Past exam of the particle physics course of the University of Oxford / 2019 / 5 / d / Solution

Take $p_z>0$ and $E\gg m$. Then $\sqrt{E+p_z}\simeq \sqrt{2E}$ while $\sqrt{E-p_z}=m/\sqrt{E+p_z}\simeq m/\sqrt{2E}$. HenceThe helicity operator is $\hat{h}=\frac{1}{2}\mathrm{diag}({\sigma }_3,{\sigma }_3)$ for momentum along $+z$. It givesThe positive-helicity state becomes entirely right chiral, while the negative-helicity state becomes entirely left chiral. The suppressed opposite-chirality amplitude is of relative order $m/(2E)$; at finite mass chirality is not the same as helicity. If $p_z<0$, helicity is measured along the reversed momentum and the labels of the two spin-axis states interchange.