= Chiral-superfield component expansion
With <left Grassmann derivatives>, a <chiral superfield> is a finite expansion in chiral coordinates $y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$:
$$
\Phi(y,\theta)=\phi(y)+\sqrt2\theta\psi(y)+\theta^2F(y).
$$
Here $\phi$ is a <complex scalar field>, $\psi$ a <Weyl spinor> and $F$ an <auxiliary field>. The translation form of its ordinary-coordinate expansion is
$$
\Phi(x,\theta,\bar\theta)=\exp\bigl(i\theta\sigma^\mu\bar\theta\,\partial_\mu\bigr)\bigl(\phi(x)+\sqrt2\theta\psi(x)+\theta^2F(x)\bigr).
$$
The exponential terminates in the <Grassmann algebra>. To reduce its contractions, choose signature $(+---)$, $\sigma^\mu=(I,\boldsymbol\sigma)$, left differentiation, and $\epsilon_{12}=1$ for both lower spinor epsilon tensors. Define $\theta^2=\theta^\alpha\theta_\alpha$ and $\bar\theta^2=\bar\theta_{\dot\alpha}\bar\theta^{\dot\alpha}$. Then $v^\mu=\theta\sigma^\mu\bar\theta$ satisfies $v^\mu v^\nu=\tfrac12\theta^2\bar\theta^2\eta^{\mu\nu}$. Expanding the translation gives
$$
\Phi=\phi+\sqrt2\theta\psi+\theta^2F+i\theta\sigma^\mu\bar\theta\,\partial_\mu\phi-\frac{i}{\sqrt2}\theta^2\partial_\mu\psi\sigma^\mu\bar\theta-\frac14\theta^2\bar\theta^2\Box\phi,
\qquad\Box=\eta^{\mu\nu}\partial_\mu\partial_\nu.
$$
The last sign follows from $i^2/2=-1/2$ in the translation expansion. Defining the barred square in the reverse order, $\bar\theta^{\dot\alpha}\bar\theta_{\dot\alpha}$, negates it and writes the same term with a positive quarter coefficient. Spinor contractions and the <Grassmann algebra> convention must therefore accompany the reduced formula.
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