One-Qubit Euler Decomposition

Direct multiplication of the matrices in [F8] gives $Y^2=Z^2=I_2$. Grouping the even and odd terms of the exponential series in [F7] then gives $R_z(t)=\mathrm{diag}(e^{-it\mathrm{/}2},e^{it\mathrm{/}2})$ and $R_y(t)=\left(\begin{array}{cc}\cos (t\mathrm{/}2)& -\sin (t\mathrm{/}2)\\ \sin (t\mathrm{/}2)& \cos (t\mathrm{/}2)\end{array}\right)$. Thus the matrices used in the Statement match the QC7 convention exactly.

The Leibniz formula [F4] gives $\det (\bar{U})=\overline{\det U}$; by [F5], $\det (U^{\ast })=\det ({\bar{U}}^{\mathsf{T}})=\overline{\det U}$. Thus [F6] and $U^{\ast }U=I$ give $\mathrm{\mid }\det U{\mathrm{\mid }}^2=\det (U^{\ast }U)=1$.

By [F1], choose $\delta \in \mathbb{R}$ with $\det U=e^{2i\delta }$ and put $W=e^{-i\delta }U$. The scalar has modulus one and [F4],[F6] give $\det W=e^{-2i\delta }\det U=1$, so $W$ is unitary with determinant one.

Write the first column of $W$ as $(a,c{)}^T$. Its norm is one, so $(\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }c\mathrm{\mid })$ is on the unit circle in the closed first quadrant. By [F2], choose $t\in [0,2\pi )$ with $\mathrm{\mid }a\mathrm{\mid }=\cos t$ and $\mathrm{\mid }c\mathrm{\mid }=\sin t$. By [F3], sine is positive on $(0,\pi )$: it increases from $0$ to $1$ on $[0,\pi \mathrm{/}2]$ and decreases to $0$ on $[\pi \mathrm{/}2,\pi ]$. Its $\pi$-shift makes it negative on $(\pi ,2\pi )$, so $\sin t\ge 0$ forces $t\in [0,\pi ]$. Cosine decreases strictly on $[0,\pi ]$ and vanishes at $\pi \mathrm{/}2$, so $\cos t\ge 0$ then forces $t\in [0,\pi \mathrm{/}2]$. Set $\theta =2t$. This also covers either zero coordinate.

The unit vector orthogonal to $(a,c{)}^T$ is a unit scalar multiple of $(-\bar{c},\bar{a}{)}^T$. If its scalar is $s$, the determinant of the matrix with these columns is $s(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2)=s$; since $\det W=1$, the second column is exactly $(-\bar{c},\bar{a}{)}^T$. Choose phases $\gamma ,\eta$ so $a=e^{-i\gamma }\cos t$ and $c=e^{i\eta }\sin t$; when one modulus is zero choose its phase arbitrarily.

Put $\alpha =\gamma +\eta$ and $\beta =\gamma -\eta$. Direct multiplication of the matrices obtained in step 1.1 gives $R_z(\alpha )R_y(\theta )R_z(\beta )=\left(\begin{array}{cc}{e}^{-i\gamma }\cos t& -e^{-i\eta }\sin t\\ e^{i\eta }\sin t& e^{i\gamma }\cos t\end{array}\right)=W\mathrm{.}$ The phase choices in step 4.1 make this exactly the two columns of $W$, including the cases $t=0$ and $t=\pi \mathrm{/}2$.

Restoring the scalar from step 2.1 proves the formula for arbitrary $U$. If $\det U=1$, choose $\delta =0$ in step 2.1, so $W=U$, and apply steps 3.1–5.1.