= 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 written
$$
V=\begin{pmatrix}u&v\\-v^*&u^*\end{pmatrix},\qquad |u|^2+|v|^2=1,
$$
and consequently
$$
V=q_0 I-i(q_x\sigma_x+q_y\sigma_y+q_z\sigma_z),\qquad q_0,q_x,q_y,q_z\in\mathbb R,\quad q_0^2+q_x^2+q_y^2+q_z^2=1.
$$
Choose $\theta$ with $q_0=\cos(\theta/2)$ and $\|q\|=\sin(\theta/2)$, and set $n=q/\|q\|$ when $q\ne0$. The <Pauli matrix multiplication law> shows $(n\cdot\sigma)^2=\|n\|^2I=I$. Applying part (a) proves
$$
\boxed{U=e^{i\chi}e^{-i\theta(n\cdot\sigma)/2},\qquad \|n\|=1.}
$$
For $q=0$, $V=\pm I$ 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>.
Back to article page
