fields the chiral superfield $\Phi$ is given by $$\begin{aligned}
\Phi(x, \theta, \bar{\theta}) &=& A(x) 
+ \sqrt{2}\theta^\alpha\psi_\alpha(x)
+ \theta\theta F(x) + i\theta\sigma^l\bar{\theta}\p_l A(x) \nonumber\\
&& -\frac{i}{\sqrt{2}}\theta\theta\p_m\psi^\alpha(x)
\sigma^m_{\ \alpha\dot{\alpha}}\bar{\theta}^{\dot{\alpha}}
+ \frac{1}{4}\theta\theta\bar{\theta}\bar{\theta}\Box A(x).\label{chiral}\end{aligned}$$ It is easy to calculate the $\star$-product of two chiral fields from (\[star\]). It is given by $$\begin{aligned}
\Phi\star\Phi &=& A^2 - \frac{C^2}{2}F^2 +
\frac{1}{4}C^{\alpha\beta}\bar{C}^{\dot{\alpha}\dot{\beta}}
\sigma^m_{\ \alpha\dot{\alpha}}\sigma^l_{\ \beta\dot{\beta}}(\p_m
A)(\p_l A)
+\frac{1}{64}C^2\bar{C}^2 (\Box A)^2\nonumber\\
&& + \theta^\alpha\Big( 2\sqrt{2}\psi_\alpha A 
-\frac{1}{\sqrt{2}}C^{\gamma\beta}\bar{C}^{\dot{\alpha}\dot{\beta}}
\varepsilon_{\gamma\alpha}
(\p_m\psi^\rho)\sigma^m_{\ \rho\dot{\beta}}\sigma^l_{\ \beta\dot{\alpha}}
(\p_l A) \Big)
\nonumber\\
&& -\frac{i}{\sqrt{2}}C^2 \bar{\theta}_{\dot{\alpha}}
\bar{\sigma}^{m\dot{\alpha}\alpha}(\p_m\psi_\alpha)F
+ \theta\theta \Big( 2AF - \psi\psi\Big) \nonumber\\
&& + \bar{\theta}\bar{\theta}\Big( -\frac{C^2}{4}\big( F\Box A
- \frac{1}{2}(\p_m\psi)\sigma^m\bar{\sigma}^l(\p_l\psi) \big) \Big) 
\nonumber\\
&& + i\theta\sigma^m\bar{\theta}\Big( (\p_m A^2) 
+ \frac{1}{4}C^{\alpha\beta}\bar{C}^{\dot{\alpha}\dot{\beta}}
\sigma_{m\alpha\dot{\alpha}}\sigma^l_{\ \beta\dot{\beta}}(\Box A)
(\p_l A)\Big) \nonumber\\
&& +  i\sqrt{2}\theta\theta\bar{\theta}_{\dot{\alpha}}
\bar{\sigma}^{m\dot{\alpha}\alpha}
\big( \p_m(\psi_\alpha A)\big) + \frac{1}{4}\theta\theta\bar{\theta}
\bar{\theta} (\Box A^2) , \label{phistarphi}\end{aligned}$$ where $C^2 =
C^{\alpha\beta}C^{\gamma\delta}\varepsilon_{\alpha\gamma}\varepsilon_{\beta\delta}$ and $\bar{C}^2 =
\bar{C}_{\dot{\alpha}\dot{\beta}}\bar{C}_{\dot{\gamma}\dot{\delta}}
\varepsilon^{\dot{\alpha}\dot{\gamma}}\varepsilon^{\dot{\beta}\dot{\delta}}$. One sees that due to the $\bar{\theta}$-term and the $\bar{\theta}\bar{\theta}$-term (\[phistarphi\]) is not a chiral field. But, in order to write an action invariant under the deformed SUSY transformations (\[defsusytr\]) we need to preserve the notion of chirality. This can be done in different ways. One possibility is to use a different $\star$-product, the one which preserves chirality [@Seiberg]. However, chirality-preserving $\star$-product implies working in 