Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 307 / 3

With ${ϵ}_{0123}=+1$, define the Pauli-Lubanski pseudovector byFor a translation $a^{\mu }$ and a Lorentz transformation $\Lambda =\exp (\omega )$, a left-handed Weyl spinor transforms asEquivalently, the active field at a fixed point uses ${\psi }^{\prime }(x)=S(\Lambda )\psi ({\Lambda }^{-1}(x-a))$.

The Clebsch-Gordan decomposition isIndeed,The symmetric spinor is the $(1,0)$ representation, while its antisymmetric part is the Lorentz scalar $(0,0)$.

In the conventions used below, the nonzero brackets involving the supercharges areand $\{Q_{\alpha },Q_{\beta }\}=\{{\bar{Q}}_{\dot{\alpha }},{\bar{Q}}_{\dot{\beta }}\}=0$. These equations, together with the Poincare algebra, are the four-dimensional $\mathcal{N}=1$ Super-Poincaré algebra without central charges.

Both terms in $B_{\mu }=W_{\mu }-\frac{1}{4}\bar{Q}{\bar{\sigma }}_{\mu }Q$ commute with momentum. For the first,because the two resulting products of momenta are symmetric in indices contracted with the Levi-Civita symbol. For the second, $[P_{\rho },Q]=[P_{\rho },\bar{Q}]=0$. Therefore

Using part i and $[P_{\mu },P_{\rho }]=0$,Thus

Move $Q_{\alpha }$ through the two odd supercharges and use $\{Q_{\alpha },Q_{\beta }\}=0$:Consequently,This unsimplified form keeps all signs transparent; changing the placement convention for the conjugate supercharge changes the displayed intermediate sign consistently with the definition of $B_{\mu }$.

The Pauli-Lubanski pseudovector and the spinor Lorentz transformation giveCombining this with part iii and the Pauli matrix identity ${\sigma }^{\nu }{\bar{\sigma }}_{\mu }={\eta }^{\nu }{}_{\mu }1+2i{\sigma }^{\nu }{}_{\mu }$ cancels the term containing ${\sigma }_{\mu \nu }{P}^{\nu }$. In these conventions,Simultaneously reversing the convention for $W_{\mu }$ reverses the final sign, but the proportionality to $P_{\mu }{Q}_{\alpha }$ is invariant and is what the next part needs.

Since momentum commutes with the supercharge, part iv givesThereforeHermitian conjugation also gives $[C_{\mu \nu },{\bar{Q}}_{\dot{\alpha }}]=0$.

The antisymmetric $C_{\mu \nu }$ transforms as a Lorentz tensor, so its complete contractionis a Lorentz scalar. HenceParts ii and v give $[P_{\rho },{\tilde{C}}_2]=[Q_{\alpha },{\tilde{C}}_2]=[{\bar{Q}}_{\dot{\alpha }},{\tilde{C}}_2]=0$. It therefore commutes with every generator of the Super-Poincaré algebra and is a Casimir element, called the superspin Casimir.