Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 43 / 2

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.

The defining restriction is the reality conditionFor a non-Abelian gauge group, write the vector superfield in a Hermitian generator basis with real superfield coefficients. This is a condition on the full superfield, not a chirality constraint. It relates conjugate component coefficients and makes the vector and $D$ auxiliary field real. Wess-Zumino gauge is a further supergauge transformation choice removing redundant components; it is not the defining condition for a vector superfield.

Introduce $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and retain the same Grassmann variables. The supersymmetric derivatives in chiral coordinates follow from the left Grassmann derivative chain rule. Differentiating $y$ givesThe minus sign in the second relation comes from moving the odd Grassmann derivative past $\theta$. Hence, as operators on a superfield expressed in $(y,\theta ,\bar{\theta })$,Substitution into the two supersymmetric covariant derivatives adds the two unbarred spacetime terms and cancels the two barred ones:These operator equalities prove both requested actions on $V$. In particular, a chiral superfield becomes independent of $\bar{\theta }$ at fixed $y$. The TeX aid corrupts the second formula by replacing its ordinary barred derivative with a covariant one; the original PDF has the ordinary $-{\bar{\partial }}_{\dot{\alpha }}$.

The Abelian field-strength chiral projection is linear in the vector superfield, so isolate the terms containing the gaugino $\lambda$ and the $D$ auxiliary field. Replacing $x$ by $y-i\theta \sigma \bar{\theta }$ leaves these terms unchanged: the shift of the gaugino term would contain three barred Grassmann variables, and the shift of the $D$ term would contain three of each chirality, hence both vanish. Their contribution is thereforeThe $2i({\sigma }^{\mu }\bar{\theta }{)}_{\alpha }{\partial }_{\mu }$ part of the supersymmetric covariant derivative also adds a third barred factor, so it vanishes on these two terms. The ordinary left Grassmann derivative, with ${\partial }_{\alpha }(\theta \theta )=2{\theta }_{\alpha }$, givesAt fixed $y$, $\bar{D}=-\bar{\partial }$ and ${\bar{D}}^2(\bar{\theta }\bar{\theta })=-4$. Thus the chiral field-strength superfield hasThe remaining components come from the other independent terms in the vector superfield; linearity ensures they cannot alter the two coefficients just calculated. This proves the requested components without computing the omitted terms. The gaugino phase and the sign of the vector component are those in the printed Wess-Zumino gauge expansion.

$W_{\alpha }$ is a fermionic chiral spinor superfield. Its chirality follows directly from the Abelian field-strength chiral projection:There are only two independent barred supersymmetric covariant derivatives, and their equal-chirality anticommutators vanish. Every product of three barred derivatives therefore vanishes. Equivalently, the previous calculation has no independent barred Grassmann variable at fixed $y$. The free undotted index makes this a chiral spinor superfield, rather than a scalar chiral superfield; its lowest component is the odd gaugino ${\lambda }_{\alpha }$.