrow E-\frac{1}{4}\Box(\phi+\bar{\phi}) \end{align}

From the above transformations, one finds the following gauge invariant real fields

$$\lambda_{\alpha}(x)\equiv\eta_{\alpha}(x)-\frac{i}{2}(\sigma^{\mu})_{\alpha\dot{\beta}}\partial_{\mu}\bar{\xi}^{\dot{\beta}}(x),\quad D(x)\equiv E(x)-\frac{1}{4}\Box C(x).$$

In Wess-Zumino gauge, $C(x)=0$, $G(x)=0$ and $\xi(x)=0$, and the vector superfield can be written as

$$V(x,\theta,\bar{\theta})=\bar{\theta}\bar{\sigma}^{\mu}\theta v_{\mu}(x)+\bar{\theta}^{2}\theta\lambda(x)+\theta^{2}\bar{\theta}\bar{\lambda}(x)+\theta^{2}\bar{\theta}^{2}D(x).$$

One can define the following spinorial superfield

$$W_{\alpha}=-\frac{1}{4}\bar{D}^{2}D_{\alpha}V,\quad\bar{W}_{\dot{\alpha}}=-\frac{1}{4}D^{2}\bar{D}_{\dot{\alpha}}V,$$

where $D_{\alpha}=\partial_{\alpha}-i(\sigma^{\mu})_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\partial_{\mu}$, $\bar{D}_{\dot{\alpha}}=\bar{\partial}_{\dot{\alpha}}-i(\bar{\sigma})_{\dot{\alpha}\beta}\theta^{\beta}\partial_{\mu}$ and $D^{2}=D^{\alpha}D_{\alpha}$, and $\bar{D}^{2}=\bar{D}^{\dot{\alpha}}\bar{D}_{\dot{\alpha}}$.

It is easy to check that in Wess-Zumino gauge, the vector superfield can be written as

$$V(x,\theta,\bar{\theta})=e^{-iU}[\bar{\theta}\bar{\sigma}^{\mu}\theta v_{\mu}(x)+\bar{\theta}^{2}\theta\lambda(x)+\theta^{2}\bar{\theta}\bar{\lambda}+\theta^{2}\bar{\theta}^{2}(D(x)+\frac{i}{2}\partial^{\mu}v_{\mu}(x))],$$

where $U=\theta\sigma^{\mu}\bar{\theta}\partial_{\mu}$.

My question is exercise V.4.

It's easy to calculate $\bar{D}^{\dot{\alpha}}V$ and $\bar{D}^{2}V$, but this doesn't help anything in the calculation of $W_{\alpha}$ in equation (80).

Is there any trick to show the following identity (equation (80) on page 46)?

$$W_{\alpha}=e^{-iU}[\lambda_{\alpha}+2\theta_{\alpha}D+\frac{i}{2}\theta_{\beta}(\sigma^{\mu\nu})^{\beta}_{\,\,\,\alpha}f_{\mu\nu},+i\theta^{2}(\bar{\sigma})_{\dot{\beta}\alpha}\partial_{\mu}\bar{\lambda}^{\dot{\beta}}],$$

where $f_{\mu\nu}=\partial_{\mu}v_{\nu}-\partial_{\nu}v_{\mu}$.

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