 Solution

The transformed components are $a^{\prime }=e^{i{\alpha }_3/2}a$, $b^{\prime }=e^{-i{\alpha }_3/2}b$. Hence $(a^{\prime }{)}^{\ast }{b}^{\prime }=e^{-i{\alpha }_3}{a}^{\ast }b$, while the two moduli are unchanged. Taking real and imaginary parts in the expression from part (ii) givesThis is a right-handed active rotation through $-{\alpha }_3$ about the third axis. The minus sign follows from the positive sign in $U=e^{+iT/2}$; changing that convention reverses the rotation. A $2\pi$ spatial rotation changes the spinor sign while leaving the three-vector unchanged.