# Possible typo in Weinberg's QFT parity for massless particles (p78)

A massless particle state with the standard momentum $$k^\mu=(\kappa,0,0,\kappa)$$ and helicity $$\sigma$$ is denoted by $$\Psi_{k,\sigma}$$, Weinberg defines the parity phase $$\eta_\sigma$$ for the parity operator $$P$$ \begin{align} U(R_2^{-1}) P \Psi_{k,\sigma} = \eta_\sigma \Psi_{k,-\sigma} \end{align} where $$R_2$$ is the rotation of the reference frame around 2 axis by angle $$\pi$$ (or in the "positive" point of view I prefer, rotation of the physical system around 2 axis by angle $$-\pi$$).

To show how parity acts on states with general momentum, I would have \begin{align} P \Psi_{p,\sigma} &= P U(R(\hat{p}))U(B(\frac{|\vec{p}|}{\kappa}))\Psi_{k,\sigma} = P U(R(\hat{p}))U(B(\frac{|\vec{p}|}{\kappa}))P^{-1}U(R_2^{-1})^{-1}U(R_2^{-1}) P \Psi_{k,\sigma} \\ &=\eta_\sigma P U(R(\hat{p}))U(B(\frac{|\vec{p}|}{\kappa}))P^{-1}U(R_2^{-1})^{-1} \Psi_{k,-\sigma} \\ &=\eta_\sigma U(R(\hat{p}))U(R_2^{-1})^{-1}U(B(\frac{|\vec{p}|}{\kappa})) \Psi_{k,-\sigma} \end{align} where we have used \begin{align} P U(B(\frac{|\vec{p}|}{\kappa})) P^{-1} &= U(R_2^{-1})^{-1} U(B(\frac{|\vec{p}|}{\kappa})) U(R_2^{-1}) \\ P U(R(\hat{p})) P^{-1} &= U(R(\hat{p})) \end{align} Our result differs from the one in the book because $$U(R_2^{-1})^{-1}\neq U(R_2)$$ or $$e^{i 2\pi J_2}\neq 1$$ for particles with half-integer spin. Is that a typo of the book and we should define $$U(R_2)^{-1} P \Psi_{k,\sigma} = \eta_\sigma \Psi_{k,-\sigma}$$?

That is not a typo. Since $$R_2$$ is a rotation by $$180°$$ about the two-axis, $$R_2^{-1}$$ is a rotation by $$180°$$ about the two-axis in the opposite sense. So if $$U(R_2)=e^{i \pi J_2 }$$, then $$U(R_2^{-1})=e^{-i \pi J_2}$$. This gives $$(U(R_2^{-1}))^{-1} = (e^{-i \pi J_2})^{-1}=e^{i \pi J_2} = U(R_2)$$.