Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 50 / 1 / d

Both blocks of the given rotation matrix carry the same spin one-half representation, so a stationary spin-up state along the third axis uses the $+1$ eigenspinor ${\xi }_3=(1,0{)}^T$ of ${\sigma }_3$. At rest $A=B=m{I}_2$, givingFor a massless particle moving along the positive third axis, $p=(E,0,0,E)$ with $E>0$, andKeeping the same two-spinor label givesIt has positive helicity $+1/2$ and positive chirality in the Weyl representation.

For spin-up along the first axis, take the normalized $+1$ eigenspinor of ${\sigma }_1$, namely ${\xi }_1=(1,1{)}^T/\sqrt{2}$. At rest this givesFor momentum $(E,0,0,E)$, using the square roots displayed in part (i) with the same label givesThis is the massless limit of a transverse canonical-spin label: it is the equal superposition of the negative-helicity left-chiral mode and the positive-helicity right-chiral mode. It solves the massless Dirac equation and has norm squared $2E$.

A qualification is essential if “spin along the first axis” is interpreted literally at zero mass. The fixed transverse rotation generator is ${\Sigma }_1/2$, with ${\Sigma }_1=\mathrm{diag}({\sigma }_1,{\sigma }_1)$, and the massless Hamiltonian at this momentum is $H=E{\gamma }^0{\gamma }^3$. They anticommute. If $Hu=Eu$ and ${\Sigma }_{1}u=u$ both held, then $H{\Sigma }_{1}u=-{\Sigma }_{1}Hu=-Eu$, whereas $H{\Sigma }_{1}u=Hu=Eu$, forcing $u=0$. Thus there is no nonzero positive-energy massless solution with that literal transverse-spin eigenvalue. The displayed nonzero spinor is the transported rest-spin label requested by the massive-to-massless construction, rather than a transverse-spin eigenstate.