Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 6

The defining condition isThe entries are complex and the matrices are invertible; products and inverses retain determinant one. There is no unitarity condition $S^{†}S=I$ in general. Four complex entries subject to one complex determinant equation give three complex, or six real, local group parameters.

The association between four real coordinates and Hermitian two-by-two matrices is a real-linear bijection. Congruence by $S$ preserves Hermiticity, so $X^{\prime }=SX{S}^{†}$ defines another real four-vector depending linearly on the original. Its determinant isThus the induced invertible linear map preserves the Minkowski quadratic form:The maps obtained continuously from the identity are proper and orthochronous. For example, a future timelike vector gives a positive-definite $X$, which remains positive definite under invertible congruence; its time direction is preserved. The matrices $S$ and $-S$ induce the same four-vector transformation, illustrating the Lorentz spinor double cover.

The source uses $x+iy$ in the upper-right entry, corresponding to $X=ctI+x{\sigma }_x-y{\sigma }_y+z{\sigma }_z$ for the standard Pauli matrices. This is a permissible basis convention, but its minus sign in the $y$ basis must also be retained in the spinor identification.

For nonzero $\chi$, the outer product $X=\chi {\chi }^{†}$ is rank-one Hermitian positive semidefinite, so $\det X=0$ and $ct={\chi }^{†}\chi /2>0$. It therefore represents a future-directed null vector. Conversely any nonzero future-null Hermitian $X$ has eigenvalues $2ct$ and zero, so its positive eigenvector multiplied by $\sqrt{2ct}$ supplies $\chi$. The phase $\chi \mapsto e^{i\zeta }\chi$ leaves $X$ unchanged. A past-directed null vector needs $X=-\chi {\chi }^{†}$, rather than the positive outer product alone.

For the printed normalized spinor, direct multiplication yieldsComparing with the printed $x+iy$ convention givesIts usual spatial azimuth is $-\varphi$, not $\varphi$. The source's asserted usual-angle identification has a sign inconsistency. Either conjugating the two spinor phases or replacing the upper-right matrix entry by $x-iy$ restores the usual azimuth $\varphi$. Neither alteration is made silently here. A further positive scale factor sets the magnitude of an arbitrary future-null vector; the displayed normalized spinor fixes $ct=1/2$ in the chosen coordinate units.

The spinor transformation is ${\chi }^{\prime }=S\chi$, with $S=\mathrm{diag}(e^{\kappa /2},e^{-\kappa /2})$. Its squared norm becomesDividing by $\sqrt{\mathcal{R}}$ isolates the new direction without discarding the physical scale factor in $X^{\prime }$. The lower-to-upper component ratio givesEquivalently,This is relativistic aberration for the active positive-$z$ boost specified in the question. At the poles, azimuth is undefined rather than a measurable transformed variable. Under the literal matrix convention from (c), the actual azimuth $-\varphi$ is likewise unchanged. The time coordinate is multiplied by $\mathcal{R}$; normalizing a spinor for its direction must not erase that boost-induced change of null-vector scale.

The generator acts only in the $ct,z$ plane, with its square equal to the identity on that two-dimensional subspace. Its exponential isFor the vector in (c), this giveswhile $x^{\prime }=x$ and $y^{\prime }=y$. Since the vector is future-null, its spatial magnitude is $c{t}^{\prime }$. ThereforeThe transverse azimuth does not change. These are exactly the direction and scale obtained from the spinor congruence. In particular $c{t}^{\prime }=\mathcal{R}ct$ and $\tan ({\theta }^{\prime }/2)=e^{-\kappa }\tan (\theta /2)$. The azimuth-sign qualification in (c) is compatible with this check because a boost along $z$ leaves both transverse coordinates unchanged.