 Solution

Use left Grassmann derivatives and the usual contractions ${\partial }^2={\partial }^{\alpha }{\partial }_{\alpha }$ and ${\bar{\partial }}^2={\bar{\partial }}_{\dot{\alpha }}{\bar{\partial }}^{\dot{\alpha }}$. Set $a={\theta }^1,b={\theta }^2,c={\bar{\theta }}^{\dot{1}},d={\bar{\theta }}^{\dot{2}}$. The printed epsilon convention givesThe reversal of the barred contraction is essential. The left Grassmann derivative obeys the graded Leibniz rule, so ${\partial }_a(ab)=b$ but ${\partial }_b(ab)=-a$. With the index-raising convention,Applying these to the displayed Grassmann algebra monomials gives $A=B=-4$. Since every antisymmetric two-index product is proportional to epsilon, the $12$ and $\dot{1}\dot{2}$ components giveThus $C=-2$ and $D=2$.

In the remaining contraction, moving the first barred Grassmann variable past the second unbarred one introduces a minus sign. Inserting the two spinor identities then givesThe trace normalization is the one explicitly supplied in the paper. These two-component superspace contraction signs therefore giveAs a direct check in the mostly-minus convention, ${\sigma }^0=I$ gives $(ac+bd{)}^2=-2abcd$, whereas $(\theta \theta )(\bar{\theta }\bar{\theta })=-4abcd$. Their ratio is $+1/2$. A convention with $\mathrm{Tr}({\sigma }^{\mu }{\bar{\sigma }}^{\nu })=-2{\eta }^{\mu \nu }$ would change this last metric-relative sign; it is not the printed trace convention.