 Solution

A general single-qubit quantum gate belongs to $U(2)$. Remove its global phase to obtain $V\in SU(2)$; the displayed expression in the question is literally a special-unitary gate and represents a general physical gate up to that phase. Every $V\in SU(2)$ can be writtenand consequentlyChoose $\theta$ with $q_0=\cos (\theta /2)$ and $\Vert q\Vert =\sin (\theta /2)$, and set $n=q/\Vert q\Vert$ when $q\ne 0$. The Pauli matrix multiplication law shows $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I=I$. Applying part (a) provesFor $q=0$, $V=\pm I$ and any axis can be used with angle zero or $2\pi$. Thus the phase-free gate is a rotation gate by angle $\theta$ about the unit axis $n$. An arbitrary exact $U(2)$ operator also needs the scalar phase $e^{i\chi }$, which cannot in general be produced by a traceless Quantum Hamiltonian.