In the chiral coordinate $y^\mu=x^\mu+i\bar\theta\bar\sigma^\mu\theta$, a <chiral superfield> is
$$
\Phi(y,\theta)=\phi(y)+\sqrt2\,\theta^\alpha\psi_\alpha(y)+\theta^\alpha\theta_\alpha F(y).
$$
Here $\phi$ is a complex scalar, $\psi_\alpha$ is a two-component <Weyl spinor>, and $F$ is a complex <auxiliary field>. Off shell, $\phi$ and $F$ contribute two real bosonic components each, while $\psi$ has two complex Grassmann components. Thus there are four real bosonic and four real fermionic degrees of freedom.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 307 / 1 / Solution

{wiki}

A chiral superfield obeys $\bar D_{\dot\alpha}\Phi=0$. In chiral coordinates $y^\mu=x^\mu+i\bar\theta\bar\sigma^\mu\theta$ it expands as
$$
\Phi(y,\theta)=\phi(y)+\sqrt2\,\theta\psi(y)+\theta^2F(y).
$$


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