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

The ordering of Weyl components depends on the supplied basis transformation. In this paper use $u_W=U{u}_D=(\xi ,\eta {)}^{\mathsf{T}}$. Applying the printed $U$ to the printed Dirac matrices givesThus $\xi$ is right chiral and $\eta$ left chiral in this ordering. The free massless Weyl equations areThey describe spin-one-half massless fields in the two inequivalent chiral spinor representations of the Lorentz group. Their rotation generators agree, but their boost generators have opposite signs; spatial parity interchanges the two. A positive-energy particle in the $\xi$ field has helicity $+1/2$, and one in the $\eta$ field has helicity $-1/2$. Their antiparticles have the opposite helicities, so a Weyl field is not a field containing only one helicity for both particles and antiparticles.

Neutrinos provide the observed ultrarelativistic approximation to a left-chiral particle and right-helicity antiparticle. However, neutrino oscillation establishes nonzero neutrino mass differences, so physical neutrinos are not all exactly massless Weyl particles. Chiral Weyl fields occur in the Standard Model, but an isolated exactly massless elementary Weyl particle is not established by neutrino observations. Whether neutrino mass eigenstates are Dirac or Majorana remains a separate question.^[ref]

With the gamma matrices derived in a, the momentum-space Dirac equation readsFor a positive-energy particle at rest, $E=m>0$ and $\mathbf{p}=0$, so $\xi =\eta$. The Dirac adjoint in this basis is $\bar{u}=({\eta }^{†},{\xi }^{†})$, and the specified normalization givesTherefore, for any normalized two-component spinor $\chi$,Equivalently, choosing ${\chi }_{+}=(1,0{)}^{\mathsf{T}}$ and ${\chi }_{-}=(0,1{)}^{\mathsf{T}}$ gives the two independent rest-frame spin states. A strictly massless particle has no rest frame; this step requires $m>0$.

Choose the active boost producing $p_z=m\sinh \rho$ and $E=m\cosh \rho$. The supplied two-component boost $L=\exp (-\rho {\sigma }_3/2)$ acts on the left-chiral block $\eta$; the right-chiral block $\xi$ transforms by its inverse. Applying them to b givesUse $m{e}^{\pm \rho }=E\pm p_z$ and ${\sigma }_3{\chi }_{\pm }=\pm {\chi }_{\pm }$ to obtainBoth are particle spinors, with spin respectively parallel and antiparallel to the positive $z$ axis. A compact equivalent expression is $\xi =\sqrt{E+p_z{\sigma }_3}\chi$, $\eta =\sqrt{E-p_z{\sigma }_3}\chi$, using the positive matrix square roots.

The products of corresponding nonzero upper and lower components equal $\sqrt{E^2-p_z^2}=m$, so ${\bar{u}}_{\pm }{u}_{\pm }=2m$ and $u_{\pm }^{†}{u}_{\pm }=2E$. Substitution also verifies $(E-p_z{\sigma }_3)\xi =m\eta$ and $(E+p_z{\sigma }_3)\eta =m\xi$. These checks fix which boost belongs to which component and prevent silently exchanging the chiral ordering defined by the source's $U$.

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.