Past exam of the symmetry and relativity course of the University of Oxford / 2019 / 4 / d / Solution

Write $u=(a,b{)}^T$. The Pauli matrices giveHenceTherefore $\boxed{U_{\mu }{U}^{\mu }=0}$, with $U^0>0$ for $u\ne 0$: the bilinear is a future null Lorentz four-vector, and $u=0$ gives the zero vector.

For $S=e^{\rho {\sigma }_z/2}=\mathrm{diag}(e^{\rho /2},e^{-\rho /2})$, the transformed components are $a^{\prime }=e^{\rho /2}a$ and $b^{\prime }=e^{-\rho /2}b$. Thus $a^{\prime \ast }{b}^{\prime }=a^{\ast }b$, while the two squared magnitudes acquire factors $e^{\rho }$ and $e^{-\rho }$. Substitution givesThe matrix preserves the Minkowski metric and is an active boost along $+z$ of rapidity $\rho$, with velocity $c\tanh \rho$. Relative to the passive standard-boost convention of part i, its velocity parameter has the opposite sign.

For $S=e^{i\theta {\sigma }_z}$, $a^{\prime }=e^{i\theta }a$ and $b^{\prime }=e^{-i\theta }b$, so $a^{\prime \ast }{b}^{\prime }=e^{-2i\theta }{a}^{\ast }b$. ConsequentlyThis is an active right-handed spatial rotation through $-2\theta$ about $z$. The minus sign follows from the specified positive $i\theta {\sigma }_z$ exponent; the familiar positive-angle spin rotation uses $e^{-i\phi {\sigma }_z/2}$. The doubled angle exhibits the Lorentz spinor double cover: $u$ and $-u$ give the same four-vector.